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Homework Help: The probability that spin will be in -x direction!

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    A spin 1/2 particle is in the state [tex]\left| \Psi \right\rangle[/tex] = [tex]\sqrt{2/3}\left|\uparrow\right\rangle + i\sqrt{1/3}\left|\downarrow\right\rangle[/tex]

    A measurement is made of the x-component of the spin. What is the probability that the spin will be in the -x direction?

    2. Relevant equations

    Spin states are represented as linear combinations of the spin in the +z direction and -z direction, since these spins form an orthonormal basis set:

    In this case, spin in -x direction is represented by

    [tex]\sqrt{1/2} \[
    \left( {\begin{array}{cc}
    1 \\
    -1 \\
    \end{array} } \right)

    3. The attempt at a solution

    P = [tex]\left|\left\langle\leftarrow\left|\Psi\right\rangle\left|^{2}[/tex]
    = the transpose of the [tex]\leftarrow[/tex] matrix, times the [tex]\Psi[/tex] matrix, squared.

    When calculating this straightforwardly, I will end up with a complex probability because of the i term in the [tex]\Psi[/tex] matrix. That doesn't make sense!

    So, do I just take this complex number and find its magnitude in the complex plane, and then square that?

    Or something else?

  2. jcsd
  3. Jul 19, 2010 #2


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    The method is correct.
    To find the probability you multiply the projection by its complex conjugate. Did you do that or did you just square it?
  4. Jul 19, 2010 #3
    That is a fantastic idea!

    P = [tex]\left(\sqrt{1/2}*\sqrt{2/3}-i\sqrt{1/2}*\sqrt{1/3}\right)*\left(\sqrt{1/2}*\sqrt{2/3}+i\sqrt{1/2}*\sqrt{1/3}\right)[/tex]

    = 1/2

    Is this correct?

    Actually, this is the same answer I got when I used the magnitude-in-the-complex-plane method. Hrm. Coincidence, or no... I will look into it.
  5. Jul 19, 2010 #4


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    It is correct.
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