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Time dependent quantum state probability calculation

  1. May 27, 2014 #1
    For part a I have (H0-ω[itex]\hbar[/itex]m)|nlm>, which I think the (H0-ω[itex]\hbar[/itex]m) part is the eigenvalue of the Hamiltonian, also is the energies?

    And mainly, I am not sure how to approach part b, the time variable is not in any of the states. I saw this in our lecture notes: ψ(r,t)=∑Cnψn(r) e-iEnt/[itex]\hbar[/itex]. Do I simply add the e-iEnt/[itex]\hbar[/itex] term after each ψ state? And then what? Please help me with this, thanks
     

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  3. May 28, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    ##H_0## is the Hamiltonian for the hydrogen atom in the absence of the external magnetic field. You should be able to find its eigenvalues.

    Basically, yes. The time evolution of a state ##\psi## (with a time-independent Hamiltonian ##H##) is given by
    $$
    \psi(t) = e^{-i H t / \hbar} \psi(0)
    $$
    For ##\phi_n## and eigenstate of ##H## with eigenvalue ##E_n##,
    $$
    e^{-i H t / \hbar} \phi_n = e^{-i E_n t / \hbar} \phi_n
    $$
    Therefore, if ##\psi## is a superposition of eigenstates ##\phi_n##,
    $$
    \psi(0) = \sum_n c_n \phi_n
    $$
    then
    $$
    \psi(t) = \sum_n c_n e^{-i E_n t / \hbar} \phi_n
    $$

    When you've done a correctly, you'll have ##E_{nlm}## for each state ##| n l m \rangle##, and therefore can get ##|\Psi(t)\rangle## for any time ##t##. You will then have to calculate projections like ##\langle \psi_1 | \Psi(t) \rangle##, from which you can calculate the probabilities.
     
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