# Spectrum of bosonic Hamiltonian in 2nd quantization

I have the hamiltonian

$$H=\varepsilon(a_1^\dagger a_1 + a_2^\dagger a_2) + g(a_1^\dagger a_2 + a_2^\dagger a_1)$$

with $$\varepsilon>g\ge 0$$, $$[a_1,a_1^\dagger]=[a_2,a_2^\dagger]=1$$ and all other commutators equal to zero.

What is the spectrum of the Hamiltonian?

In the exercise there is a given hint that it is a possible solution strategy to write the hamiltonian in terms of ladder operators $$L_+^\dagger$$ and $$L_-^\dagger$$.

So far I have tried to act with the hamiltonian on a state $$|n_1,n_2\rangle$$:

$$H|n_1,n_2\rangle =\varepsilon(n_1+n_2)|n_1,n_2\rangle + g\left(\sqrt{(n_1+1)n_2}|n_1+1,n_2-1\rangle + \sqrt{n_1(n_2+1)}|n_1-1,n_2+1\rangle\right)\,.$$

So we see that the hamiltonian is not diagonal in the $$|n_1,n_2\rangle$$ basis.

I also did not see a way to factorize the hamiltonian in some way to construct ladder operators.
Any idea how to proceed?

It can be diagonalized in the subspace of states $$\{\vert n_1n_2\rangle\, , n_1+n_2=N\}$$. Indeed, in terms of $$\hat L_\pm$$ and the total number operator $$\hat N$$, your Hamiltonian is just \begin{align} \hat H=\epsilon \hat N + 2g\hat L_x \end{align} with $$\hat L_x$$ connecting states with the same total $$N=n_1+n_2$$ (as your expression suggests).
Thus the spectrum will be $$\epsilon N+2g m$$, where $$-\frac{N}{2}\le m\le \frac{N}{2}$$ since the eigenvalues of $$\hat L_x$$ are the same as those of $$\hat L_z$$. You can work out the expression for $$\hat L_z$$ in terms of $$a_1,a_1^\dagger, a_2, a_2^\dagger$$ by yourself to confirm the connection between the $$m$$ values and $$N$$.