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Given many copies of some unknown quantum state $\rho$, I would like to compute its von Neumann entropy $S(\rho)$. What algorithm could be used for this that minimizes the number of copies required? We require that the estimate of the entropy has to be $\varepsilon-$close and one will need more copies as $\varepsilon\rightarrow 0$.

The naive solution is to do tomography and obtain a classical description of the state. This would require exponentially many copies as we increase the dimension of $\rho$. But the classical description of the state has a lot more information so perhaps there is a smarter way?

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    $\begingroup$ My guess is to look at whether there are methods to estimate the spectrum of the quantum state that are more efficient than full tomography. Actually, after a quick search I found this work Measuring Quantum Entropy which deals with the exact problem you state. $\endgroup$
    – Rammus
    Commented Sep 1, 2020 at 16:34
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    $\begingroup$ That's a tricky question, and it depends also whether you want epsilon to scale with N etc.. You could think e.g. about a series expansion of the entropy function in rho. On the other hand, measuring tr(rho^k) is not very efficient since the relative error is typically large. Generally, the entropy is not so easy to estimate well, even given many copies ... $\endgroup$
    – Norbert Schuch
    Commented Sep 1, 2020 at 19:36

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Similar ideas with Quantum PCA may be useful. Meaning, apply Quantum Phase Estimation on unitary $e^{-i\rho t}$ to obtain estimates of the eigenvalues of $\rho$ and finally estimate von Neumann entropy as $S(\rho) = -\sum \tilde{\lambda_i} \text{log}\tilde{\lambda_i}$. In the original paper there is the claim that you obtain estimates of the eigenvalues with accuracy $O(\epsilon)$ from $O(\frac{1}{\epsilon^3})$ copies of $\rho$.

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    $\begingroup$ That's pretty likely not the best way to do it; obtaining the spectrum is far too much information. (And if there are many small eigenvalues - which is the case if the entropy is large - you are pretty much lost. You can't even store that amount of data.) $\endgroup$
    – Norbert Schuch
    Commented Sep 1, 2020 at 19:34

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