How do I show that a given Hamiltonian does not affect the overall number of particles in a given state?

Question

I'm struggling with the following problem:

Consider a system of an arbitrary number of indistinguishable bosonic particles. The system has two sites and $a_i^{\dagger}$ and $a_i$ are the corresponding creating and annihiliation operators. If $$\displaystyle \hat{H}=-a_1^{\dagger}a_2-a_2^{\dagger}a_1+\sum_{i=1,2}a_i^{\dagger}a_i(a_i^{\dagger}a_i-1),$$ show that applying this Hamiltonian to state does not affect the overall number of particles in that state.

I use occupation number representation and that

$$a_i|n_1,\dots,n_i,\dots ⟩=\sqrt{n_i}|n_1,\dots,n_i-1,\dots⟩ $$ and $$ a_i^{\dagger}|n_1,\dots,n_i,\dots ⟩=\sqrt{n_i+1}|n_1,\dots,n_i+1,\dots⟩$$

So, \begin{align} \hat{H}|n_1,n_2⟩ &= \left[ -a_1^{\dagger}a_2-a_2^{\dagger}a_1+\sum_{i=1,2}a_i^{\dagger}a_i(a_i^{\dagger}a_i-1) \right] |n_1,n_2⟩ \\ &= -a_1^{\dagger}a_2 |n_1,n_2⟩ -a_2^{\dagger}a_1 |n_1,n_2⟩ +\sum_{i=1,2}a_i^{\dagger}a_i(a_i^{\dagger}a_i-1) |n_1,n_2⟩ \\&= -\sqrt{n_1+1}\sqrt{n_2}|n_1+1,n_2-1⟩ -\sqrt{n_1}\sqrt{n_2+1} |n_1-1,n_2+1⟩ \\&\quad + \frac{1}{2}n_1^2|n_1,n_2⟩ +\frac{1}{2}n_2^2|n_1,n_2⟩-\frac{1}{2}n_1|n_1,n_2⟩-\frac{1}{2}n_2|n_1,n_2⟩ \end{align}

Am I able to just say $|n_1+1,n_2-1⟩ + |n_1-1,n_2+1⟩ =|n_1,n_2⟩$? Im not sure how to interpret the answer I found.

Is anyone able to do this by computing the action of $\hat{H}$ on an arbitrary occupation number representation state?

Answer 1

The correct statement is that the hamiltonian does not affect the total number of particles in a given state. This means that you do cannot equate the state $|n_1,n_2⟩$ with $|n_1\pm1,n_2\mp1⟩$, because they are different states, but they are all eigenstates of the total particle number operator, $$ \hat N=a_1^\dagger a_1+a_2^\dagger a_2. $$ Your calculation is enough to show that the total particle number is conserved, because $\hat H|n_1,n_2⟩$ is a linear combination of eigensstates of $\hat N$ with the same eigenvalue, which means that $\hat H|n_1,n_2⟩$ is also an eigenstate of $\hat N$ with eigenvalue $n_1+n_2$.

The classical proof, though, is rather simpler (and it is really the argument you were asked to produce). The operators $\hat O$ which are conserved by $\hat H$ are exactly those that commute with it, so that's what you need to show. This is easy: \begin{align} [\hat H,\hat N] & = \left[ a_1^{\dagger}a_2-a_2^{\dagger}a_1 +\sum_{i=1,2}a_i^{\dagger}a_i(a_i^{\dagger}a_i-1),a_1^\dagger a_1+a_2^\dagger a_2 \right] \\ & = \left[ a_1^{\dagger}a_2-a_2^{\dagger}a_1,a_1^\dagger a_1+a_2^\dagger a_2 \right] + \left[ \sum_{i=1,2}n_i(n_i-1),n_1+n_2 \right] \\ & = \left[ a_1^{\dagger}a_2-a_2^{\dagger}a_1,n_1+n_2 \right]. \end{align} I won't spoil the fun, because this is the crucial step of the calculation, but you can and should show using the bosonic algebra that this is zero.

With commuting $\hat H$ and $\hat N$, then, you know that $\hat H$ will preserve the eigenstate status, and eigenvalue, of any eigenstate of $\hat N$. This is what "conserve total particle number" means.