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Electric field perturbation HELP PLEASE!

  1. Apr 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A plane (x-y plane) rotor has moment of inertia I and electric dipole moment P.

    A uniform electric field is applied along the x-direction (Ex, so the interaction

    energy can be written as

    H'=P*E cosΦ, where phi is the angle between the E-field and the dipole P

    Find the eigenenergies of the rotor for arbitary eigenstate up to 1st order perturbation

    2. Relevant equations

    torque of dipole in unifrom electric field
    τ=p*E

    dipole
    P=Q*D

    S=I*omega (spin angular momentum)

    3. The attempt at a solution
    from E&M
    A dipole placed in a uniform electric field has no force on it however a torque IS induced that tends to align the dipole up parallel to the E-field

    how do i calculate the hamiltonian? i am given the perturbation but do not have the wavefunctions for this system.

    any help would be impressive as i am not sure how to even approach the problem
    cheers
    nate
     
  2. jcsd
  3. Apr 26, 2007 #2

    nrqed

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    The unperturbed hamiltonian is simply the energy of a spinning rotor with moment of inertia I. So it's basically just rotational kinetic energy. You may write this as something proportional to [itex]{\vec L}^2[/itex] (I will let you figure out the constant, which will of course contain I). This is your unpertubed hamiltonian. You then know the eigenstates and eigenvalues since you surely already know the eigenstates and eigenvalues of [itex]{\vec L}^2[/itex] .
    Then you apply perturbation theory with your H' (notice that there is degeneracy in the unperturbed solution)


    Patrick
     
    Last edited: Apr 26, 2007
  4. Apr 26, 2007 #3
    ok patrick i thank u. i am still i tad confused...

    i know the rotational K.E. is 1/2*I*ω^2 and the potential is just from the E-field = q*φ


    so, the unperturbed Hamiltonian is T+V = 1/2*I*ω^2 +q*φ? i made a mistake on the perturbed hamiltonian, there should be a negative sign

    H'=-P*E*cosφ

    <ψn|H'|ψm>

    what are the ψ's for this?
     
  5. Apr 26, 2007 #4

    nrqed

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    The E field here is treated as a perturbation so there is no electric field contribution to the unperturbed hamiltonian.

    You have the right kinetic energy. Now, recalling that the magnitude of the angular momentum is [itex] L = I \omega [/itex], you can rewrite [itex] H_0 [/itex] in terms of the angular momentum (which will be squared, obviously). Since you know the eigenvalues and eigenstates of [itex] \vec{L}^2 [/itex], you basically have solved the unperturbed hamiltonian.


    Patrick
     
    Last edited: Apr 26, 2007
  6. Apr 26, 2007 #5
    i see u r saying that u JUST need the rotational kinetic energy and no potential energy V=0
    H=T+V = 1/2*I*omega^2
    I=1/3ML^2 (for rotation about end of rod of length L)

    now i need the eigenvalues, but i again have a question!

    this is a quantum test but do i don't need quantum mechanics to solve this?

    this is just perturbation theory, the Hamiltonian still being the total energy E of the system in question.

    for 1st order correction, i use power series to the 2nd term, Ao + εA1?
     
  7. Apr 26, 2007 #6

    nrqed

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    There is no need to introduce an explicit form for I. Just leave it as "I" (we don't know the exact mass distribution so we can't use any specific formula)


    What you need now is to rewrite the unperturbed Hamiltonian in terms of the angular momentum L. see my previous post.
    You do. You will use that to identify the unperturbed eigenstates and eigenvalues.
    Yes...
     
    Last edited: Apr 26, 2007
  8. Apr 26, 2007 #7
    H=Ho+H'

    Ho=1/2*I*ω^2, but L=I*ω, so i can rewrite into
    Ho=1/2*(I*ω)*ω=1/2*L∙ω

    now
    Lz|lm>=mhbar|lm>
    and
    L^2|lm>=l(l+1)*hbar^2|lm>

    so the eigenvalues and eigenstates of L^2 are l(l+1)hbar^2 with (2l+1) possible eigenvalues for Lz

    m=-l,-l+1...,0,...l-1,l

    so the total Hamiltonian is

    H=1/2*L*ω-P*E*cosφ
     
  9. Apr 26, 2007 #8

    nrqed

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    Wait...


