I have $N$ pure, but nonorthogonal, states $|\psi_n\rangle$ with density matrix $\rho_n=|\psi_n\rangle\langle\psi_n|$.

Say we call the the total density matrix $\rho=\frac{1}{N}\sum_n \rho_n$.

Are there any formulas to calculate $S_{vn}[\rho]=-\mathrm{Tr}[\rho\log{\rho}]$ solely from the overlaps $\langle\psi_n|\psi_m\rangle$ (and the fact that all individual states are pure)? Gram-Schmidt orthogonalization might be a possibility but perhaps there is an easier way/existing result?

Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?

  • $\begingroup$ Be so kind and give a definition for $S_{vn}$. $\endgroup$ – flippiefanus Aug 26 at 12:52
  • $\begingroup$ @flippiefanus done, thanks $\endgroup$ – Wouter Aug 26 at 13:37

Let us denote the overlap matrix by $O$, this is, $O_{nm}=\langle\psi_n\vert\psi_m\rangle$. Then, $$ S_{\mathrm{vN}}(\rho)=S_\mathrm{vN}(O)\ . $$ More generally, $O$ has the same non-zero eigenvalues as $\rho$, so any function of the eigenvalues (which is insensitive to zero eigenvalues) can be evaluated on $O$ instead of $\rho$.

This can be seen by defining a matrix $X$ whose columns are the states $\vert\psi_n\rangle$, i.e. $X_{kn}=\langle k\vert\psi_n\rangle$. Then, $\rho=XX^\dagger$, while $O=X^\dagger X$. However, two matrices $AB$ and $BA$ always have the same non-zero eigenvalues, and thus, the non-zero eigenvalues of $\rho$ and $O$ are the same.

  • $\begingroup$ Very interesting, thanks! Is there no way to get rid of the phases in O then? $\endgroup$ – Wouter Aug 26 at 16:57
  • $\begingroup$ @Wouter Why should there? E.g. for 3 states, you can transform away the relative phase of 12 and 13, but the phase of 23 cannot be changed at will. $\endgroup$ – Norbert Schuch Aug 26 at 21:41
  • $\begingroup$ ... It shouldn't be too hard to construct an example on those grounds. $\endgroup$ – Norbert Schuch Aug 26 at 21:47
  • $\begingroup$ Ok, my question has arisen in the context where $\ket{\psi_n}$ are all manymode Gaussian states, and there seem to be formulas much more readily available for Tr[\rho_n \rho_m] =|<\psi_n|\psi_m>|^2 than <\psi_n|\psi_m> itself; although density matrices should in principle contain all relavant information (also corresponding to eg. a W-function). But I suspect the issue is that the vn-entropy depends also on correlators with more than two density matrices... $\endgroup$ – Wouter Aug 27 at 7:06
  • $\begingroup$ It turns out that all phasefactors cán be transformed away, this is just how addition works: the difference in phase between 1 and 3 is the difference between 1 and 2 plus the difference between 2 and 3. The short answer, therefore, is that these phasefactors are not relevant, and the spectrum and hence $S_{vN}$ remain unaffected by replacing $\langle\psi_n|\psi_m\rangle\rightarrow|\langle\psi_n|\psi_m\rangle|$ in every entry of $O_{nm}$. If you agree with this point (which I tested for some random examples), perhaps it would be useful to add it to your answer for visibility. $\endgroup$ – Wouter Sep 10 at 12:38

In such situations it is common to calculate the linear entropy instead of the full VN entropy. The linear entropy comes from the first term approximation in the expansion of $\log(\rho)$ as Taylor series around $\rho= I$, which is $I-\rho$ (I might be wrong about the sign here). Then the first term approximation to the VN entropy is $$Tr[\rho\log\rho]\longmapsto Tr[\rho (I-\rho)]=1-Tr[ \rho^2]$$ where $Tr[ \rho^2]$ is commonly referred to as purity. The purity is 1 for pure states and $\frac{1}{d}$ for completely mixed states (here $d$ is the dimension of the Hilbert space).

Back to your problem. The purity of $\rho$ is $$Tr[\rho^2]=Tr[(\frac{1}{N}\sum_n \rho_n)^2]=\frac{1}{N^2}\sum_{n,m}Tr[\rho_n\rho_m] $$ which, as you have anticipated, depends on the overlaps.

  • $\begingroup$ Why should that be common in "such" situations? Also, the linear entropy is not a "VN" entropy -- it is the 2-Renyi entropy (or, in your normalization, maybe rather a Tsallis entropy). $\endgroup$ – Norbert Schuch Aug 26 at 23:01
  • $\begingroup$ It should be noted that the formula you get is exactly $\mathrm{tr}[O^2]$, which is consistent with the fact that you can replace $\rho$ by $O$ for any quantity which only depends on the eigenvalues. (As it happens, this is the only entropic quantity which only depends on the absolute values of $O_{nm}$.) $\endgroup$ – Norbert Schuch Aug 26 at 23:32
  • $\begingroup$ @Norbert Schuch Thank you for pointing out my confusion of the nomenclature -- I edited out the VN out of linear entropy. The context of this problem is not clear here so the "such" refers to the part that asks about how the absolute values of overlaps can play a part. Since the very common notion of purity is clearly related to this intuition I decided to point that out (it is not given in opposition or contradiction to your answer). $\endgroup$ – oleg Aug 27 at 1:39
  • $\begingroup$ Indeed, the latter is a good point -- which (see my comment above) comes from the fact that the Hilbert-Schmidt-norm has the property that it only depends on the absolute value. $\endgroup$ – Norbert Schuch Aug 27 at 1:58
  • $\begingroup$ Thanks, perhaps I will try this one first and see how far it gets me $\endgroup$ – Wouter Aug 27 at 7:08

Since part of the questions is

Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?

let me provide an explicit counterexample:


$$\newcommand{\ket}[1]{\vert#1\rangle}\ket{\psi_1}=\frac{\ket0+\ket1}{\sqrt{2}}, \ket{\psi_2}=\frac{\ket1+\ket2}{\sqrt{2}}, \ket{\psi_3}=\frac{\ket0+\ket2}{\sqrt{2}}\ . $$ The mixing entropy of $ \newcommand{\kb}[1]{\vert#1\rangle\langle#1\vert} \rho=\frac13(\kb{\psi_1}+\kb{\psi_2}+\kb{\psi_3})$ is then $S(\rho)=\log(3)-\tfrac13\log(2)\approx0.87$.

On the other hand, if we change $\ket{\psi_3}$ to $$\ket{\psi_3}=\frac{\ket0-\ket2}{\sqrt{2}}\ ,$$ we obtain $S(\rho)=\log(2)\approx0.69$.


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