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Finding eigenstates and eigenvalues of hamiltonian

  1. Mar 12, 2013 #1
    Hey there, the question i'm working on is written below:-

    Let |a'> and |a''> be eigenstates of a Hermitian operator A with eigenvalues a' and a'' respectively. (a'≠a'') The Hamiltonian operator is given by:

    H = |a'>∂<a''| + |a''>∂<a'|

    where ∂ is just a real number.

    Write down the eigenstates of the hamiltonian. what are their energy eigenvalues?




    Was feeling a bit confused by the question at first, and was just wondering if someone could let me know if my thoughts so far are on the right track?

    Essentially, I just wrote the eigenstates as the kets: |E[itex]_{a'}[/itex]> and |E[itex]_{a''}[/itex]>. In order to find the eigenvalues of the energy, I constructed the matrix:

    H = [tex]
    \begin{pmatrix}
    0 & ∂\\
    ∂ & 0
    \end{pmatrix}
    [/tex]


    so that I could use the general det(H - λI) = 0 to find the eigenvalues. is this reasoning vaguely in the correct direction?

    Also, was just wondering, if in the original hamiltonian equation, i'm allowed to take the ∂ symbol out (written below) because it is just a real number?

    H = ∂(|a'><a''| + |a''><a'|)
     
  2. jcsd
  3. Mar 12, 2013 #2
    Yes to both, you are headed in the right direction
     
  4. Mar 14, 2013 #3
    thanks for that. i have actually continued on with this problem, and i've come across another question.

    i found the eigenvalues to be ±∂. the problem then asks

    b) suppose the system is known to be in state |a'> at t=0. write down the schrodinger picture for t>0.

    Which my working out is:|α, t[itex]_{0}[/itex]=0> = |a'>

    |α, t[itex]_{0}[/itex]=0; t> = U(t,t[itex]_{0}[/itex])|a'>
    =U(t,0)|a'>
    =exp( (-iE[itex]_{a'}[/itex]t) / h-bar )|a'>
    =exp( -i∂t / h-bar )|a'>

    question c)
    what is the probability for finding the system in state |a''> for t>0, if the system is known to be in state |a'> at t=0?

    for this, i was thinking that for |a''>= U(t,t[itex]_{0}[/itex])|a'> = =exp( -i∂t / h-bar )|a'>

    and for the probability, i need to calculate |<a''|a'>|[itex]^{2}[/itex] = |(exp( -i∂t / h-bar ) <a'| )(|a'>)|[itex]^{2}[/itex], where <a'|a'> = 1

    am i allowed to do this??? i'm probably ignoring something important....
     
  5. Mar 15, 2013 #4

    vela

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    The state ##\lvert a'\rangle## isn't an eigenstate of the Hamiltonian, so it's not correct to say it has an energy ##E_{a'}##. You need to express ##\lvert a'\rangle## in terms of the eigenstates of H.
     
  6. Mar 15, 2013 #5
    oh ok then. would this be the right plan of action then?

    using the eigenvectors i found for the hamiltonian|E[itex]_{1}[/itex]> = (1,1) and |E[itex]_{2}[/itex]> = (1,-1). i then constructed:

    ||E[itex]_{1}[/itex]> = |a'> + |a''> and
    |E[itex]_{2}[/itex]> = |a'> - |a''> ( i have left out the normalization constant here)

    then i can have |a'> and |a''> written in terms of the eigenstates of H. back in the schrodinger picture, can i then substitute |a'> for |E[itex]_{1}[/itex]>+|E[itex]_{2}[/itex]>, and then use E[itex]_{1}[/itex] & E[itex]_{1}[/itex] in the exponential? ==>

    |α, t[itex]_{0}[/itex]=0; t> = U(t,t[itex]_{0}[/itex])|a'>
    =U(t,0)|a'>
    =exp((-iE[itex]_{a}[/itex]′t) / h-bar)|a'>
    =(exp(-i∂t/h-bar)||E[itex]_{1}[/itex]> + exp(i∂t/h-bar)||E[itex]_{2}[/itex]>

    then sub |a'> and |a''> back into this equation?
     
  7. Mar 15, 2013 #6

    vela

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    Yes, that's exactly what you want to do.
     
  8. Mar 15, 2013 #7
    yay! thanks for your help :)
     
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