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Spin state function of a beam of particles in terms of eigenfunctions.

  1. Jan 7, 2012 #1
    Hi there, I apologise that I should probably know this/its a stupid question but I seem to have forgotten all physics over the holiday and so any help would be great!

    I have been told that there is a beam of atoms with spin quantum number 1/2 and zero orbital angular momentum, with spin +1/2 along the x axis. I am then asked what the spin state function of this beam is in terms of eigenfunctions of [itex]\hat{S}^2[/itex] and [itex]\hat{S}_z[/itex], being the kets |[itex]1/2 , m_{s_z} = 1/2>[/itex] and |[itex]1/2 , m_{s_z} = -1/2>[/itex]

    Would I be right in assuming that the answer is simply
    [itex]\phi_s = \frac{1}{\sqrt{2}}|1/2 , m_{s_z} = 1/2> + \frac{1}{\sqrt{2}}|1/2 , m_{s_z} = 1/2> [/itex]
    because knowing the spin in the x direction doesn't tell you about the z direction (as they are incompatible observables) or is it more complex than that? Any help/advice greatly appreciated, thanks.
     
  2. jcsd
  3. Jan 8, 2012 #2

    vela

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    You want to find the eigenstates of Sx in terms of the eigenstates of Sz. Your answer is a good guess, but there could be a different relative phase between the two terms.
     
  4. Jan 8, 2012 #3
    Thanks for the help :).

    I got the eigenstates of [itex]S_x[/itex] in terms of [itex]S_z[/itex] as

    [itex] \frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> + |\frac{1}{2},\frac{-1}{2}> [/itex]
    and
    [itex] \frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> - |\frac{1}{2},\frac{-1}{2}> [/itex]

    With the first one corresponding to the same eigenvalue as the positive spin in the x direction given by [itex]S_x[/itex]. Is that anything like what the correct answer would be? (I'm guessing that it should be similar to if the beam is not polarised as the following questions suggest that they would not be differentiated by a Stern-Gerlach experiment with the magnetic field along z.).

    I don't know how much you're actually allowed to say "yes that's the correct answer" but thanks for the help anyway ;).
     
  5. Jan 8, 2012 #4

    vela

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    Yup, that's it.
     
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