# Coupled Oscillators

This is an exercise of my last exam. Since I couldn't find anybody who solved it or knows how to, it would be really nice if somebody could tell me if my thoughts on it go into the right direction.

Two harmonic Oscillators can be described by the Hamiltonian

$\hat H_a + \hat H_b + \hat H_c$

where

$\hat H_a = \hbar\omega(\hat a^\dagger \hat a + \frac{1}{2})$
$\hat H_b = \hbar\omega(\hat b^\dagger \hat b + \frac{1}{2})$

and the coupling Hamiltonian is

$\hat H_c = \hbar\Omega(\hat a^\dagger \hat b + \hat b^\dagger \hat a)$

with $\Omega \ll \omega$

a) Evaluate the commutation relation between $\hat H_c$ and the operators $\hat a^\dagger \hat a$ and $\hat b^\dagger \hat b$.

This is not a problem at all, at least when you know about the canonical commutation relation. $[\hat H_c,\hat a^\dagger \hat a] = -[\hat H_c,\hat b^\dagger \hat b] = \hbar\Omega(\hat b^\dagger \hat a - \hat a^\dagger \hat b)$

b) Given that the two oscillators are degenerate, what is the significance of the four operators $\hat a^\dagger \hat a,\ \hat b^\dagger \hat b,\ \hat a^\dagger \hat a + \hat b^\dagger \hat b$ and $\hat a^\dagger \hat a - \hat b^\dagger \hat b$ in evaluation the energy level shift resulting from the coupling.

Ok, if things commute that means that they have simultaneous eigenstates. The operators do all commute with $\hat H_a$ and $\hat H_b$ which makes totally sense because they are just the solutions of two different oscillators. To evaluate the energy shift occurring due to the coupling Hamiltonian I need to use an operator that commutes with every Hamiltonian. In this case it would be the one for the total quantum number $n_{tot} = n_a + n_b$ which is $\hat a^\dagger \hat a + \hat b^\dagger \hat b$ because from (a) I know that it commutes. So, if this is right, with part (c) comes my problem.

c) Evaluate the first order energy shift for the states

$\varphi_+ = (\varphi_{n_a}\varphi_{n_b - 1} + \varphi_{n_a-1}\varphi_{n_b})/\sqrt{2}$
$\varphi_- = (\varphi_{n_a}\varphi_{n_b - 1} - \varphi_{n_a-1}\varphi_{n_b})/\sqrt{2}$

where $\varphi_{n_a},\varphi_{n_b}$ are the eigenstates of $\hat H_a$ and $\hat H_b$ respectively.

To get the energyshift I need to send $\hat H_c$ onto my states. But to use these eigenstates with my coupling Hamiltonian I would have to rewrite it ($\hat H_c$) in terms of the total energy because this operator does commute and would have simultaneous eigenstates. But this doesn't work. I thought that it might be some expansion I need to do because one should calculate the 'first order' energy shift, but this could be because the coupling Hamiltonian is a simplified one and I didn't figure out how to expand the total energy operator.

Did I get this right? Maybe someone could give me a hint how to proceed.
I got the ideas of how it could work from perturbation theory.

After part (b) you write

"To evaluate the energy shift occurring due to the coupling Hamiltonian I need to use an operator that commutes with every Hamiltonian."

This is sort of true, but a better way to say it is that you need to use STATES that are eigenvalues of operators that commute with every Hamiltonian.

It's important to keep in mind the physical picture that goes on in degenerate perturbation theory. Let's say there's an $N$ fold degeneracy. The unperturbed Hamiltonian has an $N$ dimensional linear subspace of states (ie, an infinite number of states) that all have the energy $E$. All linear combinations of these states have the same energy before turning on the perturbation. The perturbation then picks out $N$ special linear combinations of the states, those special linear combinations end up having definite energies, but the $N$ different special linear combinations will have different energies than each other (there will be $N$ energy eigenvalues after turning on the perturbation). All other linear combinations of the states, which previously were energy eigenstates, will no longer be energy eigenstates after turning on the perturbation. So it's crucial to determine the $N$ special linear combinations that remain energy eigenstates after turning on the perturbation. How do we do it? Well, say you find an operator that commutes with the free and perturbing hamiltonian. An eigenstate of this new operator will be a particular linear combination of the energy eigenstates of the free hamiltonian, since this operator commutes with the free hamiltonian. But this linear combination of states will also be an energy eigenstate of the free + peturbing hamiltonian, because the operator also commutes with the perturbation. So the eigenstates of this new operator are the special linear combinations you are looking for.

The point isn't to express the perturbing hamiltonian in terms of the unperturbed hamiltonian. The point is to evaluate the basic first order perturbation equation

$$\delta E^{(1)} = \langle \varphi | V^{(1)} | \varphi \rangle$$

using STATES that are eigenvalues of operators that commute with $V^{(1)}$ and $H_0$.

You have identified an operator that commutes with $V^{(1)}$ and $H_0$ in your problem (namely $a^\dagger a + b^\dagger b$). So, are the states in part (c) eigenvalues of that operator? Are they eigenstates of $a^\dagger a$ or $b^\dagger b$?

I would also try expressing the states $\varphi_{n_a}\varphi_{n_b}$ in terms of creation and annhilation operators. That might also help suggest how to evaluate the energy shift.

• Well in retrospect I'm not actually sure how helpful my answer is but I'm glad you got it. The key point here is that the perturbation commutes with the total free hamiltonian, as you said, not with a or b alone. So your new eigenstates will be combinations of a's and b's eigenstates corresponding to eigenstates of the total free hamiltonian. But it doesn't break the degeneracy of the total free Hamiltonian. A good computational trick here is to rewrite $\varphi_{n_a}=\frac{1}{\sqrt{n_a!}}(a^{\dagger})^{n_a} |0 \rangle$, then the energy shift reduces to evaluating commutation relations – Andrew Aug 3 '13 at 17:36