- #1
DocZaius
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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?):
[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 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?
[itex]X=X^{(z)}[/itex] ?
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?):
[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 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?
[itex]X=X^{(z)}[/itex] ?
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