Phase-space QM

Quantum Brownian motion: Definition

By Jess Riedel October 10, 2014 June 7, 2016 Physics

In this post I’m going to give a clean definition of idealized quantum Brownian motion and give a few entry points into the literature surrounding its abstract formulation. A follow-up post will give an interpretation to the components in the corresponding dynamical equation, and some discussion of how the model can be generalized to take into account the ways the idealization may break down in the real world.

I needed to learn this background for a paper I am working on, and I was motivated to compile it here because the idiosyncratic results returned by Google searches, and especially this MathOverflow question (which I’ve answered), made it clear that a bird’s eye view is not easy to find. All of the material below is available in the work of other authors, but not logically developed in the way I would prefer.

Preliminaries

Quantum Brownian motion (QBM) is a prototypical and idealized case of a quantum system $\mathcal{S}$ , consisting of a continuous degree of freedom, that is interacting with a large multi-partite environment $\mathcal{E}$ , in general leading to varying degrees of dissipation, dispersion, and decoherence of the system. Intuitively, the distinguishing characteristics of QBM is Markovian dynamics induced by the cumulative effect of an environment with many independent, individually weak, and (crucially) “phase-space local” components. We will defined QBM as a particular class of ways that a density matrix may evolve, which may be realized (or approximately realized) by many possible system-environment models. There is a more-or-less precise sense in which QBM is the simplest quantum model capable of reproducing classical Brownian motion in a $\hbar \to 0$ limit.

In words to be explained: QBM is a class of possible dynamics for an open, quantum, continuous degree of freedom in which the evolution is specified by a quadratic Hamiltonian and linear Lindblad operators.… [continue reading]

4 Comments

In what sense is the Wigner function a quasiprobability distribution?

By Jess Riedel September 22, 2014 May 22, 2020 Physics

For the upteenth time I have read a paper introducing the Wigner function essentially like this:

The Wigner-representation of a quantum state $\rho$ is a real-valued function on phase space defined^a (with $\hbar=1$ ) as
(1) $\begin{align*} W_\rho(x,p) \equiv \int \! \mathrm{d}\Delta x \, e^{i p \Delta x} \langle x+\Delta x /2 \vert \rho \vert x-\Delta x /2 \rangle. \end{align*}$
It’s sort of like a probability distribution because the marginals reproduce the probabilities for position and momentum measurements:
(2) $\begin{align*} P(x) \equiv \langle x \vert \rho \vert x \rangle = \int \! \mathrm{d}p \, W_\rho(x,p) \end{align*}$
and
(3) $\begin{align*} P(p) \equiv \langle p\vert \rho \vert p \rangle = \int \! \mathrm{d}x \, W_\rho(x,p). \end{align*}$
But the reason it’s not a real probability distribution is that it can be negative.

The fact that $W_\rho(x,p)$ can be negative is obviously a reason you can’t think about it as a true PDF, but the marginals property is a terribly weak justification for thinking about $W_\rho$ as a “quasi-PDF”. There are all sorts of functions one could write down that would have this same property but wouldn’t encode much information about actual phase space structure, e.g., the Jigner^b function

$J_\rho(x,p) \equiv P(x)P(p) = \langle x \vert \rho \vert x \rangle \langle p \vert \rho \vert p \rangle,$

which tells as nothing whatsoever about how position relates to momentum.

Here is the real reason you should think the Wigner function $W_\rho$ is almost, but not quite, a phase-space PDF for a state $\rho$ :

Consider an arbitrary length scale $\sigma_x$ , which determines a corresponding momentum scale $\sigma_p = 1/2\sigma_x$ and a corresponding set^c of coherent states $\{ \vert \alpha \rangle \}$ .
If a measurement is performed on $\rho$ with the POVM of coherent states $\{ \vert \alpha \rangle \langle \alpha \vert \}$ , then the probability of obtaining outcome $\alpha$ is given by the Husimi Q function representation of $\rho$ :
(4) $\begin{align*} Q_\rho(\alpha) = \langle \alpha \vert \rho \vert \alpha \rangle. \end{align*}$
If $\rho$ can be constructed as a mixture of the coherent states $\{ \vert \alpha \rangle \}$ , then

… [continue reading]

No comments

Diagonal operators in the coherent state basis

By Jess Riedel June 13, 2014 October 11, 2014 Physics

I asked a question back in November on Physics.StackExchange, but that didn’t attract any interest from anyone. I started thinking about it again recently and figured out a good solution. The question and answer are explained below.^a

Q: Is there a good notion of a “diagonal” operator with respect the overcomplete basis of coherent states?
A: Yes. The operators that are “coherent-state diagonal” are those that have a smooth Glauber–Sudarshan P transform.

