# Von Neumann entropy of mixtures of coherent states

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 $$\rho = \sum_i p_i |\alpha_i\rangle\langle\alpha_i|~?$$

• 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? – Jess Riedel May 23 '16 at 16:04
• 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. – Ilya May 23 '16 at 21:08
• 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. – Jess Riedel May 23 '16 at 22:11
• 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 :) – Jess Riedel May 23 '16 at 23:22
• 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. – 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.

• 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.) – Norbert Schuch May 25 '16 at 22:56
• Thank you for pointing out the mistake. Corrected that and edited the post. – Ilya Jun 3 '16 at 9:16
• 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? – Rococo Jun 3 '16 at 15:16
• @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 $|α\rangle±|β\rangle$ – Frédéric Grosshans Apr 15 '18 at 15:01