Expectation value of Pauli string for VQE

Question

One approach to get the expectation value $\langle\psi|P\psi\rangle$ of a pauli string $P\in \{I, X, Y, X\}^{\otimes n}$ is the following.

1. Let $(a_i, |\lambda_i\rangle)$ be eigenvalue-eigenvector pair of $P$ where $a_i\in \{0, 1\}$ as $P$ is unitary. We may write $$P = \sum_i a_i|\lambda_i\rangle\langle\lambda_i|$$ so it follows that $$\langle\psi|P\psi\rangle = \sum_i a_i|\langle\lambda_i|\psi\rangle|^2.$$
2. We can thus perform a change of basis measure in the $|\lambda_i\rangle$ basis to recover the coefficients $|\langle\lambda_i|\psi\rangle|^2$ via repeated sampling.
3. Evaluate the sum in step 1.

My questions about this are:

1. How are we sampling? ie. Do we measure each of the $n$ qubits "separately" and estimate $p_0, p_1$ in $p_0|0\rangle + p_1|1\rangle$? This would lead to us adding $2^n$ numbers in step 3, so I don't think this is the case. Do we measure each outcome $|x_1\dots x_n\rangle$ and estimate its probability $|\langle\lambda_i|\psi\rangle|^2$ directly?
2. If we measure each outcome and estimate $|\langle\lambda_i|\psi\rangle|^2$ directly, how many repeated measurements do we need to obtain accuracy within $\epsilon$? If we perform a polynomial number of measurements, most of our estimates for $|\langle\lambda_i|\psi\rangle|^2$ will be $0.$

Answer 1

I think we could get the estimate of $\langle \psi | P | \psi \rangle $ using your method in the following way.

Let's assume that we know the eigenvalue decomposition (https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix) of the Pauli chain, i.e we know $Q$ and $\Lambda$ in :

$$ P = Q \Lambda Q^{\dagger}$$

where $\Lambda$ is a diagonal matrix with the eigenvalues and $Q$ is a unitary matrix with the eigenvectors. The expected value over the state $|\psi\rangle$ becomes:

$$ \langle \psi | P | \psi \rangle = \langle \psi | Q \Lambda Q^{\dagger} | \psi \rangle $$

This is what I think you have written in equation $⟨ψ|P|ψ⟩=\sum_i a_i|⟨\lambda_i|ψ⟩|^2$. In order to answer your questions:

1. Let's define $|\varphi \rangle = Q^\dagger |\psi\rangle $, the expected value becomes:

$$ \langle \psi | Q \Lambda Q^{\dagger} | \psi \rangle = \langle \varphi |\Lambda|\varphi \rangle$$

Now, given that we know $Q$ we can create a circuit that applies $Q^\dagger$ to the state $|\psi\rangle$. Then we should sample the outcome of this circuit, i.e measure all the n qubits and estimate the probability of outcome of each state $|i\rangle$ with $i = 0, 1, \dots, 2^n - 1$. Each probability would be in fact the coefficient $|⟨\lambda_i|ψ⟩|^2$ that we were looking for. You would get indeed $2^n$ numbers. Finally, it would be enough to compute the sum $\sum_i a_i|⟨\lambda_i|ψ⟩|^2$ to get the expected value.

2. You can find information about the precision in the estimation of these coefficients as a function of the number of samples in these questions:

When increase the shot, why the result is different?

How many shots should one take to get a reliable estimate in a quantum program?

Now, having said this. There is a more efficient method to estimate the expected value of a unitary operator. The Hadamard test (https://en.wikipedia.org/wiki/Hadamard_test_(quantum_computation) ). It uses this circuit:

In short, it allows you to compute the real part of the expected value $\langle \psi |U | \psi \rangle$ using the difference between the probabilities of measurement of the ancilla qubit (the one at the top), i.e:

$$ \mathcal{Re}{\langle \psi |U|\psi \rangle } = p_0 - p_1$$

It is also possible to compute the imaginary part by adding a phase gate with angle $-\pi/2$ in the ancilla qubit. So, if you replace U in the circuit with your Pauli chain P, you can compute $ \langle \psi | P | \psi \rangle$ without sampling $2^n$ probabilities. You would only have to sample 4, 2 for the real part, and 2 for the imaginary part.