The primary motivation for this question is to get a clean mathematical condition for diagonality (presumably with a notion of “approximately diagonal”) for the density matrix of a system of a continuous degree of freedom being decohered. More generally, one might like to know the intuitive sense in which $X$ , $P$ , and $X+P$ are all approximately diagonal in the basis of wavepackets, but $RXR^\dagger$ is not, where $R$ is the unitary operator which maps

(1) $\begin{align*} \vert x \rangle \to (\vert x \rangle + \mathrm{sign}(x) \vert - x \rangle) / \sqrt{2}. \end{align*}$

(This operator creates a Schrodinger’s cat state by reflecting about $x=0$ .)

For two different coherent states $\vert \alpha \rangle$ and $\vert \beta \rangle$ , we want to require an approximately diagonal operator $A$ to satisfy $\langle \alpha \vert A \vert \beta \rangle \approx 0$ , but we only want to do this if $\langle \alpha \vert \beta \rangle \approx 0$ . For $\langle \alpha \vert \beta \rangle \approx 1$ , we sensibly expect $\langle \alpha \vert A \vert \beta \rangle$ to be within the eigenspectrum of $A$ .

One might consider the negativity of the Wigner-Weyl transform ^b of the density matrix (i.e. the Wigner phase-space quasi-probability distribution aka the Wigner function) as a sign of quantum coherence, since it is known that coherent superpositions (which are clearly not diagonal in the coherent state basis) have negative oscillations that mark the superposition, and also that these oscillations are destroyed by decoherence.… [continue reading]

6 Comments

Wigner function = Fourier transform + Coordinate rotation

By Jess Riedel April 1, 2014 January 26, 2017 Physics

[Follow-up post: In what sense is the Wigner function a quasiprobability distribution?]

I’ve never liked how people introduce the Wigner function (aka the Wigner quasi-probability distribution). Usually, they just write down a definition like

(1) $\begin{align*} W(x,p) = \frac{1}{\pi \hbar} \int \mathrm{d}y \rho(x+y, x-y) e^{-2 i p y/\hbar} \end{align*}$

and say that it’s the “closest phase-space representation” of a quantum state. One immediately wonders: What’s with the weird factor of $2$ , and what the heck is $y$ ? Usually, the only justification given for the probability interpretation is that integrating over one of the variables recovers the probability distribution for the other (if it were measured):

(2) $\begin{align*} \int \! \mathrm{d}p \, W(x,p) = \vert \rho(x,x) \vert^2 , \\ \int \! \mathrm{d}x \, W(x,p) = \vert \hat{\rho}(p,p) \vert^2 , \end{align*}$

where $\hat{\rho}(p,p')$ is just the density matrix in the momentum basis. But of course, that doesn’t really tell us why we should think of $W(x,p)$ , as having anything to do with the (rough) value of $x$ conditional on a (rough) value of $p$ .

Well now I have a much better idea of what the Wigner function actually is and how to interpret it. We start by writing it down in sane variables (and suppress $\hbar$ ):

(3) $\begin{align*} W(\bar{x},\bar{p}) = \frac{1}{2 \pi} \int \! \mathrm{d}\Delta x \,\rho \left(\bar{x}+\frac{\Delta x}{2}, \bar{x}-\frac{\Delta x}{2} \right) e^{-i \bar{p} \Delta x}. \end{align*}$

So the first step in the interpretation is to consider the function

(4) $\begin{align*} M(\bar{x},\Delta x) \equiv \rho \left(\bar{x}+\frac{\Delta x}{2}, \bar{x}-\frac{\Delta x}{2} \right) , \end{align*}$

which appears in the integrand. This is just the (position-space) density matrix in rotated coordinates $\bar{x} \equiv (x+x')/2$ and $\Delta x = x-x'$ . There is a strong sense in which the off-diagonal terms of the density matrix represent the quantum coherence of the state between different positions, so $\Delta x$ indexes how far this coherence extends; large values of $\Delta x$ indicate large spatial coherence. On the other hand, $\bar{x}$ indexes how far down the diagonal of the density matrix we move; it’s the average position of the two points between which the off-diagonal terms of the density matrix measures coherence. (See the figure below.)… [continue reading]