So I was trying to solve the Liouville equation, which for the classical harmonic oscillator in one dimension looks like this:

$$\frac{\partial \rho }{\partial t} + \frac{p}{m} \frac{\partial \rho }{\partial x} - m \omega^2 x \frac{\partial \rho }{\partial p} = 0$$

Specifically, I'm interested in stationary solutions, where the phase space distribution function $\rho$ is independent of time, so that $\partial \rho/ \partial t=0$. Well, I noticed that by scaling the position coordinate with the new variable $q=m \omega x$, we obtain:

$$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0$$

Now, dividing by $pq$ yields:

$$\frac{1}{q} \frac{\partial \rho }{\partial q} - \frac{1}{p} \frac{\partial \rho }{\partial p} = \frac{\partial \ln \rho }{\partial q} - \frac{\partial \ln \rho }{\partial p}= 0$$

This equation has the general solution $\ln \rho(x,p) = f(q +p) = f(m \omega x +p)$, and so $\rho(x,p) = g(m \omega x +p)$, where $f(s)$ and $g(s)=\exp(f(s))$ are arbitrary functions. However, it appears that this solution is not normalizable as a probability density function over phase space (assuming that both $x$ and $p$ can take on any real value). This is because the solution $\rho(x,p) = g(m \omega x +p)$ is constant for fixed values of $m \omega x +p$, so $\rho(x,p)$ does not decay to zero for arbitrarily large values of $x$ and $p$ along these lines of constant $\rho(x,p)$, unless the function $g(s)$ is zero everywhere.

So these are my questions: Is this analysis correct, implying that there are no stationary physical solutions to the Liouville equation for the harmonic oscillator, or is there some class of solutions that I've missed? And if so, is this just a peculiarity of the harmonic oscillator, or is there some general class of problems for which no stationary solutions to the Liouville equation can be found?

  • $\begingroup$ While I do not know the answer and it's an interesting question to me, is it wise to divide by $pq$ (which might well be zero)? Further, I do not like your argument that the solution is not normalisable because you only show that $\rho$ does not decay along lines - which have Lebesgue measure zero. $\endgroup$
    – Sanya
    Jul 19, 2016 at 21:16
  • $\begingroup$ possibly related: physics.stackexchange.com/questions/140968/… and physics.stackexchange.com/questions/145566/… $\endgroup$
    – Sanya
    Jul 19, 2016 at 21:18
  • 1
    $\begingroup$ @Sanya you're right, I didn't include a rigorous explanation of why my solution (which turns out to be wrong anyway) is non-normalizable. to be exact, we can see that the solution isn't normalizable by integrating over all $x$ and $p$, and then making a change of variables to variables of the form $y=m \omega x + p$, $z = m \omega x - p$. Then the integral separates into the product of an integral over $y$ and another over $z$. The integral over $z$ looks like $dz$ integrated from negative to positive infinity, which is undefined. $\endgroup$ Jul 19, 2016 at 21:32

1 Answer 1


I think you made an mistake in the algebra. Let's focus on the PDE

$$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0.$$

You claim that $\rho(p,q) = \exp^{\, f(p+q)}$ is a solution to this equation for any function $f$, but that's simply not true. If you plug that Ansatz in the PDE, you get

$$ (p-q)\; f'(p+q) = 0 $$ which is very constraining.

As a matter of fact, the general solution appears to be (don't have a proof)

$$\rho(p,q) = f(p^2 + q^2).$$

  • $\begingroup$ Ok I see the mistake! $(1/q)*(\partial \rho / \partial q)$ is not $\partial \ln \rho / \partial q$. Rather, $(1/\rho)*(\partial \rho / \partial q) =\partial \ln \rho / \partial q$. And similarly for the momentum term. Thank you! $\endgroup$ Jul 19, 2016 at 21:23

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