(This post from the lecture by Yueqi Sheng)

In this post, we will talk about detecting phase transitions using
Approximate-Message-Passing (AMP), which is an extension of
Belief-Propagation to “dense” models. We will also discuss the Replica
Symmetric trick, which is a heuristic method of analyzing phase
transitions. We focus on the Rademacher spiked Wigner model (defined
below), and show how both these methods yield the same phrase transition
in this setting.

The Rademacher spiked Wigner model (RSW) is the following. We are given
observations $Y = \frac{\lambda}{n}xx^T + \frac{1}{\sqrt{n}}W$ where
$x \in \{\pm 1\}^n$ (sampled uniformly) is the true signal and $W$ is a
Gaussian-Orthogonal-Ensemble (GOE) matrix:
$W_{i, j} \sim \mathbb{N}(0, 1)$ for $i \neq j$ and
$W_{i, i} \sim \mathbb{N}(0, 2)$. Here $\lambda$ is the signal to noise
ratio. The goal is to approximately recover $x$.

The question here is: how small can $\lambda$ be such that it is
impossible to recover anything reasonably correlated with the
ground-truth $x$? And what do the approximate-message-passing algorithm

To answer the first question, one can think of the task here is to
distinguish $Y \sim \frac{\lambda}{n}xx^T + \frac{1}{\sqrt{n}}W$ vs
$Y \sim W$. One approach to distinguishing these distributions is to
look at the spectrum of the observation matrix $Y$. (In fact, it turns
out that this is an asymptotically optimal distinguisher [1]). The spectrum of $Y$ behaves as ([2]):

• When $\lambda \leq 1$, the empirical distribution of eigenvalues in
spiked model still follows the semicircle law, with the top
eigenvalues $\approx 2$

• When $\lambda > 1$, we start to see an eigenvalue $> 2$ in the
planted model.

Approximate message passing

This section approximately follows the exposition in [3].

First, note that in the Rademacher spiked Wigner model, the posterior
distribution of the signal $\sigma$ conditioned on the observation $Y$
is: $\Pr[\sigma | Y] \propto \Pr[Y | \sigma] \propto \prod_{i \neq j} \exp(\lambda Y_{i, j} \sigma_i \sigma_j /2 )$ This
defines a graphical-model (or “factor-graph”), over which we can perform
Belief-Propogation to infer the posterior distribution of $\sigma$.
However, in this case the factor-graph is dense (the distribution is a
product of potentials $\exp(\lambda Y_{i, j} \sigma_i\sigma_j)$ for all
pairs of $i, j$).

In the previous blog post, we saw belief propagation works great when the underlying interaction
graph is sparse. Intuitively, this is because $G$ is locally tree like,
which allows us to assume each messages are independent random
variables. In dense model, this no longer holds. One can think of dense
model as each node receive a weak signal from all its neighbors.

In the dense model setting, a class of algorithms called Approximate
message passing (AMP) is proposed as an alternative of BP. We will
define AMP for RWM in terms of its state evolution.

State evolution of AMP for Rademacher spiked Wigner model

Recall that in BP, we wish to infer the posterior distributon of
$\sigma$, and the messages we pass between nodes correspond to marginal
probability distribution over values on nodes. In our setting, since the
distributions are over $\{\pm 1\}$, we can represent distributions by
their expected values. Let $m^t_{u \to v} \in [-1, 1]$ denote the
message from $u$ to $v$ at time $t$. That is, $m_{u \to v}$ corresponds
to the expected value ${{\mathbb{E}}}[\sigma_u]$.

To derive the BP update rules, we want to compute the expectation
${{\mathbb{E}}}[\sigma_v]$ of a node $v$, given the
messages ${{\mathbb{E}}}[\sigma_u]$ for $u \neq v$. We can
do this using the posterior distribution of the RWM, $\Pr[\sigma | Y]$,
which we computed above.
$\displaystyle \Pr[\sigma_v = 1 | Y, \{\sigma_u\}_{u \neq v}] = \frac{ \prod_u \exp(\lambda Y_{u, v} \sigma_u) - \prod_u \exp(-\lambda Y_{u, v} \sigma_u) }{ \prod_u \exp(\lambda Y_{u, v} \sigma_u) + \prod_u \exp(-\lambda Y_{u, v} \sigma_u) }$

And similarly for $\Pr[\sigma_v = -1 | Y, \{\sigma_u\}_{u \neq v}]$.
From the above, we can take expectations over $\sigma_u$, and express
${{\mathbb{E}}}[\sigma_v]$ in terms of
$\{{{\mathbb{E}}}[\sigma_u]\}_{u \neq v}$. Doing this (and
using the heuristic assumption that the distribution of $\sigma$ is a
product distribution), we find that the BP state update can be written
as:
$m^{t}_{u \to v} = f(\sum_{w \neq v}f^{-1}(A_{w, u} m^{t - 1}_{w \to u}))$
where the interaction matrix $A_{w, u} = \lambda Y_{w, u}$, and
$f(x) = tanh(x) = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(x)}$.

Now, Taylor expanding $f^{-1}$ around $0$, we find
$m^{t}_{u \to v} = f\left( (\sum_{w \neq v} A_{w, u} m^{t - 1}_{w \to u}) + O(1/\sqrt{n}) \right)$
since the terms $A_{w, u}$ are of order $O(1/\sqrt{n})$.

At this point, we could try dropping the “non-backtracking” condition
$w \neq v$ from the above sum (since the node $v$ contributes at most
$O(1/\sqrt{n})$ to the sum anyway), to get the state update:
$m^{t}_{u} = f\left( \sum_{w} A_{w, u} m^{t - 1}_{w}) \right)$ (note the messages no longer
depend on receiver – so we write $m_u$ in place of $m_{u \to v}$).
However, this simplification turns out not to work for estimating the
signal. The problem is that the “backtracking” terms which we added
amplify over two iterations.

In AMP, we simply perform the above procedure, except we add a
correction term to account for the backtracking issue above. Given $u$,
for all $v$, the AMP update is:
$m^{t}_{u \to v} = m^{t}_u = f(\sum_{w}A_{w, u} m^{t - 1}_{w}) + [\text{some correction term}]$

The correction term corresponds to error introduced by the backtracking
terms. Suppose everything is good until step $t - 2$. We will examine
the influence of backtracking term to a node $v$ through length 2 loops.
At time $t - 1$, $v$ exert $Y_{v, u}m^{t - 2}_v$ additional influence to
each of it’s neighbor $u$. At time $t$, $v$ receive roughly
$Y_{u, v}^2m^{t - 2}_v$. Since $Y_{u, v}^2$ has magnitude
$\approx \frac{1}{n}$ and we need to sum over all of $v$’s neighbors,
this error term is to large to ignore. To characterize the exact form of
correction, we simply do a taylor expansion

$m^{t}_v = \sum_{u}f(Y_{u, v}m^{t - 1}_u) = \sum_{u}f(Y_{u, v} \left(\sum_{w}f(Y_{w, u}m^{t - 2}_w) - f(Y_{u, v}m^{t - 2}_w)\right) )\\ \approx \sum_u f(Y_{u, v} m^{t - 1}_u) - Y_{u, v}f'(m^{t - 1}_u)m^{t - 2}_v\\ \approx \sum_u f(Y_{u, v} m^{t - 1}_u) - \frac{1}{n}\sum_{u}f'(m^{t - 1}_u)m^{t - 2}_v$

State evolution of AMP

In this section we attempt to obtain the phase transition of Rademacher
spiked Wigner model via looking at $m^{\infty}$.

We assume that each message could be written as a sum of signal term and
noise term. $m^t = \mu_t x + \sigma_t g$ where
$g \sim \mathbb{N}(0, I)$. To the dynamics of AMP (and find its phase
transition), we need to look at how the signal $\mu_t$ and noise
$\sigma_t$ evolves with $t$.

We do the following simplification: ignore the correction term and
assume each time we obtain an independent noise $g$.

$m^{t} = Yf(m^{t - 1}) = (\frac{\lambda}{n}x^Tx + \frac{1}{\sqrt{n}}W)f(m^{t - 1}) = \frac{\lambda}{n} < f(m^{t - 1}), x > x + \frac{1}{\sqrt{n}} Wf(m^{t - 1})$

Here, we see that $\mu_t = \frac{\lambda}{n}< f(m^{t - 1}), x>$
and $\sigma_t = \frac{1}{\sqrt{n}}Wf(m^{t - 1})$.

Note that $\mu_{t}$ is essentially proportional to overlap between
ground truth and current belief
, since the function $f$ keeps the
magnitude of the current beliefs bounded.

$\frac{\lambda}{n} = \frac{\lambda}{n} \approx\lambda {{\mathbb{E}}}_{X \sim unif(\pm 1), G\sim \mathbb{N}(0, 1)}[X f(\mu_{t - 1}X + \sigma_{t - 1}G)] = \lambda {{\mathbb{E}}}_G[f(\mu_{t - 1} + \sigma_{t - 1}G)]$

For the noise term, each coordinate of $\sigma_t$ is a gaussian random
variable with $0$ mean and variance

$\frac{1}{n} \sum_v f(m^{t - 1})_v^2 \approx {{\mathbb{E}}}_{X, G}[f(\mu_{t - 1}X + \sigma_{t - 1}G)^2] = {{\mathbb{E}}}_{G}[f(\mu_{t - 1} + \sigma_{t - 1}G)^2]$

It was shown in [4] that we can introduce a new
parameter $\gamma_t$ s.t.
$\gamma_t = \lambda^2 {{\mathbb{E}}}[f(\gamma_{t - 1} + \sqrt{\gamma_{t - 1}}G)]$
As $t \to \infty$, turns out $\mu_t = \frac{\gamma_t}{\lambda}$ and
$\sigma_t^2 = \frac{\sigma_t}{\lambda^2}$. To study the behavior of
$m^t$ as $t \to \infty$, it is enough to track the evolution of
$\gamma_t$.

This heuristic analysis of AMP actually gives a phase transition at
$\lambda = 1$ (in fact, the analysis of AMP can be done rigorously as in [5]):

• For $\lambda < 1$: If $\gamma_t \approx 0$, $|\gamma_t + \sqrt{\gamma_t}G| < 1$ w.h.p., thus we have $\gamma_{t + 1} \approx \lambda^2 (\gamma_t) < \gamma_t$. Taking $t \to \infty$, we have $\gamma_{\infty} = 0$, which means there AMP solution has no overlap with the ground truth.

• For $\lambda > 1$: In this case, AMP’s solution has some correlation with the ground truth.

(Figure from [6])

Replica symmetry trick

Another way of obtaining the phase transition is via a non-rigorous
analytic method called the replica method. Although non-rigorous, this
method from statistical physics has been used to predict the fixed point
of many message passing algorithms and has the advantage of being easy
to simulate. In our case, we will see that we obtain the same phase
transition temperature as AMP above. The method is non-rigorous due to
several assumptions made during the computation.

Outline of replica method

Recall that we are interested in minizing the free energy of a given
system $f(\beta, Y) = \frac{1}{\beta n} \log Z(\beta, Y)$ where $Z$ is
the partition function as before:
$Z(\beta, Y) = \sum_{x \in \{\pm 1\}^n} exp(-\beta H(Y, x))$ and
$H(Y, x) = - = -xYx^T = -\sum_{i, j} Y_{i, j}x_ix_j$.

In replica method, $Y$ is not fixed but a random variable. The
assumption is that as $n \to \infty$, free energy doesn’t vary with $Y$
too much, so we will look at the mean of $f_Y$ to approximate free
energy of the system.

$f(\beta) = \lim_{n \to \infty}\frac{1}{\beta n}{{\mathbb{E}}}_{Y}[\log Z(\beta, Y)]$

$f(\beta)$ is called the free energy density and the goal now is to
compute the free energy density as a function of only $\beta$ , the
temperature of the system.

The replica method is first proposed as a simplification of the
computation of $f(\beta)$

It is a generally hard problem to compute $f(\beta)$ in a clear way. A
naive attempt of approximate $f(\beta)$ is to simply pull the log out
$g(\beta) = \frac{1}{\beta n}\log {{\mathbb{E}}}_Y[Z(\beta, Y)]$
Unfortunately $g(\beta)$ and $f(\beta)$ are quite different quantities,
at least when temperature is low. Intuitively, $f(\beta)$ is looking at
system with a fixed $Y$ while in $g(\beta)$, $x$ and $Y$ are allowed to
fluctuate together. When the temperature is high, $Y$ doesn’t play a big
roll in system thus they could be close. However, when temperature is
low, there could be a problems. Let $\beta \to \infty$,
$f(\beta) \approx \int_Y (\beta x_Y Y x_Y)\mu(Y) dY$,
$g(\beta) \approx \log \int_Y exp(\beta x_J Y x_Y)\mu(Y)dY \approx \beta x^* Yx^*$.

While ${{\mathbb{E}}}_X[\log(f(X))]$ is hard to compute,
${{\mathbb{E}}}[f(X)^r]$ is a much easier quantity. The
replica trick starts from rewriting $f(\beta)$ with moments of $Z$:
Recall that $x^r \approx 1 + r \log x$ for $r \approx 0$ and
$\ln(1 + x)\approx x$, using this we can rewrite $f(x)$ in the following
way:

Claim 1. Let $f_r(\beta) = \frac{1}{r \beta n}\ln[{{\mathbb{E}}}_Y[Z(\beta, Y)^r]]$
Then, $f(\beta) = \lim_{r \to 0}f_r(\beta)$

The idea of replica method is quite simple

• Define a function $f(r, \beta)$ for $r \in \mathbb{Z}_+$ s.t. $f(r, \beta) = f_r(\beta)$ for all such $r$.

• Extend $f(r, \beta)$ analytically to all $r \in {{\mathbb{R}}}_+$ and take the limit of $r \to 0$.

The second step may sound crazy, but for some unexplained reason, it has
been surprisingly effective at making correct predictions.

The term replica comes from the way used to compute
${{\mathbb{E}}}[Z^r]$ in Claim 1. We expand the $r$-th moment
in terms of $r$ replicas of the system

$Z(\beta, Y)^r = (\sum_x exp(-\beta H(Y, x)))^r = \sum_{x^1, \cdots, x^r} \Pi_{k = 1}^r exp(-\beta H(Y, x^i))$

In this section, we will see how one can apply the replica trick to
obtain phase transition in the Rademacher spiked Wigner model. Recall
that given a hidden $a \in \{\pm 1\}^n$, the observable
$Y = \frac{\lambda}{n}a^Ta + \frac{1}{\sqrt n} W$ where
$W_{i, j} \sim \mathcal{N}(0, 1)$ and $W_{i, i} \sim \mathcal{N}(0, 2)$.
We are interested in finding the smallest $\lambda$ where we can still
recover a solution with some correlation to the ground truth $a$. Note
that $\{W_{i, i}\}$ is not so important here as $x_i^2$ doesn’t carry
any information in this case.

Given by the posterior ${{\mathbb{P}}}[x|Y]$, the system we
set up corresponding to Rademacher spiked Wigner model is the following:

• the system consists of $n$ particles and the interactions between
each particle are give by $Y$

• the signal to noise ratio $\lambda$ as the inverse temperature
$\beta$.

Following the steps above, we begin by computing
$f(r, \beta) = \frac{1}{r\beta n}\ln{{\mathbb{E}}}_Y[Z^r]$
for $r \in \mathbb{Z}_+$: Denote $X^k = (x^k)^Tx^k$ where $x^k$ is the
$k$th replica of the system.

${{\mathbb{E}}}_Y[Z^r] = \int_Y \sum_{x^1, \cdots, x^r} exp(\beta \sum_k \mu(Y) dY\\ = \int_Y \sum_{x^1, \cdots, x^r} exp(\beta ) \mu(Y) dY$

We then simplify the above expression with a technical claim.

Claim 2. Let $Y = A + \frac{1}{\sqrt{n}}W$ where $A$ is a fixed matrix and
$W$ is the GOE matrix defined as above. Then,
$\int_Y exp(\beta) \mu(Y) dY = exp(\frac{\beta^2}{n}{{\|{X}\|_{F}}}^2 + \frac{\beta}{2} )$
for some constant $C$ depending on distribution of $Y$.

Denote $X = \sum_k X^k$. Apply Claim 2 with
$A = \frac{\beta}{n}a^Ta$, we have
${{\mathbb{E}}}_Y[Z^r] = \sum_{x^1, \cdots, x^r} exp(\frac{\beta^2}{n}{{\|{X}\|_{F}}}^2 + \frac{\beta^2}{2n} )$
To understand the term inside exponent better, we can rewrite the inner
sum in terms of overlap between replicas:

${{\|{X}\|_{F}}}^2 = \sum_{i, j}X_{i, j}^2 = \sum_{i, j}(\sum_{k = 1}^r x^k_ix^k_j)^2 =\sum_{i, j}(\sum_{k = 1}^r x^k_ix^k_j)(\sum_{l = 1}^r x^l_ix^l_j)\\ = \sum_{k, l} (\sum_{i = 1}^n x^k_ix^{l}_i)^2 = \sum_{k, l} ^2$

where the last equality follows from rearranging and switch the inner
and outer summations.

Using a similar trick, we can view the other term as

$ = \sum_{i, j}\sum_{k = 1}^rx^k_ix^k_ja_ia_j = \sum_{k = 1}^r (\sum_{i = 1}^n a_ix^k_i)^2 = \sum_{k}^2$

Note that $Q_{k, l} = $ represents overlaps between the
$k$ and $l$th replicas and $Q_k = $ represents the
overlaps between the $k$th replica and the ground truth vector.

In the end, we get for any integer $r$, (Equation 1):

$\displaystyle f(r, \beta) = \frac{1}{r\beta n}\ln(\sum_{x^1, \cdots, x^r} exp(\frac{\beta^2}{n}\sum_{k, l}Q_{k, l}^2 + \frac{\beta^2}{2n}\sum_k Q_k^2)) \label{e:1}\\ = \frac{1}{r\beta n} \ln(\sum_{Q}\nu_{x^k}(Q)exp(\frac{\beta^2}{n}\sum_{k, l}Q_{k, l}^2 + \frac{\beta^2}{2n}\sum_k Q_k^2))$

Our goal becomes to approximate this quantity. Intuitively, if we think
of $Q_{k, l}$ as indices on a $(r + 1) \times (r + 1)$ matrices, $Q$,
with $Q(i,i) = 1$, then $Q$ is the average of $n$ i.i.d matrices. So we
expect $Q_{j, k} \in [\pm \frac{1}{n}]$ for $j \neq k$ w.h.p. In the
remaining part, We find the correct $Q$ via rewriting Equation 1.

Observe that by introducing a new variable $Z_{k, l}$ for $k \neq l$ and
using the property of gaussian intergal (Equation 4):

$\label{e:4} exp(\frac{\beta^2}{n}Q_{k, l}^2) = \sqrt{\frac{n}{4\pi}}\int_{Z_{k, l}} exp(-\frac{n}{4}Z_{k, l}^2 + \beta Q_{k, l}Z_{k, l})dZ_k$

$\exp(\frac{\beta^2}{2n}Q_k^2) = \sqrt{\frac{1}{8\pi n}}\int_{Z_k}exp(-(2n)Z_k^2 + 2\beta Q_{k}Z_k)dZ_k$
Replace each $exp(\frac{\beta^2}{n}Q_{k, l}^2)$ by a such integral, we
have (Equation 2):

$\begin{gathered} {{\mathbb{E}}}[Z^r] = \sum_{x^1, \cdots, x^r} exp(\frac{\beta^2}{n}\sum_{k, l}Q_{k, l}^2 + \frac{\beta^2}{2n}\sum_k Q_k^2) \label{e:2}\\ = C\sum_{x^1, \cdots, x^r} \exp(\beta^2 n)\int_{Z_{k, l}}exp(-\frac{n}{4}\sum_{k \neq l}Z_{k, l}^2 - \frac{n}{2}\sum_k Z_k^2 + \beta \sum_{k \neq l}Y_{k, l}Q_{k, l} + 2\beta\sum_kZ_k Q_k) dZ \\ =C\exp(\beta^n) \int_{Y_{k, l}}exp(-\frac{n}{4}\sum_{k \neq l}Y_{k, l}^2 - \frac{n}{2}\sum_k Z_k^2 + \ln(\sum_{x_1,\cdots, x_r}exp(\beta \sum_{k\neq l}Y_{k, l}Q_{k, l} + 2\beta\sum_kY_k Q_k)) dY \label{e:2}\end{gathered}$

where $C$ is the constant given by introducing gaussian intergals.

To compute the integral in (Equation 2), we need to cheat a little bit and take
$n \to \infty$ before letting $r \to 0$. Note that free energy density
is defined as
$f(\beta) = \lim_{n \to \infty}\lim_{r \to 0}\frac{1}{r\beta n}\ln {{\mathbb{E}}}_Y[Z(\beta, Y)^r]$
This is the second assumption made in the replica method and it is
commonly believed that switching the order is okay here. Physically,
this is plausible because we believe intrinsic physical quantities
should not depend on the system size.

