Spectrum of bosonic Hamiltonian in 2nd quantization

Question

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 $varepsilon>gge 0$, $[a_1,a_1^dagger]=[a_2,a_2^dagger]=1$ 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 $|n_1,n_2rangle$:
$$
H|n_1,n_2rangle =varepsilon(n_1+n_2)|n_1,n_2rangle + gleft(sqrt{(n_1+1)n_2}|n_1+1,n_2-1rangle + sqrt{n_1(n_2+1)}|n_1-1,n_2+1rangleright),.
$$
So we see that the hamiltonian is not diagonal in the $|n_1,n_2rangle$ basis.
I also did not see a way to factorize the hamiltonian in some way to construct ladder operators.
Any idea how to proceed?

ZeroTheHero · Accepted Answer

It can be diagonalized in the subspace of states ${vert n_1n_2rangle, ,
n_1+n_2=N}$.  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 + 2ghat 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 $-frac{N}{2}le mle frac{N}{2}$ 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 $N$.

Spectrum of bosonic Hamiltonian in 2nd quantization

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