# How can one estimate the von Neumann entropy of an unknown quantum state?

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?

• 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. Commented Sep 1, 2020 at 16:34
• 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 ... Commented Sep 1, 2020 at 19:36

## 1 Answer

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 $$\rho$$.

• 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.) Commented Sep 1, 2020 at 19:34