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QM: Groundstates and changing Hamiltonians

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    At time t<0 I have a Hamiltonian given by:

    [tex]
    H = \left( {\begin{array}{*{20}c}
    {\varepsilon _1 } & { - V} \\
    { - V} & {\varepsilon _2 } \\
    \end{array}} \right)
    [/tex]

    with the ground state [tex]\left| {v_ - } \right\rangle = A\left( {\sqrt 2 - 1} \right)\left| 1 \right\rangle + \left| 2 \right\rangle [/tex], where A is some constant.

    Now at t=0 the Hamiltonain changes, since [tex]\varepsilon_1=\varepsilon_2=0[/tex] for t>0.

    I have to find the probability of finding the particle in the groundstate of the new Hamiltonian at the time t=0.


    3. The attempt at a solution

    Ok, first I have found the groundstate [tex]\left| {m_0 } \right\rangle[/tex] of the new Hamiltonian. Since I have to find the probability of the particle being in the state [tex]\left| {m_0 } \right\rangle[/tex] at t=0, I still need the Hamiltonian for t<0, i.e. I need:


    [tex]
    H = \left( {\begin{array}{*{20}c}
    {\varepsilon _1 } & { - V} \\
    { - V} & {\varepsilon _2 } \\
    \end{array}} \right)
    [/tex]

    I am a little uncertain of what should be done now. Can you push me in the right direction?
     
  2. jcsd
  3. Oct 15, 2008 #2
    I'm also uncertain, but my guess is that you should first normalize your ground states and then take the square modulus of their inner product to get the probability:

    [tex]P = |\langle v_- | m_0\rangle|^2[/tex]
     
  4. Oct 15, 2008 #3
    I think we're on the right track, Irid.

    I'm quite sure I have to use [tex]P = |\langle v_- | \Psi\rangle|^2 [/tex], but how can I extract \Psi from the Hamiltonian at t=0?
     
  5. Oct 15, 2008 #4
    You said that you have Psi, only named it m_0. Anyway, I think one needs to set up an eigenvalue equation and you'll obtain eigenvectors. The one which corresponds to the lowest energy is termed "the ground state".
     
  6. Oct 15, 2008 #5
    I have already found (and normalized) the groundstate [tex]\left| {m_0 } \right\rangle[/tex]. But the groundstate is not the same as the wavefunction itself, i.e. the solution of the time-dependant Schrödinger-equation?

    Please correct me if I am wrong.
     
  7. Oct 16, 2008 #6
    Ok, I worked it out. The particle is in the groundstate, so [tex]\Psi(x,0) = \left| {v__- } \right\rangle[/tex] and the new stationary state is [tex]\psi(x) = \left| {m_0 } \right\rangle[/tex]. That is, we can find the probability as:

    [tex]
    P = |\langle m_0 | v_-\rangle|^2.
    [/tex]

    Thanks.
     
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