Now the Laplace method tells us when $n \to \infty$, the integral in (Equation 2) is dominated by the max of the exponent.

Theorem 1 (Laplace Method). Let $h(x): {{\mathbb{R}}}^n \to {{\mathbb{R}}}$then

$\int e^{nh(x)} \approx e^{nh(x^*)}(\frac{2\pi}{n})^{\frac{d}{2}}\frac{1}{\sqrt{det(H)}}$

where $x^* = argmax_x \{h(x)\}$ and $H$ is the Hessian of $h$ evaluated at the point $x^*$.

Fix a pair of $k, l$ and apply Laplace method with
$h(Z_{k, l}) = -\frac{1}{2}\sum_{0 \leq k < l \leq r}Z_{k, l}^2 + \frac{1}{n}\ln(\sum_{x_1,\cdots, x_r}exp(\beta \sum_{k \neq l}Z_{k, l}Q_{k, l} + 2\beta\sum_kZ_k Q_k))$
what’s left to do is to find the critical point of $h$. Taking the
derivatives gives
$-Y_{k, l} + \frac{A(Z_{k, l})\beta Q_{k, l}}{n A(Z_{k, l})} = 0$
where
$A(Z_{k, l}) = \sum_{x_1,\cdots, x_r}exp(\beta \sum_{k \neq l}Z_{k, l}Q_{k, l} + \beta\sum_kY_k Q_k)$.

We now need to find a saddle point of $h$ where the hessian is PSD. To
do that, we choose to assume the order of the replicas does not matter,
which is refer to as the replica symmetry case. 1 One simplest form
of $Y$ is the following: $\forall k, l > 0$, $Z_{k, l} = y$ and
$Z_{k} = y$ for some $y$. This also implies that $Q_{k, l} = q$ for some
$q$ and $y =\frac{\beta}{n} q$

Plug this back in to Equation 2 gives: (Equation 3)

$\label{e:3} {{\mathbb{E}}}[Z^r] = C\exp(\beta n)\exp(-\frac{n}{2}(\frac{r^2 - r}{2})y^2 - \frac{n^2}{2} + \ln(\sum_{x^i}\exp(y\beta\sum_{k \neq l}Q_{k, l} + 2y\beta \sum_k Q_k))$

To obtain $f(r, \beta)$, we only need to deal with the last term in
(Equation 3) as $r \to 0$. Using the fact that $Q_{k, l} = y$ for all
$k, l$ and using the same trick of introducing new gaussain integral as
in (Equation 4) we have
$\lim_{r \to 0}\frac{1}{r}\ln(\sum_{x^i}\exp(y\beta\sum_{k \neq l}Q_{k, l} + n\beta \sum_k Q_k)) = -\beta + {{\mathbb{E}}}_{z \sim \mathcal{N}(0, 1)}[\log(2cosh(y\beta + \sqrt{y\beta}z))]$

Using the fact that we want the solution to minimizes free energy,
taking the derivative of the current $f$ w.r.t. $y$ gives
$\frac{y}{\beta} = n{{\mathbb{E}}}_z[tanh(y\beta + \sqrt{y\beta}z)]$
which matches the fixed point of AMP. Plug in $q$ and $y$ will give us
$f(\beta)$. The curve of $f(\beta)$ looks like the Figure below, where
the solid line is the curve of $f(\beta)$ with the given $q$ and the
dotted line is the curve given by setting all variables $0$.

References

[1] Amelia Perry, Alexander S Wein, Afonso S Bandeira, and Ankur Moitra. Optimality and sub-optimality of pca for spiked random matrices and synchronization.
arXiv preprint arXiv:1609.05573, 2016.
[2] D. Feral and S. Pech e. The Largest Eigenvalue of Rank One Deformation of Large Wigner Matrices. Communications in Mathematical Physics, 272:185–228, May 2007.
[3] Afonso S Bandeira, Amelia Perry, and Alexander S Wein. Notes on computational-to-statistical gaps: predictions using statistical physics. arXiv preprint arXiv:1803.11132, 2018.
[4] Yash Deshpande, Emmanuel Abbe, and Andrea Montanari. Asymptotic mutual information for thebinary stochastic block model. In
Information Theory (ISIT), 2016 IEEE International Symposium on, pages 185–189. IEEE, 2016.
[5] Adel Javanmard and Andrea Montanari. State evolution for general approximate message passing algorithms, with applications to spatial coupling. Information and Inference: A Journal of the IMA, 2(2):115–144, 2013.
[6] A. Perry, A. S. Wein, and A. S. Bandeira. Statistical limits of spiked tensor models.
ArXiv e-prints, December 2016.

1. Turns out for this problem, replica symmetry is the only case. We
will not talk about replica symmetry breaking here, which
intuitively means we partition replicas into groups and re-curse.

There’s a lot of discussion and (possibly well-deserved) hype nowadays about quantum computation and its potential for computation at speeds we simply can’t reach with the classical computers we’re used to today. The excitement about this has been building for years, even decades, but it’s only very recently that we’ve really been approaching a solid proof that quantum computers do have an advantage over classical computers.

What’s rather tricky about showing such a result is that, rather than a direct argument about the capability of quantum computers, what we really need to demonstrate is the incapability of classical computers to achieve tasks that can be done with quantum computers.

One of the major leaps forward in demonstrating quantum supremacy was taken by Terhal and DiVincenzo in their 2008 paper “Adaptive quantum computation, constant depth quantum circuits and arthur-merlin games“. Their approach was to appeal to a complexity-theoretic argument: they gave evidence that there exists a certain class of quantum circuits that cannot be simulated classically by proving that if a classical simulation existed, certain complexity classes strongly believed to be distinct would collapse to the same class. While this doesn’t quite provide a proof of quantum supremacy – since the statement about the distinction between complexity classes upon which it hinges is not a proven fact – because the complexity statement appears overwhelmingly likely to be true, so too does the proposed existence of non-classically-simulatable quantum circuits. The Terhal and DiVincenzo paper is a complex and highly technical one, but in this post I hope to explain a little bit and give some intuition for the major points.

Now, let’s start at the beginning. What is a quantum circuit? I’m going to go ahead and assume you already know what a classical circuit is – the extension to a quantum circuit is rather straightforward: it’s a circuit in which all gates are quantum gates, where a quantum gate can be thought of as a classical gate whose output is, rather than a deterministic function of the inputs, instead a probability distribution over all possible outputs given the size of the inputs. For example, given two single-bit inputs, a classical AND gate outputs 0 or 1 deterministically given the inputs. A quantum AND gate on the analogous single-qubit inputs would output 0 with some probability $p$ and 1 with some probability $1-p$. Similarly, a classical AND gate on two 4-bit inputs outputs the bitwise AND, while the quantum analog has associated with it a 4-qubit output: some probability distribution over all 4-bit binary strings. A priori there is no particular string that is the “output” of the computation by the quantum gate; it’s only after taking a quantum measurement of the output that we get an actual string that we can think of as the outcome of the computation done by the gate. The actual string we “observe” upon taking the measurement follows the probability distribution computed by the gate on its inputs. In this way, a quantum circuit can then be thought of as producing, via a sequence of probabilistic classical gates (i.e., quantum gates) some probability distribution over possible outputs given the input lengths. It’s not hard to see that in this way, we can compose circuits: suppose we have a quantum circuit $c_1$ and another quantum circuit $c_2$. Let $c_1$ have an input of $n$ qubits and an output of $m$ qubits; suppose we measure $k$ of the output qubits of $c_1$ – then we can feed the remaining $m-k$ unmeasured qubits as inputs into $c_2$(assuming that those $m-k$ qubits do indeed constitute a valid input to $c_2$).

Consider, then, the following sort of quantum circuit: it’s a composition of $N$ quantum circuits, such that after each $i$-th circuit we take a measurement some of its output qubits (so that the remaining unmeasured qubits become inputs to the $(i+1)$-th circuit), and then the structure of the $(i+1)$-th circuit is dependent on this measurement. That is, it’s as though, given a quantum circuit, we’re checking every so often at intermediate layers over the course of the circuit’s computation what the value of some of the variables are (leaving the rest to keep going along through the circuit to undergo more computational processing), and based on what we measure is the current computed value, the remainder of the circuit “adapts” in a way determined by that measurement. Aptly enough, this is called an “adaptive circuit”. But since the “downstream” structure of the circuit depends on the outcomes of all the measurements made “upstream”, each adaptive circuit actually comprises a family of circuits, each of which is specified by the sequence of intermediate measurement outcomes. That is, we can alternatively characterize an adaptive circuit as a set of ordinary quantum circuits that is parameterized by a list of measurement outcomes. Terhal and DiVincenzo call this way of viewing an adaptive circuit, as a family of circuits parametrized by a sequence of measurement values, a “non-adaptive circuit” – since we replace the idea that the circuit “adapts” to intermediate measurements with the idea that there are just many regular circuits, one for each possible sequence of measurements. It’s this non-adaptive circuit concept that’ll be our main object of study going forward.

Simulating quantum circuits

Now, the result we wanted to demonstrate about quantum circuits had to do with their efficient simulatability by classical circuits – and so we should establish some notion of what we mean when we talk about an “efficient simulation”.

Terhal and DiVincenzo offer the following notion of a classical simulation – which in their paper they call an “efficient density computation”: consider a quantum circuit with some output of length $n$ qubits. Recall that to actually obtain an output value, we need to take a measurement of the circuit output – imagine doing this in disjoint subsets of cubits at a time. That is, we can break up the $n$ qubits into $k$ disjoint subsets and consider the entire output measurement as a process of taking $k$ measurements, subset by subset. An efficient density computation exists if there’s a classical procedure for computing, in time polynomial in the width and depth of the quantum circuit, the conditional probability distribution over the set of possible measurement outcomes of a particular subset of qubits, given any subset of the $k-1$ other measurement outcomes. Intuitively, this is a good notion of what a classical simulation should consist of, or at least what data it should contain, since if you know the conditional probabilities given any (possibly empty) subset of the other measurements, you can just flip coins for the outputs according to the conditional probabilities as a way of actually exhibiting a “working” simulation.

It’s with this notion of simulation, along with our concept of an adaptive quantum circuit as a family of regular circuits parameterized by a sequence of intermediate measurement outcomes, we may now arrive at the main result of Terhal and DiVincenzo’s paper. Recall that what we wanted to show from the very beginning is that there exists some quantum circuit that can’t be simulated classically. The argument for this proceeds like a proof by contradiction: suppose the contrary, and that all quantum circuits can be simulated classically. We want to show that we can find, then, a quantum circuit which, if it were possible to be simulated classically (as per our assumption), we’d wind up with some strange consequences that we believe are false, leading us to conclude that those circuits probably can’t be simulated classically.

Thus, we shall now exhibit such a quantum circuit whose classical simulatability leads (as far as we believe) to a contradiction. Consider a special case of adaptive quantum circuits, considered as a parameterized family of regular circuits, in which the circuit’s output distribution is independent of the intermediate measurement outcomes; that is, the case in which the entire family of circuits corresponding to an adaptive circuit is logically the same – that is, is the same logical circuit on input qubits independent of intermediate measurements. I’d like to point out, just for clarification’s sake, the subtlety here, which makes this consideration non-redundant, and not simply a reduction of an adaptive quantum circuit (again, thought of as a family) to a single fixed circuit (i.e., a family of one): the situation in which the family is reduced to a single fixed circuit occurs when the structure of the circuit is independent of the intermediate measurement outcomes. If the structure were independent of the measurements, then no matter what we observed in the measurements, we’d get the same circuit – hence a trivial family of one. What we’re considering instead is the case in which the structure of the circuit is still dependent on the intermediate measurements (and so the circuit is still adaptive), but where the distribution over the possible outputs of the circuit is identical no matter what the intermediate measurements are. In this case, the circuit can still be considered as a parameterized and in general non-trivial family of circuits, but for which each member produces the same distribution over outputs – hence, a family of potentially structurally different circuits, but which are logically identical.

Suppose there’s some set of such circuits that’s universal – that is, that’s sufficient to implement all polynomial-time quantum computations. (This is a reasonable assumption to make, since there do in fact exist universal quantum gate sets. But now if a simulation of the kind we defined (an efficient density computation) existed for every circuit in this set, then we could calculate the outcome probability of any polynomial-depth quantum circuit, since any polynomial-depth quantum circuit could be realized as some composition of circuits in this universal set (and in particular as a composition of particular family members of each adaptive circuit in the universal set), and an efficient density computation, as we mentioned above, precisely gives us a way to compute the output distribution.

But now here is where our believed contradiction lies:

Theorem: Suppose there exists a universal set of adaptive quantum circuits whose output distributions are independent of intermediate measurements. If there is an efficient density computation for each family member of each adaptive circuit in this universal set, then for the polynomial hierarchy PH we have PH = BPP = BQP.

The proof goes something like this: if we can do our desired efficient density computations (as we assumed, for the sake of contradiction, we could for all quantum circuits), this is equivalent to being able to determine the acceptance probability of a quantum computation, which was shown in the paper “Determining Acceptance Possibility for a Quantum Computation is Hard for the Polynomial Hierarchy” by Fenner, Green, Homer and Pruim to be equivalent to the class $\text{coC=P}$. Thus, we have that $\text{coC=P} \subseteq \text{BPP}$. But it’s known that $\text{PH} \subseteq \text{BPP}^{\text{coC=P}}$ and so $\text{PH} \subseteq \text{BPP}^{\text{BPP}} = \text{BPP}$. That is, we have $\text{PH} = \text{BPP} = \text{BQP}$, and so the polynomial hierarchy would collapse to $\Sigma^P_2$ since $\text{BPP} \subseteq \Sigma^P_2$ (for more on these more obscure complexity classes, see here). Again, this is our “contradiction”: while it hasn’t been quite proven, it is widely believed, with strong supporting evidence, that the polynomial hierarchy does not collapse as would be the case if all quantum circuits were classically simulatable. Thus this provides a strong argument that not all quantum circuits are classically simulatable, which was precisely what we were looking to demonstrate.

Conclusion

Terhal and DiVincenzo actually go even further and show that there is a certain class of constant-depth quantum circuits that are unlikely to be simulatable by classical circuits – this, indeed, seems to provide even stronger evidence for quantum supremacy. This argument, which is somewhat more complex, uses the idea of teleportation and focuses on a particular class of circuits implementable by a certain restricted set of quantum gates. If you’re interested, I highly recommend reading their paper, where this is explained.

• Terhal, Barbara and David DiVincenzo. “Adptive quantum computation, constant depth quantum circuits and arthur-merlin games.” Quantum Info. Comput. 4, 2 (2004); 134-145.
• Bravyil, Sergey, David Gosset, and Robert König. “Quantum advantage with shallow circuits.” Science 362, 6412 (2018); 308-311.
• Harrow, Aram Wettroth and Ashley Montanaro. “Quantum computational supremacy.” Nature 549 (2017): 203-209.
• Boixo, Sergio et. al. “Characterizing quantum supremacy in near-term devices.” Nature Physics 14 (2018); 595-600.

References

\begin{enumerate}

• Terhal, Barbara and David DiVincenzo. “Adptive quantum computation, constant depth quantum circuits and arthur-merlin games.” Quantum Info. Comput. 4, 2 (2004); 134-145.
• Fenner, Stephen et. al. “Determining Acceptance Possibility for a Quantum Computation is Hard for the Polynomial Hierarchy.” https://arxiv.org/abs/quant-ph/9812056. (1998).
• Bravyil, Sergey, David Gosset, and Robert König. “Quantum advantage with shallow circuits.” Science 362, 6412 (2018); 308-311.

This is the season that people are applying for postdoc positions. Unlike student and faculty hiring, which each have a fairly fixed schedule, postdoc availability can change from time to time, with new opportunities opening up all the time. So, I encourage everyone looking for a postdoc position to periodically check out https://cstheory-jobs.org/ .

(For example, a new ad was just posted on Wednesday by Venkat Guruswami and Pravesh Kothari )

tags:

[Guest post by Noah Miller – a Harvard Physics Ph.D student that took our seminar. Noah’s webpage contains wonderful and extensive notes that can be of interest to computer scientists. –Boaz]

(The following blog post serves as an introduction to the following notes:)

Black Holes, Hawking Radiation, and the Firewall

There are many different types of “theoretical physicists.” There are theoretical astrophysicists, theoretical condensed matter physicists, and even theoretical biophysicists. However, the general public seems to be most interested in the exploits of what you might call “theoretical high energy theorists.” (Think Stephen Hawking.)

The holy grail for theoretical high energy physicists (who represent only a small fraction of all physicists) would be to find a theory of quantum gravity. As it stands now, physicists have two theories of nature: quantum field theory (or, more specifically, the “Standard Model”) and Einstein’s theory of general relativity. Quantum field theory describes elementary particles, like electrons, photons, quarks, gluons, etc. General relativity describes the force of gravity, which is really just a consequence of the curvature of spacetimes.

Sometimes people like to say that quantum field theory describes “small stuff” like particles, while general relativity describes “big stuff” like planets and stars. This is maybe not the best way to think about it, though, because planets and stars are ultimately just made out of a bunch of quantum particles.

Theoretical physicists are unhappy with having two theories of nature. In order to describe phenomena that depend on both quantum field theory and general relativity, the two theories must be combined in an “ad hoc” way. A so-called “theory of everything,” another name for the currently unknown theory of “quantum gravity,” would hypothetically be able to describe all the phenomena we know about. Just so we’re all on the same page, they don’t even have a fuly worked out hypothesis. (“String theory,” a popular candidate, is still not even a complete “theory” in the normal sense of the word, although it could become one eventually.)

So what should these high energy theoretical physicists do if they want to discover what this theory of quantum gravity is? For the time being, nobody can think up an experiment that would be sensitive to quantum gravitation effects which is feasible with current technology. We are limited to so-called “thought experiments.”

This brings us to Hawking radiation. In the 1970’s, Stephen Hawking considered what would happen to a black hole once quantum field theory was properly taken into account. (Of course, this involved a bit of “ad hoc” reasoning, as mentioned previously.) Hawking found that, much to everybody’s surprise, the black hole evaporated, realeasing energy in the form of “Hawking radiation” (mostly low energy photons). More strangely, this radiation comes out exactly in the spectrum you would expect from something “hot.” For example, imagine heating a piece of metal. At low temperatures, it emits low energy photons invisible to the human eye. Once it gets hotter, it glows red, then yellow, then perhaps eventually blue. The spectrum of light emitted follows a very specific pattern. Amazingly, Hawking found that the radiation which black holes emit follow the exact same pattern. By analogy, they have a temperature too!

This is more profound than you might realize. This is because things which have an temperature should also have an “entropy.” You see, there are two notions of “states” in physics: “microstates” and “macrostates.” A microstate gives you the complete physical information of what comprises a physical system. For example, imagine you have a box of gas, which contains many particles moving in a seemingly random manner. A “microstate” of this box would be a list of all the positions and momenta of every last particle in that box. This would be impossible to measure in pratice. A “macrostate,” on the other hand, is a set of microstates. You may not know what the exact microstate your box of gas is in, but you can measure macroscopic quantities (like the total internal energy, volume, particle number) and consider the set of all possible microstates with those measured quantities.

The “entropy” of a macrostate is the logarithm of the number of possible microstates. If black holes truly are thermodynamic systems with some entropy, that means there should be some “hidden variables” or microstates to the black hole that we currently don’t understand. Perhaps if we understood the microstates of the black hole, we would be much closer to understanding quantum gravity!

However, Hawking also discovered something else. Because the black hole is radiating out energy, its mass will actually decrease as time goes on. Eventually, it should disappear entirely. This means that the information of what went into the black hole will be lost forever.

Physicists did not like this, however, because it seemed to them that the information of what went into the black hole should never be lost. Many physicists believe that the information of what went into the black hole should somehow be contained in the outgoing Hawking radiation, although they do not currently understand how. According to Hawking’s original calculation, the Hawking radiation only depends on a parameters of the black hole (like its mass) and have nothing to do with the many finer details on what went in, the exact “microstate” of what went in.

However, physicists eventually realized a problem with the idea that the black hole releases its information in the form of outgoing Hawking radiation. The problem has to do with quantum mechanics. In quantum mechanics, it is impossible to clone a qubit. That means that if you threw a qubit in a black hole and then waited for it to eventually come out in the form of Hawking radiation, then the qubit could no longer be “inside” the black hole. However, if Einstein is to be believed, you should also be able to jump into the black hole and see the qubit on the inside. This seems to imply that the qubit is cloned, as it is present on both the inside and outside of the black hole.

