# 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? Sep 6, 2015 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. Sep 6, 2015 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. 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.
• I think you should say "performing the sum" rather than "canceling the $\delta$". Sep 6, 2015 at 19:57