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

cfitz
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Homework Statement


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, \chi in the z basis, and then asked "in what direction is the spin state pointing?".


Homework 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:

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



The Attempt at a Solution



My goal is to find the angles theta and phi such that S applied to \chi just gives me back \chi. 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 \chi 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 :)
 
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Check your eigenvector.
The one I usually work with is
|S_{r};+> = cos (\theta/2) |S_{z};+> + e^{i \phi}sin (\theta/2) |S_{z};->

Another recommendation is to cast ##\chi## in complex exponential form. This allows you to eliminate overall phase factors.
 
Last edited:
Fightfish said:
Check your eigenvector.
The one I usually work with is
|S_{r};+> = cos (\theta/2) |S_{z};+> + e^{i \phi}sin (\theta/2) |S_{z};->

Another recommendation is to cast ##\chi## in complex exponential form. This allows you to eliminate overall phase factors.

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 |S_{r};+>.

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