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Two-Level Quantum System, Need help Finding State at time t

  1. Aug 27, 2016 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    [itex]|1>[/itex] and [itex]|2>[/itex] form an orthonormal basis for a two-level system. The Hamiltonian of this system is given by:
    [tex]
    \hat{H} = \epsilon
    \begin{pmatrix}
    1 & i \\
    -i & 1
    \end{pmatrix}
    [/tex]

    a.) Is this Hamiltonian hermitian? What is the significance of a hermitian operator?
    b.) Find the eigenvalues and eigenvectors of this hamiltonian.
    c.) Suppose a particle is in the state [itex]|1>[/itex]. An energy measurement is performed on the particle. What are the possible outcomes of such a measurement, and what are the probabilities?
    2. Relevant equations
    All that should be necessary is the Schrodinger equation.
    [tex]\hat{H} |\psi> = E|\psi>[/tex]

    3. The attempt at a solution
    For part a I checked that the eigenvalues of the hamiltonian are real. This kills two birds with one stone because part b asks for the eigenvectors. They were real, and given by 2ε and 0. The significance of an operator being hermitian is that hermitian operators correspond to observables.

    For part b I just went through the rigor of finding the eigenvectors.
    They were
    [itex] |1> =
    \begin{pmatrix}
    i \\
    1
    \end{pmatrix}
    [/itex] and
    [itex]
    |2> =
    \begin{pmatrix}
    -i \\
    1
    \end{pmatrix}
    [/itex] The first corresponds to the eigenvalue 2ε and the second corresponds to the eigenvalue 0.

    For part C I used Schrodinger's equation and said that since |1> is an eigenvector of the hamiltonian it has a constant definite energy given by 2ε with 100% probability. This is where I feel like I am wrong, because I think this violates an uncertainty principle.

    For part D I am not sure what to do. I know I should multiple by some exponent with time, but I'm not sure how to find this.
     
  2. jcsd
  3. Aug 27, 2016 #2

    blue_leaf77

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    The problem doesn't say that ##|1\rangle## and ##|2\rangle## are energy eigenvectors and you shouldn't assume this way.
    If in part b) you don't assume that ##|1\rangle## and ##|2\rangle## are energy eigenvectors, these vectors will be some superposition states of the energy eigenvectors and thus the energy measurement will yield a multiple possibilities.
     
  4. Aug 27, 2016 #3

    B3NR4Y

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    Yeah my assumption about the basis was stupid. I did some extra reading to refresh my memory from my first quantum course and realized that.

    Since |1> and |2> are any two linearly independent vectors, then an arbitrary state |ψ> can be written c1 |1> + c2 |2>. In the case that part c says, c2 = 0 and c1=1. However I am not sure how to write the |1> state in terms of the eigenvectors which I think is necessary. I was thinking that if the basis vectors are given by
    [itex] |1> =
    \begin{pmatrix}
    1 \\
    0
    \end{pmatrix}
    [/itex] and
    [itex]
    |2> =
    \begin{pmatrix}
    0 \\
    1
    \end{pmatrix}
    [/itex]

    Writing |1> in terms of the eigenvectors is difficult to me.
     
  5. Aug 27, 2016 #4

    blue_leaf77

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    Yes that's right.
    You actually have a good start already since you have calculated the eigenvectors of ##H## in the basis ##\{| i \rangle\}## with ##i = 1,2##. However, let's rename the two eigenvectors using different notation, let's say ##|a\rangle## and ##|b\rangle##. Written in vector notation, these eigenvectors of ##H## will become
    $$
    |a\rangle =
    \begin{pmatrix}
    i \\ 1
    \end{pmatrix} = i
    \begin{pmatrix}
    1\\ 0
    \end{pmatrix}+
    \begin{pmatrix}
    0\\ 1
    \end{pmatrix} = i|1\rangle + |2\rangle
    $$
    and
    $$
    |b\rangle =
    \begin{pmatrix}
    -i \\ 1
    \end{pmatrix} = -i
    \begin{pmatrix}
    1\\ 0
    \end{pmatrix}+
    \begin{pmatrix}
    0\\ 1
    \end{pmatrix} = -i|1\rangle + |2\rangle
    $$
    Your task is to invert these two equations such that ##|1\rangle## and ##|2\rangle## are explicitly written in terms of ##|a\rangle## and ##|b\rangle##. After that, you should normalize them.
     
  6. Aug 28, 2016 #5

    B3NR4Y

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    I see, that is shockingly easy.

    So once I have that I should compute [itex]<1|\hat{H}|1>[/itex] and the coefficient of each term is the probability of that energy?
     
  7. Aug 28, 2016 #6

    blue_leaf77

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    You are not required to calculate the average of the measurement results, only the possible outcomes and the corresponding probability are asked.
     
  8. Aug 28, 2016 #7

    B3NR4Y

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    Okay I have that |1> can be written:
    [tex]|1\rangle = \frac{1}{\sqrt{2}i} \left(|a\rangle-|b\rangle\right)[/tex]
    With this I apply H, which I can use the schrodinger equation to find. The only thing I am uneasy on now is that I think the coefficients will be imaginary, and I am not sure how to reconcile that. I remember in the past we had problems with clebsch-gordan tables and if the coefficients were imaginary we just took the length of that vector (i.e. -i/2 would just be given 1/2).

    Sorry if all of these are stupid questions, I haven't done any of this all summer and am trying to get back in the swing of it.
     
    Last edited: Aug 28, 2016
  9. Aug 28, 2016 #8

    blue_leaf77

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    Yes that's right.
    The coefficients are generally complex numbers, it's the probabilities which equal the modulus square of those coefficients that are real.
    I don't remember one has to deal with imaginary coefficients in calculating CG coefficients. The proportionality constant between the states ##|s_1s_2;SS\rangle## and ## |s_1s_2;s_1s_2\rangle## is agreed to be 1 and also the action of the lowering and raising operators gives only real coefficient, so the CG coefficients must be real.
     
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