Degenerate perturbation theory for harmonic oscillator

[CAF123]

Homework Statement

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The isotropic harmonic oscillator in 2 dimensions is described by the Hamiltonian $$\hat H_0 = \sum_i \left\{\frac{\hat{p_i}^2}{ 2m} + \frac{1}{2} m\omega^2 \hat{q_i}^2 \right\} ,$$ for $i = 1, 2$ and has energy eigenvalues $E_n = (n + 1)\hbar \omega \equiv (n_1 + n_2 + 1)\hbar \omega, n = 0, 1, 2, \dots$. What is the degeneracy of the first excited level? Use degenerate perturbation theory to determine the splitting induced by the perturbation $\hat H' = K\hat q_1 \hat q_2$ where $K$ is a constant.

Homework Equations

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Raising and lowering operators

The Attempt at a Solution

The degeneracy of the first excited state is 2 so the splitting of the energy to first order is given by the expression $$\sum_{n=1}^2 b_n (\hat H'_{kn} - E^{(1)} \delta_{kn}) = 0$$. The matrix element I need to compute is $$\hat H'_{kn} = \langle E_k^{(0)} | K \hat q_1 \hat q_2 | E_n^{(0)} \rangle = \frac{K \hbar}{2 m \omega} \langle E_k^{(0)} | (\hat a_1 + a_1^{\dagger}) (\hat a_2 + a_2^{\dagger})| E_n^{(0)} \rangle$$ I then thought if the $\hat a_i$ raise the $n_i$ or $k_i$, I could write this like $$\frac{K \hbar}{2 m \omega} \langle k_1, k_2^{(0)}| (\hat a_1 + a_1^{\dagger}) (\hat a_2 + a_2^{\dagger})| n_1, n_2^{(0)} \rangle$$ and then for example, $\hat a_1 \hat a_2 |n_1, n_2 \rangle = \hat a_1 \sqrt{n_2} | n_1, n_2 - 1 \rangle = \sqrt{n_1} \sqrt{n_2} | n_1 - 1, n_2 - 1 \rangle?$. Would this be right? Many thanks.

 

[TSny]

I could write this like $$\frac{K \hbar}{2 m \omega} \langle k_1, k_2^{(0)}| (\hat a_1 + a_1^{\dagger}) (\hat a_2 + a_2^{\dagger})| n_1, n_2^{(0)} \rangle$$ and then for example, $\hat a_1 \hat a_2 |n_1, n_2 \rangle = \hat a_1 \sqrt{n_2} | n_1, n_2 - 1 \rangle = \sqrt{n_1} \sqrt{n_2} | n_1 - 1, n_2 - 1 \rangle?$. Would this be right? Many thanks.

Yes, that looks right to me. [I'm not sure what the superscripts (0) refer to in your state vectors]

 

[Jufro]

If you look at acting on one of your states |0,1> for example, then you cannot lower both. So these states would not contribute to the splitting.

 

[CAF123]

Hi TSny,

Yes, that looks right to me. [I'm not sure what the superscripts (0) refer to in your state vectors]

Ok thanks, the (0) was to indicate the zero used in the notation $|E_n^{(0)}\rangle$, but yes I'll suppress it from now on. The question boils down to solving a 2x2 determinant system where I need to compute the perturbation matrix elements $H'_{11}, H'_{12}, H'_{21}$ and $H'_{22}$, where the labels are the $k$ and $n$ indices. Can I just check some of my reasoning for these matrix elements?

Consider one of the terms in the expansion before using the raising and lowering operators. E.g $\langle k_1, k_2 | n_1 - 1, n_2 - 1 \rangle.$ I think initially I can maybe write this as a sum of two deltas, so that this term is equal to $\delta_{k_1, n_1 - 1} \delta_{k_2, n_2 - 1} + \delta_{k_1, n_2-1}\delta_{k_2, n_1-1}$. My reasoning is e.g the states |0,1> and |1,0> correspond to same e.value. However, in practice, this might imply the values of $n_i$ might go below 0 for any terms to survive. (E.g in the matrix element $H'_{11}$, if $k,n=1$ then if we suppose $k_1 = 0, k_2 = 1$ then for this inner product not to vanish must have $n_1 - 1 = 0, n_2 - 1 = 1$ or $n_2 - 1 = 0, n_1 - 1 = 0$. But this set of equations does not give an n corresponding to 1, so it must be neglected. Am I thinking about this in the right way? Thanks!

 

[CAF123]

Hi TSny,

Ok thanks, the (0) was to indicate the zero used in the notation $|E_n^{(0)}\rangle$, but yes I'll suppress it from now on. The question boils down to solving a 2x2 determinant system where I need to compute the perturbation matrix elements $H'_{11}, H'_{12}, H'_{21}$ and $H'_{22}$, where the labels are the $k$ and $n$ indices. Can I just check some of my reasoning for these matrix elements?

