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

Finding expectation value using Heisenberg picture

  1. Sep 12, 2005 #1
    We have a particle in a harmonic oscillator potential. The eigenstates are denoted {|0>,|1>,...,|n>,...}. Initially the particle is in the state |s> = exp(-ipa)|0>, where p is the momentum operator.

    I need to find <x> as a function of time using the Heisenberg picture. The problem is, how do I find an expression for |s>? My guess is to express p as p = (some constant)(a- - a+) (where a-/a+ are the ladder operators) and then:

    |s> = exp(-i(some constant)(a- - a+)a)|0> = exp(-i(some constant)a)|1>.

    But I know this is wrong... What am I supposed to do??
     
  2. jcsd
  3. Sep 12, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well, you've got [itex] |s\rangle [/itex] in the Schrödinger picture. Use the

    [tex] |state\rangle_{Heisenberg}\left(t_{0}\right) = \hat{U}^{\dagger} \left(t,t_{0}\right)|state\rangle_{Schrödinger}(t) \hat{U}\left(t,t_{0}\right) [/tex]

    thingy and the expression of the time evolution operator involving the Hamiltonian.

    Daniel.
     
  4. Sep 12, 2005 #3

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    dextercioby,

    You can't multiply a ket on both sides with an operator. You are thinking of how operators transform:

    [tex] A_H(t) = U^\dagger(t) A_S U(t) [/tex]

    where [tex] U(t) = e^{-i H t /\hbar} [/tex]

    The connection between the Schrodinger state and Heisenberg state is:

    [tex] | \psi \rangle_H = U^\dagger (t) | \psi \rangle_S [/tex]

    broegger,

    The state you've got there is called a coherent state. You can do the problem using the raising and lowering operators, but your manipulation of the raising and lowering operators is not quite right. You need to be careful with your exponentials. Rememeber that [tex] e^{A+B} = e^A e^B [/tex] only if A and B commute. Also I'm not sure how you got just the |1> state in your final step. There is a theorem called the Baker-Campbell-Hausdorff theorem that you might find useful for writing a coherent state in terms of number states.

    There is also a more elegant way to do this problem using the Baker-Campbell-Hausdorff theorem (or a version of it) where you need to compute the two objects:

    [tex] e^{i \hat{p}_S a} \hat{p}_S e^{- i \hat{p}_S a} [/tex]
    [tex] e^{i \hat{p}_S a} \hat{x}_S e^{- i \hat{p}_S a} [/tex]

    See if you can make progress with these hints.
     
    Last edited: Sep 12, 2005
  5. Sep 12, 2005 #4

    reilly

    User Avatar
    Science Advisor

    The Heisenberg equations of motion for x are coupled with those for p -- you get them from the commutators, [H,x] and [H,p], and they are directly solvable. The only slightly tricky point is to do everything in harmonic oscillator eigenstates. The integrals involved are standard Gaussian integrals. The rest is standard QM in the Heisenberg Picture as explained in many text books. I would highly recommend doing the problem the old fashioned way, described here, before going on to coherent states, which although very elegant, are not orthogonal and form a so-called over-complete basis.\
    Simple is good.
    Regards,
    Reilly Atkinson
     
  6. Sep 12, 2005 #5

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Simple is good. Nevertheless, I think it is possible to mistake familiar methods for simple methods. You can certainly do the whole problem with good ol' position dependent wavefunctions, etc. This is definitely the most familiar method, but I don't think it is most physical approach to the problem. As a physicist I tend to regard the most physical approach as the simplest one. That being said, I think if someone is planning on a future without much quantum mechanics, then the familiar wavefunction approach is fine. On the other hand, if someone is planning to use their QM knowledge again, then I say the quicker you learn the easy physical ways to do problems (even if they involve more complicated mathematics) the better. Physicists would do the problem the way I suggested (for the most part). This could just be the physicist/theorist in my talking though :smile: . Any thoughts?
     
  7. Sep 12, 2005 #6

    reilly

    User Avatar
    Science Advisor

    Physics Monkey--
    I think we are talking matters of taste and style. To me my approach is eminentnly physical -- I was trained roughly 40 years ago, and my profs were steeped in classical mechanics -- so I like x's and p's and wave functions. I'm also a theoretician, but clearly of an older generation -- the simpler the math, the better.

    If coherent states work for you in this problem, that's fine. In the early days of QED, after WWII, some prefered Schwinger's approach, many others preferred Feynman's approach.

    And, by the way, I think coherent states provide a very powerful tool. But, I think it's best to get the basic groundwork before moving onto more sophisticated topics, including coherent states. To me, the notion of overcompleteness is not an easy one.

    I'm happy to state. that both our approaches are equally valid.

    Regards,
    Reilly Atkinson
     
  8. Sep 13, 2005 #7
    Thanks everybody. I'm still having problems, though.

    I've found the Baker-Hausdorff theorem in my book and I'll try to apply that (I don't have my papers here right now), but I'm also interested in the "wave mechanics" approach. My problem is, I'm bad at working with exponentials of operators and the like.

