Degeneracies of Eigenvalues for H0 with Operators a and b | QM Homework

[Tangent87]

Homework Statement

The unperturbed Hamiltonian H₀ of two independent one-dimensional operators is

$$H_0=a^{\dagger}a+2b^{\dagger}b$$

where a and b are operators such that $$[a,a^{\dagger}]=1=[b,b^{\dagger}]$$

Find the degeneracies of the eigenvalues of H₀ with energies E₀ = 0, 1, 2, 3, 4.

The Attempt at a Solution

As I understand it, the eigenvalues of H₀ ARE the energies E₀ = 0, 1, 2, 3, 4. So that we have the equation H₀|n>=E₀|n>. But I'm not sure how to evaluate H₀|n> as all we know about the operators a and b is the commutation relations they satisfy.

 

[Chopin]

This is a ladder operator problem, so in addition to the operators a and b, you must also have a ground state, $$|0\rangle$$, right? This state is defined as the state which is annihilated by the lowering operators, so $$a|0\rangle = b|0\rangle = 0$$. Now, given that state, try to work out what the hamiltonian does to states like $$a^\dagger|0\rangle$$, $$b^\dagger|0\rangle$$, $$a^\dagger b^\dagger|0\rangle$$, etc.

 

[Tangent87]

I've had a brainwave:

Can we call our states |n m>=|n>_a|m>_b so that
$$H_0|n m>=(a^{\dagger}a|n>_a)|m>_b+2|n>_a(b^{\dagger}b|m>_b)=(n+2m)|n m>$$?

 

[Chopin]

I'm not exactly sure what you mean by your notation (what are the subscripts?) If I understand you right, though, then your end result is correct, but your reasoning for getting there is slightly wrong. You can indeed label your states as $$|nm\rangle$$, and they will have the eigenvalues that you state. That should make it fairly straightforward to find the degenerate states that the problem asks for. What I don't think you have quite right is how those states are defined.

I assume you have learned about how ladder operators work with a ground state? Like I mentioned in my previous post, you start by assuming there is a ground state which is annihilated by both lowering operators. By definition, then, this state is an eigenstate of the hamiltonian, with eigenvalue 0. You then apply various combinations of raising and lowering operators to that state to form new states which are also eigenstates of the hamiltonian.

For instance, take the state $$|\Psi\rangle = a^\dagger|0\rangle$$:

$$H_0|\Psi\rangle = (a^\dagger a + 2b^\dagger b)a^\dagger|0\rangle$$
$$= a^\dagger a a^\dagger |0\rangle + 2b^\dagger b a^\dagger | 0\rangle$$

Now, since a and b commute, the second term is (by the definition of $$|0\rangle$$) equal to:

$$2b^\dagger b a^\dagger | 0\rangle = 2b^\dagger a^\dagger b |0\rangle = 0$$

The first term does not equal 0, because $$a$$ and $$a^\dagger$$ don't commute.

$$a^\dagger a a^\dagger |0\rangle = a^\dagger (a^\dagger a + [a, a^\dagger])|0\rangle$$
$$= a^\dagger a^\dagger a|0\rangle + a^\dagger |0\rangle = a^\dagger|0\rangle$$

In the last step, the first term again vanishes due to the definition of $$|0\rangle$$. In this case, then, we have $$H_0|\Psi\rangle = 1|\Psi\rangle$$, so $$\Psi$$ is an eigenstate of the hamiltonian with eigenvalue 1.

By similar logic, you should be able to show that $$b^\dagger|0\rangle$$, $$a^\dagger b^\dagger|0\rangle$$, or any other combination of $$a^\dagger$$ and $$b^\dagger$$ in front of $$|0\rangle$$ is an eigenstate of the hamiltonian. Each of these states can then be labeled according to your $$|mn\rangle$$ notation, and from there it should be fairly easy to see how to compute the eigenvalue for any such state.

Apologies if you had already worked all of this out, I just wasn't quite sure I understood what you meant in your last post.

 

[Tangent87]

I'm not exactly sure what you mean by your notation (what are the subscripts?) If I understand you right, though, then your end result is correct, but your reasoning for getting there is slightly wrong. You can indeed label your states as $$|nm\rangle$$, and they will have the eigenvalues that you state. That should make it fairly straightforward to find the degenerate states that the problem asks for. What I don't think you have quite right is how those states are defined.

I assume you have learned about how ladder operators work with a ground state? Like I mentioned in my previous post, you start by assuming there is a ground state which is annihilated by both lowering operators. By definition, then, this state is an eigenstate of the hamiltonian, with eigenvalue 0. You then apply various combinations of raising and lowering operators to that state to form new states which are also eigenstates of the hamiltonian.

For instance, take the state $$|\Psi\rangle = a^\dagger|0\rangle$$:

$$H_0|\Psi\rangle = (a^\dagger a + 2b^\dagger b)a^\dagger|0\rangle$$
$$= a^\dagger a a^\dagger |0\rangle + 2b^\dagger b a^\dagger | 0\rangle$$

Now, since a and b commute, the second term is (by the definition of $$|0\rangle$$) equal to:

$$2b^\dagger b a^\dagger | 0\rangle = 2b^\dagger a^\dagger b |0\rangle = 0$$

The first term does not equal 0, because $$a$$ and $$a^\dagger$$ don't commute.

$$a^\dagger a a^\dagger |0\rangle = a^\dagger (a^\dagger a + [a, a^\dagger])|0\rangle$$
$$= a^\dagger a^\dagger a|0\rangle + a^\dagger |0\rangle = a^\dagger|0\rangle$$

In the last step, the first term again vanishes due to the definition of $$|0\rangle$$. In this case, then, we have $$H_0|\Psi\rangle = 1|\Psi\rangle$$, so $$\Psi$$ is an eigenstate of the hamiltonian with eigenvalue 1.

By similar logic, you should be able to show that $$b^\dagger|0\rangle$$, $$a^\dagger b^\dagger|0\rangle$$, or any other combination of $$a^\dagger$$ and $$b^\dagger$$ in front of $$|0\rangle$$ is an eigenstate of the hamiltonian. Each of these states can then be labeled according to your $$|mn\rangle$$ notation, and from there it should be fairly easy to see how to compute the eigenvalue for any such state.

Apologies if you had already worked all of this out, I just wasn't quite sure I understood what you meant in your last post.

Thank you for helping me out with this. What I basically meant by my subscripts a and b was that we can form our eigenstates as the direct product of the eigenstates of $$a^{\dagger}a$$ and $$b^{\dagger}b$$ and just use the usual results of ladder operators as you said. I.e. we get simultaneous eigenstates of n,m.

I'm a little uncertain by your method because as you've shown it takes quite a bit of work to show possible combinations of a and b are eigenstates and to get their eigenvalues, and that's just for E=0, to do the same for E=1,2,3 and 4 would take forever!

 

[Chopin]

Ok, yeah, you're basically on the right track. You can think of the states as a direct product if you want, but if you do it that way you won't be able to describe how to actually construct them. If you do it according to the procedure I outlined, you can actually express the states in terms of the ladder operators, which is the whole goal of this problem.

It's definitely a pain in the butt to run through all of the commutators and such to show how the hamiltonian acts on these states, but it's really important to thoroughly understand how this works. The ladder operator formalism features very heavily later on in quantum theory, especially for QFT. It's worth it to go through the pain once, to be sure you get what's actually going on. Once you do it for n=1 and n=2, it should be clear how it generalizes to higher eigenvalues, and then you won't have to do it explicitly anymore.