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Sx acting on up spin particle confusion 
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#1
Nov2412, 04:31 PM

P: 296

I am reading Griffiths and I am having trouble interpreting the results of measuring the x component of spin on a spinup 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 hbar/2 and half of negative hbar/2. But what I get is hbar/2 times spin down (down in z axis right?): [itex]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[/itex] I don't see why this would make any sense. You measure the x component of the spin of a spinup particle and get hbar 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 spinup spinor without a (z) superscript implies it is a spinor of the z axis right? [itex]X=X^{(z)}[/itex] ? 


#2
Nov2412, 04:44 PM

Sci Advisor
Thanks
P: 4,160

You didn't measure anything. Applied to eigenstates of S_{z}, the S_{x} operator acts as a stepping operator (it's equal to ½(S_{+} + S_{})) The S_{+} gives zero, and the S_{} steps you down. So no wonder you got the spin down state!: wink:
What you want to do is find the overlap between the spin up state and the eigenstates of S_{x}. This will give you the probability amplitude of measuring each value of S_{x}. From the matrix you wrote, the eigenstates of S_{x} are (1/√2)(1, ±1). 


#3
Nov2412, 05:33 PM

P: 296

So if I wish to measure the x component of the spinup particle, I first project spinup onto Sx's eigenvectors, thereby expressing it as a linear combination of them, then Sx that vector?
1) V = <Sx's first eigenvectorSpin up vector> (Sx's first eigenvector) + <Sx's second eigenvectorSpin up vector> (Sx's secondeigenvector) 2) (Sx)V Is that right? 


#4
Nov2512, 01:20 AM

P: 1,020

Sx acting on up spin particle confusion
yes,you will have to first define up and down with respect to x axis.



#5
Nov2512, 11:25 AM

P: 362

?? why would you apply the spinx operator to spin up state



#6
Nov2512, 01:32 PM

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 [itex]\langle \uparrow_{x}\uparrow_{z}\rangle [/itex] or something like that depending what you want to know.



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