Degenerate perturbation theory for harmonic oscillator

In summary, the degeneracy of the first excited state in the isotropic harmonic oscillator in 2 dimensions is 2. To determine the splitting induced by the perturbation, degenerate perturbation theory is used to compute the matrix elements. These matrix elements are found to be 2Cℏ/mω for H'11, 0 for H'12 and H'21, and 5√2Cℏ/mω for H'22. Non-diagonal elements were found to be 0.
  • #1
CAF123
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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.
 
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  • #2
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]
 
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  • #3
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.
 
  • #4
Hi TSny,
TSny said:
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!
 
  • #5
CAF123 said:
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.
 
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  • #6
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?
 
  • #7
TSny said:
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.
 
  • #8
CAF123 said:
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##?
 
  • #9
TSny said:
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##
 
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  • #10
That looks right to me, with C = K. Good.
 
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What is degenerate perturbation theory?

Degenerate perturbation theory is a mathematical method used in quantum mechanics to calculate the energy levels of a system that has multiple degenerate (or equally spaced) energy states.

How does degenerate perturbation theory work?

In degenerate perturbation theory, the Hamiltonian of the system is divided into two parts: the unperturbed Hamiltonian, which describes the system without any external influence, and the perturbation, which represents the small external influence on the system. The perturbed Hamiltonian is then diagonalized to obtain the perturbed energy levels.

Why is degenerate perturbation theory important?

Degenerate perturbation theory is important because it allows us to accurately calculate the energy levels of a system that has degenerate energy states. This is particularly useful in quantum systems, where degeneracy is a common phenomenon.

What is the difference between non-degenerate and degenerate perturbation theory?

In non-degenerate perturbation theory, the energy levels of the system are not degenerate, meaning they are not equally spaced. In this case, the perturbed energy levels can be calculated without any additional mathematical techniques. In degenerate perturbation theory, however, the energy levels are degenerate and require special techniques to accurately calculate them.

What are some applications of degenerate perturbation theory?

Degenerate perturbation theory is commonly used in various fields of physics, including quantum mechanics, atomic and molecular physics, and condensed matter physics. It is also used in chemistry to study molecular energy levels and spectroscopy. Additionally, it has applications in engineering, such as in designing electronic devices and materials with desired properties.

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