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

• Tangent87
So I wondered if there was a more direct way of finding the eigenstates and eigenvalues for each E. I realise that this is a slightly different question to the one I asked earlier, but while we're discussing the problem I wouldn't mind knowing if there was an easier way of solving it, as I'm sure there must be.In summary, the problem involves finding the degeneracies of the eigenvalues of the unperturbed Hamiltonian H0 with energies E0 = 0, 1, 2, 3, 4. This can be done by using ladder operators and a ground state to create various combinations of eigenstates, which can then be labeled according to the notation |nm\rangle.

## 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.

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.

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

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.

Last edited:
Chopin said:
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!

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.

## 1. What are degenerate states in quantum mechanics?

Degenerate states in quantum mechanics refer to energy levels that have the same value. This means that electrons in these states have the same energy, but they may have different wave functions.

## 2. How do degenerate states affect the properties of a system?

Degenerate states can affect the properties of a system by allowing for multiple possible outcomes for a measurement. This can lead to phenomena such as electron spin and the Zeeman effect.

## 3. Can degenerate states exist in all quantum systems?

Yes, degenerate states can exist in all quantum systems, including atoms, molecules, and subatomic particles. They are a fundamental aspect of quantum mechanics and are essential for understanding the behavior of these systems.

## 4. How are degenerate states resolved in quantum mechanics?

In quantum mechanics, degenerate states are resolved through the use of perturbation theory. This involves introducing small changes to the system, which can lift the degeneracy and allow for a more accurate understanding of the system's properties.

## 5. What is the significance of degenerate states in quantum computing?

Degenerate states play a crucial role in quantum computing as they can be used for quantum information processing and quantum algorithms. They allow for a larger number of possible states and operations, making quantum computers more powerful than classical computers.