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Hamiltonian eigenstate problem

  1. Apr 14, 2013 #1
    Hi PF members,

    I am stuck with a problem about larmor precession. I cannot find the eigenstates of the hamiltonian given as

    [itex]H = \frac{\hbar}{2}\begin{pmatrix} \omega_{0} & \omega_{1}\delta(t-t') \\ \omega_{1}\delta(t-t') & \omega_{0} \end{pmatrix}[/itex]

    Can anyone help me? Since it has time dependence I cannot figure out how to solve this problem.

    Thank you.
     
  2. jcsd
  3. Apr 14, 2013 #2
    Any advice :(
     
  4. Apr 14, 2013 #3
    I'm confused: are those ##\delta##'s Dirac delta functions? If so, then ##\hat{H}(t)## is ill-defined when ##t=t'##. (The integral ##\int \hat{H}(t) dt## can still be defined using step functions.)

    I think there might be a typo in there, too. The usual way to write ##\hat{H}## for a spin-1/2 in a ##\mathbf{B}## field is:
    ##
    \hat{H}
    = - \tfrac{\hbar\omega_{0}}{2}\hat{\sigma}_z
    = \frac{\hbar \omega_{0}}{2}\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}
    ##
     
  5. Apr 14, 2013 #4
    Yes, they are dirac-delta functions. Those dirac-deltas stand there for representing a magnetic field pulse in x direction at t = t'
     
    Last edited: Apr 14, 2013
  6. Apr 14, 2013 #5
    Any ideas?
     
  7. Apr 14, 2013 #6
    When ##t \neq t'##, ##\hat{H}## is ##\tfrac{1}{2}\hbar \omega_0## times the identity matrix. Then every vector is an eigenvector with eigenvalue ##\tfrac{1}{2}\hbar \omega_0##.

    When ##t = t'##, ##\hat{H}## is
    ##
    \hat{H} = \frac{\hbar}{2}
    \begin{bmatrix}
    \omega_0 & \infty \\
    \infty & \omega_0
    \end{bmatrix}
    ##
    Then ##\hat{H}## is ill-defined. This is why I think the problem has somehow been miscommunicated.
     
  8. Apr 14, 2013 #7
    Yes I understand your point. But problem says that we apply an instantaneous magnetic field in x direction on the precessing particle at time t'. I tried to solve this like the dirac-delta potential problem but it is different I think. Because in this case I cannot find a time dependent wave function for the initial hamiltonian since now potential is time dependent. So we cannot seperate time dependent solution and time-independent solution.
     
  9. Apr 14, 2013 #8
    When the Hamiltonian is time-dependent, there aren't really "eigenstates." Or rather, for a given t you can find the eigenstates of ##H(t)##, but the eigenstates for different t will in general be different.

    However, note that for t < t' and t > t', the Hamiltonian is time-independent. You can find the eigenstates of the time-independent Hamiltonian that governs the system for these times. Then to connect t < t' to t > t' you need to integrate the Schrodinger equation

    ##i \hbar \frac{d}{dt} \psi(t) = H(t) \psi(t)##

    over the infinitesimal interval ##t' - \epsilon < t < t' + \epsilon##. If you take ##\epsilon \to 0## this will give a simple relationship between ##\psi(t - \epsilon)## and ##\psi(t + \epsilon)##.

    Mathematically, this procedure is much like the way in which you solve the time-independent Schrodinger equation for the potential ##V(x) = -V_0 \delta(x-x_0)##.
     
  10. Apr 14, 2013 #9
    But in the larmor precession case, eigenvectors of the hamiltonian is all time-independent. How can I know the before and after wave functions. Before and after time t' wave functions are same.
     
  11. Apr 14, 2013 #10
    Here is the main nontrivial part of the problem:

    Suppose the state of the system at ##t = t' - \epsilon## is ##\psi_0##. What is the state at ##t = t' + \epsilon##? Assume ##\epsilon## is infinitesimal.

    You can get the answer to this sub-problem by integrating the Schrodinger equation over the time interval in question.

    Once you have solved this sub-problem, you should be able to extend it to answer the general question: given the state of the system at any time ##t_1##, what is the state at any other time ##t_2##?
     
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