Can QAOA be considered as simulation of a quantum annealer on a gate-based quantum computer?

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Quantum annealers are single purpose machines allowing to solve quadratic unconstrained binary optimization (QUBO) problems. QUBO problems have following objective function: $$ F=-\sum_{i<j}J_{ij}x_ix_j-\sum_{i=1}^N h_ix_i, $$ where $x_i$ is a binary varibale and $h_i$ and $J_{ij}$ are coefficients. Such objective function is equivalent to Ising Hamiltonian $$ H_{\text{ISING}}=-\sum_{i<j}J_{ij}(\sigma^z_i\otimes\sigma^z_j)-\sum_{i=1}^N h_i\sigma^z_i, $$ where $\sigma^z_i$ is Pauli Z gate acting on $i$th qubit and there are identity operators on other qubits, tensor product $\sigma^z_i\otimes\sigma^z_j$ means that Z gates act on $i$th and $j$th qubits and there are identity operators on other qubits.

Quantum annealers physically implements simulation of Hamiltonian $$ H(t)=\Big(1-\frac{t}{T}\Big)\sum_{i=1}^N h_i\sigma^x_i+\frac{t}{T}H_{\text{ISING}}, $$ where $t$ is a time, $T$ total time of simulation and $\sigma^x_i$ is Pauli X gate acting on $i$th qubit. Initial state of a quantum annealer is equal superposition of all qubits which is ground state of the Hamiltonian $H(0)$.

Quantum Approximate Optimization Algorithm (QAOA) is described by an operator $$ U(\beta, \gamma) = \prod_{i=1}^{p}U_B(\beta_i)U_C(\gamma_i), $$ where $p$ is number of iteration of QAOA, $$ U_B(\beta) = \mathrm{e}^{-i\beta\sum_{i=1}^N \sigma^x_i}, $$ and $$ U_C(\gamma) = \mathrm{e}^{-i\gamma(\sum_{i,j=1}J_{ij}(\sigma^z_i\otimes\sigma^z_j)+\sum_{i=1}^N h_i\sigma^z_i)}. $$ Initial state for QAOA is $H^{\otimes n}|0\rangle ^{\otimes n}$, i.e. equally distributed superposition as in case of the quantum annealer.

Since time evolution of quantum system described by Hamiltonian $H$ from state $|\psi(0)\rangle$ to state $|\psi(t)\rangle$ is expressed by $$ |\psi(t)\rangle = \mathrm{e}^{-iHt}|\psi(0)\rangle, $$ it seems that operator $U(\beta, \gamma)$ from QAOA is simply simulation of Hamiltonian $H(t)$ describing quantum annealer beacause exponents of $\mathrm{e}$ are sums in Hamiltonian $H(t)$.

However, $H(t)$ is composed of two term containing Pauli matrices X and Z and $\mathrm{e}^{A+B}=\mathrm{e}^A\mathrm{e}^B$ is valid only for commuting matrices $[A,B]=O$. But Pauli matrices X and Z fulfil anti-commutation relation $\{X,Z\}=O$, not the commutation one.

So my questions are these:

  1. Can QAOA be realy considered as a simulation of quantum annealer on gate-based universal quantum computer?
  2. What I am missing in discussion above concerning commutation of Pauli matrices? Or is there any condition for matrices $A$ and $B$ allowing equality $\mathrm{e}^{A+B}=\mathrm{e}^A\mathrm{e}^B$?

Martin Vesely

Posted 2020-10-06T14:04:29.787

Reputation: 7 763

Answers

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  1. If you use infinite depth then QAOA can be consider as quantum annealer on gate-based. The authors of QAOA original paper probably deduce it from quantum annealing. What I mean by infinite depth is you take $p \to \infty$ in the operator

$$U(\beta, \gamma) = \Pi_{i=1}^p U_B(\beta_i)U_C(\gamma_i) $$

  1. Your are right about the commutation problem. However, remember that

$$ \lim_{p \to \infty} \big(e^{A/p}e^{B/p} \big)^p = e^{A+B} $$

so in a sense, at infinite limit depth, it is not a problem. So you recover the quantum annealing approach.

KAJ226

Posted 2020-10-06T14:04:29.787

Reputation: 6 322

Thank you, now everything is clear. – Martin Vesely – 2020-10-07T09:02:38.130