Spin 1/2 System: Eigenstates of Sz and Probability of Measurement

[unscientific]

Homework Statement

For a spin 1/2 system, the eigenstates of z-component of the angular-momentum operator S_z 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 S_z is measured, what are the possible results of measurement and their probabilities?

Homework Equations

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?

 

[Goddar]

Hi.
You should see that there's something wrong when you give a probability in terms of h, since it should be dimensionless... Look at this thread for reference:
https://www.physicsforums.com/showthread.php?t=737947

 

[unscientific]

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

 

[George Jones]

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?

 

[unscientific]

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?

I hope I got this right.

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

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

The observables of S_z 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.

 

[Goddar]

An observable Q, such as S_z can take upon real values q₁, q₂, ...

Yes. (upon measurement of course)

For each possible value, the system can be in a state |q₁>, |q₂>, ...

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.

Possible states of system are |q₁>, |q₂>, ... and in this case it would be $\frac{\hbar}{2}|\pm>$

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

Expected value ≠ Probability

Yes.

Probability of a measurement = possibility of finding observable in a state |q_i> which means $<q_i|\psi>^2 = (\frac{\hbar}{2})^2<\pm|\psi>^2$

Probability of a measurement =<q_i|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.

Expected value of observable A = $<\psi|A|\psi>$ ≠ Probability of measurement

Yep.