# Can the Wigner function be described using coherent states?

The Wigner function for a wave function $$\Psi(\vec{r})$$ is $$W(\vec{r},\vec{k}) = \frac{1}{2\pi} \int dy e^{-i \vec{k} \cdot \vec{y}} \Psi^{*}(\vec{r}-\vec{y}/2) \Psi(\vec{r}+\vec{y}/2) . \tag{1}$$ Recently, I've come to learn of the generalized Cahill phase space representation $$F^{s}(\alpha) = \frac{1}{\pi^{2}} \int d^{2}\xi e^{\alpha\xi^{*}-\alpha^{*}\xi+\frac{s}{2}|\xi|^{2}} \mathrm{Tr}D(\xi)\rho , \tag{2}$$ which is a continuous family of quasiprobability distribution functions with respect to the parameter $$s\in\left[-1,1\right]$$. Here, $$D(\xi) = e^{\xi a^{\dagger} - \xi^{*} a}$$ is the displacement operator, where $$a$$ and $$a^{\dagger}$$ are the annihilation and creation operators satisfying $$[a, a^{\dagger}] = 1$$. As $$s$$ interpolates from $$-1$$ to $$0$$ to $$1$$, it is able to take the form of the Glauber–Sudarshan $$P$$ function, the Wigner Function and the Husimi $$Q$$ function.

This means to me that the Wigner function can be represented in a coherent state expansion $$W(\alpha)$$, but to my knowledge not all states admit to a quantum optics description (like maybe a simple hydrogen atom), therefore can they be presented as such? Is it realizable only mathematically, or does it possess physical meaning in coherent state space?

• The answer to your question is "of course", and you provide the answer, to boot. Your question has little to do with Wigner functions, though. You are seeking to represent the generic density matrix, so, really, state, in terms of the complete basis of coherent states. Write down the wave function, e.g. of hydrogen, in the coherent state basis, and the rest, based on your prescription, is just traces and integrals. That is the question you should be asking and answering, first. – Cosmas Zachos Aug 20 '17 at 13:27
• @CosmasZachos: OK, so it is realizable mathematically, much as I surmised, since the real and imaginary part of a coherent state could be used to correspond to position and momentum respectively, thank you for the insight, much appreciated. – 2ub Aug 25 '17 at 12:11

## 1 Answer

Coherent states exist for a large number of algebraic systems so it's not hard to define a $Q$-function. The difficulty is that, for other values of $s$, the displacement operator takes on much more complicated expression than the Heisenberg-Weyl algebra, meaning that the "formula" you suggest are not terribly convenient.

There is this recent review (open access) that gives some details on the $SU(2)$ case and the Stratonovich approach to Wigner functions, which would be applicable to $SU(n)$. (The work that "resurrected" this Stratonovich approach is Várilly J C and Gracia-Bondía J M 1989 Ann. Phys. 190 107; see also Dowling, Jonathan P., Girish S. Agarwal, and Wolfgang P. Schleich. "Wigner distribution of a general angular-momentum state: Applications to a collection of two-level atoms." Physical Review A 49.5 (1994): 4101.)

Additional examples of Wigner functions have been published for various types of potential (Morse, Poeshl-Teller etc): arXiv is full of examples.

So in short, no the WF cannot always be defined as a coherent state although the so-called Wigner kernel is basically just that. There are other bits and pieces to the kernel when you want anything but the $s=1$ quasiprobability distribution.

• So when can and can't a Wigner function be represented in coherent state space? Say, for the ground state hydrogen atom, can it be expressed as such? – 2ub Aug 20 '17 at 6:26
• Given the symmetry group of hydrogenis $SO(4)$, how do you define a coherent state? You can do a Wigner transform (something like it has been done for the Morse potential for instance) but the result is not a function that satisfies the basic properties of the axiomatic formulation of Stratanovich. – ZeroTheHero Aug 20 '17 at 11:05