# I Eigenstates of two Coupled Harmonic Oscillator

1. Sep 27, 2016

### Esquer

Hello everyone! For my quantum mechanics class I have to study the problem of two quantum oscillator coupled to each other and in particular to find the eigenstates and eigenergies for a subspace of the Fock space.
I know that, in general, to solve this kind of problem I have to diagonalize the hamiltonian of the system that in this case is the following one:
$$H=\hbar\omega_0 (a^+a+b^+b)+\hbar J(a^+b+b^+a)$$
with a and b bosonic creation and annhilation operator for the two harmonic oscillator.
What I do not understand is how to write the matrix in a given subspace. For example in the case of one quanta of energy present in the oscillators my two eigenstate would be: |00> a superposition of |01> and |10> (one quanta of energy in the first oscillator and 0 in the second one and viceversa). Is that correct?
I don't know how to set the problem,
thank to everyone for the help.

2. Sep 27, 2016

### Lucas SV

What you said is correct on the state with one quanta of energy.

In order to diagonalize the operator, you need to find the eigenvalues and eigenstates of the operator. To do this write a general state, which would be a superposition of $|m n\rangle$ states, then act upon this state by $H$, and require it to be an eigenstate. Since $H$ contains creation and annihilation operator, the computation is straightforward. I am wondering, however, what is $J$ in your equation?

3. Sep 27, 2016

### Esquer

Sorry I forgot to specify, J is the coupling coefficient between the two cavities

4. Sep 27, 2016

### vanhees71

Another trick is to find a canonical transformation (Bogoliubov transformation) of the annihilation operators,
$$\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$$
such that your Hamiltonian becomes "diagonalized",
$$H=\hbar (\omega_1 \alpha^{\dagger} \alpha + \omega_2 \beta^{\dagger} \beta).$$

5. Sep 27, 2016

### Esquer

With that it would be actually preatty easy. But if I want to find the matrix form such $$H \begin{pmatrix} n_a \\ n_b \end{pmatrix} = E \begin{pmatrix} n_a \\ n_b \end{pmatrix}$$ what should I do?

6. Sep 28, 2016

### stevendaryl

Staff Emeritus
You're not going to find that, because there will be infinitely many eigenstates of $H$, and the way you've written it assumes that there are just two. As vanhees71 suggested, you have to find two operators $\alpha$ and $\beta$ such that:
1. $H = \hbar \omega_1 \alpha^\dagger \alpha + \hbar \omega_2 \beta^\dagger \beta + C$ (for some constants $\omega_1, \omega_2, C$).
2. $\alpha \alpha^\dagger - \alpha^\dagger \alpha = 1$
3. $\beta \beta^\dagger - \beta^\dagger \beta= 1$
4. $\alpha \beta - \beta \alpha = \alpha^\dagger \beta - \beta \alpha^\dagger = 0$
Then you can let $|0\rangle$ be some state such that $\alpha|0\rangle = \beta|0\rangle = 0$. Then every state of the form $(\alpha^\dagger)^n (\beta^\dagger)^m |0\rangle$, which we can write as $C_{nm} |n, m\rangle$, for some normalization constant $C_{nm}$, will be an eigenstate of $H$, whose eigenvalues you can work out using the example of the simple harmonic oscillator. You can't represent $|n,m\rangle$ as a column matrix unless you use infinitely many rows.