What is the value of the second term in the commutator for an N particle system?

[B4cklfip]

Homework Statement

Given is the Hamiltonian H of a n particle system and it should be shown that H and P commucate, thus [H,P] = 0.
Where P is given as $$P = \sum_{n=1}^N p_n.$$

Relevant Equations

$$H = \sum_{n=1}^N \frac{p_n^2}{2m_n} +\frac{1}{2}\sum_{n,n'}^N V(|x_n-x_n'|)$$

I have insertet the equations for H and P in the relation for the commutator which gives

$$[H,P] = [\sum_{n=1}^N \frac{p_n^2}{2m_n} +\frac{1}{2}\sum_{n,n'}^N V(|x_n-x_n'|),\sum_{n=1}^N p_n]
\\ = [\sum_{n=1}^N \frac{p_n^2}{2m_n},\sum_{n=1}^N p_n]+\frac{1}{2}[\sum_{n,n'}^N V(|x_n-x_n'|),\sum_{n=1}^N p_n]$$

The first term should become zero, but what about the second term? I don't really know how to go on here.

 

[Abhishek11235]

The potential is the function of only relative coordinates. If I translate all of them simultaneously,the potencial won't change i.e it has translation symmetry. It is well known to people in physics that translation symmetry implies conservation of momentum (Noether Theorem). So, the commutator becomes zero.
If you are not satisfied(Since you didn't see Mathematics), then try Taylor's expansion of $H(x_i+a)$ around $x_i=a$ to first order.
Reference: Modern Quantum Mechanics,J.J Sakurai.

 

[TSny]

To see how it works, try writing out explicitly the expression $\frac{1}{2} \left[ \sum_{m,n}^N V(|x_m-x_n|),\sum_{n=1}^N p_n \right]$ for the case N = 2.

 

[Cem]

We have $\frac{\partial}{\partial p_i}\sum_{n=1}^N p_n = 1$ for all $i=1,2,...,N$. If you show that the terms in $\sum_i \frac{\partial}{\partial x_i} \sum_{n,n'}^N V(|x_n-x_{n'}|)$ cancel each other out, then you get $\left[\sum_{n,n'}^N V(|x_n-x_{n'}|),\sum_{n=1}^N p_n\right]=0$. You can look at the sum of the two terms $\frac{\partial V(|x_k-x_m|)}{\partial x_k}+\frac{\partial V(|x_k-x_m|)}{\partial x_m}$ for some $k\neq m$ and try to show that this is 0.