# What's the intuitive reason that phase space flow is incompressible in Classical Mechanics but compressible in Quantum Mechanics?

One of the most important results of Classical Mechanics is Liouville's theorem, which tells us that the flow in phase space is like an incompressible fluid.

However, in the phase space formulation of Quantum Mechanics, one of the main results due to Moyal is that quantum flows are compressible.

So what's the intuitive reason for this difference?

Formulated a bit differently: What exactly is the assumption used in the derivation of Liouville's theorem that is no longer valid in Quantum Mechanics?

• Quantum mechanics is so fundamentally different from classical mechanics that I would reverse the question and ask you: Why would you even think that both theories behave the same in this respect? Quantum mechanics uses a wave function on configuration space in its usual formulation, so phase space does not have any fundamental role any more. – Luke Jul 4 '18 at 11:31
• @luke sry but this is wrong. Pls check out the wiki article I linked to above. The phase space formulation of QM works perfectly and is in fact analogous to CM in phase space. There are only a few crucial differences which I'm trying to understand here. And btw. wave functions do not live in configuration space but Hilbert space. We can also formulate QM in configuration space and this is known as path integral formulation. – jak Jul 4 '18 at 11:49
• – Qmechanic Jul 4 '18 at 12:06
• Just a guess but I would think the uncertainty principle is the reason. – Lewis Miller Jul 4 '18 at 12:07
• @LewisMiller Yes that's probably it. But so far I was unable to spell it out explicitly. The incompressibility in Classical Mechanics means that there are no sources or sinks of trajectories in phase space. So somehow the uncertainty principle must lead to such sources and sinks in phase space... – jak Jul 4 '18 at 13:17

1. The so-called (generic) failure of the quantum Liouville theorem, i.e. the (generic) violation of the continuity equation $$\rho~{\rm div}_{\rho} X^Q_{-H} + \frac{\partial \rho}{\partial t}~\neq~0 \tag{1}$$ $$\qquad\Leftrightarrow\qquad \frac{\partial \rho}{\partial t} ~\stackrel{(1)}{\neq}~\rho~{\rm div}_{\rho} X^Q_H ~\stackrel{(6)+(7)}{=}~\rho~{\rm div}_{\rho} X_H ~\stackrel{\text{Leibniz}}{=}~ X_H[z^I] ~\frac{\partial\rho}{\partial z^I} ~+~ \rho~\underbrace{ {\rm div}_1 X_H}_{=0}\tag{2}$$ for the quantum flow on a $2n$-dimensional phase space, can intuitively be understood as the appearance of higher-order differential operators $X$ in the $\star$-product, which do not (necessarily) obey Leibniz rule $$X[fg] ~\neq~X[f]g+f X[g] .\tag{3}$$

2. In eq. (1) we have defined the quantum ($Q$) version $$X^Q_H~:=~\frac{1}{i\hbar}[H\stackrel{\star}{,}\cdot] ~=~ \frac{2}{i\hbar}H\sinh\left(\frac{i\hbar}{2}\stackrel{\leftarrow}{\partial} \wedge\stackrel{\rightarrow}{\partial}\right) ~=~X_H + {\cal O}(\partial^3) \tag{4}$$ of a Hamiltonian vector field $$X_H ~:=~ \{H, \cdot \}~=~H\stackrel{\leftarrow}{\partial} \wedge\stackrel{\rightarrow}{\partial} .\tag{5}$$ Note that the coordinate components are the same $$X^Q_H[z^I]~\stackrel{(4)}{=}~X_H[z^I],\tag{6}$$ which is part of the problem. Also in eq. (1) we have for convenience defined a divergence $${\rm div}_{\rho} X~:=~ \rho^{-1}\frac{\partial(\rho X[z^I])}{\partial z^I}\tag{7}$$ of a possibly higher-order differential operator $X$. Eq. (7) is not a geometric object, which foretells the doom of eq. (1).

