# Expectation value of Pauli string for VQE

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.

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 $$0.$$

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.

1. 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.

• Is doing a Hadamard test more efficient? Wouldn't doing a controlled-P operation be messy and use a lot more gates? May 20, 2022 at 0:29
• In order to implement the controlled-P operator, you would only have to apply a control Pauli gate for each qubit. So $n$ controlled Pauli gates for each operator control-P. The other option is to apply the circuit $Q^\dagger$ to the state $\psi$, this presents a couple of problems: 1. You would have to do classically the decomposition $P = Q\Lambda Q^\dagger$ and then find a circuit that implements $Q^\dagger$ which could be quite expensive to implement. 2. The bigger the $n$ the more measurements you would have to do in order to obtain a meaningful estimation of the probabilities. May 20, 2022 at 8:45