# How do you prove that the number operator commutes with a general Hamiltonian?

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 $[a^\dagger_i,a^\dagger_j]=0$.

• What if you assume there is no potential? – Luka Milic Sep 6 '15 at 18:25
• $N$ and $H$ do commute. Hint: $[N,\cdot]$ counts the number of creation operators minus the number of annihilation operators. And each term in $H$ has one of each. – Qmechanic Sep 6 '15 at 18:26

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. Canceling the $\delta$ with the sum 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.
• I think you should say "performing the sum" rather than "canceling the $\delta$". – march Sep 6 '15 at 19:57