Physicists eventually came up with a strange fix called “Black Hole Complementarity” (BHC). According to BHC, according to people outside the black hole, the interior does not exist. Also, according to people who have entered the black hole, the outside ceases to exist. Both descriptions of the world are “correct” because once someone has entered the black hole, they will be unable to escape and compare notes with the person on the outside.

Of course, it must be emphasized that BHC remains highly hypothetical. People have been trying to poke holes in it for a long time. The largest hole is the so called “Firewall Paradox,” first proposed in 2012. Essentially, the Firewall paradox tries to show that the paradigm of BHC is self-contradictory. In fact, it was able to use basic quantum mechanics to show that, under some reasonable assumptions, the interior of the black hole truly doesn’t exist, and that anyone who tries to enter would be fried at an extremely hot “firewall!” Now, I don’t think most physicists actually believe that black holes really have a firewall (although this might depend on what day of the week you ask them). The interesting thing about the Firewall paradox is that it derives a seemingly crazy result from seemingly harmless starting suppositions. So these suppositions would have to be tweaked in the theory of quantum gravity order to get rid of the firewall.

This is all to say that all this thinking about black holes really might help physicists figure out something about quantum gravity. (Then again, who can really say for sure.)

If you would like to know more about the Firewall paradox, I suggest you read my notes, pasted at the top of this post!

The goal of the notes was to write an introduction to the Black Hole Information Paradox and Firewall Paradox that could be read by computer scientists with no physics background.

The structure of the notes goes as follows:

1. Special relativity
2. General relativity
3. Quantum Field Theory (in which I ambitiously tell you what QFT actually is!)
4. Statistical Mechanics (this is the best part)

Because the information paradox touches on all areas of physics, I thought it was necessary to take “zero background, infinite intelligence” approach, introducing the all the necessary branches of physics (GR, QFT, Stat Mech) in order to understand what the deal with Hawking radiation really is, and why physicists think it is so important. I think it is safe to say that if you read these notes, you’ll learn a non-trivial amount of physics.

Nilin Abrahamsen nilin@mit.edu

Daniel Alabi alabid@g.harvard.edu

Mitali Bafna mitalibafna@g.harvard.edu

Emil Khabiboulline ekhabiboulline@g.harvard.edu

Juspreet Sandhu jus065@g.harvard.edu

Two-prover one-round (2P-1R) games have been the subject of intensive study in classical complexity theory and quantum information theory. In a 2P-1R game, a verifier sends questions privately to each of two collaborating provers , who then aim to respond with a compatible pair of answers without communicating with each other. Sharing quantum entanglement allows the provers to improve their strategy without any communication, illustrating an apparent paradox of the quantum postulates. These notes aim to give an introduction to the role of entanglement in nonlocal games, as they are called in the quantum literature. We see how nonlocal games have rich connections within computer science and quantum physics, giving rise to theorems ranging from hardness of approximation to the resource theory of entanglement.

Introduction

In these notes we discuss 2-prover 1-round games and the classical complexity of approximating the value of such games in the setting where the provers can share entanglement. That is, given the description of a game, we ask how hard it is to estimate the winning probability of the best winning strategy of the entangled provers. Let us first formally define games and its relation to the label cover problem. We write ${ [n]=\{1,\ldots,n\}}$.

Definition (Label cover)

A label cover instance ${ I=(S,T,\Sigma,\Pi)}$ consists of variable sets ${ S=\{s_i\}_{i\in[n]},T=\{t_j\}_{j\in[n]}}$, alphabet set ${ \Sigma}$, and a collection ${ \Pi}$ of constraints of the form ${ t_j=f_{i,j}(s_i)}$. Given an assignment (or coloring) ${ c : S\cup T \rightarrow \Sigma}$ we define its value to be ${ \omega(c)=\mathbb P_{f_{ij}\sim\Pi}\big(c(t_j)=f_{ij}(c(s_i))\big)}$. Define the value of ${ I}$ to be the maximum over all possible assignments, i.e. ${ \omega(I) = \max_{c} \omega(c)}$. Many familiar computational problems can be formulated as a label cover, such as ${ \textsc{3SAT}}$, ${ \textsc{3Lin}}$, and ${ \textsc{MaxCut}}$.

Definition (2-prover 1-round game)

Let ${ I = (S,T,\Sigma,\Pi)}$ be a label cover instance. We can then associate the following two-prover one-round game ${ G(I)}$ with ${ I}$. Let ${ P_1,P_2}$ be two provers who cannot communicate, and let ${ V}$ be the verifier. Given the label cover instance ${ I}$, the verifier uniformly samples a constraint ${ f_{i,j} \in \Pi}$ and sends ${ s_i}$ to ${ P_1}$ and ${ t_j}$ to ${ P_2}$. The provers then reply with ${ a \in \Sigma}$ and ${ b\in\Sigma}$ respectively to ${ V}$. Finally, ${ V}$ outputs ${ 1}$ if and only if ${ b = f_{i,j}(a)}$. Any coloring of the label cover instance corresponds to a deterministic strategy for the corresponding game. Therefore, with an optimal strategy the provers win the game associated to label cover instance ${ I}$ with probability ${ \omega(I)}$. That is, the value of the game equals that of the label cover instance. However, this is with the assumption that provers can only use deterministic strategies or convex combinations of these (that is, using shared randomness). If the provers share an entangled quantum state, then the provers (who still cannot communicate) can enjoy correlations that allow them to win with a higher probability than classically. In the quantum literature, these 2P-1R games are known as nonlocal games referring to the fact that the correlations arise without signaling between the provers. We are concerned with the complexity of approximating the winning probability of this strategy. We refer to the optimal winning probability within some class (classical or entangled) of strategies as the classical and quantum value of the game, respectively, and we use the terms quantum strategy and entangled strategy interchangeably. Fixing different constraint families in the label cover game changes the complexity of finding the (classical and entangled) values of the game. We will show that approximating the entangled value of XOR games, or more generally unique games (to be defined later on), is possible in polynomial time. This is remarkable because a famous conjecture known as the unique games conjecture says that approximating the classical value of unique games is NP-hard. In contrast, we will see that for unrestricted edge constraints, it is NP-hard to approximate the entangled value of a nonlocal game. Thus, hardness of approximation of the game’s value, established by the celebrated PCP theorem , still applies in the presence of entanglement. In the quantum world, we have new complexity classes such as QMA (which can be regarded as “quantum NP”), so one may conjecture whether approximating the entangled value of a general game is QMA-hard (the games formulation of the quantum PCP conjecture ). We will indicate progress in this direction but will explicitly demonstrate the NP-hardness result. Entanglement is often regarded as an expensive resource in quantum information because it is difficult to produce and maintain. Hence, even if sharing entanglement can improve the success probability of winning a game, the resource consumption may be costly. We will conclude by discussing lower bounds on the number of shared entangled bits required to achieve the optimal value of a game.

Notation and quantum postulates

Let us first establish notation and define what is meant by an entangled strategy. In keeping with physics notation we write a column vector as ${ |{v}\rangle \in\mathbb C^d}$ and its conjugate-transpose (a row vector) as ${ \langle{v}| }$ . More generally the conjugate-transpose (Hermitian conjugate) of a matrix ${ A}$ is written ${ A^\dag}$. Then ${ \langle{v}| w\rangle}$ is the inner product of two vectors (a scalar) and ${ |{v}\rangle \langle{w}| }$ the outer product (a rank-1 matrix). A matrix ${ A}$ is said to be Hermitian if ${ A^\dag=A}$. A matrix ${ A}$ is positive semidefinite , written ${ A\succeq 0}$, if ${ A=B^\dag B}$ for some matrix ${ B}$. We write the identity matrix as ${ {\mathbb{I}}}$, denote by ${ Herm(\mathbb{C}^d)}$ the set of ${ d}$-by-${ d}$ Hermitian matrices.

Observables, states, and entanglement

In a quantum theory the observables are Hermitian operators ${ A\in Herm(\mathbb C^d)}$ on a vector space ${ \mathbb C^d}$. It then makes sense to say that a state is a functional ${ \varphi}$ on the set of observables. That is, to specify the state of a physical system means giving a (expected) value for each observable. It turns out states are linear functionals of the observables, and such functionals can be written ${ \varphi(A)=\langle A,\rho\rangle}$ for some ${ d}$-by-${ d}$ matrix ${ \rho}$. We call ${ \rho}$ the density matrix and require moreover that ${ \rho}$ is positive semidefinite and has trace ${ 1}$. Every density matrix is a convex combination of rank-one projections ${ |\psi\rangle\langle\psi|}$ known as pure states. The unit vectors ${ |\psi\rangle\in\mathbb C^d}$ are also themselves known as pure states. If the state of one particle is described by a vector in ${ \mathbb C^d}$ (referring here to pure states), then two particles are described by a vector in the tensor product ${ \mathbb C^d\otimes\mathbb C^d}$. The two particles are entangled if their state is not in the form of a pure tensor ${ |\phi\rangle\otimes|\psi\rangle}$. We also write product states as ${ |\phi\rangle|\psi\rangle}$, omitting the tensor symbol.

Quantum measurements

A quantum measurement can be described in terms of a projection-valued measure (PVM). Definition 1 (PVM) A projection-valued measure on vector space ${ \mathbb C^d}$ (where the quantum states live) is a list of projection matrices ${ A_1,\ldots,A_k}$ on ${ \mathbb C^d}$ such that ${ A_iA_j=0}$ for ${ i\neq j}$ and ${ \sum_i A_i={\mathbb{I}}}$. The PVM describes a measurement which, on state ${ |\psi\rangle\in\mathbb C^d}$ outputs measurement outcome ${ i}$ with probability ${ \langle\psi| A_i|\psi\rangle}$. The quantum state after obtaining outcome ${ i}$ is ${ \frac1{\|A_i|\psi\rangle\|}A_i|\psi\rangle}$. When the projections are rank-one projections ${ A_i= |{\beta_i}\rangle \langle{\beta_i}| }$ we say that we measure in the basis ${ \{ |{\beta_i}\rangle \}_i}$. In this case the probability of outcome ${ i}$ in state ${ |\psi\rangle}$ is ${ |\langle\beta_i |{\psi}\rangle |^2}$, and the post-measurement state is simply ${ |{\beta_i}\rangle}$ . Applying the measurement ${ (A_i)_i}$ on the left half of a two-particle state ${ |{\Psi}\rangle \in\mathbb C^d\otimes\mathbb C^d}$ means applying the PVM ${ (A_i\otimes {\mathbb{I}})_{i}}$ on the two-particle state.

Quantum strategies for nonlocal games

We now introduce the notion of a quantum strategy for a nonlocal game. Each prover holds a particle, say with state space ${ \mathbb C^d}$, and Alices particle may be entangled with Bob’s. The global state is ${ |{\phi}\rangle _{AB}\in\mathbb C^d\otimes\mathbb C^d}$. Each player receives a question from the verifier and then chooses a measurement (a PVM) depending on the question. The player applies the measurement to their own particle and responds to the verifier with their measurement outcome. Hence for Alice we specify ${ n}$ PVM’s ${ A^s}$ where ${ s\in S}$ is a question, and each PVM is a list ${ A^s_{a=1},\ldots,A^s_k}$. By definition 1, given questions ${ s,t}$ the probability that Alice outputs ${ a}$ and Bob outputs ${ b}$ is

$\displaystyle P(a,b|s,t)=\|({\mathbb{I}}\otimes B^t_b)(A^s_a\otimes {\mathbb{I}}) |{\phi}\rangle _{AB}\|^2= \langle{\phi}| A^s_a\otimes B^t_b|\phi\rangle, \ \ \ \ \ (1)$

where we have used that squaring a projection leaves it unchanged.

Quantum strategies beat classical ones

For any 2P-1R game ${ G}$, let ${ \omega(G)}$ be the maximum probability — over the players’ classical strategies — that the verifier accepts and ${ \omega^*(G)}$ the maximum probability that the verifier accepts when the provers use qubits such that player 1’s qubits are entangled with those of player 2. The game of Clauser, Horne, Shimony, and Holt (CHSH) has the property that the provers can increase their chance of winning by sharing an entangled pair of qubits, even when no messages are exchanged between the players. We show that there’s a characterization of the CHSH game’s value ${ \omega^*(G)=\cos^2(\frac{\pi}{8}) = \frac{1}{2}+\frac{\sqrt{2}}{4}\approx 0.85}$ which is better than the classical value ${ \omega(G)\leq \frac{3}{4}}$. Let us first define XOR games, of which the CHSH game is a special case.

Definition (XOR game)

An XOR game is a 2-player classical game (the questions and answers are classical bits) where:
1. Questions ${ (s, t)\in\{0, 1, \ldots, n-1\}^2}$ are asked according to some distribution ${ \Pi}$ (e.g. uniform).
2. Answers ${ a, b\in\{0, 1\}}$ are provided by players (call them Alice and Bob).
3. The verifier computes a predicate ${ V(a, b|s, t) = f_{s, t}(a\oplus b)}$ used to decide acceptance/rejection.

Definition (CHSH Game)

An XOR game with ${ n=2}$ where ${ s, t\in\{0, 1\}}$ are independent random bits and ${ V(a, b | s, t)=1 \Longleftrightarrow a\oplus b = s\wedge t}$. To win the CHSH game, Alice and Bob need to output bits ${ a}$ (from Alice) and ${ b}$ (from Bob) that disagree if ${ s=t=1}$ and agree otherwise. If Alice and Bob are classical then they can do no better than by always outputting ${ 0}$, say, in which case they win in the three out of four cases when one of the questions is ${ 0}$. Equivalently, ${ \omega(G)=\frac34}$ where ${ G}$ is the CHSH game. This is the content of Bell’s inequality :

Lemma (Bell’s Inequality)

For any two functions ${ g, h: \{0, 1\}\rightarrow\{0, 1\}}$, we have ${ {\mathop{\mathbb{P}}}_{x, y\in\{0, 1\}}\left[g(x)\oplus h(y) = x\wedge y\right] \leq \frac{3}{4} }$ where ${ x}$ and ${ y}$ are independent uniformly random bits.
Proof.
The probability of any event is a multiple of ${ 1/4}$ so it suffices to show that ${ {\mathop{\mathbb{P}}}_{x, y\in\{0, 1\}}\left[g(x)\oplus h(y) = x\wedge y\right]\neq1}$. So assume for contradiction that ${ g(x)\oplus h(y) = x\wedge y}$ for all pairs ${ x,y}$. Then we have that ${ g(0)\oplus h(0) = 0}$ and ${ g(0)\oplus h(1) = 0}$ which implies that ${ h(0)=h(1)}$. But then ${ 0=g(1)\oplus h(0) =g(1)\oplus h(1) = 1}$ which is a contraction.

The strategy

The entangled strategy for the CHSH game requires that Alice and Bob each hold a qubit, so that their two qubits together are described by a vector in ${ \mathbb C^2\otimes\mathbb C^2}$. The two qubits together are in the state ${ |{\phi}\rangle = \frac{1}{\sqrt{2}}\left( |{0}\rangle _A |{0}\rangle _B + |{1}\rangle _A |{1}\rangle _B\right), }$ forming what is known as an EPR (Einstein-Podolsky-Rosen) pair. Now for ${ \theta\in[-\pi, \pi]}$ define ${ |{\beta_0(\theta)}\rangle = \cos\theta |{0}\rangle + \sin\theta |{1}\rangle }$ and ${ |{\beta_1(\theta)}\rangle = -\sin\theta |{0}\rangle + \cos\theta |{1}\rangle }$ . Now we describe a (quantum) strategy with winning probability ${ \cos^2(\frac{\pi}{8})}$. In each case Alice and Bob respond with their measurement outcome, where subscripts of the measurement basis vectors correspond to the answer to be sent back to the verifier.
• If ${ s=0}$, Alice measures in basis ${ \{ |{\beta_0(0)}\rangle , |{\beta_1(0)}\rangle \} = \{ |{0}\rangle , |{1}\rangle \}}$. If ${ s=1}$, Alice measures in ${ \{ |{\beta_0(\frac{\pi}{4})}\rangle , |{\beta_1(\frac{\pi}{4})}\rangle \}}$. Alice answers bit ${ a = 0}$ if outcome is ${ \beta_0(\cdot)}$ and answers ${ a = 1}$ otherwise.
• If ${ t=0}$, Bob measures in basis ${ \{ |{\beta_0(\frac{\pi}{8})}\rangle , |{\beta_1(\frac{\pi}{8})}\rangle\}}$. If ${ t=1}$, Bob measures in ${ \{ |{\beta_0(-\frac{\pi}{8})}\rangle , |{\beta_1(-\frac{\pi}{8})}\rangle \}}$.
• Each player responds with their respective measurement outcome.
Lemma 2 Alice and Bob win the CHSH game with probability ${ \cos^2(\frac{\pi}{8})}$.
Proof.
We will show that for each pair of questions ${ s,t}$ the pair of answers ${ a,b}$ is correct with probability ${ \cos^2(\pi/8)}$. We can split the pairs of questions into the two cases ${ s\wedge t=0}$ and ${ s\wedge t=1}$:
• (${ s\wedge t=0}$) The three cases ${ (s,t)=(0,0),(0,1),(1,0)}$ are all analogous: in each case Alice an Bob must output the same answer, and in each case Bob’s measurement basis is almost the same as Alice’s except rotated by a small angle ${ \pm\pi/8}$. Of the three above cases we consider the one where ${ s,t=0}$ and check that indeed the two measurement outcomes agree with probability ${ \cos^2(\pi/8)}$: When Alice measures her qubit and obtains some bit ${ a}$, the shared pair ${ |{\beta}\rangle}$ collapses to ${ |{a}\rangle _A |{a}\rangle _B}$. Indeed, since the question was ${ s=0}$, Alice measures her qubit in the basis ${ \{|0\rangle,|1\rangle\}}$. This means that Alice applies the measurement ${ \{|0\rangle\langle 0|\otimes {\mathbb{I}},|1\rangle\langle 1|\otimes {\mathbb{I}}\}}$ on the global state. The post-measurement state is the normalization of ${ ( |{a}\rangle \langle{a}| \otimes {\mathbb{I}}) |{\phi}\rangle _{AB}=\frac{1}{\sqrt2} |{a}\rangle \overbrace{\langle a |{0}\rangle }^{\delta_{a,0}}\otimes | 0\rangle + |{a}\rangle \overbrace{\langle a |{1}\rangle }^{\delta_{a,1}}\otimes |{1}\rangle =\frac {1}{\sqrt2} |{a}\rangle _A |{a}\rangle _B }$ because ${ \langle a |{a'}\rangle}$ can be viewed as a Kronecker delta of ${ a}$ and ${ a'}$. In particular, Bob is now in the pure state ${ |{a}\rangle}$ . Because Bob received question ${ t=0}$ he measures in the basis ${ \{ |{\beta_b(\pi/8)}\rangle \}_b}$ Therefore his probability of correctly outputting ${ b=a}$ is ${ {\mathop{\mathbb{P}}}[\text{Bob gets outcome }a] = |\langle\beta_a(\tfrac{\pi}{8}) |{a}\rangle |^2 = \cos^2\left(\frac{\pi}{8}\right) }$
• (${ s\wedge t=1}$) Now consider the case ${ s=1,t=1}$ where Alice and Bob are supposed to give different answers. Alice measures in basis consisting of ${ |{\beta_0(\pi/4)}\rangle =\frac{|0\rangle+|1\rangle}{\sqrt2}}$ and ${ |{\beta_1(\pi/4)}\rangle =\frac{-|0\rangle+|1\rangle}{\sqrt2}}$. If Alice gets outcome ${ a}$ then the post-measurement global state is ${ |{\beta_a(\frac\pi4)}\rangle |{\beta_a(\frac\pi4)}\rangle}$ . Therefore when Bob applies the measurement in basis ${ \{ |{\beta_a(-\frac\pi8)}\rangle \}_a}$ he mistakenly outputs ${ a}$ only with probability ${ |\langle\beta_a(\frac\pi4) |{\beta_a(-\frac\pi8)}\rangle |^2=\sin^2(\frac\pi8)=1-\cos^2(\frac\pi8)}$.
Lemma 2 implies a lower bound on the value of the CHSH game.

Corollary

${ \omega^*(G)\ge\cos^2(\frac\pi8)}$
It turns out that this lower bound is sharp, that is, the strategy just described is optimal.

