How to compute the energy of a non-eigenstate in a Heisenberg limited form?

Question

Let us have a Hamiltonian $H$ and a state $|\psi\rangle = \sum_i a_i |E_i\rangle$, a linear combination of eigenstates $|E_i\rangle$ of $H$ with eigenvalues $E_i$. What is the best way to achieve a Heisenberg-limited measurement of the energy of $\psi$, $E_\psi = \sum_i a_i E_i$.

If $\psi$ were an eigenstate, Quantum Phase Estimation provides an answer. However, if it were not we would have to sample the result of QPE subsequently. I have also considered an old Grover paper to compute the mean of a probability distribution, with an additional $O(1/\epsilon)$ overhead over QPE (already $O(1/\epsilon)$). Is there a way to do it better?

Answer 1

Here is a suggestion how to obtain a Heisenberg-limited energy estimation -- there's several details to be filled in, though.

You can run phase estimation on $N$ copies of the state, $|\Psi\rangle :=|\psi\rangle^{\otimes N}$, for $$ \mathcal H = \tfrac1N\sum H_i\ , $$ where $H_i$ is the Hamiltonian $H$ acting on the $i$'th copy. Note that the energy expectation value of $\mathcal H$ in $|\Psi\rangle$ is the same as that of $H$ in $|\psi\rangle$.

For each energy eigenstate of $\mathcal H$, you will get Heisenberg-limited accuracy for the energy. On the other hand, as the number of copies increases, the probability to be in an eigenstate with an energy $\delta$-close to the average energy (that is, in a $\delta$-typical subspace) will converge exponentially due to the central limit theorem -- which is essence means that with correspondingly high probability, the phase estimation will return a value $\delta$-close to the desired average. Choosing $\delta$ sufficiently smaller than the desired accuracy $\epsilon$ (e.g. $\epsilon/2$, and run phase estimation with accuracy $\epsilon/2$) and then $N$ sufficiently large should give the desired result.

Answer 2

I may be wrong about this but when you run the QPE on some state $|\psi\rangle$ by simulating some Hamiltonian $H$, your precision is limited at least by:

1. The error in the simulation of $U=\exp(-iHt)$;
2. The precision you decide to choose for the phase estimation with the (inverse) Fourier transform; and
3. The number of times you have to sample from the QPE using multiple copies of $|\psi\rangle$.

You can increase the precision with the first factor by using a larger Trotter factor or using more sophisticated Hamiltonian simulation. You can increase the precision of the second by adding an additional qubit of precision but you're still limited by the No Fast-Forwarding Theorem and you'd need to run your circuit for twice as long for each qubit of precision:

As for the third factor, if we are promised that $|\psi\rangle$ is an eigenstate then we only need to run the circuit once, because we'll get the phase right away. If $|\psi\rangle$ is in a superposition of two or more eigenstates and we don't know how much each contributes to $|\psi\rangle$, then I think you're right that statistics tells us the number of times we need to call the QPE grows quadratically.

But I think it's somewhat similar to a classical test. If we are given an audio signal that we are promised is a pure tone, and we have magical Born-rule box that will let us sample from the signal to return its frequency, it only takes one sample for us to know what the tone is. If we are given a signal that's a sum of two or more tones and we don't know the weight of each pure tone with respect to our signal, then I'm still pretty sure we'd have to run our Born box for a quadratic number of times to get a good estimte of $E_\psi$.

Answer 3

I think some of the methods discussed in https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.129.240501 would work.