# Homework Help: Measuring Spin - wave function collapse

1. Apr 26, 2010

### benbenny

It is my understanding that a measurement of $$S_z$$ followed by a measurement of $$S_y$$ will result in a particle which is in an eigenstate of $$S_y$$. But it appears that a measurement of say $$S_y$$ followed by a measurement of $$S_x$$ results in zero. I see this from a question in which im asked to find the expectation value of $$S_x$$ for a particle in an eigenstate of $$S_y$$ and my result is zero. But for $$S_z$$ it is non zero.
I dont understand why this is so. Could someone help me understand why we find non zero $$S_z$$ values but zero $$S_x$$ values for a partilce in an $$S_y$$ eigenstate.

B

2. Apr 26, 2010

### vela

Staff Emeritus
Taking a measurement is not the same as finding the expectation value. If you take a measurement of Sx or Sz, you'll always get 1/2 or -1/2 (in units of h-bar). The expectation value is what you'd expect if you took many measurements and averaged the results.

3. Apr 30, 2010

### benbenny

Ok. I understand now. Thank you Vela.

It makes sense that the expectation value in the $$S_z$$ direction is NOT zero even if we are in a quantum state of $$S_x$$ or $$S_y$$ since it is the "preferred" direction I suppose, while $$S_x$$ and $$S_y$$ should exhibit zero expectation if we are in the $$S_z$$ eigenstate as they sort of rotate around $$S_z$$.
or something like that. cheers.

4. Apr 30, 2010

### vela

Staff Emeritus
If you're in an eigenstate of Sx, the preferred direction is along the x-axis. You should find the expectation values of Sy and Sz to be zero.

Keep in mind that x, y, and z are just labels we use to keep things straight. There's nothing intrinsically special about the z-axis.

5. Apr 30, 2010

### benbenny

oh ok. now I think I really do understand. Source of confusion: hypothetical triple stern-gerlach experiment which exhibits that taking a measurement of Sx on an Sz eigenstate beam would produce 2 more beams in eigenstates of Sx. Obviously 2 resulting beams in the up and down eigenstates of Sx still amount to a zero expectation value.

Thanks again Vela, and if your interested in helping me with another unrelated question regarding GR and the cylinder condition that would be great as well : https://www.physicsforums.com/showthread.php?t=399738

cheers.

B