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In my quantum theory course, there is a question ask for checking whether the expectations in quantum and classical Liouville theory are identical.

Here is the original version:

"Assume the system is classical but that your initial knowledge of the system is described by a probability density function $\rho$(q,p,t). Use the Liouville equation to obtain a pair of differential equations describing the time-evolution of the classical expectation values" $$\langle q\rangle=\int dq \int dp\,\rho(p,q,t)q\quad\langle p\rangle=\int dq\int dp \,\rho(p,q,t)p$$ Which means to derive: $$\frac{d}{dt}\langle q\rangle=\langle p\rangle\quad \frac{d}{dt}\langle p\rangle=\langle F\rangle$$ with $$\frac{dq}{dt}=p \quad \frac{dp}{dt}=F$$and$$ \frac{d\rho}{dt}=0\quad(\text{By Liouville's Thm})$$ as known.

However, one will have one problem:

Notice that: $$ \frac{d}{dt}\int f(x,t)\,dx =\int \frac{\partial f(x,t)}{\partial t}\, dx $$ not $$\frac{d}{dt}\int f(x,t)\,dx =\int \frac{d f(x,t)}{dt}\, dx $$ One will obtain: $$\frac{d\langle q\rangle}{dt}=\frac{d}{dt}\int dq \int dp\,\rho(p,q,t)q=\int dq \int dp\,\frac{\partial}{\partial t}(\rho(p,q,t)q)\neq \int dq \int dp\,\frac{d}{d t}(\rho(p,q,t)q) $$ However, the latter is what expected.

What is wrong here?

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Liouville's theorem says that $\frac{d\rho}{dt} = 0$ along the motion.

Expanding, the theorem is that $$0 = \frac{\partial \rho}{\partial q} \frac{dq}{dt} + \frac{\partial \rho}{\partial p}\frac{dp}{dt} + \frac{\partial \rho}{ \partial t}.$$

Substitute from this for $\partial \rho / \partial t$, and integrate by parts to find your desired result.

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  • $\begingroup$ Yes, you are right. I will post the proof to complete the question. It is really an interesting coincidence. : ) $\endgroup$ – Hamio Jiang Jan 27 '17 at 0:36

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