Lemma

The value of the CHSH game using a quantum strategy is at most ${ \frac{1}{2} + \frac{\sqrt{2}}{4} = \cos^2\frac{\pi}{8}\approx 0.85}$. Proof. We can describe the quantum strategy of Alice and Bob in an XOR game by (i) a shared quantum state ${ |{\phi}\rangle _{AB}\in\mathbb{C}^d\otimes\mathbb{C}^d}$ (note that for the CHSH game, ${ d=2}$); (ii) measurements ${ \{A^0_s, A^1_s\}}$ for every question ${ s}$ sent to Alice; (iii) measurements ${ \{B^0_t, B^1_t\}}$ for every question ${ t}$ sent to Bob. The probability of answering ${ (a, b)}$ given questions ${ (s, t)}$ is ${ \langle{\phi}| A^a_s\otimes B^b_t |{\phi}\rangle}$ . Now let us write ${ A_s=A^0_s - A^1_s}$ and ${ B_t=B^0_t - B^1_t}$ so that for any ${ a, b\in\{0, 1\}}$, we can write ${ A_s^a = \frac{{\mathbb{I}} + (-1)^aA_s}{2}, B_t^b = \frac{{\mathbb{I}} + (-1)^bB_t}{2} }$ Note that since the possible outcomes here are finite, ${ A_s, B_t}$ are Hermitian and we may assume have bounded norm of 1. Furthermore, we assume that ${ A_s, B_t}$ are observables so that ${ A_s^2 = B_t^2 = {\mathbb{I}}}$ . Now denoting ${ f_{s, t}(a\oplus b)}$ as the XOR predicate to be computed, we can write the quantum game value as $\omega^*(G) = \sup_{ |{\phi}\rangle , \{A_s^a\}, \{B_t^b\}} {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}f_{s, t}(a\oplus b) \langle{\phi}| A^a_s\otimes B^b_t |{\phi}\rangle \\ = \sup_{ |{\phi}\rangle , \{A_s\}, \{B_t\}} {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}f_{s, t}(a\oplus b) \langle{\phi}| \frac{{\mathbb{I}} + (-1)^aA_s}{2}\otimes \frac{{\mathbb{I}} + (-1)^bB_t}{2} |{\phi}\rangle \\ = {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}\frac{f_{s, t}(a\oplus b)}{4} + \sup_{ |{\phi}\rangle , \{A_s\}, \{B_t\}} {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}f_{s, t}(a\oplus b) \langle{\phi}| \frac{{\mathbb{I}} + (-1)^aA_s}{2}\otimes \frac{{\mathbb{I}} + (-1)^bB_t}{2} |{\phi}\rangle \\ = {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}\frac{f_{s, t}(a\oplus b)}{4} + \sup_{ |{\phi}\rangle , \{A_s\}, \{B_t\}} {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}\frac{f_{s, t}(a\oplus b)(-1)^{ab}}{4} \langle{\phi}| A_s\otimes B_t |{\phi}\rangle \\ = {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\sum_{a, b}\frac{f_{s, t}(a\oplus b)}{4} + \sup_{ |{\phi}\rangle , \{A_s\}, \{B_t\}} {\mathop{\mathbb{E}}}_{(s, t)\sim\Pi}\frac{f_{s, t}(0) - f_{s,t}(1)}{2} \langle{\phi}| A_s\otimes B_t |{\phi}\rangle \\$ where the summation ${ \sum_{a, b\in\{0, 1\}}\left(\cdot\right)}$ has been evaluated in the last line. Now note that the first term is independent of the quantum strategy and as a result equals the value of the uniformly random strategy which is 1/2. So we proceed to focus on the second term. Note that for CHSH ${ f_{s, t}(0) - f_{s,t}(1) = (-1)^{st}}$ simplifying the second term to ${ \frac{1}{8}( \langle{\phi}| A_0\otimes B_0 |{\phi}\rangle + \langle{\phi}| A_1\otimes B_0 |{\phi}\rangle + \langle{\phi}| A_0\otimes B_1 |{\phi}\rangle - \langle{\phi}| A_1\otimes B_1 |{\phi}\rangle ). }$ Next, we invoke Tsirelson’s Theorem (See Theorem 3) to bound this second term as $= \sup_{{\textbf{x}}_s, {\textbf{y}}_t}\frac{1}{8}({\textbf{x}}_0\cdot {\textbf{y}}_0 + {\textbf{x}}_0\cdot {\textbf{y}}_1 + {\textbf{x}}_1\cdot {\textbf{y}}_0 - {\textbf{x}}_1\cdot {\textbf{y}}_1) \\ \leq \sup_{{\textbf{x}}_s, {\textbf{y}}_t}\frac{1}{8}(\|{\textbf{x}}_0\|\|{\textbf{y}}_0 + {\textbf{y}}_1\| + \|{\textbf{x}}_1\|\|{\textbf{y}}_0 - {\textbf{y}}_1\|) \\ \leq \sup_{{\textbf{x}}_s, {\textbf{y}}_t}\frac{\sqrt{2}}{8}\sqrt{2\|{\textbf{y}}_0\|^2 + 2\|{\textbf{y}}_1\|^2} \leq \frac{\sqrt{2}}{4}$ where we have used Cauchy-Schwartz and the concavity of the ${ \sqrt{\cdot}}$ function. This completes our proof showing the exact characterization of the value (${ =\cos^2(\frac{\pi}{8})=\frac{1}{2}+\frac{\sqrt{2}}{4}}$) of the CHSH game using a quantum strategy. This proof is an adaptation of the one in [12].
Theorem 3 (Tsirelson’s Theorem [1]) For any ${ n\times n}$ matrix ${ C = (C_{s, t})}$, the following are equivalent:
1. There exist ${ d\in\mathbb{N}, |{\phi}\rangle \in\mathbb{C}^{d}\otimes\mathbb{C}^{d}, A_s, B_t\in\text{Herm}(\mathbb{C}^d), A_s^2 = B_t^2 = I}$ such that for any ${ s, t\in[n]}$ ${ C_{s, t} = \langle{\phi}| A_s\otimes B_t |{\phi}\rangle}$ . Further this would imply that ${ d\leq 2^{\lceil\frac{n+2}{2}\rceil}}$;
2. There exist real unit vectors ${ { {\textbf{x}}}_s, { {\textbf{y}}}_t\in{\mathbb{R}}^{n+2}}$ for ${ s, t\in [n]}$ such that ${ C_{s, t} = { {\textbf{x}}}_s\cdot { {\textbf{y}}}_t}$;

Entangled unique games are easy

The CHSH game provides the first example that the entangled value ${ \omega^*(G)}$ of a nonlocal game can exceed the classical value ${ \omega(G)}$. XOR-games like the CHSH game are the special case corresponding to alphabet size ${ k=2}$ of the class of unique games :

Definition (Unique Games)

A 2-prover 1-round game is called a unique game if its constraints are of the form ${ b=\pi_{ij}(a)}$ where ${ \pi_{ij}}$ is a permutation of the alphabet ${ \Sigma}$ for each edge ${ i\sim j}$. The famous unique games conjecture (UGC) by Khot says that ${ \omega(G)}$ is NP-hard to approximate for unique games. Surprisingly, Kempe et al. showed that a natural semidefinite relaxation for unique games yields an approximation to the entangled value ${ \omega^*(G)}$ which can be computed in polynomial time. In other words the UGC is false for entangled provers, in contrast to the classical case where the conjecture is open.
Theorem 4 There is an efficient (classical) algorithm which takes a description of a nonlocal game ${ G}$ as its input and outputs ${ \hat\omega(G)}$ such that ${ 1-6(1-\hat\omega(G))\le\omega^*(G)\le\hat\omega(G) }$ Put differently, if ${ \omega^*(G)=1-\varepsilon^*}$ and ${ \hat\omega(G)=1-\hat\varepsilon}$, then ${ \varepsilon^*\in[\hat\varepsilon,6\hat\varepsilon] }$
The algorithm of theorem 4 proceeds by relaxing the set of quantum strategies to a larger convex set ${ \mathcal S}$ of pseudo-strategies and maximizing over ${ \mathcal S}$ instead of actual strategies, a much easier task. In approximation theory one often encounters a collection of hypothetical moments not arising from a distribution, known as a pseudo-distribution. In contrast, our pseudo-strategies are actual conditional probability distributions on answers (conditional on the questions). What makes ${ \mathcal S}$ a set of “pseudo”-strategies rather than actual strategies is that they may enjoy correlations which cannot be achieved without communication.

Convex relaxation of quantum strategies

We will define ${ \mathcal S}$ to be a class of conditional probability distributions ${ \tilde{P}(a,b|s,t)}$ on answers given questions. We will require that the pseudo-strategies satisfy a positive semidefinite constraint when arranged in matrix form. In particular this matrix has to be symmetric, so we symmetrize the conditional probability ${ \tilde{P}(a,b|s,t)}$ by allowing each of ${ s}$ and ${ t}$ to be either a question for Alice or for Bob. That is, we extend the domain of definition for ${ \tilde{P}(a,b|s,t)}$ from ${ \Sigma^2\times S\times T}$ to ${ \Sigma^2\times (S\cup T)^2}$ where ${ S}$ and ${ T}$ are the question sets. So each question ${ s}$ and ${ t}$ and answer ${ a}$ and ${ b}$ can be either for Alice or Bob — we indicate this by changing notation from ${ (s,t)\in S\times T}$ to ${ q,q'\in S\cup T}$ and for the answers replacing ${ a,b}$ by ${ c,c'}$.
Definition 5 (Block-matrix form) Given a function ${ \tilde{P}=\tilde{P}(\cdot\cdot|\cdot\cdot)}$ defined on ${ (S\cup T)^2\times \Sigma^2}$ with ${ |S|=|T|=n}$ and ${ |\Sigma|=k}$, define a ${ 2nk}$-by-${ 2nk}$ matrix ${ M^{\tilde{P}}}$ whose rows are indexed by pairs ${ (q,c)\in(S\cup T)\times\Sigma}$ and columns by pairs ${ (q',c')\in (S\cup T)\times\Sigma}$, and whose entries are ${ M^{\tilde{P}}_{(q,c),(q',c')}=\tilde{P}(c,c'|q,q') }$ In other words ${ M^{\tilde{P}}}$ consists of ${ k}$-by-${ k}$ blocks where the block ${ M^{\tilde{P}}_{q,q'}}$ at position ${ q,q'}$ contains the entries ${ \tilde{P}(\cdot\cdot|q,q')}$.
Definition 5 is simply a convenient change of notation and we identify ${ \tilde{P}}$ with ${ M^{\tilde{P}}}$, using either notation depending on the context.
Definition 6 (Pseudo-strategies) Let ${ S}$ and ${ T}$ be the question sets for Alice and Bob, respectively. We say that ${ \tilde{P}=\tilde{P}(\cdot\cdot|\cdot\cdot):\Sigma^2\times (S\cup T)^2\rightarrow[0,1]}$ (or its matrix form ${ M^{\tilde{P}}}$) is a pseudo-strategy if:
1. ${ M^{\tilde{P}}}$ is positive semidefinite.
2. For any pair of questions ${ q,q'\in S\cup T}$, ${ \sum_{c,c'=1}^k \tilde{P}(c,c'|q,q')=1}$.
3. The blocks ${ M^{\tilde{P}}_{q,q'}}$ on the diagonal are themselves diagonal ${ k}$-by-${ k}$ matrices.
Define the winning probability or value of a pseudo-strategy as: ${ \omega^{\tilde{P}}(G)=\mathbb E_{(s,t)\sim \Pi} \sum_{a,b} \tilde{P}(a,b|s,t)V(a,b|s,t) }$ The algorithm outputs the maximum winning probability: ${ \hat\omega(G)=\max_{\tilde{P}\in\mathcal S}\omega^{\tilde{P}}(G) }$ over pseudo-strategies ${ \tilde{P}\in \mathcal S}$. This maximum is efficiently computable using standard semidefinite programming algorithms. As we will see, actual quantum strategies are in ${ \mathcal S}$ which immediately implies ${ \hat\omega(G)\ge\omega^*(G)}$. It then remains to show that the optimal pseudo-strategy can be approximated by an actual entangled strategy, thus bounding the gap from ${ \omega^*(G)}$ to ${ \hat\omega(G)}$.

Quantum strategies are in ${ \mathcal S}$

Let us establish that ${ \mathcal S}$ is indeed a relaxation of the class of quantum strategies, that is, it contains the quantum strategies. So suppose we are given a quantum strategy. By equation 1 the probability of answers ${ a}$ and ${ b}$ given questions ${ s}$ and ${ t}$ is of the form ${ \langle\phi|A^s_a\otimes B^t_b |\phi\rangle=\langle\phi|(A^s_a\otimes {\mathbb{I}})({\mathbb{I}}\otimes B^t_b )|\phi\rangle }$ for some PVM’s ${ A^s}$ and ${ B^t}$ and some ${ |\phi\rangle\in\mathbb C^d\otimes\mathbb C^d}$. This conditional probability distibution is not immediately in the form of a pseudo-strategy because we cannot evaluate it on pairs of Alice-questions or pairs of Bob-questions. We therefore have to extend it, as follows: Place all the column vectors ${ A^s_a\otimes {\mathbb{I}}|\phi\rangle}$, ${ (s,a)\in S\times \Sigma}$ side by side, and then append the vectors ${ I\otimes B^t_b|\phi\rangle}$, resulting in a ${ d^2}$-by-${ 2nk}$ matrix ${ R}$. We then define ${ {P}(\cdot\cdot|\cdot\cdot):\Sigma^2\times(S\cup T)^2\rightarrow[0,1]}$ through its matrix form (see the comment below definition 5):

$\displaystyle M^{P}:=R^\dag R=\begin{pmatrix} (\langle\phi|A^s_a A^{s'}_{a'}\otimes {\mathbb{I}} |\phi\rangle)_{(s,a),(s',a')} (\langle\phi|A^s_a\otimes B^t_b |\phi\rangle)_{(s,a),(t,b)} \\\\ (\langle\phi|B^t_b\otimes A^s_a |\phi\rangle)_{(t,b),(s,a)} (\langle\phi|{\mathbb{I}}\otimes B^t_bB^{t'}_{b'} |\phi\rangle)_{(t,b),(t',b')} \end{pmatrix} \ \ \ \ \ (2)$

Lemma 7 ${ M^{P}}$ defined in 2 is a pseudo-strategy, that is, ${ M^{P}\in\mathcal S}$.
Proof.
We verify the conditions in definition 6. Condition 1 (${ M^{P}\succeq0}$) holds because it is of the form ${ R^\dag R}$. Condition 2 (Each block ${ M^{P}_{q,q'}}$ sums to ${ 1}$) holds because PVM’s sum to the identity. Condition 3 (Diagonal blocks ${ \tilde M=M^{P}_{qq}}$ are diagonal) holds because the projections in the PVM ${ A^q}$ are mutually orthogonal, hence ${ \langle{\phi}| A^q_cA^q_{c'} |{\phi}\rangle =0}$ if ${ c\neq c'}$.
Lemma 7 means ${ \hat\omega(G)=\max_{{P}\in \mathcal S}\omega^{P}(G)}$ is an (efficiently computable) upper bound for ${ \omega^*(G)}$:

Corollary

${ \hat\omega\ge \omega^*(G)}$.
To finish the proof of theorem 4 we need to show that any pseudo-strategy ${ \tilde{P}}$ can be rounded to an actual quantum strategy with answer probabilities ${ {P}}$ such that

$\displaystyle 1-\omega^{P}\le 6(1-\omega^{\tilde P}) \ \ \ \ \ (3)$

Applying this rounding to the optimal pseudo-strategy implies that ${ 1-\omega^*\le 6(1-\hat\omega) }$ or ${ \omega^*\ge 1-6(1-\hat\omega)}$, which will finish the proof of theorem 4.
Proof.
[Proof of theorem 4] Let ${ \tilde{P}\in\mathcal S}$ be a pseudo-strategy. We construct a quantum strategy ${ {P}}$ approximating ${ \tilde{P}}$. Since ${ M^{\tilde{P}}}$ is positive semidefinite we can write ${ \tilde M=R^\dag R }$ for some matrix ${ R\in \mathbb C^{r\times 2nk}}$ and ${ r\leq 2nk}$. Now let us define ${ 2nk}$ vectors ${ |\tilde u^s_a\rangle}$ and ${ |\tilde v^t_b\rangle}$ in ${ \mathbb C^{r}}$, and let ${ |{u^s_a}\rangle}$ and ${ |{v^t_b}\rangle}$ be the same vectors normalized. The strategy is constructed as follows. Alice and Bob share the maximally entangled state ${ |\phi\rangle=\frac1{\sqrt r}\sum_{i=1}^r|i\rangle|i\rangle\in\mathbb C^r\otimes\mathbb C^r }$ Before deciding on Alice and Bob’s PVM’s ${ A^s}$ and ${ B^t}$ let us see what this choice of shared state means for the conditional distribution on answers (see equation (1)).

$\langle\phi| A^s_a\otimes B^t_b|\phi\rangle =\frac1r\sum_{i,j=1}^r\langle i | A^s_a| j\rangle\langle i | B^t_b| j\rangle =\frac1r A^s_a\cdot B^t_b=\frac1r\langle A^s_a,\overline{B^t_b}\rangle$, (*)

where the bar represents entrywise complex conjugation, ${ \cdot}$ is the entrywise dot product of matrices, and ${ \langle\:,\rangle}$ the entrywise complex inner product (Hilbert-Schmidt inner product). We now choose the measurements. Given question ${ s}$, Alice measures in the PVM ${ A^s=(A^s_a)_{a=0}^k}$ with ${ A^s_a= |{u^s_a}\rangle \langle{u^s_a}| \text{ for }a=1,\ldots,k\text{, and }A^s_0={\mathbb{I}}-\sum_{i=1}^k |{u^s_a}\rangle \langle{u^s_a}| }$ Similarly, Bob on question ${ t}$ applies the PVM ${ B^t}$ with ${ B^t_b= |{v^t_b}\rangle \langle{v^t_b}| \text{ for }b=1,\ldots,k\text{, and }B^t_0={\mathbb{I}}-\sum_{i=1}^k |{v^t_b}\rangle \langle{v^t_b}| }$ The condition 3 in definition 6 ensures that for any question ${ s}$, the vectors ${ |{u^s_1}\rangle ,\ldots, |{u^s_1}\rangle}$ are orthogonal so that this is a valid PVM. The measurement outcome “$0$” is interpreted as “fail”, and upon getting this outcome the player attempts the measurement again on their share of a fresh copy of ${ |\phi\rangle_{AB}}$. This means that the strategy requires many copies of the entangled state to be shared before the game starts. It also leads to the complication of ensuring that with high probability the players measure the same number of times before outputting their measurement, so that the outputs come from measuring the same entangled state. By (*), at a given round of measurements the conditional distribution of answers is given by ${ \langle\phi|A^s_a\otimes B^t_b|\phi\rangle=\frac1r\Big\langle |{u^s_a}\rangle \langle{ u^s_a}| \:,\: |{{v^t_b}\rangle } \langle{ {v^t_b}| }\Big\rangle=\frac1r|\langle {u^s_a}|v^t_b\rangle|^2=\frac1{r|\tilde u^s_a|^2|\tilde v^t_b|^2}\Big(M^{\tilde{P}}_{(s,a),(t,b)}\Big)^2, }$ We wish to relate the LHS to ${ M^{\tilde{P}}_{(s,a),(t,b)}}$, so to handle the factor ${ \frac1r}$ each prover performs repeated measurements, each time on a fresh copy of ${ |\phi\rangle_{AB}}$, until getting an outcome ${ \neq0}$. Moreover, to handle the factor ${ \frac1{|u^s_a|^2|v^t_b|^2}}$, each prover consults public randomness and accepts the answer ${ a\in[k]}$ with probability ${ |u^a_s|^2}$ and ${ |v^b_t|^2}$ respectively, or rejects and start over depending on the public randomness. Under a few simplifying conditions (more precisely, assuming that the game is uniform meaning that an optimal strategy exists where the marginal distribution on each prover’s answers is uniform), we can let ${ M^{\tilde{P}}_{(s,a),(t,b)}\le 1/k}$ for all ${ s,a,t,b}$, and one can ensure that the conditional probabilities ${ {P}}$ of the final answers satisfy

$\displaystyle 1/k-P(a,b|s,t)\le 3\big(1/k-k(M^{\tilde P}_{(s,a),(t,b)})^2\big),\ \ \ \ \ (4)$ At this stage it is important that we are dealing with a unique game . Indeed, by (4) we have for every ${ s}$ and ${ t}$, $1-\sum_{a=1}^k P(a,\pi_{st}(a)|s,t)=\sum_a \Big(\frac{1}{k}-P(a,\pi_{st}(a)|s,t) \Big) \\ \leq 3\sum_{b=\pi_{st}(a)} \Big(\frac{1}{k}-k(M^{\tilde{P}}_{(s,a),(t,b)})^2 \Big) \\ \leq 6\sum_{b=\pi_{st}(a)}\big(\frac{1}{k}-M^{\tilde{P}}_{(s,a),(t,b)}\big)=6\Big(1-\sum_{b=\pi_{st}(a)} M^{\tilde{P}}_{(s,a),(t,b)}\Big)$ where the last inequality follows from concavity. Taking the expectation over ${ s}$ and ${ t}$ implies the bound (3), thus concluding the proof of theorem 4.