3. The quantum Liouville theorem (1) is replaced by the quantum Liouville equation $$0~=~\frac{d\rho}{dt}~=~X^Q_{-H}[\rho]+\frac{\partial \rho}{\partial t}\tag{8}$$ $$\qquad\Leftrightarrow\qquad\frac{\partial \rho}{\partial t} ~\stackrel{(8)}{=}~X^Q_{H}[\rho] ~\stackrel{(3)}{\neq}~ X^Q_H[z^I] ~\frac{\partial\rho}{\partial z^I} ~\stackrel{(6)}{=}~ X_H[z^I] ~\frac{\partial\rho}{\partial z^I} .\tag{9}$$ The ineq. (9) is precisely the ineq. (2). $\Box$

4. See also this related Phys.SE post.

Apologies for my inability to share intuition, a frequently subjective issue... I have learned a lot by reading the Steuernagel group numerical flows and topological features of such flows, in practice. For a recent discussion/proof of the zeros, singularities,and negative probability density features, hence your source-sink query in anharmonic quantum systems see Kakofengitis, Oliva & Steuernagel, 2017. Basically all bets are off when you (a point in phase space) and neighbors enter a phase-space cell of order $$\hbar$$, by dint of the uncertainty principle, and that includes a definition of what a trajectory is.

If you watch the nifty movies of Cabrera and Bondar in the WP article you are linking to for the Morse and quartic potentials, you actually see this in real time, as a lump (you) spread all over the phase space in a highly organized way... I defy you to discern trajectories there! There is powerful topology at work, but I'd defer to Steuernagel for that.

As a practical reassurance, I'll work out a trivial exercise from our book, on compressibility of Euler flows. For a Hamiltonian $$H=p^2/(2m)+V(x)$$, the Moyal evolution equation amounts to an Eulerian probability transport continuity equation, $$\frac{\partial f(x,p)}{\partial t} +\partial_x J_x + \partial_p J_p=0~,$$ where, for $$\mathrm{sinc}(z)\equiv \sin z/~ z$$ , the phase-space flux is $$J_x=pf/m~ ,\\ J_p= -f \mathrm{sinc} \left( {\hbar \over 2} \overleftarrow {\partial _p} \overrightarrow {\partial _x} \right)~~ \partial_x V(x).$$
Classical mechanics is crucially different, in that the phase-space current is always $${\bf J}=(p/m,-\partial_xV(x))f$$, and the velocity $${\bf v}=(p/m,-\partial_xV(x))$$, manifestly divergenceless in phase space.

Now note for the oscillator, $$V_1= x^2/2$$, $$J_p=-f x$$, so the phase-space velocity $${\bf v}=(mp,- x)$$ and $$\nabla \cdot {\bf v}=0$$, incompressibility. This is a reminder that the quantum oscillator is basically classical, and its wave packets do not spread, as iconically pointed out by Schroedinger... coherent states. But this is a crying exception.

For a more generic potential, like the quartic, $$V_2= x^4/4$$, $$v_p=J_p/f= -x^3 +\hbar^2 x ~\partial_p^2f /f , \\ \nabla \cdot {\bf v}= \hbar^2 x ~\partial_p (\partial_p^2 f(x,p) /f(x,p))\neq 0,$$ so the flow is modified by $$O(\hbar^2)$$ to compressible.

So the strictly quantum difference between the quantum Moyal bracket and the classical Poisson bracket is the crucial element in increasing or decreasing the amount of (quasi)probability in a comoving phase-space region $$\Omega$$, since $${d \over dt}\! \int_{\Omega}\! \! dx dp ~f= \int_{\Omega}\!\! dx dp \left ({\partial f \over \partial t}+ \partial_x (\dot{x} f) + \partial_p (\dot{p} f ) \right) = \int_{\Omega}\! \!\! dx dp~ (\{\!\!\{ H,f\}\!\!\}-\{H,f\})\neq 0 ~.$$

