Ansatz state for finding the lowest eigenvalue of a $2^ntimes 2^n$ real matrix using VQE

Question

I would like to find the lowest eigenvalue of a $2^ntimes 2^n$ real matrix $H$ using the VQE procedure. The measurement part is simple — I just expand $H$ in a sum of all possible $n$-qubit Pauli terms.

But I have not yet come up with a nice way of preparing an arbitrary $n$-qubit state with real amplitudes. What would be a good ansatz operator $U(theta_1,ldotstheta_{2^n-1})$ to do the job?

(If I were solving the problem in the second-quantized formalism using $2^n$ qubits, I would simply use the Unitary Coupled Cluster with $(2^n - 1)$ single amplitudes — does it have an analogue in such a 'binary' encoding??)

Arthur Pesah · Answer

If you don't have any prior on your matrix, you can use the Strongly Entangling Circuit as a very generic ansatz for qubits.

KAJ226 · Answer

Giving a general $2^n times 2^n $ Hermitian matrix $H$, it's not guarantee that you can decompose/expand $H$ in polynomial strings of $n$-qubit Pauli matrices. So I don't know if VQE is a good method to find lowest eigenvalues of a matrix... unless you know that you can sum $H$ in polynomial terms of Pauli matrices. Even then, in general, VQE only gonna give you a good answer if you have some intuition about the problem. If you pick a variational form that is hardware efficient, using polynomial number of gates, then you can't expect it explore the entire Hilbert space...

Answered by KAJ226 on January 31, 2021

Ansatz state for finding the lowest eigenvalue of a $2^ntimes 2^n$ real matrix using VQE

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