Measuring Spin - wave function collapse...

[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.

Thanks in advance.

B

[vela]

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.

[benbenny]

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.

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.

[vela]

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.

[benbenny]

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.

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 : http://www.physicsforums.com/showthread.php?t=399738

cheers.

B