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I am having a little trouble understanding the meaning of the fact that the convolution of two Liouville densities $\rho_{1}(p,q,t)$ and $\rho_{2}(p,q,t)$ (i.e. the classical probability overlap) remains constant in phase-space as the densities evolve over time, under the same Hamiltonian:

$$\frac{d}{dt} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\rho_{1}(p,q,t) \rho_{2}(p,q,t)\, dq \, dp = 0.$$

What is the physical consequence of this result? Does it imply that if there are two ensembles of particles such that some of them share the same set of starting coordinates in phase-space, the number of such particles remains constant as the two ensembles evolve over time?

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    $\begingroup$ Maybe I'm missing something, but how could this be true if $\rho_1$ and $\rho_2$ are evolving under different Hamiltonians? For example, a small region near $p = q = 0$ will be driven in opposite directions in phase space by the Hamiltonians $H_1 = p^2 + q$ and $H_2 = p^2 - q$; so regions in phase space that overlap initially will have no overlap at later times. $\endgroup$ – Michael Seifert Mar 15 at 15:09
  • $\begingroup$ Hmm you might be right, I will update the question. Assuming the Hamiltonian is the same for both densities, what does the overlap integral being constant in time physically mean? $\endgroup$ – Yejus Mar 15 at 15:18
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    $\begingroup$ If the hamiltonian is the same, the product of two densities is a density. What does Liouville's theorem tell you? $\endgroup$ – Cosmas Zachos Mar 15 at 15:32
  • $\begingroup$ @CosmasZachos The integrand evaluates to zero per Liouville's theorem, but what does that physically mean? $\endgroup$ – Yejus Mar 15 at 15:44
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    $\begingroup$ Conservation of probability, mass, particles? $\endgroup$ – Cosmas Zachos Mar 15 at 16:02

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