Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Perturbation theory

  1. Jun 1, 2005 #1
    Hi,

    I'm new to this subject, so bear with me. We consider the harmonic oscillator with a pertubation:

    [tex]\hat{H}' = \alpha\hat{p}.[/tex]​

    (What kind of a perturbation is that anyway, it's not a disturbance in the potential, what does it correspond to physically.)

    Now I have to calculate the first and second order energy corrections. I express p by the ladder operators:

    [tex]p=i\sqrt{\tfrac{\hslash m\omega}2}(a_+-a_-).[/tex]​

    I find that the first order correction is 0, and that the second order correction is [tex]E_n^2=m\alpha^2/2[/tex], that is, the energy shift is INDEPENDENT of n (i.e. it is the same for all excited states of the oscillator), but this can't be true, can it?
     
    Last edited: Jun 1, 2005
  2. jcsd
  3. Jun 2, 2005 #2
    Hint 1: H_harm= k1^2.p^2+k2^2.q^2
    If you add this potential (alpha.p) what do you obtain?

    Hint 2: what is the interaction of a spinless charged particle with a magnetic field?

    Seratend.
     
  4. Jun 2, 2005 #3
    Well, it could be solved analytically, but I don't know how. I'm not particularly familiar with solving such differential equations. I've tried solving it numerically in Matlab, and this confirms my result, but it's not quite satisfying.

    I don't know much about magnetic interactions in QM, haven't touched the subject in our intro course, unfortunately.
     
  5. Jun 2, 2005 #4
    Hint 3: a^2+ka= (a+k/2)^2 - k^2

    Hint 4: do you known the classical hamiltonian of a particle with an electromagnetic field?

    Seratend.
     
  6. Jun 2, 2005 #5
    I've tried completing the square, but I don't know how to go on.

    The energy of a classical particle in an electromagnetic field is just the sum of the energy associated with the electric and magnetic field, respectively, right?

    [tex]E = \frac{q_1q_2}{4\pi\epsilon_0r^2}-\mu\cdot B[/tex]​
     
  7. Jun 2, 2005 #6
    Ok,

    First, Let't take the harmonic oscillator hamiltonian:
    H= p^2+k.q^2
    You know the solutions of such an hamiltonian. Therefore I suppose you know the solution of H'=(p+a)^2+k.q^2, don't you?

    (k>0, I have choosen the units to got simple equations, you are free to choose your own units)

    If I had the interaction hamiltonian, I get:

    choosing alpha= 2a

    H=p^2+k.q^2+2a.p= (p+a)^2 + k.q^2 -a^2

    We just recover another harmonic oscillator with p'=p+a:
    [p,q]=[p+a,q]=-ihbar.

    The constant -a^2 does not change the spectrum (when you write the hamiltonian it is defined up to a constant).
    Or if you prefer a more mathematical way:
    the commutator [p'^2 + k.q^2 -a^2, p'^2 + k.q^2]=0 => we have the same eigenvalues. Or just solve the equation with an additional constant (not very difficult)

    Second: Classical hamiltonian: H= (P - qA)^2/2m + eV. Where A and V are the electromagnetic field potentials.
    Now, develop the square and assume A^2 is very very small, what do you obtain?

    Seratend.

    Edit: correction of minor typo errors + others : )
     
    Last edited: Jun 2, 2005
  8. Jun 2, 2005 #7
    I don't know if I'm doing it right, but I get the same result as perturbation theory - E = (n+1/2)hw - ma^2/2 - (that's a good sign, isn't it :)), so perturbation theory yields the exact result in this case?
     
  9. Jun 2, 2005 #8
    For the sign, I think it's ok.
    bravo!

    Seratend
     
  10. Jun 2, 2005 #9
    Weeh, thanks for helping!
     
  11. Jun 2, 2005 #10
    Additionnal question for you: why does the pertubation theory give the same result in this case?

    Seratend.
     
  12. Jun 2, 2005 #11
    Hmm, I think it's because we were able to factor the Hamiltonian perfectly, but I don't know, I was wondering the same thing...

    I also have a quick question for you (I asked this in a another thread, but I'm still thinking about it). Suppose you have a hydrogen electron in the mixed state:
    [tex]|\psi\rangle=\frac1{\sqrt2}(|210\rangle + |211\rangle)[/tex]​
    (|nlm> denotes an eigenstate of H, L^2 and L_z - n is the main quantum number and l and m is the quantum numbers associated with L^2 and L_z, respectively).

    Now, I find that L_x and L_y has different expectation values (h/sqrt(2) and 0, respectively), but how can this be, when there is nothing to distinguish the x- and y-directions in this case??
     
  13. Jun 2, 2005 #12
    Apply a rotation (z axis) to the state, does it change?

    Seratend.
     
  14. Jun 2, 2005 #13
    Think on the functions: f(x) and f(x+h)~f(x)+hf'(x). When do you have the equality?
    Think now on the expansion of the hamiltonian, with the pertubation theory. It just allows more possibilities.


    Seratend.

    EDIT change = into ~to avoid confusion : )
     
  15. Jun 2, 2005 #14
    No, I don't think so. It's symmetric with respect to x and y, that's the problem. I might be wrong, though :)
     
  16. Jun 2, 2005 #15
    I am not sure, but you should try to work a little : )

    here is the rotation matrix for an angle phi:
    Code (Text):
    (exp(-i.phi/2)      0)  
    (0      exp(iphi/2))   
    Do you keep saying the state does not change?

    Seratend.

    EDIT: corrected a missing i in the roation matrix. : )
     
    Last edited: Jun 2, 2005
  17. Jun 2, 2005 #16
    Hehe, you're right, I should...

    Well no, I just thought "geometrically", you know :) I'll try working it out from the start again...
     
  18. Jun 2, 2005 #17
    Good. Progression in the knowledge requires somtimes some efforts : )

    Seratend.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Perturbation theory
  1. Perturbation Theory (Replies: 1)

  2. Perturbation Theory (Replies: 5)

  3. Perturbation theory. (Replies: 0)

  4. Perturbation Theory (Replies: 3)

Loading...