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

1. May 1, 2013

### cfitz

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, $\chi$ 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:

$\chi$ = (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 $\chi$ 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 :)

2. May 2, 2013

### Fightfish

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: May 2, 2013
3. May 3, 2013

### cfitz

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))