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?


1 Answer 1


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 $N$.

  • 2
    $\begingroup$ Maybe it is worth mentioning that this is called Schwinger boson transformation / representation... $\endgroup$ Commented Apr 6, 2021 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.