General games are hard

We just saw that a specific class of games becomes easy in the presence of shared entanglement, in that semidefinite programming allows the entangled value ${ \omega^*}$ to be approximated to within exponential precision in polynomial time. Does this phenomenon hold more generally, so that the value of entangled games can always be efficiently approximated? We answer in the negative, by constructing a game where ${ \omega^*}$ is NP-hard to approximate to within inverse-polynomial factors. The complexity for 2P-1R entangled games can be strengthened to constant-factor NP-hardness, putting it on par with the classical PCP theorem. This result is used to prove (with some conditions) the games formulation of the quantum PCP theorem, which states that the entangled value of general games is QMA-hard to approximate within a constant factor.

Formulation of game

Given any instance ${ \phi}$ of a ${ k}$-CSP (constraint satisfaction problem, where ${ k}$ is the number of literals), we can define a clause-vs-variable game ${ G_\phi}$ (see clause-vs-variable figure):
1. The referee (verifier) randomly sends a clause to Alice (first prover) and a variable to Bob (second prover).
2. Alice and Bob reply with assignments.
3. The referee accepts if Alice’s assignment satisfies the clause and Bob’s answer is consistent with Alice’s.
To show hardness of approximation, we need to go beyond the usual 2-player construction. In particular, in our game [3] one of the players receives an extra dummy question (see subfigure (a)). Mathematically, the result is very similar to introducing another player and having the referee play the 2-player game with two players chosen randomly [4] (see subfigure (b)). In either variation, the quantum phenomenon of monogamy of entanglement , imposing that only two parties can be maximally entangled to one another, is key to establishing hardness. The players do not know where to use their entanglement, which prevents them from coordinating as well as they could in the standard game.

NP-hardness of approximating the entangled value

To prove hardness, we rely on several results from classical complexity theory.
Theorem 8 ([6]) Given an instance of 1-in-3 3SAT (a CSP), it is NP-hard to distinguish whether it is satisfiable or no assignments satisfy more than a constant fraction of clauses.
Theorem 9 ([2]) For a PCP game ${G}$ (emulating the CSP) and its oracularization ${G'}$ (transformation to a 2P-1R game),

$\displaystyle \omega(G)\leq \omega(G')\leq 1-\frac{1-\omega(G)}{3}\,. \ \ \ \ \ (5)$

Theorem 8 establishes the CSP variant of the classical PCP theorem: distinguishing between ${ \omega(\phi)=1}$ and ${ \omega(\phi)\leq 1/2}$ is NP-hard for some ${ k}$-CSP. Here, ${ \omega(\phi)}$ denotes the maximum fraction of clauses that are simultaneously satisfiable. Theorem 9 relates the general game ${ G}$ obtained from the CSP to a two-player one-round game ${ G'}$, in terms of the value (probability of winning) the game. The first inequality, equivalently saying ${ \omega(\phi)\leq \omega(G_\phi)}$, is achieved since the players can answer the questions in the game ${ G_\phi}$ to satisfy the clauses in ${ \phi}$. These theorems together imply that ${ \omega(G_\phi)}$ is NP-hard to approximate to within constant factors. Allowing the two players to share entanglement can increase the game value to ${ \omega^*(G_\phi)\geq \omega(G_\phi)}$. Classical results do not necessarily carry over, but exploiting monogamy of entanglement allows us to limit the power of entangled strategies. One can show the following lemma, which is weaker than what we have classically.
Lemma 10 ([3]) There exists a constant ${c>0}$ such that for a CSP ${\phi}$,

$\displaystyle \omega^*(G_\phi)\leq 1 - \frac{c(1-\omega(\phi))^2}{n^2}\,, \ \ \ \ \ (6)$

where ${n}$ is the number of variables. Combining Theorem 9 and Lemma 10, we have ${ \omega(\phi)\leq \omega^*(G_\phi)\leq 1 - \frac{c(1-\omega(\phi))^2}{n^2}\,. }$ Using Theorem 8, approximating ${ \omega^*(G_\phi)}$ is NP-hard to within inverse polynomial factors. Proving Lemma 10 takes some work in keeping track of approximations. For simplicity, we will show a less quantitative statement and indicate where the approximations come in.
Proposition 11 (adapted from [13]) ${\phi}$ is satisfiable iff ${\omega^*(G_\phi)=1}$.

Proof of Proposition 11

The forward direction is straightforward: If ${ \phi}$ is satisfiable, then there exists a perfect winning strategy where the questions are answered according to the satisfying assignment. For the reverse direction, suppose there exists a strategy that succeeds with probability 1, specified by a shared entangled state ${ |{\psi}\rangle \in\mathbb{C}^d\otimes\mathbb{C}^d}$ and measurements ${ (A^j_{a_1,\ldots,a_k})_{a_1,\ldots,a_k}}$ for Alice and ${ (B^{t,l}_{b,b'})_{b,b'}}$ for Bob, where the questions ${ j\in[m]}$, ${ t,l\in[k]}$ and the answers ${ a,b}$ are from the CSP’s alphabet. Since one of the questions/answers for Bob corresponds to a dummy variable that is irrelevant to the game, trace over the dummy variable to define a new measurement operator ${ \tilde{B}^t_b=\frac{1}{n}\sum_{l,b'} B^{t,l}_{b,b'}}$. We can introduce a distribution on assignments to the ${ n}$ relevant variables,

$\displaystyle p(a_1,\ldots,a_n)=\lVert \mathbb{I} \otimes \tilde{B}^1_{a_1}\cdots\tilde{B}^n_{a_n} |\psi\rangle \rVert^2\,. \ \ \ \ \ (7)$

If we show that the distribution for assignments ${a_{i_1},\ldots,a_{i_k}}$ on variables ${x_{i_1},\ldots,x_{i_k}}$ in any clause ${C_j}$ is

$\displaystyle p(a_{i_1},\ldots,a_{i_k})= \lVert A^j_{a_{i_1},\ldots,a_{i_k}} \otimes \mathbb{I} |\psi\rangle \rVert^2\,, \ \ \ \ \ (8)$

then, since the players win with certainty, ${\phi}$ has a satisfying assignment. To transform Eq. 7 to Eq. 8, we need a relation between the ${A}$ and ${\tilde{B}}$ measurement operators and a way to commute the ${\tilde{B}}$ operators. The success probability of the players’ strategy is expressed as ${ P=\frac{1}{m}\sum_{j\in[m]}\frac{1}{k}\sum_{i\in C_j} \sum_{(a_1,\ldots,a_k)\vdash C_j} \langle{\psi}| A^j_{a_1,\ldots,a_k}\otimes \tilde{B}^i_{a_i} |{\psi}\rangle \,, }$ where ${ i}$ is the index of one of the ${ k}$ variables on which ${ C_j}$ acts, and ${ (a_1,\ldots,a_k)\vdash C_j}$ indicates that the assignment satisfies the clause. By positivity and summation to identity of the measurement operators ${ A}$ and ${ \tilde{B}}$, each term is at most 1; for our hypothesis ${ P=1}$, each has to be 1. Hence, using orthogonality of the vectors ${ {\mathbb{I}}\otimes\tilde{B}^i_{a_i} |{\psi}\rangle}$ for different ${ a_i}$, we have

$\displaystyle \sum_{\substack{(a_{i_1},\ldots,a_{i_k})\vdash C_j \\ a_i=b}} A^j_{a_{i_1},\ldots,a_{i_k}}\otimes \mathbb{I} |\psi\rangle = \mathbb{I} \otimes \tilde{B}^i_{a_i}|\psi\rangle\,, \ \ \ \ \ (9)$

for any ${ j\in[m]}$ and ${ i\in C_j}$. We now demonstrate that two different ${ \tilde{B}^{t_1}_{b_1}}$, ${ \tilde{B}^{t_2}_{b_2}}$ commute, so that Bob can match any satisfied clause/variable. ${\mathbb{I}} \otimes \tilde{B}^{t_1}_{b_1} \tilde{B}^{t_2}_{b_2} |{\psi}\rangle = {\mathbb{I}} \otimes (\frac{1}{n}\sum_{l_1,b_1'} B^{t_1,l_1}_{b_1,b_1'}) (\frac{1}{n}\sum_{l_2,b_2'} B^{t_2,l_2}_{b_2,b_2'}) |{\psi}\rangle \\ = {\mathbb{I}} \otimes (\frac{1}{n}\sum_{t_2,b_1'} B^{t_1,t_2}_{b_1,b_1'}) (\frac{1}{n}\sum_{t_1,b_2'} B^{t_2,t_1}_{b_2,b_2'}) |{\psi}\rangle \\ = {\mathbb{I}} \otimes \frac{1}{n^2}\sum_{t_1,t_2} B^{t_1,t_2}_{b_1,b_2} |{\psi}\rangle \\ = {\mathbb{I}} \otimes \frac{1}{n^2}\sum_{t_1,t_2} B^{t_2,t_1}_{b_2,b_1} |{\psi}\rangle \\ = {\mathbb{I}} \otimes \tilde{B}^{t_2}_{b_2} \tilde{B}^{t_1}_{b_1} |{\psi}\rangle$ In the second line, we used (9) to relate the measurements. The third line follows by the orthogonality of ${ B^{t_1,t_2}_{b_1,b_2}}$ for different ${ b_1,b_2}$. For the fourth equation, we simply swap ${ t_1,t_2}$ since the questions are indistinguishable to Bob. Thus, we can see how the dummy variable comes into play. If we had assumed ${ P=1-\epsilon}$ and kept track of approximations, we would find

$\displaystyle \frac{1}{n^2}\sum_{t_1,b_1}\sum_{t_2,b_2}\lVert \mathbb{I} \otimes (\tilde{B}^{t_1}_{b_1} \tilde{B}^{t_2}_{b_2} - \tilde{B}^{t_2}_{b_2} \tilde{B}^{t_1}_{b_1}) |\psi\rangle \rVert^2 = O(\epsilon)\,. \ \ \ \ \ (10)$

This approximate commutativity results in the hardness of approximation holding only to within inverse poly${ (n)}$ factors. Now we are ready to transform Eq. 7 to Eq. 8 to conclude the proof. $p(a_1,\ldots,a_n)=\lVert {\mathbb{I}} \otimes \tilde{B}^1_{a_1} \ldots \tilde{B}^n_{a_n} |{\psi}\rangle \rVert^2 \\ =\lVert {\mathbb{I}} \otimes \tilde{B}^{i_1}_{a_{i_1}} \ldots \tilde{B}^{i_k}_{a_{i_k}} |{\psi}\rangle \rVert^2 \\ = \lVert A^j_{a_{i_1},\ldots,a_{i_k}} \otimes {\mathbb{I}} |{\psi}\rangle \rVert^2 \\ =p(a_{i_1},\ldots,a_{i_k})\,.$ In the second line, we used (10) to commute the measurement operators, along with their properties of orthogonality and summation to identity. For the third equality, we used (9) to relate Bob’s measurements to Alice’s, along with orthogonality of ${ A^j_{a_{i_1,\ldots,i_k}}}$ for different ${ a_{i_1,\ldots,i_k}}$.

Constant-factor NP-hardness

The weakness in the above two-player game carries over from the original three-player variant. Thus, to achieve constant-factor NP-hardness of approximation, we could start with a different multiplayer game. Vidick [11] establishes the soundness of the “plane-vs-point” low-degree test (checking that the restriction of a low-degree polynomial to a plane matches its value at some point) in the presence of shared entanglement. Soundness , in the eponymous probabilistically checkable proof (PCP) formulation of the PCP theorem, refers to the verifier accepting a wrong proof with some bounded probability; bounding with a constant maps to constant-factor hardness of approximation. Here, soundness comes from a strong bound on error accumulation, similar to our approximate commutativity, but relies on the players’ Hilbert space being decomposable into three parts (i.e., there being three players). The particular game is constructed by combining the low-degree test with the 3-SAT test (encoding satisfying assignments in a low-degree polynomial), which can be reduced to the three-player QUADEQ test (testing satisfiability of a system of quadratic equations in binary variables, which is NP-complete). By the strong soundness result, the entangled value is NP-hard to approximate to within constant factors. Natarajan et al. [7] show that soundness holds even for two players, using a semidefinite program. They then construct a two-player game in a way similar to what we demonstrated.

Constant-factor QMA-hardness

The above can be thought of as the games formulation of the classical PCP theorem holding under shared entanglement. A true quantum PCP theorem states that the entangled value of general games is QMA-hard to approximate to within constant factors. Natarajan et al. [8] establish such a theorem, but under randomized reductions. This requirement stems from the lack of a sufficiently strong QMA-hardness result for local Hamiltonians (the quantum analog of CSPs). The soundness of the two-player low-degree test above is one instrumental component in the proof.

How much entanglement is needed?

We now focus on the question of quantifying exactly how much entanglement is needed to play XOR games optimally. As we shall see, the answer depends on the size of the question sets posed to Alice & Bob in the game. The previous bound given by Tsirelson [10] (see table below) is tight for certain families of games, but is not tight for other families of games (such as a generalization of the CHSH game). The reason for this discrepancy is closely tied in with the the properties of the representation of the Observables that form the Optimal Strategy (${ |{\psi}\rangle , \{A_{i}\}_{i \in S}, \{B_{j}\}_{j \in T}}$). Slofstra [9] shows that if the Observables constitute a Clifford Algebra (that is, the solutions are pair-wise anti-commutative), then the strategy is minimally entangled (uses the least number of entangled bits) iff the strategy is a unique solution to the SDP rounding problem. As a trivial corollary, if the SDP rounding problem does not have a unique solution (and a correspondingly unique strategy), then there exists a Non-Clifford optimal strategy that uses (atleast) ${ |T|}$ bits of entanglement less than the Clifford strategy. Slofstra further states that minimally entangled ${ \epsilon}$-optimal strategies may be constructed for XOR games where the optimal strategies have ‘stable’ representations. For the purposes of this post, we will analyze the exact result and merely state the approximate result.

Main Results

EXACT

For the exact realm, the table below summarizes Slofstra and Tsirelson’s main results.
Person Strategy Bound(entangled bits)
Slofstra (Possibly) Non-Clifford ${ \log_{2}(N)}$
Tsirelson Clifford ${ \left \lfloor{\frac{r}{2}}\right \rfloor}$
Here, ${ r}$ is the largest integer such that ${ \binom{r + 1}{2} < |S| + |T|}$ and corresponds to the maximum-rank of an extremal point in the quantum correlation matrix corresponding to an optimal strategy. ${ N}$ is the minimum dimension of the representations of the Operators (${ \{B_{j}\}}$).

APPROXIMATE

In the approximate realm, the minimum entanglement dimension of the representation of the Operators from an ${ \epsilon}$-Optimal Strategy is: min(${ \mathcal{O}(\epsilon^{\frac{-1}{12}}), 2^{\left \lfloor{\frac{r}{2}}\right \rfloor}}$). As we shall see, Slofstra’s theorem allows us to recover Tsirelson’s bound easily by using a fact from Representation Theory about the irreducible representations of Clifford Algebras, but stands as a more general lower bound for solutions that aren’t Clifford.

Marginals and Solution Algebras

We’ll begin by introducing 3 key ideas: i) Degeneracy ${ \leftrightarrow}$ Non-Degeneracy ii) Existence of Marginals iii) Solution Algebra Once these ideas are defined and their notions made clear, we will be in a position to state the main result and sketch a proof for it.

Definition (Marginal Strategy)

Given an Optimal Quantum Strategy (${ |{\psi}\rangle , \{A_{i}\}_{i \in S}, \{B_{j}\}_{j \in T}}$), a marginal constitutes ${ \{B_{j}\}_{j \in T}}$, and the partial trace of ${ \psi}$ with respect to ${ H_{A}}$ (${ \rho_{B}}$). It is also possible to dualize the definition for obtaining ${ \{A_{i}\}_{i \in S}}$ and ${ \rho_{A}}$. We now define the notion of degeneracy, which is critical when proving the main theorem. The main point to drive home is that a degenerate optimal quantum strategy can be reduced to a unique, non-degenerate optimal quantum strategy.

Definition (Degenerate Quantum Strategy)

A quantum strategy (${ |{\psi}\rangle , \{A_{i}\}_{i \in S}, \{B_{j}\}_{j \in T}}$) is said to be degenerate if ${ \exists (P \in H_A)}$, ${ (Q \in H_B)}$ such that: i) ${ P}$ commutes with all ${ A_i}$ and ${ (P \otimes {\mathbb{I}}) |{\psi}\rangle = |{\psi}\rangle}$ ii) ${ Q}$ commutes with all ${ B_j}$ and ${ ({\mathbb{I}} \otimes Q) |{\psi}\rangle = |{\psi}\rangle}$ Since we can efficiently construct for any degenerate Optimal Quantum Strategy a unique, non-degenerate Optimal Quantum Strategy, we will now assume WLOG that every Optimal Quantum Strategy is non-degenerate (and unique). We now define the (unique) existence of marginal biases, which correspond to constants for the rows of the quantum correlation matrix (which is a generalization of the classical pay-off). An equivalent statement can be made for columns (${ d_{j}}$) by dualizing the existence of row marginals. These constants can be thought of as representing the (expected) optimum-payoff possible for a set of operator choices by one player, given that the other player’s choice is fixed. Intuitively, this can be seen as “collapsing” the quantum correlation matrix into a column, by summing over the rows (or collapsing into a row, by summing over the columns).
Lemma 12 (Existence of Marginals) For all ${m \times n}$ XOR games G, ${\exists \{c_{i} \geq 0 \hspace{1mm} | \hspace{1mm} i \in |S|\}}$, such that, if ${\{u_{i}\}_{i \in |S|}, \{v_{j}\}_{j \in |T|}}$ form an ${\epsilon}$-optimal vector strategy where ${\epsilon \leq \frac{1}{4(m+n)}}$,

$\displaystyle \|\sum_{j\in|T|}G_{ij}v_{j} - c_{i}u_{i}\| \leq \sqrt{10}(m + n)^{\frac{1}{4}}\epsilon^{\frac{1}{4}}, \forall i \ \ \ \ \ (11)$

If ${ \epsilon = 0}$ and our strategy is perfectly optimal, we recover an exact estimation of the marginal biases:

$\displaystyle \sum_{j\in|T|}G_{ij}v_{j} = c_{i}u_{i}, \forall i \ \ \ \ \ (12)$

The proof for the above lemma provided by Slofstra relies on using techniques to analyze the structure of the SDP program that pertains to quantum marginals. In particular, conducting trace analysis on SDP matrices that correspond to using the game matrix as off-diagonal elements leads us to the construction of the desired marginal biases. It is also critical to note that a dual statement allows us to recover the column biases ${ d_{j} \geq 0}$:

$\displaystyle \sum_{i\in[|S|]}G_{ij}u_{i} = d_{j}v_{j}, \forall j \ \ \ \ \ (13)$

We now move on to defining the notion of a solution algebra.
Definition 13 (Solution Algebra) A solution algebra ${\mathcal{A}}$ consists of self-adjoint (Hermitian) operators ${X_{j}}$ that satisfy the following predicates:

$\displaystyle X_{j}^{2} = \mathbb{I}, \forall 1 \leq j \leq n \ \ \ \ \ (14)$

$\displaystyle (\sum_{j\in[|T|]}G_{ij}X_{j})^{2} = (c_{i})^{2}\cdot\mathbb{I}, \forall i \ \ \ \ \ (15)$

The definition above merely enforces the property that our unknown marginal operators be Hermitian (14) and that they respect the optimal marginal biases (or payoffs) (15) we saw in (11), so that they correspond to being constructed from an optimal vector strategy. These unknown operators will be mapped to operators that are the marginal strategy corresponding to the optimal quantum strategy. This is at the heart of the main theorem we will now state: Theorem 14 (Slofstra, 2010) Given a ${m \times n}$ XOR game G (with no zero rows or columns) and a solution algebra ${\mathcal{A}}$, a collection of Linear Operators ${\{B_{j}\}}$ and density matrix ${\rho}$ are the marginal of an optimal strategy iff the map ${X_{j} \rightarrow B_{j}}$ induces a density-matrix representation of ${\mathcal{A}}$ and ${\rho}$ commutes with ${im(\mathcal{A})}$. Put simply, the theorem states that our unknown self-adjoint operators map to an optimal marginal strategy iff the density matrix (traced from the joint Hilbert-Space) commutes with all the mapped operators (${ \{B_{j}\}}$). The result we desire on the lower bound for the number of entangled bits, given a mapping from these indeterminate operators to the marginal of an optimal strategy, comes from a corollary to (14).
Corollary 15 Given a ${m \times n}$ XOR game G (with no zero rows or columns) and a solution algebra ${\mathcal{A}}$ with minimum dimension ${N}$ among non-zero representations, the strategy for minimum entanglement uses ${\log_{2}(N)}$ entangled bits. The proof for this corollary follows from the eigenspace decomposition of the joint Hilbert Space ${ H}$ in terms of ${ \rho}$, which is preserved by the action of ${ im(\mathcal{A})}$. As a result, each eigenspace decomposes into a finite sum of irreducible representations of ${ \mathcal{A}}$. The minimum entanglement is realized when there is exactly one invariant subspace (with one irreducible representation). The entanglement used by such a representation is ${ \log_{2}(\text{dim}\,H)}$.