Consider one of the terms in the expansion before using the raising and lowering operators. E.g $\langle k_1, k_2 | n_1 - 1, n_2 - 1 \rangle.$ I think initially I can maybe write this as a sum of two deltas, so that this term is equal to $\delta_{k_1, n_1 - 1} \delta_{k_2, n_2 - 1} + \delta_{k_1, n_2-1}\delta_{k_2, n_1-1}$. My reasoning is e.g the states |0,1> and |1,0> correspond to same e.value. However, in practice, this might imply the values of $n_i$ might go below 0 for any terms to survive. (E.g in the matrix element $H'_{11}$, if $k,n=1$ then if we suppose $k_1 = 0, k_2 = 1$ then for this inner product not to vanish must have $n_1 - 1 = 0, n_2 - 1 = 1$ or $n_2 - 1 = 0, n_1 - 1 = 0$. But this set of equations does not give an n corresponding to 1, so it must be neglected. Am I thinking about this in the right way? Thanks!

Edit: Maybe just to check my matrix elements explicitly, for $H'_{11}$ I was getting this to equal $2C\hbar/m \omega$ and $H'_{22} = \frac{5 \sqrt{2} C \hbar}{m \omega}$. Non diagonal elements I got to be 0?
E.g for $H'_{11}$, if I consider $k_1 = 0, k_2 = 1,$ then in the term $\langle 0,1 | n_1 - 1, n_2 - 1 \rangle,$ as above for $n_1 - 1$ and $n_2 - 1$ to equal 0 or 1 (so we obtain a non vanishing contribution to the inner product) we must have the $n_i^s$ summing to a value which does not equal to 1 and thus should be neglected. A surviving term is for example, $\langle 0,1 | n_1 - 1, n_2 + 1 \rangle$. Here, if we let $n_2 + 1= 1, n_1 - 1 = 0,$ we get $n_1 = 1, n_2 = 0$ which is good. (Can't have $n_2+1=0$ since this gives $n_2=-1$ and the n_i's are strictly positive).

Sorry for all my explanations but hopefully I made myself clear!
Edit2: What I said about reexpressing the terms as sums of deltas is not quite correct because for example, the state |1,1> and |0,2> correspond to same e.value and no two entries are the same across the states.

 

[TSny]

I don't agree with what you are getting for H'₁₁ and H'₂₂. Let's make sure we are thinking about this correctly. Can you describe the two degenerate states for the first excited level of the unperturbed Hamiltonian? That is, what are the values of the quantum numbers n₁ and n₂ for these two states?

 

[CAF123]

I don't agree with what you are getting for H'₁₁ and H'₂₂. Let's make sure we are thinking about this correctly. Can you describe the two degenerate states for the first excited level of the unperturbed Hamiltonian? That is, what are the values of the quantum numbers n₁ and n₂ for these two states?

If we label these states as $|E_n^{(0)}\rangle = |n_1, n_2 \rangle$ then the first excited level corresponds to the $n_i$ taking either the value 0 or 1 such that $n_1 + n_2 = n = 1$. So, e,g the two degenerate states are |0,1> and |1,0> for the first excited level.

 

[TSny]

If we label these states as $|E_n^{(0)}\rangle = |n_1, n_2 \rangle$ then the first excited level corresponds to the $n_i$ taking either the value 0 or 1 such that $n_1 + n_2 = n = 1$. So, e,g the two degenerate states are |0,1> and |1,0> for the first excited level.

OK. Let $|\phi_1\rangle =|0, 1 \rangle$ and $|\phi_2\rangle =|1 , 0 \rangle$. Then $H'_{11} = \langle \phi_1|H'|\phi_1\rangle$, $H'_{12} = \langle \phi_1|H'|\phi_2\rangle$, etc.

Can you show how you get your result for $H'_{11}$? In particular, what do you get when you apply $H'$ to $|\phi_1\rangle$?

 

[CAF123]

OK. Let $|\phi_1\rangle =|0, 1 \rangle$ and $|\phi_2\rangle =|1 , 0 \rangle$. Then $H'_{11} = \langle \phi_1|H'|\phi_1\rangle$, $H'_{12} = \langle \phi_1|H'|\phi_2\rangle$, etc.

Can you show how you get your result for $H'_{11}$? In particular, what do you get when you apply $H'$ to $|\phi_1\rangle$?

Ok, I see. I was confusing the meaning of the labels $k$ and $n$.The matrix element $H'_{11}$ for example would be $$\frac{C \hbar}{2m\omega} \langle 0,1 | \hat a_1 \hat a_2 + \hat a_1 \hat a_2^{\dagger} + \hat a_1^{\dagger} \hat a_2 + \hat a_1^{\dagger} \hat a_2^{\dagger}|0,1 \rangle = \frac{C \hbar}{2m\omega} \left\{ \langle 0,1|1,0\rangle + \sqrt{2} \langle 0,1 | 1,2 \rangle \right\} = 0$$ since the states |0,1> and |1,0> form a basis for the 2d degenerate subspace (or can be achieved by orthogonalisation). Is this right? Similarly, I have $\hat H'_{12} = \frac{C \hbar}{2 m \omega} = H'_{21}$ and $\hat H'_{22} = 0$. I then get a splitting of magnitude $C\hbar/mw$

 

[TSny]

That looks right to me, with C = K. Good.