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Time-evolution of quantum state

  1. Dec 29, 2008 #1

    KFC

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    1. The problem statement, all variables and given/known data
    A two-level system is setup so the two eigenvalues are given [tex]E_1, E_2[/tex], the Hamiltonian is also given as 2x2 matrix (not shown here). The corresponding eigenstates are easy to solved as

    [tex]
    |1\rangle =
    \left(
    \begin{matrix}
    1 \\ 0
    \end{matrix}
    \right), \qquad
    |2\rangle =
    \left(
    \begin{matrix}
    0 \\ 1
    \end{matrix}
    \right)
    [/tex]

    1) If we know initially the state of the system is in [tex]|2\rangle [/tex], what's the probability of finding the system in [tex]|1\rangle[/tex]?

    2) If now we introduce an interaction to change the Hamiltonian

    [tex]
    H' =
    \left(
    \begin{matrix}
    0 & 1 \\
    1 & 0
    \end{matrix}
    \right)
    [/tex]
    so that the total Hamiltonian is H = H0 + H'

    At time is ZERO, the initial state is also in [tex]|2\rangle[/tex], what's probability of finding the system in [tex]|1\rangle[/tex] now?

    2. The attempt at a solution
    I read some similar examples for solving time-evolution problem but I am quite confuse with the method read in those examples. Here is what I did to solve the problem

    For solving problem 1), since we already know the eigenstates and eigenvalues. In addition, the initial state of the system is one eigenstate, so the time-evolution of the system would be

    [tex]
    |\Psi(t)\rangle = \sum_i c_i \exp(-iE_it/\hbar)|i\rangle = \exp(-iE_2t/\hbar)|2\rangle
    [/tex]

    But since it doesn't contain [tex]|1\rangle[/tex], hence it is always no chance to find state 1 at time t, that is,

    [tex]
    \left|\langle 1 |\Psi(t)\rangle\right|^2 = 0
    [/tex]

    For problem 2), we first solve the eigenvectors and eigenvalues of the total Hamiltonian and get

    [tex]
    \lambda_1, \quad |\phi_1\rangle ; \qquad \qquad \qquad \lambda_2, \quad |\phi_2\rangle
    [/tex]

    Now, write [tex]|1\rangle[/tex] and [tex]|2\rangle[/tex] in terms of [tex]|\phi_1\rangle [/tex] and [tex]|\phi_1\rangle [/tex]. Later on, at time t, the state will evolute

    [tex]
    |\Psi(t)\rangle = \sum_i c_i \exp(-i\lambda_i t/\hbar)|\phi_i\rangle
    [/tex]

    so the probability to find state [tex]|1\rangle[/tex] will be of the same form [tex]
    \left|\langle 1 |\Psi(t)\rangle\right|^2
    [/tex] but there [tex]\langle 1 |[/tex] and [tex]|\Psi(t)\rangle[/tex] are all written in terms of the new eigenvectors so the probability is not more zero.

    Please tell me if my solution have any problem!

    But the way, in some book, they solve the similar problem by assuming the time-dependent wave solution is

    [tex]
    |\Psi(t)\rangle = \alpha(t) |1\rangle + \beta(t) |2\rangle
    [/tex]

    Plug these in time-dependent Schrodinger equation and find the following equations

    [tex]
    i\hbar
    \frac{d}{dt}
    \left(
    \begin{matrix}
    {\alpha} \\ {\beta}
    \end{matrix}
    \right)
    =
    H
    \left(
    \begin{matrix}
    {\alpha} \\ {\beta}
    \end{matrix}
    \right)
    [/tex]

    By solving the simultaneous equations, they find the time-dependent coefficients, so the time-dependent state could be given directly.

    I didn't solve the problem with this method, but I wonder if this method is actually equivalent to mine?
     
  2. jcsd
  3. Jan 16, 2011 #2
    Yes, the second method is identical to the first method. Or rather, the first method is derived from the second method. The exponential terms that you put in front of the new eigenstates comes from the differential equation that you wrote on the bottom of your response. Either way will work, but I think the first way is easier to do, but can be tricky remembering the details, whereas the second method is tricky to do but easy to remember where to start.
     
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