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Homework Help: Given general z-oriented spinor, determine the direction of spin

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi All, this problem is related to spin-1/2 in an arbitrary direction, in particular building off of, but going beyond, Griffiths QM 4.30.

    I am given an unnormalized general spin state, [itex]\chi[/itex] in the z basis, and then asked "in what direction is the spin state pointing?".

    2. Relevant equations

    Having solved Griffiths 4.30, I know the matrix for spin in an arbitrary direction is given by:

    S = (hbar/2)[cos(theta), e^(-i*phi)sin(theta); e^(i*phi)sin(theta), -cos(theta)]

    where the commas represent separation within a row, and the semi-colon indicates a new row (sry not to TeX it).

    The state I am given is:

    [itex]\chi[/itex] = (1+i)|up z> - (1+i*sqrt(3))|down z>

    3. The attempt at a solution

    My goal is to find the angles theta and phi such that S applied to [itex]\chi[/itex] just gives me back [itex]\chi[/itex]. That is, in the direction the spin state is pointing, I should get an eigenvalue of 1.

    From here my method is pretty straightforward (and pretty wrong, I suppose). I just apply the matrix (to the spinor form of the [itex]\chi[/itex] state, i.e. as a column vector)then set the result equal to the original and try to find theta and phi that solves the resulting set of 2 equations.

    Not only could I not solve it, I couldn't get Mathematica to solve it either, which is a red flag for me since our professor never gives overly difficult computational problems.

    I should mention that I normalized the state with a 1/sqrt(6) before proceeding.

    NOTE: I also tried setting my spinor equal to the up eigenvector of S, namely:

    |up arbitrary> = [e^(-i*phi)cos(theta/2); e^(+i*phi)sin(theta/2)]

    and had similar issues (not being able to solve).

    Is this totally the wrong approach?

    Any help much appreciated, thanks in advance :)
  2. jcsd
  3. May 2, 2013 #2
    Check your eigenvector.
    The one I usually work with is
    [tex]|S_{r};+> = cos (\theta/2) |S_{z};+> + e^{i \phi}sin (\theta/2) |S_{z};->[/tex]

    Another recommendation is to cast ##\chi## in complex exponential form. This allows you to eliminate overall phase factors.
    Last edited: May 2, 2013
  4. May 3, 2013 #3
    Thanks for the suggestions Fightfish, all is well now!

    Putting the original chi amplitudes into polar form was the trick, exactly for the reason you said. I was able to factor out an overall phase and them simply match coefficients with [tex]|S_{r};+>[/tex].

    If anyone finds this and wants to try, I got, using the [tex]|S_{r};+>[/tex] Fightfish posted, that phi= 13pi/12 and theta = 2*cos^-1(1/sqrt(3)) = 2*sin^-1(sqrt(2/3))
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