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Spin 1/2 particle in B field

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider a spin 1/2 particle placed in a magnetic field [tex]\vec{B_0}[/tex] with components:

    [tex] B_x = \frac{1}{\sqrt{2}} B_0 [/tex]
    [tex] B_y = 0 [/tex]
    [tex] B_z = \frac{1}{\sqrt{2}} B_0 [/tex]

    a) Calculate the matrix representing, in the {| + >, | - >} basis, the operator H, the Hamiltonian of the system.
    b) Calculate the eigenvalues and the eigenvectors of H.
    c) The system at time t = 0 is in the state | - >. What values can be found if the energy is measured, and with what probabilities?


    2. Relevant equations
    [tex] \omega_0 = - \gamma B_0 [/tex]
    [tex] H = \omega_0 S_z [/tex]
    [tex] S_z = \frac{\hbar}{2} \[ \left( \begin{array}{cc}
    1 & 0 \\
    0 & -1 \\ \end{array} \right)\] [/tex]

    3. The attempt at a solution

    I'm stuck on part a).

    My initial instinct is to do this:

    [tex] H = \omega_0 S_z [/tex]
    [tex] H = - \gamma \vec{B_0} S_z [/tex]
    [tex] H = - \gamma \vec{B_0} \frac{\hbar}{2} \[ \left( \begin{array}{cc}
    1 & 0 \\
    0 & -1 \\ \end{array} \right)\] [/tex]

    But [tex]\vec{B_0}[/tex] is a 3D column vector, and I can't multiply that into a 2x2 matrix. And I have to somehow express that with | + > and | - >... I have a feeling I'm on the wrong track.
     
  2. jcsd
  3. Nov 4, 2007 #2

    malawi_glenn

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    The basical hamiltonian is on the form:

    [tex] H = \vec{B} \cdot \vec{S} = B_x \cdot S_x + B_y \cdot S_y + B_z \cdot S_z [/tex]

    And the [tex] S_x = \frac{1}{2} \sigma _x [/tex] pauli matrix, etc

    (I use natural units, so dont bother)¨

    I hope my hint helped you anyway to solve a)
     
    Last edited: Nov 4, 2007
  4. Nov 4, 2007 #3
    So with this I get:

    [tex] H = \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} \[ \left( \begin{array}{cc}
    0 & 1 \\
    1 & 0 \\ \end{array} \right)\] [/tex]
    [tex]+ \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} \[ \left( \begin{array}{cc}
    1 & 0 \\
    0 & -1 \\ \end{array} \right)\] [/tex]

    [tex] H = B_0 \hbar \[ \left( \begin{array}{cc}
    \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \\
    \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} \\ \end{array} \right)\] [/tex]

    Right?
     
  5. Nov 4, 2007 #4

    malawi_glenn

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    Yes, that looks correct. Now find eigenvalue and eigenvector of this matrix.
     
  6. Nov 4, 2007 #5
    For the eigenvalue:

    [tex] (\frac{B_0 \hbar}{2\sqrt2})^2 (1-\lambda)(1-\lambda) - (\frac{B_0 \hbar}{2\sqrt2})^2 = 0 [/tex]

    [tex](1-\lambda)(1-\lambda) = 0[/tex]
    [tex] \lambda = 1 [/tex]

    And the eigenvector:

    [tex]\frac{B_0 \hbar} {2\sqrt2} \[ \left( \begin{array}{cc}
    0 & 1 \\
    1 & -2 \\ \end{array} \right)\] \times \[ \left(\begin{array}{c}
    c_1 \\
    c_2 \\ \end{array} \right)\] = 0 [/tex]

    This gives me [tex] c_2 = 0 [/tex] and [tex] c_1 = \frac{B_0 \hbar} {2\sqrt2} [/tex]. I think I did something wrong.
     
  7. Nov 4, 2007 #6

    malawi_glenn

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    for matrix:

    [tex] A = \[ \left( \begin{array}{cc} a & a\\ a & -a \\ \end{array} \right)\] [/tex]

    The secular eq is [tex] (a- \lambda )(-a- \lambda ) - a^2 = 0 [/tex]

    if lambda is the eigenvalue.
     
  8. Nov 4, 2007 #7
    Thanks, I found the problem, I should have had [tex] (1-\lambda)(-1-\lambda) = 1 [/tex]
     
  9. Nov 4, 2007 #8
    Ok so I get the following system of equations for the case of the eigenvalue [tex]+\sqrt2[/tex]

    [tex](1-\sqrt2)c_1 + c_2 = 0[/tex]
    [tex]c_1 + (-1-\sqrt2)c_2 = 0 [/tex]

    which according to myself and my calculator has no solution....
     
  10. Nov 4, 2007 #9

    malawi_glenn

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    If you can get eigenvalues to a matrix, then there exists corresponding eigenvectors.

    Eigen vectors are, by using Matlab:

    for sqrt2 = (-0.92388,-0.38268)
    for -sqrt2 = (0.38268,-0.92388)
     
    Last edited: Nov 4, 2007
  11. Nov 4, 2007 #10
    Ok I found the two eigenvectors using Matlab. I'm not sure how to write them when applied to the system so that it makes sense.

    [tex] |\psi(t)> = 0.92388 | + > + 0.382683 | - > [/tex]
    [tex] |\psi(t)> = 0.382683 | + > - 0.92388 | - > [/tex]

    Is this correct? And for part c), which one do I use to find the probability?
     
  12. Nov 4, 2007 #11

    malawi_glenn

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    But in order to get them analytically, just do substituion
    [tex]c_1 = (1+\sqrt2)c_2[/tex]
    From your second equation and put in into the first one and solve for c_2

    Now how does a state evolve with time? Ever heard of "Time evolution operator" or similar?

    Time evolution of a ket is
    [tex] |a(t) \rangle = \exp (-i E_a t/\hbar)|a(0) \rangle [/tex]
    where E_a is the energy eigenvalue of that ket.

    So your egeinvectors are:
    [tex] |\psi +> = 0.92388 | + > + 0.382683 | - > [/tex]
    [tex] |\psi -> = 0.382683 | + > - 0.92388 | - > [/tex]

    Dont use time, as you did, it is not correct.
    The psi + has eigenvalue +sqrt2 etc.
    Now I have helped you very much.
     
    Last edited: Nov 4, 2007
  13. Nov 4, 2007 #12
    Yes I've seen it. So the eigenvectors are just for [tex] |\psi(0)>[/tex] ?
     
  14. Nov 4, 2007 #13

    malawi_glenn

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    no, just for the hamiltonian.

    at time = 0; the state is in |->

    Then you must find out what just |-> is in superposition of the eigenvectors to the hamiltonian, in order to get the time evolution.

    i.e you should first write |-> = a|phi + > + b|phi ->
     
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