# Angular Momentum

1. Apr 14, 2010

### Nusc

S_z = hbar/2 [ (|+><+|)-(|-><-|) ]

Normally we're given these relations, how does one derive them?

2. Apr 14, 2010

### kanato

The general form of any operator in terms of its eigenvalues and eigenvectors is
$$A = \sum_a |a\rangle a \langle a |$$
where |a> is the eigenvector with eigenvalue a. This is simply like expressing an operator in its diagonal basis, and the operator $$|a\rangle \langle a |$$ is the projection operator for that eigenvector. That's what the S_z operator you've written is.

3. Apr 14, 2010

### Nusc

Sorry. I don't see how that is derived. Can you show me explicitly?

What does that say about S_x?

S_x = hbar/2 [ (|+><-| + |-><+|)]

4. Apr 14, 2010

### SpectraCat

Do you understand how the resolution of the identity works? The expression you have in square brackets is equal to 1 (more specifically the 2x2 identity matrix) ... do you understand why? That is a good place to start here ...

5. Apr 14, 2010

### Nusc

There's a negative sign in front of |-><-| so its not the identity.

But (|+><+|)+(|-><-|) = 1 is.

Actually for an arbitrary set $$\sum_a |a\rangle \langle a | = 1$$

|+> = ( 1 0 )^T
|-> = ( 0 1 )^T

Last edited: Apr 14, 2010
6. Apr 14, 2010

### SpectraCat

Sorry ... you are of course correct (it's late and I am not paying close enough attention). However, the equation you gave was generated from the resolution of the identity as follows:

$$S_z = S_z \{|+\rangle\langle+|\:+\:|-\rangle\langle-|\}$$

But of course,

$$S_z|+\rangle=\frac{\hbar}{2}|+\rangle;\:\:S_z|-\rangle=-\frac{\hbar}{2}|-\rangle$$

So we have (parentheses used to group terms for clarity):

$$S_z = (S_z|+\rangle)\langle+|\:+\:(S_z|-\rangle)\langle-| = \frac{\hbar}{2}[\:|+\rangle\langle+|\:-\:|-\rangle\langle-|\:]$$

As you correctly wrote. As for Sx, the treatment is basically the same, although you have to be a little bit careful at the beginning. I find it clearest to do the resolution of the identity in terms of the eigenstates of Sx (as opposed to |+> and |->, which are the eigenstates of Sz by convention). Then use the expressions for the Sx eigenstates in the basis of |+> and |-> to simplify the result.

EDIT: Ok, that TeX code came out pretty ugly, but it is correct .. let me know if you have any more questions. Bedtime now.

7. Apr 16, 2010

### Nusc

What's S_x |+> and S_x |->?

8. Apr 16, 2010

### SpectraCat

What do you think? Are you familiar with the Stern-Gerlach experiment? The |+> and |-> are eigenstates of Sz, so what should happen when the Sx operator is applied to them? Are Sz and Sx commuting operators?

Last edited: Apr 16, 2010
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