# Short spin 1/2 question

1. Feb 18, 2014

### unscientific

1. The problem statement, all variables and given/known data

For a spin 1/2 system, the eigenstates of z-component of the angular-momentum operator Sz are given by:

$$S_z |\pm> = \pm \frac{\hbar}{2}|\pm>$$

Suppose at time t, the state of the system is given by:

$$|\psi> = a|+> + b|->$$

If Sz is measured, what are the possible results of measurement and their probabilities?

2. Relevant equations

3. The attempt at a solution

Possibilities: State |+> or |->.

$$<+|S_z|\psi> = \frac{\hbar}{2} <+|\psi> = \frac{\hbar}{2}a$$

Probability of |+> state = $\frac{\hbar ^2 a^2}{4}$

Similarly,

Probability for |-> state = $\frac{\hbar ^2 b^2}{4}$

This is weird, as the probability should just be $a^2$ and $b^2$ correspondingly?

2. Feb 18, 2014

### Goddar

3. Feb 19, 2014

### unscientific

But what's wrong with this? It is literally the "expected value of Sz".

4. Feb 19, 2014

### George Jones

Staff Emeritus
Is an expected value the same thing as a probability?

How is the probability of a measurement calculated?

How is the expected value of an observable calculated?

5. Feb 19, 2014

### unscientific

I hope I got this right.

Explanation
The eigenvalue of the operator Sz is the observable, $\pm\frac{\hbar}{2}$.

The observable corresponding to operator Sz is angular momentum, $\pm\frac{\hbar}{2}$.

The observables of Sz can take upon real values q1, q2, .... which are $\pm\frac{\hbar}{2}$.

For each possible value, the system can be in a state |q1>, |q2>, ... which are for $+\frac{\hbar}{2}$ is $|+>$ and for $-\frac{\hbar}{2}$ is $|->$

Possible states of system are |+>, |->

Expected value ≠ Probability

Expected value of observable A = sandwich of operator that gives observable A = $<\psi|\hat{A}|\psi>$ ≠ Probability of measurement

Probability of a measurement = possibility of finding observable in a state $\|q_i>$ which is $|\pm>$ which means $<q_i|\psi>^2 = <\pm|\psi>^2$.

This comes out as unitless, which looks right.

Last edited: Feb 19, 2014
6. Feb 19, 2014

### Goddar

Yes. (upon measurement of course)

Upon measurement returning a certain eigenvalue, the wave function collapses into the
corresponding eigenstate, yes, but before measurement it could have been in any linear
combination including this eigenstate. That's where probabilities come into play.

The eigenstates are |±>, with Sz|±> = ±$\frac{\hbar}{2}|\pm>$

Yes.

Probability of a measurement =<qi|ψ>2=$<\pm|\psi>^2$≠ $(\frac{\hbar}{2})^2<\pm|\psi>^2$
Also, the rest is bad wording: the probability of a measurement is self explanatory, it's the
probability that a measurement returns a certain value. You don't observe a state.

Yep.