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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?

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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 $N$.

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    $\begingroup$ Maybe it is worth mentioning that this is called Schwinger boson transformation / representation... $\endgroup$
    – Jakob
    Apr 6 at 17:27

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