As far as I know after minimization I have to obtain a value $E_{0}\le \frac{\langle \psi (\theta)|H|\psi (\theta)\rangle}{\langle \psi (\theta)|\psi (\theta)\rangle}$, where $E_{0}$ - eigenvalue of ground state for hamiltonian $H$. Sometimes the algorithm give the value close to $E_{0}$, but far more often I get values lower than that.

I use hardwave efficient ansatz for initial state generation.

Hamiltonian consists of Pauli-strings $H=\sum_{ijkl}\sigma_i\sigma_j\sigma_k\sigma_l$.

For parameters optimization I use "COBYLA" and "Nelder-Mead" methods.

Could it be that the ansatz produce a state space which is not large enough?

  • $\begingroup$ hard to know the exact reason without knowing the details. It's probably due to numerical errors if you get values lower than $E_0$ but not by much. Isolate the state found by the optimizer that gives the problem and try computing the expectation values in different ways. $\endgroup$
    – glS
    Dec 9 '19 at 17:12
  • $\begingroup$ I tried to use the algorithm for less number of Pauli-strings in Hamiltonian, I found out that the simpler the Hamiltonian the more correct the result.Could you explain what you mean by "Isolate the state", please? $\endgroup$
    – C-Roux
    Dec 9 '19 at 19:14
  • $\begingroup$ The most likely reason in practice (assuming your implementation is error-free) is noise. I'll write a more detailed response tomorrow. $\endgroup$
    – Arthur-1
    Dec 10 '19 at 4:39
  • $\begingroup$ I mean that if the algorithm is finding the value of $\theta$ corresponding to what it thinks is the minimum, you can investigate directly what is going wrong in the numerics using that value $\endgroup$
    – glS
    Dec 10 '19 at 21:07

The proof of the variational theorem (the theorem that the ground state energy is the lowest possible energy you can get from $\frac{\langle \psi|H|\psi\rangle}{ \langle \psi | \psi \rangle}$) is very simple: https://en.wikipedia.org/wiki/Variational_method_(quantum_mechanics)

If you get a lower energy, it means you don't actually have $\frac{\langle \psi|H|\psi\rangle}{ \langle \psi | \psi \rangle}$. For example if the quantum hardware doesn't give you $H|\psi\rangle$ but instead gives you $H|\psi\rangle + \epsilon |\psi\rangle $ where $\epsilon$ is some non-zero error, then when you plug everything into the proof of the variational theorem you may find that you are no longer guaranteed to always get energies equal to or higher than the ground state energy.


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