Sampling Advantage (ANSCSE2023)

Quantum Sampling Advantage

How to prove it and does it matter?

The 26th International Annual Symposium on Computational Science and Engineering

20 Jul 2023

Ninnat Dangniam

The Institute for Fundamental Study (IF)

Naresuan University

PC: Google Quantum AI

Extra Slides

50-100 qubits

1,000

10,000

1,000,000

...

Fault-tolerant, universal quantum computing

Gil Kalai, modified from Devoret and Schoelkpf, Science (2013)

Noisy, intermediate-scale quantum (NISQ) devices

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Commercially-relevant applications, unambiguous quantum advantage

10 years?

A Quantum Future

BBC

Codebreaking

iStock/sergunt

Quantum chemistry simulation

Modified from Bremner, QIP2018

Fermionic Linear Optics (FLO)

X = \begin{pmatrix}0&1\\1&0 \end{pmatrix}

X = \begin{pmatrix}0&1\\1&0 \end{pmatrix}

Y = \begin{pmatrix}0&\!\!\!-i\\i&0 \end{pmatrix}

Y = \begin{pmatrix}0&\!\!\!-i\\i&0 \end{pmatrix}

Z = \begin{pmatrix}1&0\\0&\!\!\!-1 \end{pmatrix}

Z = \begin{pmatrix}1&0\\0&\!\!\!-1 \end{pmatrix}

Reminder

\sigma^{\dagger} = X-iY = \begin{pmatrix}0&0\\1&0 \end{pmatrix}

\sigma^{\dagger} = X-iY = \begin{pmatrix}0&0\\1&0 \end{pmatrix}

\sigma^{\dagger}|0\rangle = |1\rangle

\sigma^{\dagger}|0\rangle = |1\rangle

Identify computational basis states with occupation number states

Jordan-Wigner

f_j f^{\dagger}_k +f^{\dagger}_k f_j = \delta_{jk}\mathbb{I}

f_j f^{\dagger}_k +f^{\dagger}_k f_j = \delta_{jk}\mathbb{I}

(f^{\dagger}_j)^2 = 0

(f^{\dagger}_j)^2 = 0

CAR

f^{\dagger}_j \coloneqq \sigma_j^{\dagger} \displaystyle{\prod_{k=1}^{j-1}} Z_k

f^{\dagger}_j \coloneqq \sigma_j^{\dagger} \displaystyle{\prod_{k=1}^{j-1}} Z_k

$f_j^{\dagger}$ creates a fermion at site $j$

Pauli exclusion

Fermionic Linear Optics (FLO)

m_j^X \coloneqq X_j \displaystyle{\prod_{k=1}^{j-1}} Z_k

m_j^X \coloneqq X_j \displaystyle{\prod_{k=1}^{j-1}} Z_k

m_j^Y \coloneqq Y_j \displaystyle{\prod_{k=1}^{j-1}} Z_k

m_j^Y \coloneqq Y_j \displaystyle{\prod_{k=1}^{j-1}} Z_k

cf. Dirac's gamma matrices

m_{\alpha}m_{\beta} + m_{\beta}m_{\alpha} = 2 \delta_{\alpha\beta} \mathbb{I}

m_{\alpha}m_{\beta} + m_{\beta}m_{\alpha} = 2 \delta_{\alpha\beta} \mathbb{I}

\alpha = (j,X\,\mathrm{or}\,Y)

\alpha = (j,X\,\mathrm{or}\,Y)

Majoranas

\begin{pmatrix} 1 & & 0 \\ & \!\!\!\!\!\Large{-iX}\!\!\!\! \\ 0 & & 1 \end{pmatrix}

\begin{pmatrix} 1 & & 0 \\ & \!\!\!\!\!\Large{-iX}\!\!\!\! \\ 0 & & 1 \end{pmatrix}

Non-example

\mathrm{SWAP}= \mathbb{I}\!\otimes\!\mathbb{I}\!+\!X\!\otimes\! X \!+\! Y\!\otimes \!Y \!+\! Z\!\otimes \!Z

