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Degenerate perturbation theory degeneracy not lifted by perturbation

  1. May 22, 2013 #1
    Hi,

    I have an equation of the form

    [tex](-i \lambda \frac{d}{dr}\sigma_z+\Delta(r)\sigma_x) g =(\epsilon + \frac{\mu \hbar^2}{2mr^2}) g[/tex]

    where [itex]\sigma[/itex] refers to the Pauli matrices, g is a two component complex vector and the term on the right hand side of the equation is small compared to the other terms. The authors of the paper where i found this equation (http://www.sciencedirect.com/science/article/pii/0031916364903750) say that this can be handled using perturbation theory.

    Ignoring the right hand side gives the possible solutions of g as

    [tex]g=const (1,\pm i)^T e^{\pm K}[/tex]

    where

    [tex]K=\frac{1}{\lambda}\int_0^r \Delta dr[/tex].

    Since the known behaviour of [itex]\Delta[/itex] is that it goes to a constant for large r, only the negative sign gives a well behaved function for large r.

    Now as I see it both these states are zero energy solutions to the Hamiltonian

    [tex]H_0= -i \lambda \frac{d}{dr} \sigma_z +\Delta(r) \sigma_x[/tex]

    and the perturbation

    [tex]V=\epsilon +\frac{\mu \hbar^2}{2mr^2}[/tex]

    being a scalar does not remove the degeneracy between the states. I'm clueless how to proceed here...
     
    Last edited: May 22, 2013
  2. jcsd
  3. May 22, 2013 #2

    Bill_K

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    No, the perturbation is off-diagonal. If you let g± = g1 ± ig2, then you have roughly

    λ dg±/dr ± iΔ g± = iε g
     
  4. May 23, 2013 #3
    i'm sorry but i don't understand how that helps... the perturbation is ε+[itex]\frac{\mu\hbar^2}{2mr^2}[/itex] and that term as I see it has the same eigenvalue/ expectation value for both [itex]g_+ = (1,1)^T (1+i)[/itex] and [itex]g_- = (1,-1)^T (1-i)[/itex]. Did I misunderstand you?
     
  5. May 23, 2013 #4

    DrDu

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    Are you sure that your expression for the unperturbed g is correct?
     
  6. May 23, 2013 #5
    Yes DrDu, I would be very surprised if it turns out to be incorrect...

    the solution after treating it to first order, claim the authors, is
    [tex]g=const (e^{i\psi/2}, -ie^{-i\psi/2})^T e^{-K}[/tex]
    where
    [tex]\psi(r)=-\int_r^{\infty}e^{2K(r)-2K(r')} \frac{2}{\lambda}\left(\epsilon +\frac{\mu\hbar^2}{2mr'^2}\right) dr'[/tex].

    As you can see this reduces to the solution above as the perturbation is turned off.
     
  7. May 23, 2013 #6

    DrDu

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    Yes, I was wrong here. You can easily see that the unperturbed eigenfunction must be an eigenstate of sigma_y by multiplying the whole equation by sigma_z and using that sigma_z^2=1 and sigma_y sigma_x is proportional to sigma_y. Then you also see that the right hand side becomes proportional to sigma_z which is not diagonal in the basis of the eigenstates of sigma_y.

    The structure becomes even clearer upon the cyclic substitution sigma_z->sigma_x, sigma_y-> sigma_z, and sigma_x-> sigma_y. The resulting equations then look basically like the ones given by Bill_K.
     
  8. May 23, 2013 #7

    Bill_K

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    The perturbation does not have the same eigenvectors. It is an off-diagonal term that connects one of the eigenvectors g± with the other. Consequently its effect is to split the degeneracy.

    You are apparently thinking of it as proportional to the identity matrix, and therefore the same in every direction, but it is not. Write the matrix equation out as a pair of coupled equations for the components of g, and you'll see your error.
     
    Last edited: May 23, 2013
  9. May 23, 2013 #8
    Thanks, DrDu. That does seem to be a good idea. And your explanation explains Bill_K's suggestion.
    If i label the unperturbed solutions as |g_1> and |g_2> then the vectors
    [tex] g_{\pm}=e^{-K}|g_1> \pm e^{K}|g_2> [/tex]
    are proportional to [itex] \sigma_z [/itex] eigenstates and are therefore the perturbation is indeed diagonal in this basis and off diagonal in the original one. So in this basis the first order shifts in energy are given by [itex] \pm (\epsilon + ..) [/itex].

    However, there is still the problem that the two states thus found are degenerate eigenstates of the hamiltonian [itex] H_0 [/itex]. From what i know of degenerate perturbation theory, the corrections to the eigenvectors actually arise from coupling to the non-degenerate subspace of solutions. It seems to me that in this problem the only states are degenerate. In this situation Sakurai says "Note also that if the degenerate subspace were the whole space, we would have solved the problem exactly in this manner." But its obvious that the sigma_z eigenstates are not the solutions to the full problem. How do I get the first order correction to the eigenvector |g_2>?
     
  10. May 25, 2013 #9
    All right, i did manage to did it down eventually by writing
    [tex]g=e^{-(K+\delta K)}(e^{i\psi/2},-i e^{-i\psi/2})^T[/tex]
    and solving for [itex]\delta K[/itex] and [itex]\psi [/itex] to first order.
    The resulting differential equations for [itex]\delta K[/itex] and [itex]\psi [/itex] show that the first correction to K is second order and [itex]\psi[/itex] comes out as given earlier.

    Possibly this is what the authors meant by treating the right hand side as a perturbation, and not the whole machinery of perturbation theory.
     
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