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Non-degenerate and degenerate perturbation theory

  1. Jul 30, 2009 #1
    Consider a system of a rigid rotator together with a uniform E-field directing along z-axis. So to calculate the perturbed energy and wavefunction we have to use perturbation theory. But the book said we can use non-degenerate one to calculate the result. I wonder why. It is because the original hamitonian and the perturbed one commute with Lz so that we are sure that the first term of the perturbed wavefunction must be the original unperturbed eigenfunction? Does it mean that there are no more physical quantities except Lz which could be certain if we are certain of its energy?
    Also, why do we use a linear combination of the degenerate eigenfunction for the first term of the perturbed wavefunction? I think it is becos we are not sure of how other physical quantities behave under the perturbed situation. What do you guys think?:smile:
     
  2. jcsd
  3. Jul 31, 2009 #2
    Sorry lucien, your post is a bit hard to follow. Can you give us a better description of your rotator- perhaps a mathematical statement?
     
  4. Aug 1, 2009 #3
    Sorry for poor presentation
    The hamitonian of the system is [tex]\frac{L^2}{2m} - kBz[/tex]

    and the unperturbed eigenfunction is [tex]Y_{l,m}(\theta,\phi)[/tex]

    As you know, the unperturbed eigenfunction is degenerate with respect to m (where m =-l,-l+1,...,l ). But the book said we can use non-degenerate perturbation theory to handle this case. I wonder why.

    And I have to make it clear that the second paragraph is an independent question and not related to the rotator system. It is just my thought.
     
    Last edited: Aug 1, 2009
  5. Aug 1, 2009 #4
    Thanks for the clarification. I think you're right that the relevant point is that the commutation relations associated with Lz, although in general "A commutes with B" and "B commutes with C" does not mean that "A commutes with C", so you have to be careful about how you structure the argument.
    Physically, I don't think it's quite as strange as you first think. The case m= +/-l means that the motion is in a plane parallel to the xy plane, so z is constant; as the modulus of m gets smaller this plane is rotated, but the expectation- the average- value of z won't change, and the 1st order correction to the energy is directly proportional to the expectation value of z. It's not like the zeeman effect in atoms, where the fact that electrons are charged leads to additional interaction between their magnetic moment and the applied field that is dependent upon m.

    As for your second comment, I think I understand you now. The whole point of perturbation theory is that you assume that your system is "close to" one that you can solve exactly- if it's not, then the whole thing blows up in your face. In the derivation of these results it's this closeness that you use to simplify expressions. Perhaps it's the presentation in my lecture notes, but the whole thing really seems like a natural, sensible way to proceed.
     
  6. Aug 2, 2009 #5
    I can't quite understand the first paragraph. Can you explain it more?

    Yeah my thought is kind of related to the closeness you mentioned. Thanks for reminding me this. I feel I understand it more now.
     
  7. Aug 2, 2009 #6
    I should really have put my first paragraph as two separate points, sorry.

    The first point- you have to be careful about what conclusions you draw from the commutation relations. For example, Lz commutes with L^2 and L^2 commutes with Lx, but Lx and Lz do not commute. The fact that old and new hamiltonians commute with Lz means that old and new energy eigenstates are both eigenstates of Lz, but it doesn't mean that they are the same eigenstates of Lz. And even if they were, it wouldn't necessarily follow that an eigenstate of Lz had the same energy eigenvalues under the two hamiltonians.
    In this example, we are sure that the first term of the perturbed wavefunction is the unperturbed one, but not because of the commutation relations- it's because that's the way perturbation theory works! Once you start thinking about the corrections to the unperturbed states, however, you're in dangerous territory -they're neither a basis set nor orthogonal.

    The second point is the actual reason we can use non-degenerate perturbation theory, which is because all of the eigenstates of m experience the same shift in their energies
    to first order. The simplest way of seeing this is to examine the 1st order correction:
    [tex]<\psi|H'|\psi>=-kB<\psi|z|\psi>[/tex]
    -it's completely independent of the angular momentum about the z axis. Here you can say that the commutator of z, Lz is 0 in order to support this statement.

    However, I don't see how you could possibly compute the new state without using degenerate perturbation theory- does the book do actually do this, or does it just compute the correction to the energy?

    One final aside:
    Not quite- we've taken the radius to be fixed, I think?
     
  8. Aug 2, 2009 #7
    Yeah I know if A,B commute and B,C commute doesn't mean A,C commute.
    Actually I just think of a good way to present my thought. Can you comment on this?

    So both unperturbed hamilotonian and perturbed one commute with Lz. Therefore, under both unperturbed and perturbed situations the energy and the angular momentum about z axis can be determined simultaneously.
    Let the perturbed wavefunction be
    [tex]\phi^{(0)}+\lambda\phi^{(1)}+\lambda^{2}\phi^{(2)}+...[/tex]
    and this wavefunction corresponds to mh as its angular momentum about z axis
    So when we substitute [tex]\lambda=0[/tex] into the equation, the resulting wavefunction, which is the unperturbed wavefunction, also correspond to mh as its angular momentum about z axis. So the first term of the series is exactly [tex]Y_{m,l}(\theta,\phi)[/tex]
     
  9. Aug 4, 2009 #8
    As I mentioned above, this is true but it's not interesting- as it's just because you're doing perturbation theory. The first term in your expansion is ALWAYS the unperturbed term; what you're interested in is the 1st order correction(s) to this term, i.e. [tex]\lambda=1[/tex].

    That the 1st order correction keeps the degeneracy is because of the specific form of this perturbation, and isn't generally true. If your rotator was charged, for example, then you'd have an additional term as a result of its magnetic moment interacting with the B-field, and hence the different values of m would no longer be degenerate.
     
  10. Aug 5, 2009 #9
    Do you mean that since the perturbed wavefunction to the first order of [tex]\lambda[/tex] is degenerate to m so that non-degenerate wavefunction can be applied? Why is it true? Can you explain a bit more? Thank you:smile:
     
  11. Aug 5, 2009 #10
    I'm not convinced that you actually can use non-degenerate perturbation theory to compute the new wavefunction. What I've tried to explain above is why you can use non-degenerate perturbation theory to compute the corrections to the energy. Can you give us a reference so I can have a look at this book please?
     
  12. Aug 5, 2009 #11
    Oh you are only talking about the energy shift, I am so careless to take this important point. Sorry for that. I understand all you said now.
    Well my book is Quantum Mechanics by B.H.Bransden and C.J.Joachain(Second Edition).
    That part is CH12.1 "The Stark effect and the rigid rotator".
    Sorry I try to find you a ebook for quick look but fail.

    The book only calculates the first and second order correction of energy by non-degenerate perturbation theory. But it said," Since H0 and H commute with Lz, non-degenerate perturbation theory is applicable."
     
  13. Aug 5, 2009 #12
    Can I use these to say that we can use non-degenerate perturbation theory to calculate the new wavefunction
     
  14. Aug 5, 2009 #13
    As it happens I own the book! :smile:
    I'll try and dig it out tomorrow, but I'm fairly certain you can't compute the correction to the actual wavefunction without using degenerate perturbation theory- the commutation relations wouldn't stop you dividing by zero!

    Also, I'm off to Germany tomorrow evening, so if I don't reply for a few days to any further responses that's why.
     
  15. Aug 6, 2009 #14
    If you use the non-degenerate perturbation theory, it happens that the correction to wavefunction has undetermined coefficients. I guess maybe these coefficients are not physically important and hence can be arbitrarily chosen. What do you think?
     
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