# Spin average value

## 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

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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

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?

TSny
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Gold Member
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?

Yes.

TSny
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Gold Member
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?

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 $\langle \hat{S}_n \rangle =0$.

TSny
Homework Helper
Gold Member
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.