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Spin average value

  • #1

Homework Statement


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.


Homework 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.


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!
 

Answers and Replies

  • #2
TSny
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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!
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
 
  • #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?
 
  • #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?
 
  • #5
Yes.
 
  • #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?
 
  • #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##.
 
  • #8
TSny
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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##.
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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