\mathrm{SWAP}= \mathbb{I}\!\otimes\!\mathbb{I}\!+\!X\!\otimes\! X \!+\! Y\!\otimes \!Y \!+\! Z\!\otimes \!Z

$Z\!\otimes\! Z$ interaction is not FLO

FLO Hamiltonians are quadratic

Antisymmetric

H_{\mathrm{FLO}} = \displaystyle{\frac{i}{4}} \sum_{\substack{\alpha,\beta \\ \alpha>\beta}} h_{\alpha\beta} [m_{\alpha},m_{\beta}]

H_{\mathrm{FLO}} = \displaystyle{\frac{i}{4}} \sum_{\substack{\alpha,\beta \\ \alpha>\beta}} h_{\alpha\beta} [m_{\alpha},m_{\beta}]

Fermionic Linear Optics (FLO)

\vec{m}(t) = V^{\dagger} \vec{m} V = e^{{\bf h}t} \vec{m}

\vec{m}(t) = V^{\dagger} \vec{m} V = e^{{\bf h}t} \vec{m}

Generators of rotations SO( $2n$ )

V_{\mathrm{FLO}} = e^{iH_{\mathrm{FLO}}t}

V_{\mathrm{FLO}} = e^{iH_{\mathrm{FLO}}t}

(Defining representation)

Antisymmetric

Manipulating $2n\times 2n$ matrices instead of $2^n\times 2^n$ matrices!

Only need to specify the correlation matrix (2nd moments)
All higher moments can be extracted from $\Sigma$ via Wick's theorem

\Sigma_{\alpha\beta} \coloneqq \frac{i}{2}\mathrm{Tr}(\rho[m_{\alpha},m_{\beta}])

\Sigma_{\alpha\beta} \coloneqq \frac{i}{2}\mathrm{Tr}(\rho[m_{\alpha},m_{\beta}])

p_{\bf x} = \frac{1}{i^{2^n}} \mathrm{Pf} (\Sigma)

p_{\bf x} = \frac{1}{i^{2^n}} \mathrm{Pf} (\Sigma)

$\rho = (|0\rangle\langle 0|)^{\otimes n}$ is a fermionic Gaussian state

$l^2$ -distance to the uniform distribution:

$\Vert p-p_{\mathrm{unif}}\Vert^2 = Z - \displaystyle{\frac{1}{2^n}} = Z-Z_{\mathrm{unif}}$

Z = \mathbb{E}_U \mathbb{E}_{\bf x}\, [ p^2_{\bf x}(U)]

Z = \mathbb{E}_U \mathbb{E}_{\bf x}\, [ p^2_{\bf x}(U)]

Z \le \displaystyle{\frac{1}{\alpha 2^n}}

Z \le \displaystyle{\frac{1}{\alpha 2^n}}

$p_{\bf{x}}(U)$ anti-concentrates if

Z_{\mathrm{unif}} = \displaystyle{\frac{1}{2^n}}

Z_{\mathrm{unif}} = \displaystyle{\frac{1}{2^n}}

Implies $\Pr\left(p_{{\bf x}} \ge \displaystyle{\frac{\beta}{2^n}}\right) \ge \alpha(1-\beta)^2$ via the Paley-Zygmund inequality

\Pr\left( X\ge \beta\mathbb{E}[X] \right) \ge (1-\beta)^2 \displaystyle{\frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}}

\Pr\left( X\ge \beta\mathbb{E}[X] \right) \ge (1-\beta)^2 \displaystyle{\frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}}

2

2

2

Anti-concentration

Modified from Dalzell et al., PRX Quantum 2022

\mathbb{E}_V [p^2_{\bf 0}] = \int dV\, \mathrm{Tr}(V|\Psi\rangle \langle \Psi| V^{\dagger}|{\bf 0}\rangle\langle {\bf x}|)^2

