Sx acting on up spin particle confusion


by DocZaius
Tags: acting, confusion, particle, spin
DocZaius
DocZaius is offline
#1
Nov24-12, 04:31 PM
P: 287
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] ?
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
Bill_K
Bill_K is online now
#2
Nov24-12, 04:44 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 3,853
You didn't measure anything. Applied to eigenstates of Sz, the Sx operator acts as a stepping operator (it's equal to (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 Sx. This will give you the probability amplitude of measuring each value of Sx. From the matrix you wrote, the eigenstates of Sx are (1/√2)(1, 1).
DocZaius
DocZaius is offline
#3
Nov24-12, 05:33 PM
P: 287
So if I wish to measure the x component of the spin-up particle, I first project spin-up onto Sx's eigenvectors, thereby expressing it as a linear combination of them, then Sx that vector?

1) V = <Sx's first eigenvector|Spin up vector> (Sx's first eigenvector) + <Sx's second eigenvector|Spin up vector> (Sx's secondeigenvector)
2) (Sx)V

Is that right?

andrien
andrien is offline
#4
Nov25-12, 01:20 AM
P: 977

Sx acting on up spin particle confusion


yes,you will have to first define up and down with respect to x axis.
HomogenousCow
HomogenousCow is offline
#5
Nov25-12, 11:25 AM
P: 315
?? why would you apply the spinx operator to spin up state
jfy4
jfy4 is offline
#6
Nov25-12, 01:32 PM
jfy4's Avatar
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.


Register to reply

Related Discussions
how spinor vector envolves in describing the atom Quantum Physics 6
Dirac Spinor, Weyl Spinor, Majorana Spinor Quantum Physics 0
why spinor bec is called ''spinor bec''? Quantum Physics 1
spinor Advanced Physics Homework 0
Spinor fields and spinor wave functions Quantum Physics 1