# Sx acting on up spin particle confusion

by DocZaius
Tags: acting, confusion, particle, spin
 P: 296 I am reading Griffiths and I am having trouble interpreting the results of measuring the x component of spin on a spin-up particle. If you have a spin up particle, my understanding is that it is assumed to be up in the preferred axis, z. I would think that measuring its x component should give half probability of h-bar/2 and half of negative h-bar/2. But what I get is h-bar/2 times spin down (down in z axis right?): $Sx\uparrow=\frac{\hbar}{2} \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{ccc} 1 \\ 0 \end{array} \right) =\frac{\hbar}{2}\downarrow$ I don't see why this would make any sense. You measure the x component of the spin of a spin-up particle and get h-bar over 2 times spin down? Does that mean the particle is now spin down on the z axis? Shouldn't it be on the x axis now that we measured it in respect to the x component? Also, just to be clear: the generic spin-up spinor without a (z) superscript implies it is a spinor of the z axis right? $X=X^{(z)}$ ?
 P: 647 One cannot make a measurement in quantum mechanics. One can only predict probabilities for numerous measurements. There is no mathematical operation as we know it now that one can use in quantum mechanics were one applies something to a state vector, and gets back a result. This can't happen since the results are random, and you never hit a state vector with something and get a random result back. You can calculate the probability of obtaining a result in a lab where you do an actual mesearment. For your case it would be $\langle \uparrow_{x}|\uparrow_{z}\rangle$ or something like that depending what you want to know.