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A question with a time dependent Hamiltonian

  1. Nov 24, 2012 #1
    To cut to the chase, I have to solve for the evolution of a two-state system where the system's state at time t satisfies the equation
    [tex]\mathrm{i}\hbar\left(
    \begin{array}{cc}
    \dot{c}_1(t)\\
    \dot{c}_2(t)
    \end{array}
    \right)=\left(
    \begin{array}{cc}
    0 & \gamma \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\
    \gamma \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 0
    \end{array}
    \right)\left(
    \begin{array}{cc}
    c_1(t)\\
    c_2(t)
    \end{array}
    \right)[/tex]
    with initial conditions [itex]c_1(0)=1[/itex] and [itex]c_2(0)=0[/itex].
    I decided to solve this by diagonalization. The eigenvectors of the Hamiltonian are
    [tex]\frac{1}{\sqrt{2}}\left(
    \begin{array}{cc}
    \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\
    1
    \end{array}
    \right)[/tex] and [tex]\frac{1}{\sqrt{2}}\left(
    \begin{array}{cc}
    \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\
    -1
    \end{array}
    \right),[/tex] with eigenvalues [itex]\gamma[/itex] and [itex]-\gamma[/itex], respectively.

    The change-of-basis matrix then is just
    [tex]U^\dagger = \frac{1}{\sqrt{2}}\left(
    \begin{array}{cc}
    \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\
    \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1
    \end{array}
    \right).[/tex] So in the new basis, the state vector becomes
    [tex]\frac{1}{\sqrt{2}}\left(
    \begin{array}{cc}
    \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\
    \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1
    \end{array}
    \right)\left(
    \begin{array}{cc}
    c_1(t)\\
    c_2(t)
    \end{array}
    \right):=\left(
    \begin{array}{cc}
    d_1(t)\\
    d_2(t)
    \end{array}
    \right).[/tex]

    Therefore, we have
    [tex]\mathrm{i}\hbar\left(
    \begin{array}{cc}
    \dot{d}_1(t)\\
    \dot{d}_2(t)
    \end{array}
    \right)=\left(
    \begin{array}{cc}
    \gamma & 0\\
    0 & -\gamma
    \end{array}
    \right)\left(
    \begin{array}{cc}
    d_1(t)\\
    d_2(t)
    \end{array}
    \right)[/tex]
    which can be solved easily. Once the d's are found, the c's are easily computed.

    Problem is, this doesn't give the right answer. I suspect it has something to do with using a time-dependent change of basis matrix, but I'm not sure.
     
  2. jcsd
  3. Nov 25, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think your suspicions are right.

    From [tex]\mathrm{i}\hbar\left(
    \begin{array}{cc}
    \dot{c}_1(t)\\
    \dot{c}_2(t)
    \end{array}
    \right)=\left(
    \begin{array}{cc}
    0 & \gamma \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\
    \gamma \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 0
    \end{array}
    \right)\left(
    \begin{array}{cc}
    c_1(t)\\
    c_2(t)
    \end{array}
    \right)[/tex]

    You can write
    [tex]\mathrm{i}\hbar U^\dagger \left(
    \begin{array}{cc}
    \dot{c}_1(t)\\
    \dot{c}_2(t)
    \end{array}
    \right)= U^\dagger \left(
    \begin{array}{cc}
    0 & \gamma \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\
    \gamma \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 0
    \end{array}
    \right) U U^\dagger \left(
    \begin{array}{cc}
    c_1(t)\\
    c_2(t)
    \end{array}
    \right) = \left(
    \begin{array}{cc}
    \gamma & 0\\
    0 & -\gamma
    \end{array}
    \right)\left(
    \begin{array}{cc}
    d_1(t)\\
    d_2(t)
    \end{array}
    \right)[/tex]

    But the left hand side cannot be equated to ##\mathrm{i}\hbar\left(
    \begin{array}{cc}
    \dot{d}_1(t)\\
    \dot{d}_2(t)
    \end{array}
    \right)## due to the time dependence of ##U^\dagger##:

    ##\left(
    \begin{array}{cc}
    \dot{d}_1(t)\\
    \dot{d}_2(t)
    \end{array}
    \right) = \frac{d}{dt} \left(U^\dagger \left(
    \begin{array}{cc}
    c_1(t)\\
    c_2(t)
    \end{array}
    \right)\right) \neq U^\dagger \left(
    \begin{array}{cc}
    \dot{c}_1(t)\\
    \dot{c}_2(t)
    \end{array}
    \right)##
     
  4. Nov 25, 2012 #3
    you just need to write the two eqns represented by them,differentiate the second one and put the values from the first one.That's all.
     
  5. Nov 25, 2012 #4
    I understand your reasoning, and I can't find a flaw with it, but the conclusion is really bothersome. It's as if the state doesn't obey the Schrodinger equation if its matrix representation uses energy eigenkets as a basis. But how can that be right?

    I'll have to do this I guess, but I'd like to see why my approach doesn't work.
     
  6. Nov 28, 2012 #5
    Anyone?
     
  7. Nov 28, 2012 #6

    Mute

    User Avatar
    Homework Helper

    TSny showed you why it doesn't work. I'm not sure exactly what you mean when you say "It's as if the state doesn't obey the schrodinger equation if its matrix representation uses energy eigenkets as a basis". In what way does it appear to you to not obey the Schrodinger equation?

    Perhaps the issue is that your "change of basis" - your unitary transformation - is really a change of reference frame. You are moving into a rotating frame. There are no true energy eigenkets for this system because the energy is not conserved in time (though it is conserved on average). Perhaps one can say this is similar to the case in classical mechanics: when you move into a rotating frame in classical mechanics you generate "ficticious forces" into Newton's equations of motion. Similarly, your change of reference frame here is generating what look to you like new terms into Schrodinger's equation.

    I don't have a quantum book handy right now, but this is (or is pretty similar to) a fairly standard example of a spin in a rotating magnetic field. A book on NMR that treats the problem quantum mechanically should have a discussion, though I don't know if it would exactly address your philosophical problem.
     
  8. Nov 28, 2012 #7
    Thanks, now it makes sense.
     
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