    Omega is still a variable here. You want to rewrite the hamiltonian in terms of L and constants only.
     
  10. Apr 26, 2007 #9
    L^I*omega;

    L^2=I^2*omega^2
    OR
    L^2/I=I*omega^2=Ho
    yes?
     
  11. Apr 26, 2007 #10

    nrqed

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    Yes
    Not quite (that's not H_0) but you are just missing a factor of 1/2

    After that, what you said in your previous post was right...The eigenstates of your Hamiltonian will be the same as the eigenstates of [itex] L^2 [/itex] and the energies will be the eigenvalues of [itex] L^2 [/itex] multiplied by 1/(2 I). So your unperturbed system is now solved.
     
  12. Apr 26, 2007 #11
    H0=l^2/2*i?
     
  13. Apr 26, 2007 #12
    H=L^2/(2I)-PEcosφ is the correct unperturbed Hamiltonian.

    now i have H=L^2/2I-PEcosφ

    i know L^2 has l(l+1)hbar^2 for its eigevalues but what do i do with the H' term?

    again thanks for the help this problem is due Tomorrow!
     
  14. Apr 26, 2007 #13
    ok i got it!!! (i hope!)
    E(l,m)=(l(l+1)hbar^2/(2*I))-m*hbar!!!
    VIOLA!
    now i have to plug into the degenerate perturbation theory
    cheers
    nate
     
  15. Apr 26, 2007 #14

    nrqed

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    :confused: I don't think that the electric field piece shoul dbe there at all in the unperturbed energy. You should only have the first part on the left (andI don't know how you could get [itex] - m \hbar [/itex] )
     
  16. Apr 26, 2007 #15
    so its just (l(l+1)hbar^2/(2*I)?
     
  17. Apr 26, 2007 #16

    nrqed

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    Yes, this is the unperturbed energy
     
  18. Apr 28, 2007 #17
    anyone still with me on this problem?

    i need to find the solutions for 1st order perturbation for this (degenerate) system!

    ouch!

    I have as the eigenenergies E(l,m)=l(l+1)*hbar^2/(2*I)

    now how do i use denegerate perturbation theory to solve for the 1st order

    correction to an eigenstate?!?!

    i assume |lm> as the notation for eigenstate (eigenvector) for angular

    momentum operator L^2|lm>=l(l+1)*hbar^2|lm> and Lz|lm>=m*hbar|lm>

    HELP if u can, as i am spinning my wheels (rotors?!) at this point (haha, pun VERY intended lol)
     
  19. Apr 28, 2007 #18

    nrqed

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    You have to write the perturbation hamiltonian in a usable form so that the matrix element [itex] <l'm'| H_{pert} |l,m>[/itex] may be easily computed.

    Your perturbation hamiltonian is proportional to [itex] cos (\phi) [/itex] . The best thing to do probably is to rewrite this in terms of the ladder operators [itex] L_+ [/itex] and [itex] L_- [/itex]. Then the expectation values will be easy to calculate.
     
  20. Apr 30, 2007 #19
    got some of it! (i hope!)

    ok, i assume my question is a tough one, or that no one really understands

    what i am asking!

    :biggrin:

    here is what i have so far

    1st i need B in spherical coordinates

    aside from that i am using B=B0x+B0y+B0z in cartesian coordinates

    say the B-field is in the +x direction, i.e., B=B0x

    the interaction of the spin 1/2 particle( i assume electron) with the magnetic

    field is due to the magnetic

    dipole moment μ=(2*e*B0/m*c)*S ,where S is the spin vector (Sx,Sy,Sz)

    therefore μ=(2*e*B0/m*c)*Sx

    the Hamiltonian is 2*e*B0/(m*c)*Sx

    where Sx is the x-component of the spin vector

    if in the z-direction, its the same, just replace Sx with Sz etc....

    since i have to find the polarization direction <σ> (just the expectation value

    of σ) of the electron at time t>0, i assume i have to use the

    TIME_DEPENDENT S.E. -i*hbar(∂|ψ>/∂t)=H|ψ>

    hoes it looking so far? lol!
     
  21. Apr 30, 2007 #20
    oops!
    I need to move my last post to a different problem, but thank you nrqed (is

    the qed stand for what i think it does ) i thought that the ladder ops would be

    important for the states |l'm'>

    let me see what i can work out

    cheers
    nate
     
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