# Calculating the expectation value of a unitary operator on a quantum computer

What is the smartest way of calculating the expectation value of some unitary $$U$$ in some state $$|\psi\rangle$$?

There are two ways I know:

Are there more efficient techniques to do that?

• What are your conditions? Do you know what the state is? If not, how many copies of it do you have? – DaftWullie Apr 30 '19 at 5:35
• @DaftWullie The state $|\psi\rangle$ and the unitary $U$ are known, and it is assumed that we can prepare the infinite number of the copies of $|\psi\rangle$. – Andrey Kardashin Apr 30 '19 at 11:49

## 3 Answers

Since the state $$|\psi\rangle$$ is known, we can assume that we have a preparation procedure that takes some standard state (call it $$|0\rangle$$, although in general it will be composed of many qubits). Thus, there is a unitary $$V$$ such that $$V|0\rangle=|\psi\rangle.$$

In that case, to measure $$|\langle\psi|U|\psi\rangle|^2$$, we simply follow the procedure

1. Prepare $$|0\rangle$$
2. Apply $$V$$
3. Apply $$U$$
4. Apply $$V^\dagger$$
5. Measure in a basis for which $$0\rangle$$ is one of the elements
6. Repeat until you have a suitably accurate estimate of the probability of getting the answer $$|0\rangle$$.

This probability is your answer.

If you want $$\langle\psi|U|\psi\rangle$$, the Hadamard test is probably your best bet. Yes, it requires controlled-$$U$$, but you're going to need something like that. This is because there's no observable difference between $$U$$ and $$e^{i\theta}U$$, so nothing you do can access the phase of $$\langle\psi|U|\psi\rangle$$ (but there is a difference between controlled-$$U$$ and controlled-$$e^{i\theta}U$$).

To find modulus of that expectation value you can use swap test on $$|\psi\rangle \otimes U|\psi\rangle$$.
I doubt there are simpler ways since $$\langle \psi |U|\psi\rangle$$ is complex in general.

Also note, that both algorithms you mentioned require access to the controlled version of $$U$$. But you can't construct circuit for controlled $$U$$ if $$U$$ is given as blackbox. Though there are ways to physically implement controlled version. See here.

• Thanks for the answer! But as I know there is a technique to control an arbitrary unitary operation by using the qudits of dimension 4 instead of qubits. Please see this. – Andrey Kardashin Apr 30 '19 at 11:49

A bit late on this question, but for what it's worth, I recently put a paper on the arXiv which shows how to estimate $$\langle\psi|U|\psi\rangle$$ quadratically faster than the Hadamard test. https://arxiv.org/abs/2102.04975

A quick summary of the relevant part of the paper (without providing much intuition):

The algorithm uses two registers, one containing a single ancilla qubit and the other is the main quantum system. Let's define the following operations on the two-register system: $$\textbf{U} = |0\rangle\langle0| \otimes U + |1\rangle\langle1| \otimes U^\dagger$$ $$\textbf{S} = 2|+,\psi\rangle\langle+,\psi| - \textbf{I}$$ $$\textbf{Q} = [X\otimes I] \textbf{S} \textbf{U}$$ The $$\textbf{U}$$ operation acts $$U$$ on the system if the ancilla is $$0$$, and $$U^\dagger$$ if the ancilla is 1. The operation $$\textbf{Q}$$ can be implemented using $$\mathcal{O}(1)$$ calls to controlled-$$U$$ and the circuit that creates $$|\psi\rangle$$.

Let $$\cos(\theta) \equiv \mathrm{Re}[\langle\psi|U|\psi\rangle]$$. The key idea is that running QPE with $$\textbf{Q}$$ as the unitary and $$|+,\psi\rangle$$ as the input state, will return either $$\theta$$ or $$2\pi-\theta$$. SInce, $$\cos(\theta) = \cos(2\pi-\theta) = \mathrm{Re}[\langle\psi|U|\psi\rangle]$$, this completes the estimation of the real part. One can also get the imaginary part using the same algorithm by noting that $$Im[\langle\psi|U|\psi\rangle] = Re[\langle\psi|e^{-i\pi/2}U|\psi\rangle]$$.

This algorithm inherits the quadratic speedup of QPE. The error in the estimate scales as $$\mathcal{O}(1/q)$$ in the number of calls $$q$$ to the operation $$\textbf{Q}$$.