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Spin average value

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Find expectation values ##\langle \hat{S}_x \rangle##, ##\langle \hat{S}_y \rangle##, ##\langle \hat{S}_z \rangle## in state
    ##|\psi \rangle =\frac{1}{\sqrt{2}}(|+\rangle +|- \rangle)##
    ##|+\rangle## and ##|-\rangle## are normalized eigen vectors of ##z## projection of spin.


    2. Relevant equations
    ## \hat{S}_x=\frac{\hbar}{2}\sigma_x ##
    ## \hat{S}_y=\frac{\hbar}{2}\sigma_y ##
    ## \hat{S}_z=\frac{\hbar}{2}\sigma_z ##
    where sigmas are Pauli matrices.


    3. The attempt at a solution
    After calculation I get ##\langle \hat{S}_x \rangle=\frac{\hbar}{2}##, ##\langle \hat{S}_y \rangle=0##, ##\langle \hat{S}_z \rangle=0##
    Why ##\langle \hat{S}_x \rangle## isn't zero if wave function is superposition of up and down in z-direction? Tnx for the answer!
     
  2. jcsd
  3. May 29, 2013 #2

    TSny

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    The two states ##|+\rangle## and ##|-\rangle##, which are eigenfunctions of ## \hat{S}_z##, can be used as basis states for expanding any other spin state. So, a state of spin "up" in the x direction, for example, can be expanded as a superposition of the ## \hat{S}_z## eigenstates ##|+\rangle## and ##|-\rangle##.

    Can you show that the state ##|\psi \rangle =\frac{1}{\sqrt{2}}(|+\rangle +|- \rangle)## is an eigenstate of ## \hat{S}_x## with eigenvalue ##+\frac{\hbar}{2}##? If so, then ##|\psi \rangle ## represents a state with spin along the + x axis. So, it shouldn't be surprising to find that ##\langle \hat{S}_x \rangle## is nonzero
     
  4. May 29, 2013 #3
    Tnx a lot. :) And what about if I have amount of spins ##50%## in state ##|+\rangle## and ##50%## in state ##|-\rangle##. Is there some difference?
     
  5. May 29, 2013 #4

    TSny

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    Are you saying that you have 100 particles with 50 that are known to be in state ##|+\rangle## and 50 in state ##|-\rangle## and then you want to calculate the average value of the outcome of measuring the x-component of spin on all of these particles?
     
  6. May 29, 2013 #5
    Yes.
     
  7. May 29, 2013 #6

    TSny

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    What are the possible values for the outcome of measuring the spin component along the x-axis for a particle in the state ##|+\rangle## with spin up in the z direction? What is the probability for finding each of these outcomes?
     
  8. May 30, 2013 #7
    Well you want to say that if ##|\psi \rangle## is eigen function of spin projection operator ##\hat{S}_n## then ##\langle \hat{S}_n \rangle \neq 0##, otherwise ##\langle \hat{S}_n \rangle =0##.
     
  9. May 30, 2013 #8

    TSny

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    If ##|\psi \rangle## is an eigenfunction of some operator ##\hat{A}## with eigenvalue ##\lambda##, then ##\langle \hat{A}\rangle = \lambda##. So, if ##\lambda \neq 0## then ##\langle \hat{A} \rangle \neq 0##.

    You should be able to how that your ##|\psi \rangle## is an eigenfunction of ##\hat{S}_x## with eigenvalue ##\hbar/2##. So, ##\langle \hat{S}_x \rangle = \hbar/2##.

    If you have a state ##|\psi \rangle## that is not an eigenfunction of the operator ##\hat{A}##, then ##\langle \hat{A}\rangle## might or might not equal zero.
     
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