Integration of Poisson brackets by integration by parts [closed]

In the context of Statistical Mechanics I have to show that the following integral is zero:

$$\int \sum_{i=1}^{3N}(\frac{\partial O}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial O}{\partial p_i}\frac{\partial H}{\partial q_i})\rho (P,Q)d^{3N}pd^{3N}q$$

Where $$O=O(q,p);H=H(q,p)(Hamiltonian);\rho(P,Q)=$$density matrix

In the solutions it says that by using integration by parts one gets:

$$\int\sum_{i=1}^{3N} O (\frac{\partial^2 H}{\partial p_i \partial q_i}-\frac{\partial^2 H}{\partial p_i \partial q_i})\rho(P,Q) d^{3N}pd^{3N}q$$ which is obviously equal to zero. My problem is that I can't get to that expression by integrating by parts. I get some extra terms that don't cancel out in my eyes.

closed as off-topic by Aaron Stevens, Emilio Pisanty, Kyle Kanos, Jon Custer, ZeroTheHeroMar 7 at 14:17

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• I remember seeing something in Griffiths E&M I think that went something like "You can take the derivative off of one function and stick it onto another function at the cost of a minus sign and a boundary term". Maybe this qualitative explanation will help you? – Aaron Stevens Mar 6 at 19:31
• I found it but i still dont get why do the boundary terms should dissapear... In another problem it specifies that the function decays rapidly but not in this case. – Pablo Bähler Mar 6 at 19:53
• Is $O$ just some general function? – Aaron Stevens Mar 6 at 20:49
• I copy one line of my exercise:" Let$\Phi_t(P,Q):=(P(t),Q(t))$with initial condition $(P(0),Q(0)) = (P,Q)$ Show that for any observable $O(P,Q)$, the ‘time-translated’ observable $O_t(P,Q)=O(\Phi_t(P,Q))$have the same expectation value,$\langle O \rangle=\langle O_t \rangle$ for all t, where the probability distribution describing the ensemble has constant value on each energy surface." – Pablo Bähler Mar 6 at 20:59
• In this kind of integration by parts, the vanishing of the boundary terms is a little bit subtle. For the momenta, which go to $\pm \infty$ it is quite plausible that the probability density $\rho$ goes to zero (Maxwell-Boltzmann distribution). For the positions, I think one resorts to the idea that the probability density vanishes whenever a particle coordinate reaches the container walls. It's a bit hand-wavy, I know. – user197851 Mar 6 at 22:18

1. "The probability distribution describing the ensemble has constant value on each energy surface." This means that $$\rho$$ can be written as a function of $$H$$, which implies that the Poisson bracket $$\{H, \rho\} = 0$$.
2. I think we can assume that there is no probability flux through the boundaries of phase space. The probability flux is $$(\rho \dot{q}, \rho \dot{p}) = (\rho \frac{\partial H}{\partial p}, -\rho \frac{\partial H}{\partial q})$$.
After a good night of sleep everything became clearer haha. Between @Subhaneil Lahiri 'S answer, @LonelyProf comment and some re reading of the script by myself I could finally answer. The main things that I forgot were Liouville's theorem ($$\Phi_t(P,Q)$$ is area preserving, meaning $$\vert B_0 \vert= \vert B_t \vert$$ for some observable B) and the boundaries of the box. I completely forgot about the potential that limits the positions of course. Thank you all!
• The other comment to make is that the Poisson bracket $\{O,H\}$ is $\dot{O}$, the time derivative of $O(q,p)$ due to the evolution of the coordinates and momenta. So, of course, your first expression is $\langle \dot{O}\rangle$, which must be zero in an equilibrium ensemble. Possibly you already realized this. In statistical mechanics manipulations, it is often easier to bear this kind of thing in mind, rather than doing something like the integration by parts. – user197851 Mar 7 at 10:17
• I realized that indeed. I substituted the Hamilton's EOM in the time derivative of $O_t$ and got the Poisson brackets. What I feel that I didn't use is $\{H,\rho\}=0$ but I guess I used it when I treated $\rho$ as a constant when I did the time derivative of $O_t$. Thank you for your time and answers! – Pablo Bähler Mar 7 at 10:30