Basically the question is, whether you can give some estimation for the Hamiltonian of a system, given the time evolution of a density matrix $\rho$ under the assumption that it obeys the von-Neumann equation $$\frac{d}{dt} \rho = -\frac{i}{\hbar} [H,\rho]\quad .$$ If this is not the case (not even numerically) what if we make some assumptions that the Hamiltonian has some specific form, for example if you know that you only have $ZX$-coupling between some qubits, and you know their frequencies.
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2$\begingroup$ What happens in Newtonian mechanics if you only know some of the forces? $\endgroup$– By SymmetryCommented Apr 2 at 8:51
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1$\begingroup$ Are you asking for practical "estimations" or whether there is only one Hamiltonian such that the von-Neumann equation holds for a given $\rho(t)$? $\endgroup$– Tobias FünkeCommented Apr 2 at 8:58
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$\begingroup$ I mean you can do a simulation and see whether your model produces the same outcome, but I was thinking of a quicker or more straight forward way of solving this equation for H(t). I am asking for practical estimations, especially if there is a way to come up with a first guess and then do some repeated simulation/fitting $\endgroup$– JurekCommented Apr 2 at 8:58
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1$\begingroup$ Well, $H$ is certainly not unique. Assume you've found some $H$ for a given $\rho(t)$ obeying a von Neumann equation, then any Hamiltonian of the form $\tilde H=c + H$ is also a solution. With a little bit of thought, you can also see the following: Given two solutions of this problem $H_1,H_2$ then it must be the case that $H_1-H_2$ commutes with $\rho$. Also, you can add any function $f(\rho)$ to a solution and get a new solution. $\endgroup$– Tobias FünkeCommented Apr 2 at 9:01
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Found the solution myself with the help of this post https://math.stackexchange.com/questions/1307098/is-there-any-inverse-commutator-for-matrices
Essentially the equation becomes with vectorisation
$\text{Vec} \dot{\rho} = \frac{i}{\hbar} [1 \otimes \rho - \rho^T \otimes 1] \text{Vec} H$
from this we can compute a pseudo inverse. Define $A = [1 \otimes \rho - \rho^T \otimes 1]$ and let $A^*$ be the Moore-Penrose-Inverse. Then we have
$\frac{i}{\hbar} A^* A \text{Vec} H = A A^* \dot{\rho}$
However, be careful I am not yet convinced that this is a hermitian Hamiltonian.
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1$\begingroup$ As I said in the comments: There is not a unique solution; and solutions may differ by non-trivial terms. $\endgroup$ Commented Apr 2 at 19:33
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$\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$– Community BotCommented Apr 2 at 22:45