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Sx acting on up spin particle confusion

  1. Nov 24, 2012 #1
    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] ?
     
    Last edited: Nov 24, 2012
  2. jcsd
  3. Nov 24, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    You didn't measure anything. Applied to eigenstates of Sz, the Sx 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 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).
     
  4. Nov 24, 2012 #3
    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?
     
  5. Nov 25, 2012 #4
    yes,you will have to first define up and down with respect to x axis.
     
  6. Nov 25, 2012 #5
    ?? why would you apply the spinx operator to spin up state
     
  7. Nov 25, 2012 #6
    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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