Proof of Theorem 20

The rest of the section is dedicated to sketching a brief (but formal) proof for Theorem (14), and then using a simple fact about the representations of a Clifford Algebra to show how Slofstra’s result subsumes Tsirelson’s bound. For this section, ${ |{\psi}\rangle}$ refers to an arbitrary state in ${ H = H_{A} \otimes H_{B}}$ (the joint Hilbert space). We can write ${ |{\psi}\rangle = \sum_{i} |{i}\rangle \lambda |{i}\rangle}$ over some basis ${ \{{i}\}}$, where ${ \lambda}$ is a linear map. Then, the partial trace of ${ \psi}$ over ${ H_{A}}$ is given by ${ \rho = \lambda\lambda^{*}}$. Let ${ \mathcal{B}_{A}, \mathcal{B}_{B}}$ denote the algebra generated by ${ A_{1},..,A_{m}}$ and ${ B_{1},..,B_{n}}$. Here, the generating elements are the observables of an optimal quantum strategy. To arrive at a proof for the theorem, we will rely on 2 additional lemmas which we will not prove but state.
Lemma 16 Given Hermitian operators ${A}$, ${B \in H_{A}, H_{B}}$,

$\displaystyle \|(A \otimes \mathbb{I} - \mathbb{I} \otimes B)|\psi\rangle\| \leq \|\lambda\overline{A} - B\lambda\|_{F} \ \ \ \ \ (16)$

This allows us to conclude that,

$\displaystyle \|(A \otimes \mathbb{I} - \mathbb{I} \otimes B)|\psi\rangle\| \leq \epsilon \implies \|\rho(B) - B\rho\|_{F} \leq 2\epsilon \ \ \ \ \ (17)$

Lemma 17 The optimal strategy in question is non-degenerate iff

$\displaystyle closure(\mathcal{B}_{B}\lambda H_{A}) = closure(\mathcal{B}_{A}\lambda^{*}H_{B}). \\ \ \ \ \ \ (18)$

As a special case:

$\displaystyle closure(\mathcal{B}_{B}\lambda H_{A}) = H_{B} \leftrightarrow closure(\rho H_{B}) = H_{B} \ \ \ \ \ (19)$

Forward direction : We use the first lemma to prove commutativity of $\rho$ with all $B_{j}$, and we use the second lemma to show that the closure of $\rho$ is ${ H_{B}}$. We first show the forward direction: Suppose we are given an optimal quantum strategy ($|{\psi}\rangle , \{A_{i}\}_{i \in S}, \{B_{j}\}_{j \in T}$) for a game ${ G}$. Then, we fix our optimal vector strategy as:

$\displaystyle u_{i} = (A_{i} \otimes \mathbb{I})|\psi\rangle \ \ \ \ \ (20)$

$\displaystyle v_{j} = (\mathbb{I} \otimes B_{j})|\psi\rangle \ \ \ \ \ (21)$

We can now use Equations (11) and (13) to establish our optimal marginal biases to write a relationship between them and ${ |{\psi}\rangle}$ , and apply Lemma (16) to show commutativity and Lemma (17) to show that ${ im(\mathcal{A})}$ = cyclic(${ B_{j}, \rho}$). Using (13) with (20) and (21), we have:

$\displaystyle d_{j}(\mathbb{I}\otimes B_{j})|\psi\rangle = \sum_{i}G_{ij}(A_{i}\otimes\mathbb{I})|\psi\rangle \ \ \ \ \ (22)$

We can now use (17) with ${ \epsilon = 0}$ on (22) to see that ${ \rho}$ commutes with every ${ B_{j}}$. Additionally, as the terms in (22) constitute linear combinations of ${ A_{i}}$ and ${ B_{j}}$, we can compute the closure of their actions on ${ \lambda H_{A}}$ and ${ \lambda^{*} H_{B}}$, which will be equivalent. Therefore, ${ im(\rho)}$ = ${ H_{B}}$, which follows from the special case of (19). For the dual case, we substitute (20) and (21) into (12):

$\displaystyle c_{i}(A_{i}\otimes\mathbb{I})|\psi\rangle = \sum_{j}G_{ij}(\mathbb{I}\otimes B_{j})|\psi\rangle$

On taking the norm of the above on both sides and using a little algebra, we finally obtain the fact that ${ \{B_{j}\}}$ satisfy predicate (15) making them the representations of ${ X_{j}}$: ${ (\sum_{j}G_{ij}B_{j})^{2} = c_{i}^{2}{\mathbb{I}} }$ This shows that the map from ${ X_{j} \rightarrow B_{j}}$ computes a density matrix representation of ${ \mathcal{A}}$, where ${ \rho}$ commutes with all ${ B_{j}}$.
Backward Direction : The proof for the backward direction is much less involved: If we knew that ${ \{B_{j}\}, \rho}$ constituted the cyclic representation of ${ \mathcal{A}}$ with commutativity (with ${ B_{j}}$), then we can use Lemma (19) to conclude that the image of ${ \rho}$ on ${ H}$ would form a subspace of ${ \lambda H}$. We define: ${ \overline{A}_{i} = \frac{\sum_{j}G_{ij}B_{j}}{c_{i}} }$ allowing us to recover our original marginal biases ${ c_{i}}$ that satisfy (23) and therefore correspond to the optimal strategy. This shows us that ${ \{A_{i}\}}$, ${ \{B_{j}\}}$ would constitute an optimal quantum strategy.
Having proved this theorem, we now obtain Corollary 15, which is the main desired result. To see how it subsumes Tsirelson’s result as a special case, we use a simple fact from Representation Theory:
Lemma 18 For a Clifford Algebra generated by ${X_{1},..,X_{r}}$, there exist one or two irreducible representations of dimension ${2^{\left \lfloor{\frac{r}{2}}\right \rfloor}}$ Plugging Lemma 18 into Corollary 15, we simply recover the fact that the number of entangled bits of a solution algebra that is Clifford is ${ \left \lfloor{\frac{r}{2}}\right \rfloor}$. However, note that being Clifford means an extra constraint: ${ X_{i}X_{j} = -X_{j}X_{i}}$, ${ \forall i, j}$ The constraints on the Solution Algebra (14), (15) given by Slofstra do \textit{not} necessarily mean that the solution is Clifford. In fact, when an optimal quantum strategy with minimal entanglement is Clifford, ${ \{A_{i}\}}$, ${ \{B_{j}\}}$ are constructed from a unique set of ${ \{u_{i}\}}$, ${ \{v_{j}\}}$. To end, we write down a lemma that shows there exist XOR games where the optimal strategy is not unique and for minimal entanglement, a solution generated by a Non-Clifford algebra must be used:

Lemma (Existence of XOR games with Non-Clifford optimal strategies)

There exist a family of ${ m \times n}$ XOR games ${ \{G\}}$ that correspond to generalizations of the CHSH games (${CL_{n}}$), such that, the optimal strategy of minimal entanglement is Non-Clifford.

References

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author: Beatrice Nash

Abstract

In this blog post, we give a broad overview of quantum walks and some quantum walks-based algorithms, including traversal of the glued trees graph, search, and element distinctness [3; 7; 1]. Quantum walks can be viewed as a model for quantum computation, providing an advantage over classical and other non-quantum walks based algorithms for certain applications.

Continuous time quantum walks

We begin our discussion of quantum walks by introducing the quantum analog of the continuous random walk. First, we review the behavior of the classical continuous random walk in order to develop the definition of the continuous quantum walk.

Take a graph $G$ with vertices $V$ and edges $E$. The adjacency matrix $A$ of $G$ is defined as follows:

$A_{i,j} = \begin{cases} 1 \quad &\text{if } (i,j) \in E \\ 0 \quad &\text{otherwise}. \end{cases}$

And the Laplacian $L$ is given by:

$L_{i,j} = \begin{cases} -\text{degree}(i) \quad &\text{if } i = j \\ 1 \quad &\text{if } (i,j) \in E \\ 0 \quad &\text{otherwise}. \end{cases}$

The Laplacian determines the behavior of the classical continuous random walk, which is described by a length $|V|$ vector of probabilities, p(t). The $i$th entry of p(t) represents the probability of being at vertex $i$ at time $t$. p(t) is given by the following differential equation:

\begin{aligned} \frac{\text{d}}{\text{dt}} \text{p}_{i}(\text{t}) = \underset{(i,j) \in E}{\sum} L_{i,j} \text{p}_{j}(\text{t}),\end{aligned}

which gives the solution $\textbf{p}(t) = e^{Lt}\textbf{p}(0)$.

Recalling the Schrödinger equation $i \frac{\text{d}}{\text{dt}} \left|\psi\right>= H \left|\psi\right>$, one can see that by inserting a factor of $i$ on the left hand side of the equation for p(t) above, the Laplacian can be treated as a Hamiltonian. One can see that the Laplacian preserves the normalization of the state of the system. Then, the solution to the differential equation:

\begin{aligned} i \frac{\text{d}}{\text{dt}} \psi_{i}(\text{t}) = \underset{(i,j) \in E}{\sum} L_{i,j} \psi_{j}(\text{t})\end{aligned},

which is $\left|\psi(t)\right> = e^{-iLt} \left|\psi(0)\right>$, determines the behavior of the quantum analog of the continuous random walk defined previously. A general quantum walk does not necessarily have to be defined by the Laplacian; it can be defined by any operator which “respects the structure of the graph,” that is, only allows transitions to between neighboring vertices in the graph or remain stationary [7]. To get a sense of how the behavior of the quantum walk differs from the classical one, we first discuss the example of the continuous time quantum walk on the line, before moving on to the discrete case.

Continuous time quantum walk on the line

An important example of the continuous time quantum walk is that defined on the infinite line. The eigenstates of the Laplacian operator for the graph representing the infinite line are the momentum states with eigenvalues $2(\text{cos}(p) - 1)$, for $p$ in range $[-\pi,\pi]$. This can be seen by representing the momentum states in terms of the position states and applying the Laplacian operator:

\begin{aligned} \left|p\right> &= \underset{x}{\sum} e^{ipx} \left|x\right> \\ L\left|p\right> &= \underset{x}{\sum} e^{ipx} \left|x+1\right>+ e^{ipx} \left|x-1\right> - 2e^{ipx} \left|x\right> \\ &= \underset{x}{\sum} (e^{ip} + e^{-ip} - 2) e^{ipx} \left|x\right> \\ &= 2(\text{cos}(p) - 1) \left|p\right>.\end{aligned}

Hence the probability distribution at time $t$, $p(x,t) = |\left< x\right| e^{-iLt} \left|\psi(0)\right> | ^{2}$, with initial position $\left|\psi(0)\right> = \left|0\right>$ is given by:

\begin{aligned} |\left< x\right| e^{-iLt} \left|0\right> | ^{2} &= \bigg| \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-2it(\text{cos}p - 1)} \left< x|p\right> \text{d}p \bigg|^{2} \\ &= \bigg| \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-2it(\text{cos}p - 1)} e^{ipx} \text{d}p \bigg|^{2} \\ &= | J_{x}(2t) |^{2}.\end{aligned}

While the probability distribution for the classical continuous time
random walk on the same graph approaches, for large $t$, $\frac{1}{\sqrt{4\pi t}} e^{\frac{-x^{2}}{4t}}$, or a Gaussian of width $2\sqrt{t}$. One can see that the quantum walk has its largest peaks at the extrema, with oscillations in between that decrease in amplitude as one approaches the starting position at $x=0$. This is due to the destructive interference between states of different phases that does not occur in the classical case. The probability distribution of the classical walk, on the other hand, has no oscillations and instead a single peak centered at $x=0$, which widens and flattens as $t$ increases.

Walk on the glued trees graph

A glued tree is a graph obtained by taking two binary trees of equal height and connecting each of the leaves of one of the trees to exactly two leaves of the other tree so that each node that was a leaf in one of the original trees now has degree exactly $3$. An example of such a graph is shown in Figure 2.

The time for the quantum walk on this graph to reach the right root from the left one is exponentially faster than in the classical case. Consider the classical random walk on this graph. While in the left tree, the probability of transitioning to a node in the level one to the right is twice that of transitioning to a node in the level one to the left. However, while in the right tree, the opposite is true. Therefore, one can see that in the middle of the graph, the walk will get lost, as, locally, there is no way to determine which node is part of which tree. It will instead get stuck in the cycles of identical nodes and will have exponentially small probability of reaching the right node.

To construct a continuous time quantum walk on this graph, we consider the graph in terms of columns. One can visualize the columns of Figure 2 as consisting of all the nodes equidistant from the entrance and exit nodes. If each tree is height $n$, then we label the columns $0,1,\text{...},2n,2n+1$, where column $i$ contains the nodes with shortest path of length $i$ from the leftmost root node. We describe the state of each column as a superposition of the states of each node in that column. The number of nodes in column $i$, $N_{i}$, will be $2^{i}$ for $i \in [0,n]$ and $2^{2n+1-i}$ for $i \in [n+1,2n+1]$. Then, we can define the state $\left|\text{col} \; i\right>$ as:

\begin{aligned} \left|\text{col} \; i\right> = \frac{1}{\sqrt{N_{i}}} \underset{j \in \text{col} \; i}{\sum} \left|j\right>.\end{aligned}

The factor of $\frac{1}{\sqrt{N_{i}}}$latex ensures that the state is normalized. Since the adjacency matrix $A$ of the glued tree is Hermitian, then we can treat $A$ as the Hamiltonian of the system determining the behavior of the quantum walk. By acting on this state with the adjacency matrix operator $A$, we get the result (for $i \in [1,n-1]$):

\begin{aligned} A\left|\text{col} \; i\right> &= 2\frac{\sqrt{N_{i-1}}}{\sqrt{N_{i}}} \left|\text{col} \; i-1\right> + \frac{\sqrt{N_{i+1}}}{\sqrt{N_{i}}} \left|\text{col} \; i+1\right> \\ &= \sqrt{2} \left|\text{col} \; i-1\right> + \sqrt{2} \left|\text{col} \; i+1\right>.\end{aligned}

Then for $i \in [n+2,2n]$, we get the same result, because of symmetry.

For $i = n$:

\begin{aligned} A\left|\text{col} \; n\right> = \sqrt{2} \left|\text{col} \;n-1\right> + 2 \left|\text{col} \; n\right>.\end{aligned}

The case of $i = n+1$ is symmetric. One can see that the walk on this graph is equivalent to the quantum walk on the finite line with nodes corresponding to the columns. All of the edges, excluding that between columns $n$ and $n+1$, have weight $\sqrt{2}$. The edge between column $n$ and $n+1$ has weight $2$.

The probability distribution of the quantum walk on this line can be roughly approximated using the infinite line. In the case of the infinite line, the probability distribution can be seen as a wave propagating with speed linear in the time $t$. Thus, in time linear in $n$, the probability that the state is measured at distance $2n+1$ from the starting state is $\frac{1}{\text{poly} \; n}$. In [3] it is shown that the fact that the line is finite and has a single differently weighted edge from the others (that between $n$ and $n+1$) does not change the fact that in polynomial time, the quantum walk will travel from the left root node to the right one, although in this case there is no limiting distribution as the peaks oscillate. This was the first result that gives an exponential speed up over the classical case using quantum walks.

Discrete time quantum walks

In this section, we will first give an introduction to the discrete quantum walk, including the discrete quantum walk on the line and the Markov chain quantum walk, as defined in [7]. Next, we discuss how Grover search can be viewed as a quantum walk algorithm, which leads us into Ambainis’s quantum-walks based algorithm from [1] for the element distinctness problem, which gives a speed up over classical and other quantum non-walks based algorithms.

The discrete time quantum walk is defined by two operators: the coin flip operator, and the shift operator. The coin flip operator $C$ determines the direction of the walk, while the shift operator $S$ makes the transition to the new state conditioned on the result of the coin flip. The Hilbert space governing the walk is $\mathcal{H} = \mathcal{H}{C} \otimes \mathcal{H}{S}$, where $\mathcal{H}{C}$ corresponds to the space associated with the result of the coin flip operator, and $\mathcal{H}{S}$ corresponds to the locations in the graph on which the walk is defined.

For example, consider the discrete time walk on the infinite line. Since there are two possible directions (left or right), then the Hilbert space associated with the coin flip operator is two dimensional. In the unbiased case, the coin flip is the Hadamard operator,

\begin{aligned} H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix},\end{aligned}

and shift operator that produces the transition from state $\left|j\right>$ to $\left|j+1\right>$ or $\left|j-1\right>$,
conditioned on the result of the coin flip, is $S = \left|0\right>\left< 0\right| \otimes \underset{j}{\sum} \left|j+1\right> \left< j\right| + \left|1\right>\left< 1\right| \otimes \underset{j}{\sum} \left|j - 1\right> \left< j\right|$.

Each step of the walk is determined by an application of the unitary
operator $U = S \cdot (H \otimes I)$. If the walk starts at position
$\left|x\right>$, then measuring the state after one application of $U$ gives $\left|x+1\right>$ with probability $\frac{1}{2}$ and $\left|x-1\right>$ with probability $\frac{1}{2}$. This is exactly the same as the case of the classical random walk on the infinite line; the difference between the two walks becomes apparent after a few steps.

For example, the result of the walk starting at state $\left|\psi(0)\right> = \left|0\right>\left|0\right>$ after 4 steps gives:

\begin{aligned} \left|\psi(1)\right> &= \frac{1}{\sqrt{2}}\left( \left|0\right>\left|1\right> + \left|1\right>\left|-1\right> \right) \\ \left|\psi(2)\right> &= \frac{1}{2}\left( \left|0\right>\left|2\right> + \left|1\right>\left|0\right> + \left|0\right>\left|0\right> - \left|1\right>\left|-2\right> \right) \\ \left|\psi(3)\right> &= \frac{1}{2\sqrt{2}}\left(\left|0\right>\left|3\right> + \left|1\right>\left|1\right> + 2\left|0\right>\left|1\right> -\left|0\right>\left|-1\right> + \left|1\right>\left|-3\right> \right) \\ \left|\psi(4)\right> &= \frac{1}{4} (\left|0\right>\left|4\right> + \left|1\right>\left|2\right> + 3\left|0\right>\left|2\right> + \left|1\right>\left|0\right> -\left|0\right>\left|0\right> -\left|1\right>\left|-2\right> +\left|0\right>\left|-2\right>-\left|1\right>\left|-4\right>).\end{aligned}

One can see that the distribution is becoming increasingly skewed
towards the right, while in the classical case the distribution will be
symmetric around the starting position. This is due to the destructive
interference discussed earlier. The distribution after $t = 20$ time
steps is shown in Figure 4.

Now, consider the walk starting at state $\left|\psi(0)\right> = -\left|1\right>\left|0\right>$:

\begin{aligned}\left|\psi(1)\right> &= \frac{1}{\sqrt{2}}\left( -\left|0\right>\left|1\right> + \left|1\right>\left|-1\right> \right) \\ \left|\psi(2)\right> &= \frac{1}{2}\left( -\left|0\right>\left|2\right> - \left|1\right>\left|0\right> + \left|0\right>\left|0\right> - \left|1\right>\left|-2\right> \right) \\ \left|\psi(3)\right> &= \frac{1}{2\sqrt{2}}\left(-\left|0\right>\left|3\right> - \left|1\right>\left|1\right> + 2\left|1\right>\left|-1\right> - \left|0\right>\left|-1\right> + \left|1\right>\left|-3\right> \right) \\ \left|\psi(4)\right> &= \frac{1}{4} (-\left|0\right>\left|4\right> -\left|1\right>\left|2\right> - \left|0\right>\left|2\right> + \left|1\right>\left|0\right> + \left|0\right>\left|0\right> -3\left|1\right>\left|-2\right> + \left|0\right>\left|-2\right> - \left|1\right>\left|-4\right>).\end{aligned}

This distribution given by this walk is the mirror image of the first.
To generate a symmetric distribution, consider the start state $\left|\psi(0)\right> = \frac{1}{\sqrt{2}}(\left|0\right> -i\left|1\right>)\left|0\right>$. The resulting distribution after $t$ steps will be $p(x,t) = \frac{1}{2} p_{0}(x,t) + \frac{1}{2} p_{1}(x,t)$, where $p_{0}(x,t)$ is the probability distribution after $t$ steps resulting from the start state $\psi(0) = \left|0\right>\left|0\right>$ and $p_{1}(x,t)$ is the probability distribution after $t$ steps resulting from the start state $\psi(0) = -\left|1\right>\left|0\right>$. The result will be symmetric, with peaks near the extrema, as we saw in the continuous case.