• Note added: It is even odder. Quantum flows display physically significant viscosity!
• Thanks for your answer. For some reason, I wasn't able to "ping" you. So I tried to come up with a somewhat intuitive picture and would be really interested in your opinion. (See especially also the discussion in the comments below) – jak Jul 5 '18 at 9:06
• when I use your username in a comment like @username you get a little notification. However, for some reason this is currently not possible. – jak Jul 5 '18 at 14:58
• @JakobH It's because this is his post. The poster receives notifications for all untargeted comments, so the comment system removes the redundancy of addressing the poster, because it's identical to simply leaving the comment untargeted. – probably_someone Jul 5 '18 at 19:53

Here's my really naive attempt to answer my own question. Please correct me where I'm wrong.

Each point in phase space corresponds to one specific state of the system $(\vec{q}_1,\vec{q}_2,\ldots,\vec{p}_1,\vec{p}_2,\ldots)$. As time passes on this point moves and traces out an orbit in phase space. This orbit can be calculated using Hamilton's equations.

Neighboring points describe similar states. So when we are uncertain about the exact state of our system (which we always are thanks to our limited measurement accuracy) we have to take this into account by using a phase space distribution function. This function $\rho (p,q)$ determines the probability $\rho (\vec{q}_1,\vec{q}_2,\ldots,\vec{p}_1,\vec{p}_2,\ldots)\,d^{n}q\,d^{n}p$ that the system will be found in the infinitesimal phase space volume $d^{n}q\,d^{n}p$. Liouville's equation determines the "orbit" of our initial phase space distribution function. The path that is traced out this way defines a flow in phase space. The two essential ingredients in the derivation of Liouville's equation are

1. Hamilton's equations
2. The continuity equation for $\rho (p,q)$.

The second ingredient here leads us to the famous conclusion that the phase space flow is incompressible. What this means is that we can put down a pencil on each possible initial configuration (possible in a statistical sense since we are not 100% sure about the initial configuration) and then trace out the phase space flow by moving these pencils through our phase space.

Now in Quantum Mechanics, this is no longer true. Our phase space flow is compressible. In other words, the continuity equation is no longer correct since there are sources and sinks. What this means is that when we try to trace our obits using pencils we will fail. One trajectory can split into two and other trajectories possibly vanish. (There are sources for new trajectories and sinks where trajectories end.)

This is a result of the fundamental uncertainty in Quantum Mechanics. While there can also be uncertainty in Classical Mechanics (which is why we use the probability function and Liouville's equation in the first place) it is of a different kind. In Quantum Mechanics there isn't one unique orbit for each possible initial configuration. This is what we mean when we say the phase space flow in Quantum Mechanics is compressible.

• A crompressible flow still fulfills a continuity equation. What makes a flow incompressible is only the fact that the divergence of the velocity field is zero. – sagittarius_a Jul 5 '18 at 8:32
• Thanks! Maybe the terminology is somewhat unfortunate here. What I was thinking of is like the difference between water and a radioactive fluid. The number of water molecules stays the same over time and therefore the volume occupied by the water. In contrast, for the radioactive fluid molecules decay and over time a smaller and smaller volume is occupied. However, "compressible" really seems like the wrong word here. It's more like "self-compressing" (or expanding) over time. The crucial question now, of course, is what is really going on in QM. – jak Jul 5 '18 at 9:03
• So formulated differently, is the QM phase space flow compressible in the usual sense (gas vs. fluid) or more like in the radioactive fluid vs water case? I would be really interested to hear @CosmasZachos opinion – jak Jul 5 '18 at 9:04
• I'm not sure I understand radioactive fluid stuff, and I don't want to stand between you and your intuition. As WP emphasizes, compressible means failure of the material derivative to vanish because the divergence of the velocity fails to vanish. Probability is conserved, of course, but not in every comoving cell, where it may concentrate in or diffuse out of. As in my other answer linked, the efflux amounts to the integral of the difference between the MB and the PB over such a cell. – Cosmas Zachos Jul 5 '18 at 14:07