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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:

  1. quantum phase estimation algorithm;
  2. Hadamard test.

Are there more efficient techniques to do that?

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  • $\begingroup$ What are your conditions? Do you know what the state is? If not, how many copies of it do you have? $\endgroup$ – DaftWullie Apr 30 '19 at 5:35
  • $\begingroup$ @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 $|\psi\rangle$. $\endgroup$ – Andrey Kardashin Apr 30 '19 at 11:49
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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-$e^{i\theta}U$).

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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.

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  • $\begingroup$ 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. $\endgroup$ – Andrey Kardashin Apr 30 '19 at 11:49
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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 $\textbf{Q}$.

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