Markov chain quantum walk

A reversible, ergodic Markov chain with $n$ states can be represented by a $n \times n$ transition matrix $P$ with $P_{j,i}$ equal to the probability of transitioning from state $i$ to state $j$ and $P = P^{*}$. Then, $p_{0}P$, where $p_{0}$ is an initial probability distribution over the states, gives the distribution after one step.
Since $\sum_{j} P_{i,j} = 1$ for all $i$, $P$ is stochastic and thus preserves normalization.

There are multiple ways to define a discrete quantum walk, depending on the properties of the transition matrix and the graph on which it is defined (overview provided in [4]). Here we look at the quantum walk on a Markov chain as given in [2]. For the quantum walk on this graph, we define state $\left|i\right>\left|j\right>$ as the state that represents currently being at position $j$ and facing in the direction of $i$. Then, we define the state $\left|\psi_{j}\right>$ as a superposition of the states associated with position $j$:

\begin{aligned} \left|\psi_{i}\right> = \underset{j}{\sum} \sqrt{P_{j,i}} \left|i\right>\left|j\right>.\end{aligned}

The unitary operator,

$D = 2 \underset{i}{\sum} \left|\psi_{i}\right>\left< \psi_{i}\right| - I$,

acts as a coin flip for the walk on this graph. Since $P$ is reversible, we can let the shift operator be the unitary $SWAP$ operator:

$SWAP = \underset{i,j}{\sum} \left|i,j\right>\left< j,i\right|$.

A quantum walk can also be defined for a non-reversible Markov chain using a pair of reflection operators (the coin flip operator is an example of a reflection operator). This corresponds to the construction given in [7].

Search as a quantum walk algorithm

Given a black box function $f$ and a set of inputs $S$ with $|S| = N$, say we want to find whether an input $x \in S$ exists for which $f(x)$ equals some output value. We refer to the set of inputs $M$ for which this is true as marked. Classically, this requires $O(N/|M|)$ queries, for nonempty $M$. Using the Grover search algorithm, this problem requires $O(\sqrt{N/|M|})$ quantum queries. In this section, we give a quantum walks based algorithm that also solves this problem in $O(\sqrt{N/|M|})$ time. If we define a doubly stochastic matrix $P$ with uniform transitions, then we can construct a new transition matrix $P'$ from $P$ as:

$P_{i,j}' = \begin{cases} \frac{1}{N-1} \quad &\text{if } i \neq j \text{ and } i \notin M \\0 \quad &\text{if } i = j \text{ and } i \notin M \\ 1 \quad &\text{if } i = j \text{ and } i \in M \\ 0 \quad &\text{if } i \neq j \text{ and } i \in M. \end{cases}$

Then, when the state of the first register is unmarked, the operator $D$ defined in the previous section acts as a diffusion over its neighbors. When the state in the first register is marked, then $D$ will act as the operator $-I$, and the walk stops, as a marked state has been reached. This requires two queries to the black box function: one to check whether the input is marked, and then another to uncompute. By rearranging the order of the columns in $P$ so that the columns corresponding to the non-marked elements come before the columns corresponding to the marked elements, we get:

\begin{aligned} P' = \begin{pmatrix} P_{0} & 0 \\ P_{1} & I \end{pmatrix},\end{aligned}

where $P_{0}$ gives the transitions between non-marked elements and $P_{1}$ gives the transitions from non-marked to marked elements.

We now look at the hitting time of the classical random walk. Assume
that there is zero probability of starting at a marked vertex. Then, we
can write the starting distribution $p$, where the last $|M|$ elements of $p$, corresponding to the marked elements, are zero, as
$p = \underset{\lambda}{\sum} \alpha_{\lambda} \left|\lambda\right>$, where $\lambda$ are the eigenvalues of $P_{0}$, and $\left|\lambda\right>$ are the corresponding eigenvectors, with the last $|M|$ entries zero. Let $\lambda^{}$ be the principal (largest) eigenvalue. Then, the probability that, after $t$ steps, a marked element has not yet been reached will be $\sum (P_{0}^{t}p)_{i} \leq \lambda^{*t}$. Then, the
probability that a marked element has been reached in that time will be
$\geq 1 - \lambda^{t} \geq 1 - t \lambda^{*}$. Setting
$t = \frac{1}{1-\lambda^{*}}$ gives probability $\Omega(1)$ that a marked element will be reached in that time.

The eigenvalues of $P_{0}$ will be $\frac{N-|M|-1}{N-1}$ and
$\frac{-1}{N-|M|-1}$. Then, the classical hitting time will be:

\begin{aligned} t &= \frac{1}{1-\lambda^{*}} \\ &= \frac{1}{1-\frac{N-|M|-1}{N-1}} \\ &= O\left(\frac{N}{|M|}\right).\end{aligned}

It can be showed that for a walk defined by a Markov chain, the
classical hitting time will be $O(\frac{1}{\delta \epsilon})$, where $\delta = 1 - \lambda^{*}$, the spectral gap, and $\epsilon \leq \frac{|M|}{N}$ [2].

Magniez et al proved in [6] that for a reversible, ergodic
Markov chain, the quantum hitting time for a walk on this chain is
within a factor of the square root of the classical hitting time. Since
the walk on this input acts as a walk on a reversible Markov chain until
a marked element is reached, then this is also true for a walk defined
by our transition matrix $P'$. This arises from the fact that the
spectral gap of the matrix describing the quantum walk corresponding to
stochastic matrix $P$ is quadratically larger than the spectral gap of
the matrix describing the classical random walk corresponding to $P$, the proof of which is given in [2]. Thus, the quantum hitting time
is $O(\sqrt{N/|M|})$, which exactly matches the quantum query complexity of Grover search.

Element distinctness problem

Now, we describe Ambainis’s algorithm given in [1] for solving
the element distinctness problem in $O(N^{\frac{2}{3}})$ time, which
produces a speed up over the classical algorithm, which requires $O(N)$ queries, and also over other known quantum algorithms that do not make use of quantum walks, which require $O(N^{\frac{3}{4}})$ queries. The element distinctness problem is defined as follows: given a function $f$ on a size $N$ set of inputs

$S=\{x_{1}$,…,$x_{N}\}$,

determine whether there exists a pair $x_{1},\; x_{2} \in S$ for which $f(x_{1}) = f(x_{2})$. As in the search problem defined in the previous section, this is a decision problem; we are not concerned with finding the values of these pairs, only whether at least one exists.

The algorithm is similar to the search algorithm described in the previous section, except we define the walk on a Hamming graph. A Hamming graph $H(N,m)$ is defined as follows: each vertex $i$ corresponds to an $m$-tuple, ($i_{1}$,…,$i_{m}$), where $i_{k} \in S$ for all $k$ and repetition is allowed (that is, $i_{k}$ may equal $i_{j}$ for $k \neq j$), and $m$ is a parameter we will choose. Edges will exist between vertices that differ in exactly one coordinate (order matters in this graph). We describe the state of each vertex as:

$\left|i \right>=| i_{1},i_{2}$,…,$i_{m},f(i_{1})$,…,$f(i_{m})>$

Then, moving along each edge that replaces the $j$th coordinate with $x_{k}$ such that $i_{j} \neq x_{k}$ requires two queries to the black box function to erase $f(i_{j})$ and compute $f(x_{k})$. In the case, the marked vertices will be those that contain some $f(i_{k}) = f(i_{j})$ for $i_{j} \neq i_{k}$. Since the function values are stored in the description of the state, then no additional queries to the black box are required to check if in a marked state.

The transition matrix is given by $P = \frac{1}{m(n-1)} \underset{i \in [1,m]}{\sum} (J - I)^{(i)}$. $J$ is the $n \times n$ all one matrix, and the superscript $i$ denotes the operator acting on the $i$th coordinate. The factor of $\frac{1}{m(n-1)}$ normalizes the degree, since the graph is regular. We can compute the spectral gap of this graph to be $\frac{n}{m(n-1)}$ (for details of this computation, see [2]). Then, noting that that the fraction of marked vertices, $\epsilon$, is
$\geq \frac{m(m-1)(n-2)^{m-2}}{n^{m}}$, classically, the query complexity is $m + O(\frac{1}{\delta \epsilon}) = m + O(\frac{n^{2}}{m})$, where $m$ is the queries required to construct the initial state. Setting the parameters equal to minimize with respect to $m$ gives classical query complexity $O(N)$, as expected.

Then in the quantum case, $m$ queries are still required to set up the state. $O(\frac{n}{\sqrt{m}})$ queries are required to perform the walk until a marked state is reached, by [6]. Setting parameters equal gives $O(N^{\frac{2}{3}})$ queries, as desired.

[1] Ambainis, A. Quantum walk algorithm for element distinctness, SIAM Journal on Computing 37(1):210-239 (2007). arXiv:quant-ph/0311001

[2] Childs, A. Lecture Notes on Quantum Algorithms (2017). https://www.cs.umd.edu/ amchilds/qa/qa.pdf

[3] Childs, A., Farhi, E. Gutmann, S. An example of the difference between
quantum and classical random walks. Journal of Quantum Information
Processing, 1:35, 2002. Also quant-ph/0103020.

[4] Godsil, C., Hanmeng, Z. Discrete-Time Quantum Walks and Graph Structures
(2018). arXiv:1701.04474

[5] Kempe, J. Quantum random walks: an introductory overview, Contemporary
Physics, Vol. 44 (4) (2003) 307:327. arXiv:quant-ph/0303081

[6] Magniez, F., Nayak, A., Richter, P.C. et al. On the hitting times of
quantum versus random walks, Algorithmica (2012) 63:91.
https://doi.org/10.1007/s00453-011-9521-6

[7] Szegedy, M. Quantum Speed-up of Markov Chain Based Algorithms, 45th
Annual IEEE Symposium on Foundations of Computer Science (2004).
https://ieeexplore.ieee.org/abstract/document/1366222

[8] Portugal, R. Quantum Walks and Search Algorithms. Springer, New York, NY (2013).

By Abhijit Mudigonda, Richard Wang, and Lisa Yang

This is part of a series of blog posts for CS 229r: Physics and Computation. In this post, we will talk about progress made towards resolving the quantum PCP conjecture. We’ll briefly talk about the progression from the quantum PCP conjecture to the NLTS conjecture to the NLETS theorem, and then settle on providing a proof of the NLETS theorem. This new proof, due to Nirkhe, Vazirani, and Yuen, makes it clear that the Hamiltonian family used to resolve the NLETS theorem cannot help us in resolving the NLTS conjecture.

Introduction

We are all too familiar with NP problems. Consider now an upgrade to NP problems, where an omniscient prover (we’ll call this prover Merlin) can send a polynomial-sized proof to a BPP (bounded-error probabilistic polynomial-time) verifier (and we’ll call this verifier Arthur). Now, we have more decision problems in another complexity class, MA (Merlin-Arthur). Consider again, the analogue in the quantum realm where now the prover sends over qubits instead and the verifier is in BQP (bounded-error quantum polynomial-time). And now we have QMA (quantum Merlin-Arthur).

We can show that there is a hierarchy to these classes, where NP $\subseteq$ MA $\subseteq$ QMA.

Our goal is to talk about progress towards a quantum PCP theorem (and since nobody has proved it in the positive or negative, we’ll refer to it as a quantum PCP conjecture for now), so it might be a good idea to first talk about the PCP theorem. Suppose we take a Boolean formula, and we want to verify that it is satisfiable. Then someone comes along and presents us with a certificate — in this case, a satisfying assignment — and we can check in polynomial time that either this is indeed a satisfying assignment to the formula (a correct certificate) or it is not (an incorrect certificate).

But this requires that we check the entire certificate that is presented to us. Now, in comes the PCP Theorem (for probabilistically checkable proofs), which tells us that a certificate can be presented to us such that we can read a constant number of bits from the certificate, and have two things guaranteed: one, if this certificate is correct, then we will never think that it is incorrect even if we are not reading the entire certificate, and two, if we are presented with an incorrect certificate, we will reject it with high probability [1].

In short, one formulation of the PCP theorem tells us that, puzzingly, we might not need to read the entirety of a proof in order to be convinced with high probability that it is a good proof or a bad proof. But a natural question arises, which is to ask: is there a quantum analogue of the PCP theorem?

Progress

The answer is, we’re still not sure. But to make progress towards resolving this question, we will present the work of Nirkhe, Vazirani, and Yuen in providing an alternate proof of an earlier result of Eldar and Harrow on the NLETS theorem.

Before we state the quantum PCP conjecture, it would be helpful to review information about local Hamiltonians and the $k$-local Hamiltonian problem. A previous blog post by Ben Edelman covers these topics. Now, let’s state the quantum PCP conjecture:

(Quantum PCP Conjecture): It is QMA-hard to decide whether a given local Hamiltonian $H = H_{1} + ... + H_{m}$ (where each $||H_{i}|| \leq 1$) has ground state energy at most $a$ or at least $b$ when $b-a \geq c||H||$ for some universal constant $c > 0$.

Recall that MAX-$k$-SAT being NP-hard corresponds to the $k$-local Hamiltonian problem being QMA-hard when $b-a \geq 1/poly(n)$. (We can refer to Theorem 4.1 in these scribed notes of Ryan O’Donnell’s lecture, and more specifically to Kempe-Kitaev-Regev’s original paper for proof of this fact.) The quantum PCP conjecture asks if this is still the case when the gap is $c||H||$.

Going back to the PCP theorem, an implication of the PCP theorem is that it is NP-hard to approximate certain problems to within some factor. Just like its classical analogue, the qPCP conjecture can be seen as stating that it is QMA-hard to approximate the ground state energy to a factor better than $c||H||$.

Reformulation: NLTS conjecture

Let’s make the observation that, taking $a$ to be the ground state energy, the qPCP conjecture sort of says that there exists a family of Hamiltonians for which there is no trivial state (a state generated by a low depth circuit) such that the energy is at most $c||H||$ above the ground state energy.

Freedman and Hastings came up with an easier goal called the No Low-Energy Trivial States conjecture, or NLTS conjecture. We expect that ground states of local Hamiltonians are sufficiently hard to describe (if NP $\neq$ QMA). So low-energy states might not be generated by a quantum circuit of constant depth. More formally:

(NLTS Conjecture): There exists a universal constant $\epsilon > 0$ and a family of local Hamiltonians $\{H^{(n)}\}_{n=1}^{\infty}$ where $H^{(n)}$ acts on $n$ particles and consists of $m_{n}$ local terms, s.t. any family of states $\{|\psi_{n}\rangle\}$ satisfying $\langle \psi_{n} | H^{(n)} | \psi_{n}\rangle \leq \epsilon||H^{(n)}|| + \lambda_{min}(H^{(n)})$ requires circuit depth that grows faster than any constant.

To reiterate, if we did have such a family of NLTS Hamiltonians, then it we wouldn’t be able to give “easy proofs” for the minimal energy of a Hamiltonian, because we couldn’t just give a small circuit which produced a low energy state.

Progress: NLETS theorem

$\epsilon$-error states are states that differ from the ground state in at most $\epsilon n$ qubits. Now, consider $\epsilon$-error states (which “agree” with the ground state on most qubits). Then for bounded-degree local Hamiltonians (analogously in the classical case, those where each variable participates in a bounded number of clauses), these states are also low energy. So any theorem which applies to low energy states (such as the NLTS conjecture), should also apply to states with $\epsilon$-error (as in the NLETS theorem).

To define low-error states more formally:

Definition 2.1 ($\epsilon$-error states): Let $\rho, \sigma \in D((\mathbb{C}^{d})^{\otimes n})$ (the space of positive semidefinite operators of trace norm equal to 1 on $(\mathbb{C}^{d})^{\otimes n}$). Let $H$ be a local Hamiltonian acting on $(\mathbb{C}^{d})^{\otimes n}$. Then:

• $\sigma$ is an $\epsilon$-error state of $\rho$ if $\exists S \subseteq [n]$ of size at most $\epsilon n$ s.t. $\text{Tr}_{S}(\rho) = \text{Tr}_{S}(\sigma)$.
• $\sigma$ is an $\epsilon$-error state for $H$ if $\exists \rho$ s.t. $\text{Tr}(H\rho) = \lambda_{min}(H)$ and $\sigma$ is an $\epsilon$-error state for $\rho$.

Here, see that $\text{Tr}_{S}$ is just the partial trace on some subset of integers $S$, like we’re tracing out or “disregarding” some subset of $n$ qubits.

In 2017, Eldar and Harrow showed the following result which is the NLETS theorem.

Theorem 1 (NLETS Theorem): There exists a family of 16-local Hamiltonians $\{H^{(n)}\}$ s.t. any family of $\epsilon$-error states $\{|\Phi_{n}\rangle\}$ for $\{H^{(n)}\}$ requires circuit depth $\Omega(\log n)$ where $\epsilon = 10^{-9}$.

In the next two sections, we will provide background for an alternate proof of the NLETS theorem due to Nirkhe, Vazirani, and Yuen. After this, we will explain why the proof of NLETS cannot be used to prove NLTS, since the local Hamiltonian family we construct for NLETS can be linearized. Nirkhe, Vazirani, and Yuen’s proof of NLETS makes use of the Feynman-Kitaev clock Hamiltonian corresponding to the circuit generating the cat state (Eldar and Harrow make use of the Tillich-Zemor hypergraph product construction; refer to section 8 of their paper). What is this circuit? It is this one:

First, we apply the Hadamard gate (drawn as $\boxed{H}$) which maps the first qubit $|0\rangle \rightarrow \frac{|0\rangle + |1\rangle}{\sqrt{2}}$. Then we can think of the CNOT gates (drawn as $\bullet-\oplus$) as propagating whatever happens to the first qubit to the rest of the qubits. If we had the first qubit mapping to 0, then the rest of the qubits map to 0, and likewise for 1. This generates the cat state $|\textsf{CAT}_{n}\rangle = \frac{|0\rangle^{\otimes n} + |1\rangle^{\otimes n}}{\sqrt{2}}$, which is highly entangled.

Why do we want a highly entangled state? Roughly our intuition for using the cat state is this: if the ground state of a Hamiltonian is highly entangled, then any quantum circuit which generates it has non-trivial depth. So if our goal is to show the existence of local Hamiltonians which have low energy or low error states that need deep circuits to generate, it makes sense to use a highly entangled state like the cat state.

Quantum circuits

(We’ll write that the state of a qudit – a generalization of a qubit to more than two dimensions, and in this case $q$ dimensions – is a vector in $\mathbb{C}^{q}$. In our diagram above, we’ll see 4 qudits, labelled appropriately.)

Let’s briefly cover the definitions for the quantum circuits we’ll be using.

Let $U$ be a unitary operator acting on a system of $n$ qudits (in other words, acting on $(\mathbb{C}^{q})^{\otimes n}$), where $U = U_{m} \hdots U_{1}$. Here, each $U_{i}$ is a unitary operator (a gate) acting on at most two qudits, and $U$ is a product of $m$ such operators.

If there exists a partition $U$ into products of non-overlapping two-qudit unitaries (we call these layers and denote them as $L_{i} = \bigotimes_{j}U_{ij}$, where each $U_{j}$ here is in layer $L_{i}$) such that $U = L_{d} \hdots L_{1}$ then we say $U$ has $d$ layers.

In other words, $U$ has size $m$ and circuit depth $d$.

Consider $U = L_{d} \hdots L_{1}$ and $A$ an operator.

For $j < d$ define $K^{(j)}$ as the gates in layer $j$ whose supports overlap that of any gate in $K^{(j+1)}$, …, $K^{(d)}$ or with $A$.

Definition 3.1 (lightcone): The lightcone of $A$ with respect to $U$ is the union of $K^{(j)}$: $K_{U} \triangleq \bigcup_{j} K^{(j)}$.

So we can think of the lightcone as the set of gates spreading out of $A$ all the way to the first layer of the circuit. In our diagram, the lightcone of $A$ is the dash-dotted region. We have $K^{(3)} = \varnothing$, $K^{(2)} = \{U_{21}\}$, and $K^{(1)} = \{U_{11}, U_{12}\}$.

We also want a definition for what comes back from the lightcone: the set of gates from the first layer (the widest part of the cone) back to the last layer.

Define $E^{(1)} = K^{(1)}$. For $j \geq 2$, let $E^{(j)}$ be the set of gates whose supports overlap with any gate in $E^{(j-1)}$.

Definition 3.2 (effect zone): The effect zone of $A$ with respect to $U$ is the union $E_{U}(A) \triangleq \bigcup_{j} E^{(j)}$.

