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Two State Quantum System with a given Hamiltonian

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A two state system has the following hamiltonian
    [tex]

    H=E \left( \begin{array}{cc} 0 & 1 \\
    1 & 0 \end{array} \right)


    [/tex]


    The state at t = 0 is given to be

    [tex]

    \psi(0)=\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)


    [/tex]

    • Find Ψ(t).

    • What is the probability to observe the ground state energy at t = T ?

    • What is the probability that the state is

    [tex]

    \left( \begin{array}{cc} 1 \\ 0
    \end{array} \right)


    [/tex]

    at t = T ?



    2. Relevant equations
    [tex]
    i\hbar\frac{\partial\psi}{\partial t} = \frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\psi
    [/tex]
    3. The attempt at a solution

    My main problem is with the notation. I understand that this system has two states with the same energies. I think this system can be an electron in an energy level, it has up or down spin possibilities. I think since both the states have the same energy both are considered ground state therefore we have 100% possibility to observe the ground state energy at t=T and the probability that the state is [tex]

    \left( \begin{array}{cc} 1 \\ 0
    \end{array} \right)


    [/tex] at t=T seems to be 1/2 intuitively because there is no energy difference between the states. However I don't know the correct notation for Ψ(t). I am thinking of

    [tex]
    \psi(t)=\left( \begin{array}{cc} e^{(-iEt/\hbar)} \\
    e^{(-iEt/\hbar)} \end{array} \right)
    [/tex]

    but this does not define that the system is in the particular state at t=0. Any help is greatly appriciated.
     
  2. jcsd
  3. May 5, 2012 #2
    Note that [tex]\psi(0)[/tex] is not an eigenstate. So you need to solve the time dependent Schrödinger equation. The wave function at time t is given by:

    [tex]\psi (t)=\exp({-i/\hbar Ht})\psi (0)[/tex]

    You need to calculate the matrix exponential function!
     
  4. May 5, 2012 #3
    Thanks for your reply csopi. I have corrected my Ψ(t) as below, what I though of first was a fatal mistake.

    [tex]
    \psi(t)=e^{(-iEt/\hbar)}\left( \begin{array}{cc} 0 \\
    1 \end{array} \right)
    [/tex]

    Is my assumption that since both energies are the same both can be considered as ground state and probability to observe the state at t=T will always be 1 correct?

    for the next part

    What is the probability that the state is

    [tex] \left( \begin{array}{cc} 1 \\ 0
    \end{array} \right) [/tex]

    at t = T ?


    I think i will have to evaluate
    [tex]| \langle \phi | \psi \rangle |^2 [/tex]


    and i think [tex]\phi = \left( \begin{array}{cc} 1 \\ 0
    \end{array} \right)[/tex]

    and

    [tex]| \langle \phi | \psi \rangle |^2 = (\left( \begin{array}{cc} 1 & 0
    \end{array} \right)\left( \begin{array}{cc} 0 \\
    1 \end{array} \right)e^{(-iET/\hbar)})^2 =0 [/tex]

    Am I on the right track? I am not sure on how to calculate the probability of a state and an energy at a specific time, any help or tip is more than welcome.
     
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