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  Are they the same thing: Wigner distribution in quantum Boltzmann equation and Wigner function in quantum optics?

+ 4 like - 0 dislike

We know that quantum Boltzmann equation (QBE) is an equation of motion for the interacting Green's function $G^<(\vec{x}_1,t_1;\vec{x}_2,t_2)\equiv\mathrm{i}\langle \psi^\dagger(x_2)\psi(x_1)\rangle$ when transformed to center of mass and relative coordinates, also coined as Wigner distribution. Besides, in quantum optics and quantum information, we have another Wigner function for phase space description, defined as $W(x,p) = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty}{\mathrm{d} \xi \mathrm{e}^{\mathrm{i} p \xi/\hbar} \langle x-\frac{\xi}{2} \vert\psi\rangle \langle \psi\lvert x+\frac{\xi}{2}\rangle} $. (Certainly, we have more complete definition for more modes and density matrices.)

Unluckily, I happen to be that kind of person who cares about nomenclature. Both of these two Wigner distribution functions are quasiprobability distributions, i.e., not necessarily nonnegative. And I even remember that QBE shares some similar form with certain equation for Wigner function in quantum optics. However, these are no more than faint traces. I guess they are different things, but there must be some plausible relation. Can anyone shed light on this?

Update For anyone who are interested in this question, besides the answer underneath, it may also be helpful to refer to Chap. 4 & 16 of Schieve and Horwitz's book Quantum Statistical Mechanics.

This post imported from StackExchange Physics at 2014-08-22 05:05 (UCT), posted by SE-user huotuichang
asked Mar 14, 2014 in Theoretical Physics by sfman (270 points) [ no revision ]
I think they are the same. In the QBE, one works with the Wigner function of some correlation function/density matrix/wave-function. I think that in your case, it would mean to work with the fourier transform of $G$ with respect to the relative coordinate, i.e. $W(X,P)=\int d\xi e^{iP\xi}G(x_1,x_2)$ with $X=x_1+x_2$ and $\xi=x_1-x_2$.

This post imported from StackExchange Physics at 2014-08-22 05:05 (UCT), posted by SE-user Adam

1 Answer

+ 3 like - 0 dislike

I would say they are not entirely the same, but it depends on the context. First the definitions:

  • the Wigner transform of an operator $\hat{A}$ is defined as $$\tilde{W}\left[\hat{A}\right]=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{A}\left|x+z/2\right\rangle \right]$$ and this is a strange function. You see that on the left, the operator is projected onto a real-space representation, then Fourier transformed. You may find more details (especially the link with the Weyl transform) on the wonderful review by Hillery, M., O’Connel, R. F., Scully, M. O. & Wigner, E. P. Distribution functions in physics: Fundamentals, Phys. Rep. 106, 121–167 (1984) which is unfortunately beyond a paywall.

  • the Wigner transform of the density operator $\hat{\rho}=\left|\Psi\right\rangle \left\langle \Psi\right|$ is then naturally defined as the Wigner transform $$W\left(p,x\right)=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{\rho}\left|x+z/2\right\rangle \right]$$ and it is coined Wigner function in that context.

  • the Green function is not an operator, it is a correlation function, defined as $G\left(x_{1},x_{2}\right)=\left\langle \hat{T}\left[\hat{a}\left(x_{1}\right)\hat{a}^{\dagger}\left(x_{2}\right)\right]\right\rangle $ where $\hat{T}$ is the time-ordering operator, $\hat{a}$ is the (fermionic or bosonic) destruction operator, and $\left\langle \cdots\right\rangle $ represents the averaging process: it could be $\left\langle \cdots\right\rangle =\left\langle N\right|\cdots\left|N\right\rangle $ if you're working with number states $\left|N\right\rangle$, or $\left\langle \cdots\right\rangle =\text{Tr}\left\{ e^{-\beta H}\cdots\right\} /\text{Tr}\left\{ e^{-\beta H}\right\} $ if you're working with thermal averaging ($\beta=\left(k_{B}T\right)^{-1}$ is an inverse temperature in that case), ... Note there are other conventions for the Green functions, but it does not matter here. The Fourier transform of the Green function reads $$G\left(p,x\right)=\int dz\left[e^{\mathbf{i}pz/\hbar}G\left(x-z/2,x+z/2\right)\right]$$ and it looks like a Wigner transform of the Green function, but it should be more appropriate to call it a Fourier transform of the Green function when you choose $x_{1,2}=x\mp z/2$ for the components. In condensed matter theory, $G\left(p,x\right)$ is often called a mixed-Fourier Green function (the full Fourier transform would have given $G\left(p_{1},p_{2}\right)$ instead) or a quasi-classical Green function for the reason to come.

In the limit $\hbar/\tilde{p}\tilde{x}\ll1$ (called quasi-classical limit), with $\tilde{p}\tilde{x}$ the phase-space exploration of the system, the equation of motion of the quasi-classical Green function is the Boltzmann's (transport) equation. The quasi-classical Green functions are not normalised, so they can not be interpreted (whatever it means) as quasi-probability distribution.

As far as I remember, the quasi-classical equation of motion for the Wigner function is not the Boltzmann's one, but the Liouville's one: the collision term is absent, since there is no self-energy method associated with the density matrix. One needs to work with the Lindblad equation for the density matrix, whereas the self-energy method is sufficient when you work with the quasi-classical Green function. Other method to deal with open systems when working with the density matrix is the so-called stochastic method, see e.g. Walls, D. F. & Milburn, G. J. Quantum optics (Springer-Verlag, 1994).

To conclude, note I've put the time under the carpet in the above explanation. That's for a good reason: time is always more complicated to deal with in the Wigner-Weyl transform, especially in the quasi-classical limit and with the Green functions method. The use of the Wigner function is not a big problem when time is taken into account. Of course dealing with the Lindblad equation is not a simple issue... but that's an other story :-)

This post imported from StackExchange Physics at 2014-08-22 05:05 (UCT), posted by SE-user FraSchelle
answered Mar 14, 2014 by FraSchelle (390 points) [ no revision ]
Thanks a lot for your thorough answer! It's clear and very helpful. I also added an update part to my question, serving as an alternative discussion.

This post imported from StackExchange Physics at 2014-08-22 05:05 (UCT), posted by SE-user huotuichang

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