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Degenerate perturbation theory for harmonic oscillator

  1. Feb 19, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Raising and lowering operators

    3. 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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  3. Feb 19, 2015 #2

    TSny

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    Yes, that looks right to me. [I'm not sure what the superscripts (0) refer to in your state vectors]
     
    Last edited: Feb 19, 2015
  4. Feb 19, 2015 #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.
     
  5. Feb 20, 2015 #4

    CAF123

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    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!
     
  6. Feb 20, 2015 #5

    CAF123

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    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.
     
    Last edited: Feb 20, 2015
  7. Feb 20, 2015 #6

    TSny

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    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?
     
  8. Feb 20, 2015 #7

    CAF123

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    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.
     
  9. Feb 20, 2015 #8

    TSny

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    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##?
     
  10. Feb 21, 2015 #9

    CAF123

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    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##
     
    Last edited: Feb 21, 2015
  11. Feb 21, 2015 #10

    TSny

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