I'm trying to calculate the Von Neumann entropy of statistical mixtures of coherent states. The problem is that such states are in general non-Gaussian, so one cannot follow the formalism developed here: Phys. Rev. A 59, 1820 (1999). Does anybody have any hints on how to calculate the $$Tr[\rho\log[\rho]],$$ for $$\rho = \sum_i p_i |\alpha_i\rangle\langle\alpha_i|~?$$

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    $\begingroup$ Can you say more about the $p_i$? The case of two Guassians can of course be calculated exactly in terms of the overlap $\langle \alpha_1 \vert \alpha_2 \rangle$; just move to the 2-dim subspace spanned by the vectors. Likewise for a finite number of Gaussians, the answer can be computed with the Gram matrix. (Since the pairwise inner products aren't very restricted by the Gaussianity condition, I doubt the answer simplifies much from the nonGaussian case.) For a continuous distribution $p(\alpha)$, couldn't things be arbitrarily complicated? $\endgroup$ – Jess Riedel May 23 '16 at 16:04
  • $\begingroup$ I thought about $p_i$ being just classical probabilities. They are real numbers and their squares sum up to one. Yes in a continuous case as long as $p(\alpha)$ does not lead to Gaussian state, I doubt there is something general. But could you explain more in details about just two terms in the sum. I try to calculate the logarithm by the formula $\log[\rho] = \sum_n \rho^n/n!$ and things become messy quite quick. $\endgroup$ – Ilya May 23 '16 at 21:08
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    $\begingroup$ Construct a 2-dim orthonormal basis $\vert + \rangle, \vert - \rangle$ that spans the same subspace as $\vert \alpha_1 \rangle, \vert \alpha_2 \rangle$. Rewrite $\rho$ in this basis as a 2x2 matrix, which can be diagonalized, giving the spectrum of $\rho$. The entropy is a function of the spectrum. $\endgroup$ – Jess Riedel May 23 '16 at 22:11
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    $\begingroup$ Yes I know. The full Hilbert space is infinite dimensional but, if you are given only two vectors from such a Hilbert space, they span only a single, two-dimensional subspace. The density matrix, when considered as an operator, acts nontrivially only on that subspace. (It is zero in the orthogonal subspace). Ask your advisor :) $\endgroup$ – Jess Riedel May 23 '16 at 23:22
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    $\begingroup$ Maybe you can give one specific example? This would probably facilitate explaining it. (Ideally only with two non-zero $p_i$ ;-) BTW, a bit of a related discussion (compressing coherent states to a finite-dimensional space is in physics.stackexchange.com/a/208576/4888. $\endgroup$ – Norbert Schuch May 24 '16 at 7:14

It seems I have figured out an answer for 2 terms in the original state. Suppose that the state is

$$\rho = a |\alpha \rangle \langle \alpha | + (1-a) |\beta\rangle \langle \beta|$$

We need to construct an orthonormal basis to, in which this system will act as a 2-level system. One of the variants is $$|+\rangle = |\alpha\rangle; \quad |-\rangle = \frac{|\beta\rangle - k|\alpha\rangle }{\sqrt{1-k^2}}, $$ where $k=\langle \alpha |\beta \rangle$. The elements of the new density matrix $\rho^\pm$ are $$\rho_{11}=\langle+|\rho|+\rangle; \quad \rho_{12}=\langle+|\rho|-\rangle; \quad \rho_{21}=\langle-|\rho|+\rangle; \quad \rho_{22}=\langle-|\rho|-\rangle;$$ And thus it is: $$\rho^\pm=\begin{pmatrix}a+(1-a)|k|^2 & \frac{k(1-a)(1-|k|^2)}{\sqrt{1-k^2}} \\ \frac{k^*(1-a)(1-|k|^2)}{\sqrt{1-k^{*2}}} & (1-a)(1-|k|^2) \end{pmatrix}.$$

It is possible now to calculate the entropy. When $|\alpha \rangle $ and $|\beta \rangle$ have the same phase, the dependence of the entropy on the parameter $a$ and separation between states $d$, $\langle \alpha |\beta \rangle = \exp{(-d^2)}$ looks like this:


It seems reasonable as it is zero at zero separation, as the state is pure than and also goes to $0$ when $a = 1$ or $0$.

Edit: Thanks to Jess Riedel for the instructions.

  • $\begingroup$ Your basis states are not normalized properly, thus the result is likely incorrect. (In fact, depending of the phase choice of $|\alpha\rangle$ and $|\beta\rangle$, they are not even orthogonal.) $\endgroup$ – Norbert Schuch May 25 '16 at 22:56
  • $\begingroup$ Thank you for pointing out the mistake. Corrected that and edited the post. $\endgroup$ – Ilya Jun 3 '16 at 9:16
  • $\begingroup$ Nice post. So the lesson here is basically that although the full basis of coherent states is overcomplete, for a given set one can construct a orthonormal basis by the usual gram-Schmidt procedure and then calculate things normally. Is that correct? $\endgroup$ – Rococo Jun 3 '16 at 15:16
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    $\begingroup$ @Rococo : Indeed, for a discrete set, you can always use the usual Gram-Schmidt procedures. But, when the problem is nicely symmetric,there are other bases which makes the solution easier. Here, for example, I would use the basis made from the (normalized) states $|α\rangle±|β\rangle$ $\endgroup$ – Frédéric Grosshans Apr 15 '18 at 15:01

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