In our diagram, see that $E^{(1)} = \{U_{11}, U_{12}\}$, $E^{(2)} = \{U_{21}\}$, and $E^{(3)} = \{U_{31}\}$. The effect zone of $A$ is the dotted region.

Definition 3.3 (shadow of the effect zone): The shadow of the effect zone $W_{U}(A)$ of $A$ with respect to $U$ is the set of qudits acted on by the gates in the effect zone.

In our diagram, the first three qudits are effected by gates in the effect zone. So $W_{U}(A) = \{1, 2, 3\}$.

Given all of these definitions, we make the following claim which will be important later, in a proof of a generalization of NLETS.

Claim 3.1 (Disjoint lightcones): Let $U$ be a circuit and $A, B$ operators. If the qudits $B$ acts on are disjoint from $W_{U}(A)$, then the lightcones of $A$ and $B$ in $U$ are disjoint.

Toward the Feynman-Kitaev clock

Now we’ll give some definitions that will become necessary when we make use of the Feynman-Kitaev Hamiltonian in our later proofs.

Let’s define a unary clock. It will basically help us determine whatever happened at any time little $t$ along the total time big $T$. Let $|\textsf{unary}(t, T)\rangle = |0\rangle^{\otimes(T-t)} \otimes |1\rangle^{\otimes t}$. For our purposes today, we won’t worry about higher dimensional clocks. So we’ll write $|\textsf{clock}_{k}(t, T)\rangle$, but we’ll really only consider the case where $k = 1$, which corresponds to $|\textsf{unary}(t, T)\rangle$. For simplicity’s sake, we will henceforth just write $|\textsf{unary}(t)\rangle$.

Our goal is to construct something a little similar to the tableaux in the Cook-Levin theorem, so we also want to define a history state:

Definition 4.1 (History state): Let $C$ be a quantum circuit that acts on a witness register and an ancilla register. Let $C_{1}, ..., C_{T}$ denote the sequence of two-local gates in $C$. Then for all $k \in \mathbb{N}$, a state $|\Psi\rangle$ is a $k$-dimensional history state of $C$ if:

\begin{aligned}|\Psi\rangle = \frac{1}{\sqrt{T+1}}\sum_{t=0}^{T}|\textsf{clock}_{k}(t, T)\rangle \otimes |\psi_{t}\rangle\end{aligned}

where we have the clock state to keep track of time and $\psi_{t}$ is some state such that $|\psi_{t}\rangle = C_{t}|\psi_{t-1}\rangle$ and $|\psi_{0}\rangle = |\xi\rangle_{witness} \otimes |0\rangle_{ancilla}$. With this construction, we should be able to make a measurement to get back the state at time $t$.

Proof of NLETS

We provide a proof of (a simplified case of) the NLETS theorem proved by Nirkhe, Vazirani, and Yuen in [2].

Theorem 2 (NLETS): There exists a family of $3$-local Hamiltonians $\{H^{(n)}\}$ on a line (Each Hamiltonian $H^{(n)}$ can be defined on $n$ particles arranged on a line such that each local Hamiltonian acts on a particle and its two neighbors) such that for all $n \in \mathbb{N}$, the circuit depth of any $\epsilon$-error ground state for $H^{(n)}$ is at least logarithmic in $n$.

First, we’ll show the circuit lower bound. Then we’ll explain why these Hamiltonians can act on particles on a line and what this implies about the potential of these techniques for proving NLTS.

Proof: We will use the Feynman-Kitaev clock construction to construct a $5$-local Hamiltonian $\mathcal{H}^{(n)}$ for the circuit $C_n$: $|0^n\rangle \to |\textsf{CAT}_n\rangle$.

Fix $n \in \mathbb{N}$ and let $C_n$ have size $T$. The Hamiltonian $\mathcal{H}$ acts on $T+n$ qubits and consists of several local terms depending on $C_n$:

\begin{aligned}\mathcal{H} = H_{in} + \sum_{t=1}^T H_t + H_{out} + H_{stab}\end{aligned}

We can think of a $T+n$ qubit state as representing a $T$ step computation on $n$ qubits (i.e. for each time $t \in [0,T]$, we have a $n$ bit computation state $\textsf{state}_t$ of $C_n$). Intuitively, a $T+n$ qubit state has energy $0$ with respect to $\mathcal{H}$ iff it is the history state of $C_n$. This is because $H_{in}$ checks that at time $t=0$, $\textsf{state}_0$ consists of the input $|0\rangle^n$ to $C_n$. Each $H_t$ checks that $\textsf{state}_{t}$ proceed correctly from $\textsf{state}_{t-1}$ (i.e. that the $t$th gate of $C_n$ is applied correctly). Then $H_{out}$ checks that at time $t=T$, the output is $1$. Finally, $H_{stab}$ checks that the $T+n$ qubit state is a superposition only over states where the first $T$ qubits represent “correct times” (i.e. a unary clock state where time $t$ is represented by $T-t$ zeros followed by $t$ ones).

Therefore, $\mathcal{H}$ has a unique ground state, the history state of $C_n|0^n\rangle$, with energy $0$:

\begin{aligned}|\Psi\rangle = \frac{1}{\sqrt{n+1}}\sum_{t=0}^n |\textsf{unary}(t)\rangle\otimes |\psi_t\rangle = \frac{1}{\sqrt{n+1}}\sum_{t=0}^n |\textsf{unary}(t)\rangle\otimes|\textsf{CAT}_{t}\rangle\otimes |0\rangle^{\otimes (n-t)}\end{aligned}

Later we will show how to transform $\mathcal{H}$ into a Hamiltonian $H$ on $n$ qutrits on a line. Intuitively, the structure of $C_n$ allows us to fuse the $T=n$ time qubits and $n$ state qubits and represent unused state qubits by $2$. For the Hamiltonian $H$, the ground state becomes

\begin{aligned}|\Psi\rangle = \frac{1}{\sqrt{n+1}}\sum_{t=0}^n |\psi_t\rangle = \sum_{t=0}^n |\textsf{CAT}_{t}\rangle\otimes|2\rangle^{\otimes(n-t)}\end{aligned}

For the rest of this proof, we work with respect to $H$.

Let $\sigma$ be an $\epsilon$-error state and let $S \subseteq [n]$ be the subset of qutrits such that $\text{Tr}_S(\sigma) = \text{Tr}_S(|\Psi\rangle\langle\Psi|)$. We define two projection operators which, when applied to $\sigma$ alone, produce nontrivial measurements, but when applied to $\sigma$ together, produce trivial measurements.

Definition 5.1: For any $i\in[n]$, the projection operator

\begin{aligned}A_i = |0\rangle\langle 0|_i \end{aligned}

projects onto the subspace spanned by $0$ on the $i$th qutrit.

For any $j\in[n]$, the projection operator

\begin{aligned} B_j = |1\rangle\langle 1|_i\end{aligned}

projects onto the subspace spanned by $1$ on the $j$th qutrit.

Claim 5.1: For $i\not\in S$, $\text{Tr}(A_i\sigma) = \frac{1}{2} + \frac{-i}{2(n+1)}$. For $j\not\in S$, $\text{Tr}(B_j\sigma) = \frac{1}{2} + \frac{-j}{2(n+1)}$. Note that these values are positive for any $i,j\in [n]$.

Proof: If $i \not\in S$, then measurements on the $i$th qutrit are the same for $\sigma$ and $|\Psi\rangle\langle\Psi|$.

\begin{aligned} \text{Tr}(A_i\sigma) &= \text{Tr}(A_i|\Psi\rangle\langle\Psi|)\\ &= \text{Tr}\left(A_i \frac{1}{n+1}\sum_{t,t'}|\psi_t\rangle\langle\psi_{t'}|\right)\\ &= \frac{1}{n+1}\sum_{t,t'} \text{Tr}(A_i|\psi_t\rangle\langle\psi_{t'}|) \end{aligned}

If $t=t'$, then any $n$ qutrit pure state cannot have nonzero weight in both $\psi_t$ and $\psi_{t'}$ (every pure state ends in some number of $2$s which tells which $\psi_t$ (if any) it can be a part of). Therefore,

\begin{aligned} \text{Tr}(A_i\sigma) = \frac{1}{n+1}\sum_{t} \text{Tr}(A_i|\psi_t\rangle\langle\psi_{t}|) = \frac{1}{n+1}\sum_t \langle \psi_t|A_i|\psi_t\rangle \enspace. \end{aligned}

If $i \le t$, then projecting onto the $i$th qutrit gives $0$ with probability $1/2$. Therefore, $\text{Tr}(A_i\sigma) = \frac{1}{n+1}\left(\frac{n-i+1}{2}\right) = \frac{1}{2} + \frac{-i}{2(n+1)}$.

Similarly, $\text{Tr}(B_j\sigma) = \frac{1}{2} + \frac{-j}{2(n+1)}$. $\square$

Claim 5.2: For $i,j \not\in S$ such that $i < j$, $\text{Tr}(A_i \otimes B_j \sigma) = 0$.

Proof: As before, we can calculate

\begin{aligned} \text{Tr}(A_i \otimes B_j \sigma) &= \text{Tr}(A_i \otimes B_j |\Psi\rangle \langle\Psi|) = \frac{1}{n+1}\sum_t \langle\psi_t|A_i\otimes B_j|\psi_t\rangle \end{aligned}

If $j > t$, then the $j$th qutrit of $\psi_t$ is $2$ so $B_j|\psi_t\rangle = 0$. If $j \le t$, then $A_i \otimes B_j|\psi_t\rangle = 0$ because the first $t$ qutrits of $|\psi_t\rangle$ contain the $|\textsf{CAT}_{t}\rangle$ state so under any measurement, the $i$ and $j$th qutrits must be the same. $\square$

Now we use these claims to prove a circuit lower bound. Let $U$ be a circuit generating (a state with density matrix) $\sigma$. Let $d$ be the depth of $U$.

Consider some $i\not\in S$. For any operator acting on the $i$th qutrit, its lightcone consists of at most $2^d$ gates so its effect zone consists of at most $2^{2d}$ gates which act on at most $2^{2d+1}$ qudits (called the shadow of the effect zone).

Assume towards contradiction that $2^{2d+1} < n-\epsilon n$. Then the shadow of any operator acting only on the $i$th qutrit has size at most $2^{2d+1} \le n - |S|$ since $|S| \le \epsilon n$. So there is some $j$ outside of the shadow which is in the complement of $S$. By Claim 3.1, we have found two indices $i,j$ such that any pair of operators acting on $i$ and $j$ have disjoint lightcones in $U$. WLOG let $i < j$. The lightcones of $A_i,B_j$ are disjoint which implies \begin{aligned}\text{Tr}(A_i \otimes B_j \sigma) = \text{Tr}(A_i \sigma)\cdot\text{Tr}(B_j \sigma).\end{aligned}

By the two claims above, we get a contradiction.

Therefore, $2^{2d+1} \ge n-\epsilon n$. We can take any constant epsilon: letting $\epsilon = 1/2$, we get

\begin{aligned}d \ge \frac{1}{2}\left(\log \frac{n}{2} - 1\right)\end{aligned}

$\square$

This analysis relies crucially on the fact that any $\epsilon$-error state matches the groundstate on most qudits. However, NLTS is concerned with states which may differ from the groundstate on many qudits, as long as they have low energy.

Remark 2.1: The paper of Nirkhe, Vazirani, and Yuen [2] actually proves more:

• A more general lower bound: logarithmic lower bound on the circuit depth of any $\delta$-approximate ($\delta$ far in L1 norm) $\epsilon$-noisy state (probability distribution over $\epsilon$-error states).
• Assuming QCMA $\neq$ QMA (QCMA takes a $m$ bit witness string instead of a $m$ qubit state as witness), they show a superpolynomial lower bound (on the circuit depth of any $\delta$-approximate $\epsilon$-noisy state).
• “Approximate qLWC codes”, using techniques from their superpolynomial lower bound.

Back to NLTS – Tempering our Optimism

So far, we’ve shown a local Hamiltonian family for which all low-error (in “Hamming distance”) states require logarithmic quantum circuit depth to compute, thus resolving the NLETS conjecture. Now, let’s try to tie this back into the NLTS conjecture. Since it’s been a while, let’s recall the statement of the conjecture:

Conjecture (NLTS): There exists a universal constant $\epsilon > 0$ and a family of local Hamiltonians $\{H^{(n)}\}_{n=1}^{\infty}$ where $H^{(n)}$ acts on $n$ particles and consists of $m_{n}$ local terms, s.t. any family of states $\{|\psi_{n}\rangle\}$ satisfying $\langle \psi_{n} | H^{(n)} | \psi_{n}\rangle \leq \epsilon||H^{(n)}|| + \lambda_{min}(H^{(n)})$ requires circuit depth that grows faster than any constant.

In order to resolve the NLTS conjecture, it thus suffices to exhibit a local Hamiltonian family for which all low-energy states require logarithmic quantum circuit depth to compute. We might wonder if the local Hamiltonian family we used to resolve NLETS, which has “hard ground states”, might also have hard low-energy states. Unfortunately, as we shall show, this cannot be the case. We will start by showing that Hamiltonian families that lie on constant-dimensional lattices (in a sense that we will make precise momentarily) cannot possibly be used to resolve NLTS, and then show that the Hamiltonian family we used to prove NLTS can be linearized (made to lie on a one-dimensional lattice!).

The Woes of Constant-Dimensional Lattices

Definition 6.1: A local Hamiltonian $H$ acting on $n$ qubits is said to lie on a graph $G$ if there is an injection of qubits into vertices of the graph such that the set of qubits in any interaction term correspond to a connected component in the graph.

Theorem 2: If $(H^{(n)})$ is a local Hamiltonian family that lies on an $O(1)$-dimensional lattice, then $(H^{(n)})$ has a family of low-energy states with low circuit complexity. In particular, if $H$ is a local Hamiltonian on a $d$-dimensional lattice acting on $n$ qubits for large enough $n$, then for any $\epsilon$, there exists a state $|\psi\rangle$ that can be generated by a circuit of constant depth and such that $\langle \psi | H | \psi \rangle \leq H_0 + \epsilon ||H||$ where $H_0$ is the ground-state energy.

Proof: In what follows, we’ll omit some of the more annoying computational details in the interest of communicating the high-level idea.

Start by partitioning the $d$-dimensional lattice (the one that $H^(n)$ lives on) into hypercubes of side length $L$. We can “restrict” $H^{(n)}$ to a given hypercube (let’s call it $\rho$) by throwing away all local terms containing a qubit not in $\rho$. This gives us a well-defined Hamiltonian $H_{\rho}$ on the qubits in $\rho$. Define $|\phi_{\rho}\rangle$ to be the $L^d$-qubit ground state of $H_{\rho}$, and define

\begin{aligned}|\phi\rangle := \bigotimes_{\text{hypercubes } \rho} |\phi_{\rho}\rangle\end{aligned}

where $|\phi\rangle$ is an $n$-qubit state. Each $|\phi_{\rho}\rangle$ can be generated by a circuit with at most $2^{L^d}$ gates, hence at most $2^{L^d}$ depth. Then, $|\phi\rangle$ can be generated by putting all of these individual circuits in parallel – this doesn’t violate any sort of no-cloning condition because the individual circuits act on disjoint sets of qubits. Therefore, $|\phi\rangle$ can be generated by a circuit of depth at most $2^{L^d}$. $L$ and $d$ are both constants, so $|\phi\rangle$ can be generated by a constant-depth circuit.

We claim that, for the right choice of $L$, $|\phi\rangle$ is also a low-energy state. Intuitively, this is true because $\phi$ can only be “worse” than a true ground state of $H^{(n)}$ on local Hamiltonian terms that do not lie entirely within a single hypercube (i.e. the boundary terms), and by choosing $L$ appropriately we can make this a vanishingly small fraction of the local terms of $H^{(n)}$. Let’s work this out explicitly.

Each hypercube has surface area $2dL^{d-1}$, and there are $n/L^d$ hypercubes in the lattice. Thus, the total number of qubits on boundaries is at most $2d\frac{n}{L}$. The number of size $k$-connected components containing a given point in a $d$-dimensional lattice is a function of $k$ and $d$. Both of these are constants. Therefore, the number of size $k$-connected components containing a given vertex, and hence the number of local Hamiltonian terms containing a given qubit, is constant. Thus, the total number of violated local Hamiltonian terms is at most $O(\frac{n}{L})$. Taking $L$ to be $\frac{1}{\epsilon}$, we get the desired bound. Note that to be fully rigorous, we need to justify that the boundary terms don’t blow up the energy, but this is left as an exercise for the reader. $\square$

Linearizing the Hamiltonian

Now that we have shown that Hamiltonians that live on constant-dimensional lattices cannot be used to prove NLTS, we will put the final nail in the coffin by showing that our NLETS Hamiltonian (the Feynman-Kitaev clock Hamiltonian $\mathcal{H}$ on the circuit $C$) can be made to lie on a line (a $1$-dimensional lattice). To do so, we will need to understand the details of $\mathcal{H}$ a bit better.

Proposition 6.1: $\mathcal{H}$ for the circuit $C$ is $5$-local.

Proof: Recall that we defined

\begin{aligned}\mathcal{H} := H_{in} + \sum_{t=1}^T H_t+ H_{stab}\end{aligned}

Let’s go through the right-hand-side term-by-term. We will use $|\mathsf{time}(t)\rangle$ to denote the $t^{\text{th}}$ qubit of the time register and $|\mathsf{state}(s)\rangle$ to denote the $s^{\text{th}}$ qubit of the state register.

• $H_{in}$ needs to serially access the qubit pairs \begin{aligned}|\mathsf{time}(0)\rangle\otimes\textsf{state}(s) \end{aligned} for all $s$ and ensure that they are all set to $|0\rangle$. Thus, $H_{in}$ is $2$-local.
• Each $H_t$ term needs to access the states $|\textsf{time}(t-1)\rangle, |\textsf{time}(t)\rangle, |\textsf{time}(t+1)\rangle, |\textsf{state}(s)\rangle$, and $|\textsf{state}(t)\rangle$ and ensure that the state transitions are correct. Thus, $H_{t}$ is $5$-local.
• $H_{stab}$ needs to access the states \begin{aligned}|\textsf{time}(t)\rangle \otimes |\textsf{time}(t+1)\rangle \end{aligned} and ensure that the progression of the time register is correct. Thus, $H_{stab}$ is $2$-local. $\square$

Now, we follow an approach of [3] to embed $\mathcal{H}$ into a line.

Theorem 3: The Feynman-Kitaev clock Hamiltonian $H$ can be manipulated into a $3$-local Hamiltonian acting on qutrits on a line.

Proof: Rather than having $\mathcal{H}$ act on $2n$ total qubits ($n$ time qubits and $n$ state qubits), let’s fuse each $|\textsf{time}(i)\rangle$ and $|\textsf{state}(i)\rangle$ pair into a single qudit of dimension $4$. If we view $\mathcal{H}$ as acting on the space of particles $|\textsf{time}(i)\rangle \otimes |\textsf{state}(i)\rangle$, we observe that, following Proposition 6.1, each local term needs to check at most the particles corresponding to times $t-1$, $t$, and $t+1$. Therefore, $\mathcal{H}$ is $3$-local and on a line, as desired.

To see that we can have $\mathcal{H}$ act on particles of dimension $3$ (qutrits) rather than particles of dimension $4$, note that the degree of freedom corresponding to $|\textsf{time}(t)\rangle \otimes |\textsf{state}(t)\rangle = |0\rangle \otimes |1\rangle$ is unused, as the $t^{\text{th}}$ qubit of the state is never nonzero until timestamp $t$. Thus, we can take the vectors

\begin{aligned} |0\rangle := |1\rangle\otimes|0\rangle, |1\rangle := |1\rangle\otimes|1\rangle, |2\rangle := |0\rangle\otimes|0\rangle \end{aligned}

as a basis for each qutrit.

Even though we’ve shown that the clock Hamiltonian for our original circuit cannot be used to prove NLTS (which is still weaker than the original Quantum PCP conjecture) this does not necessarily rule out the use of this approach for other “hard” circuits which might then allow us to prove NLTS. Furthermore, NLETS is independently interesting, as the notion of being low “Hamming distance” away from vectors is exactly what is used in error-correcting codes.

References

• [1] Sanjeev Arora and Boaz Barak. Computational complexity: a modern approach. Cambridge University Press, 2009.
• [2] Chinmay Nirkhe, Umesh Vazirani, and Henry Yuen. Approximate low-weight check codes and circuit lower bounds for noisy ground states. arXiv preprint arXiv:1802.07419, 2018.
• [3] Dorit Aharonov, Wim van Dam, Julia Kempe, Zeph Landau, Seth Lloyd, and Oded Regev. Adiabatic quantum computation is equivalent to standard quantum computation. SIAM J. Comput., 2007.