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Solving Schrodinger Equation

  1. Nov 9, 2013 #1
    I'm working my way through some QM problems for self-study and this one has stumped me. Given the Hamiltonian as [itex]H(t) = f(t)H^0[/itex] where [itex]f(t)[/itex] is a real function and [itex]H^0[/itex] is Hermitian with a complete set of eigenstates [itex]H^0|E_n^0> = E_n^0|E_n^0>[/itex]. Time evolution is given by the Schrodinger equation [itex]i \hbar \frac{d}{dt}|\phi (t)> = H(t)|\phi (t)>[/itex]. I am supposed to write a solution to the Schrodinger equation as a linear combination of the eigenstates of [itex]H^0[/itex]. Now clearly
    [itex]|\phi (t)> = \sum\limits_{n=1}^N c_n (t)|E_n^0>[/itex]. But where do I go from there. The second part is to convert the Schrodinger equation into a first order diff eq and solve for the [itex]c_n (t)[/itex]. Any help is appreciated. Thanks.
     
  2. jcsd
  3. Nov 9, 2013 #2
    Seems pretty straight forward to me. Plug in the expressions for |phi(t)> and H(t) into the Schroedinger Equation, act on it with a bra of one of the eigenstates <En0|, use the orthonormality of those states, and solve the ODE.
     
  4. Nov 9, 2013 #3
    I'm not really sure how to deal with the sums, when I plug in and evaluate [itex]H^0[/itex]I get
    [itex]i \hbar \frac{d}{dt} \sum\limits_{n=1}^N c_n (t)|E_n^0> = f(t) \sum\limits_{n=1}^N c_n (t) E_n^0 |E_n^0> [/itex] and then if I act with a [itex]<E_n^0|[/itex] it will simply cancel out the [itex]|E_n^0>[/itex] (completeness relation), but then how do I solve it from there? Also if I am given a well-defined Hamiltonian in terms of time and a state [itex]A[/itex] with eigenvalue [itex]a[/itex] at t=0. How do I find the probability that as t->infinity a measurement of [itex]A[/itex] will give a different eigenvalue?
     
  5. Nov 9, 2013 #4
    You should apply the bra on the left side and use the orthonormality relation - not the completeness relation. There will be only one single non-vanishing term left in the summation. That term will give you an ODE.
     
  6. Nov 9, 2013 #5
    Right so because of orthogonality each term where [itex]n \neq m[/itex] will cancel leaving the left side as [itex]i \hbar \frac{d}{dt} c_n (t)[/itex] but wouldn't the same have to happen to the right side making the equation [itex]i \hbar \frac{d}{dt} c_n (t) = f(t)E_n^0 c_n (t)[/itex]. Is that right? Also what about my second question from post 3?
     
  7. Nov 9, 2013 #6
    Yes, that is the idea.
    [tex] P = \lim_{t \to +\infty}\left(1-|\langle A(t)| A(0)\rangle|^2\right),[/tex]
    where [itex]|A(t)\rangle [/itex] can be computed from Schrodinger's equation.
     
    Last edited: Nov 9, 2013
  8. Nov 9, 2013 #7
    Thanks, but how would I compute [itex]A(t)[/itex]. I'll give a little more detail. For this particular example, I am given that
    [itex]A = \left( \begin{array}{cc}
    1 & 0 \\
    0 & -1 \end{array} \right)[/itex] with eigenvalue 1 at [itex]t= 0[/itex]. Since I've now got an expression for [itex]c_n(t)[/itex] I can calculate [itex]|\phi (t)>[/itex], but what about this [itex]A(t)[/itex].
     
  9. Nov 10, 2013 #8
    I thought that [itex]A[/itex] was a quantum state.

    Do you mean that system is in one of the eigen states of operator [itex]A[/itex]?
    Also can you write down the Hamiltonian?
     
  10. Nov 10, 2013 #9
    yes at t=0


    [itex]H(t) = ae^{-t/b} \left( \begin{array}{cc}
    0 & 1 \\
    1 & 0 \end{array} \right)\[/itex]
     
  11. Nov 10, 2013 #10
    Ok.
    So the time independent part of Hamiltonian is
    [itex]H^0 = \left( \begin{array}{cc}
    0 & 1 \\
    1 & 0 \end{array} \right),[/itex]
    with eigen-vectors and eigen-values as [itex] |e_1\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}
    1 \\
    1 \end{array} \right)[/itex], [itex] |e_2\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}
    1 \\
    -1 \end{array} \right)[/itex] and [itex]1,-1[/itex] respectively.

    Now after calculating [itex]C_n(t)[/itex], you have [itex] |\phi(t)\rangle = \sum\limits_{i=1}^{2}C_i(t) |e_i \rangle [/itex].

    The operator corresponding to observable [itex]A[/itex] is
    [itex]A = \left( \begin{array}{cc}
    1 & 0 \\
    0 & -1 \end{array} \right),[/itex]
    with eigen-vectors as [itex] |a_1\rangle = \left( \begin{array}{c}
    1 \\
    0 \end{array} \right)[/itex], [itex] |a_2\rangle = \left( \begin{array}{c}
    0 \\
    1 \end{array} \right)[/itex].

    When the time approaches infinity the probability that the measurement will not result in eigen-value = 1 is given by evaluating
    [tex] P = \lim_{t \to +\infty}\left(1-|\langle a_1| \phi(t)\rangle|^2\right).[/tex]
     
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