\mathbb{E}_V [p^2_{\bf 0}] = \int dV\, \mathrm{Tr}(V|\Psi\rangle \langle \Psi| V^{\dagger}|{\bf 0}\rangle\langle {\bf x}|)^2

= \mathrm{Tr}(|\Psi\rangle \langle \Psi| \int dV\, V^{\dagger}|{\bf 0}\rangle\langle {\bf 0}|V)^2

= \mathrm{Tr}(|\Psi\rangle \langle \Psi| \int dV\, V^{\dagger}|{\bf 0}\rangle\langle {\bf 0}|V)^2

= \mathrm{Tr}\left[(|\Psi\rangle \langle \Psi|)^{\otimes 2} \underbrace{\int dV\, (V^{\dagger})^{\otimes 2}(|{\bf 0}\rangle\langle {\bf 0}|)^{\otimes 2}V^{\otimes 2}}\right]

= \mathrm{Tr}\left[(|\Psi\rangle \langle \Psi|)^{\otimes 2} \underbrace{\int dV\, (V^{\dagger})^{\otimes 2}(|{\bf 0}\rangle\langle {\bf 0}|)^{\otimes 2}V^{\otimes 2}}\right]

Sum of projectors onto irreps by Schur's lemma under nice conditions

SO( $2n$ ) for FLO
U( $n$ ) for passive (number-preserving) FLO

Anti-concentration

Numerics suggest that random FLO circuits anti-concentrate in sub-linear depth

Worst-to-average-case reduction

There exists a "worst-case circuit" $V_0$ for which $p_{\bf 0} = |\langle \mathbf{0}|V_0|\Psi\rangle|^2$ is #P-hard to approximate
Deform to average-case (Haar random) circuits where $p_{\bf 0} (\theta)= |\langle \mathbf{0}|V(\theta)|\Psi\rangle|^2$ is a polynomial of low-degree $q$
Sample $\gtrsim q$ points (some can be errorneous $\iff$ robustness) and recover $p_{\bf 0}(\theta)$ via Berlekamp–Welch
Plug in $\theta = 0$ to solve a #P-hard problem!

Unitary Paths

First idea: truncated exponential path

U(\theta) = U_0 e^{i\theta H}

U(\theta) = U_0 e^{i\theta H}

Movassagh 2019

= U_0 \left(\mathbb{I} + i\theta H + \frac{(i \theta H)^2}{2!} + \cdots \right)

= U_0 \left(\mathbb{I} + i\theta H + \frac{(i \theta H)^2}{2!} + \cdots \right)

Truncation gives rise to a non-unitary operator

f(X) = (\mathbb{I}-X)(\mathbb{I}+X)^{-1}

f(X) = (\mathbb{I}-X)(\mathbb{I}+X)^{-1}

g_{\theta} = g_0 \displaystyle{ \frac{(1-\theta) \mathbb{I} +(1+\theta) g}{ (1+\theta) \mathbb{I} +(1-\theta)g }}

g_{\theta} = g_0 \displaystyle{ \frac{(1-\theta) \mathbb{I} +(1+\theta) g}{ (1+\theta) \mathbb{I} +(1-\theta)g }}

Better idea: Cayley path

Unitary

Hermitian

Hardness Guarantees

Quantum advantage scheme	Robustness of avg-case hardness (additive error)	Anti-concentration
Boson Sampling
Random circuit sampling (Google layout)
Fermion Sampling

2^{-\tilde{O}(n)}

2^{-\tilde{O}(n)}

2^{-\tilde{O}(n^{1.5})}

2^{-\tilde{O}(n^{1.5})}

2^{-\tilde{O}(n^2)}

2^{-\tilde{O}(n^2)}

Bouland et al., FOCS 2021

Quantum Sampling Advantage How to prove it and does it matter? The 26th International Annual Symposium on Computational Science and Engineering 20 Jul 2023 Ninnat Dangniam The Institute for Fundamental Study (IF) Naresuan University PC: Google Quantum AI