TITLE: How do you prove that the number operator commutes with a general Hamiltonian?
QUESTION [3 upvotes]: If you have a hamiltonian, in the case of a bosonic system
$$
H=\sum_{ij}H_{ij}a_i^\dagger a_j,
$$
and the number operator
$$
N=\sum_{i}a_i^{\dagger}a_i.
$$
How do you show that they commute? I have tried plugging it into $[H,N]$ and using $[a_i,a^\dagger_j]=\delta_{ij}$ but have had no luck getting $[H,N]=0$. Also I used $[a_i,a_j]=0$ and $[a^\dagger_i,a^\dagger_j]=0$.

REPLY [6 votes]: Consider any particular element $[\hat a^\dagger_i ~\hat a_j, \hat N] = \sum_k \left(\hat a^\dagger_i ~\hat a_j~\hat a^\dagger_k ~\hat a_k - \hat a^\dagger_k ~\hat a_k~\hat a^\dagger_i ~\hat a_j \right).$
We know $\hat a_m~\hat a^\dagger_n = \hat a^\dagger_n~\hat a_m + \delta_{mn},$ so this is $\sum_k \left(\hat a^\dagger_i ~\delta_{jk} ~\hat a_k - \hat a^\dagger_k \delta_{ik}~\hat a_j \right),$ once you cancel out the common $\hat a^\dagger_i \hat a^\dagger_k \hat a_k \hat a_j$ term. Performing the sum over $k$ the Kronecker $\delta$ then gives $\hat a^\dagger_i \hat a_j - \hat a^\dagger_i \hat a_j = 0$ directly. The sum of zeroes is then zero, too.
Don't listen to @HChen; the commutator $[H, N]$ gives $\frac{dN}{dt}$ and all of your interactions conserve particle number, so of course this has to work out to 0.
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