    For example, Physics Monkey mentioned that exp(A+B)=exp(A)exp(B) only if [A,B]=0 and I know that if A|a>=a'|a> (|a> is an eigenket of A with eigenvalue a') then exp(A)|a>=exp(a')|a>. But obviously exp(A+B)|a> = exp(a'+B|a>) is not true. What are the arithmetics of these things??

    Update: I found [tex]\langle x \rangle(t) = a\cos(\omega t)[/tex] using the Baker-Hausdorff theorem.
     
    Last edited: Sep 13, 2005
  9. Sep 13, 2005 #8

    reilly

    User Avatar
    Science Advisor

    Actually, there's the Campbell-Baker-Hausdorf thrm (in Optical Coherence &Quantum Optics by Mandell and Wolf, P 519) which says that exp(A+B) = Exp(A)exp(B) if A and B both commute with the commutator of A and B --- [A,[A,B]] = [B,[A,B] = 0. This allows
    exp(va* - v*a) = exp(va)exp(v*a), where a and a* are creation and destruction operators -- for the harmonic oscillator/photons a= n(Wq+ip) where W is the oscillator frequency, a* =n(Wq-ip), n is a normalizing constant, and q, and p are respectively position and momentum operators. v is an arbitrary complex constant, v* is its complex conjugate.

    So, applied to a vacuum state, a|0>=0;, exp(va*-v*a)|0> = exp(va*)|0> =|v>. the coherent state with with parameter v. Note that <s|v> =/ 0, the coherent states constitute a set of non-orthogonal states. Mandel and Wolf give an excellent account of coherent states, and their use in quantum optics.(Note, [a,a*]=1, so both a and a* commute with their commutator. Their book, however, presumes familiarity with the basics of QM.

    I still suggest that the approach I outlined is straighforward, and certainly tedious. You'll find that both x and p are periodic.

    Regards,
    Reilly Atkinson
     
    Last edited: Sep 13, 2005
  10. Sep 13, 2005 #9

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    reilly,

    Let me first say your points are all well taken. Certainly if the wavefunction approach seems more physical to you, then I won't argue. It is a matter of taste and style.

    Secondly, I don't have my copy of Mandel and Wolf handy, but if you copied the theorem correctly then the book has a misprint. The correct statement of the Baker-Cambpell-Hausdorff theorem in the case when [A,[A,B]] = [B,[A,B]] = 0 is,
    [tex] e^{A+B} = e^A e^B e^{-1/2 \,[A,B]} [/tex]

    This means that the so called 'displacement' operator is,
    [tex] D(v) = e^{v a^\dagger - v^* a} = e^{v a^\dagger} e^{-v^* a} e^{-1/2\, v^* v} [/tex]
    This is important because the coherent state is really given by,
    [tex] | v \rangle = e^{-1/2\, v^* v} e^{v a^\dagger} | 0 \rangle [/tex]

    broegger,

    I'm glad to hear that you figured out the problem. Are you still interested in using the wavefunction approach as well?
     
  11. Sep 13, 2005 #10

    reilly

    User Avatar
    Science Advisor

    Physics Monkey -- You are correct. From time-to-time I blow it, and I just did. I have a bad habit of forgetting about normalizing factors. Thanks

    Broegger:
    I think the wave function approach goes something like this:
    <p|s> = const*exp(-ipa - cnst p*p), and @ t=o, x= id/dp. Get the time stuff, and then the expectation value is a gaussian integral. I think your answer is not correct, if only because no oscillator parameters are involved.

    Regards,
    Reilly Atkinson
     
    Last edited: Sep 13, 2005
  12. Sep 14, 2005 #11
    My answer, obtained by the Baker-Hausdorff theorem, was correct, it has just been confirmed :)

    Thanks, guys.

    Physicsmonkey: Is the wave mechanics approach related to the fact that exp(-ipa/h) is the translation-in-space operator?
     
  13. Sep 14, 2005 #12

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    The wavefunction approach really just means that you use the position basis (or just as well, the momentum basis) to calculate expectation values. The fact that your state is defined in terms of the translation-in-space operator and the ground state means that the wavefunction is easy to obtain: the wavefunction [tex] \psi_s(x) =\langle x | s \rangle [/tex] of the coherent state is just the ground state wavefunction displaced by an amount a. Using this fact and the formulas you derived for the time dependence of the Heisenberg operators in terms of the Schrodinger picture operators, you can calculate the expectation value by doing two integrals.
     
  14. Sep 15, 2005 #13
    Thanks. I must say I like the Baker-Hausdorff-way better, although the wave mechanics approach is more straightforward (and probably what we were supposed to do).
     
  15. Sep 19, 2005 #14

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I must have had a very bad day. :(( Sorry, it won't happen again. :approve:

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

Have something to add?



Similar Discussions: Finding expectation value using Heisenberg picture
  1. Heisenberg picture (Replies: 2)

  2. Heisenberg picture (Replies: 9)

  3. Heisenberg Picture (Replies: 3)

Loading...