# Total Angular Momentum Commutation Relations for 2 Particle

1. Jan 18, 2013

### Sekonda

Hey,

I'm not exactly sure how much this question wants, however the two in question are parts a) and b) below.

So part a) asks to write the expression for the total angular momentum J, I though this was just:

$$\hat{J}=\hat{J}^{(1)}+\hat{J}^{(2)}$$

but when we come to showing it satisfies similar commutation relations - I'm not really sure how to do this, I was thinking of expanding J like so

$$\hat{J}=\hat{J}_{x}+\hat{J}_{y}+\hat{J}_{z}=(\hat{J}^{(1)}_{x}+\hat{J}^{(2)}_{x})+(\hat{J}^{(1)}_{y}+\hat{J}^{(2)}_{y})+(\hat{J}^{(1)}_{z}+\hat{J}^{(2)}_{z})$$

but wasn't sure if this was the correct route and if the last expansion or even the first was necessary.

Part b I think relies on the commutation relations which we find in a) - I'm fairly sure I can show that they all commute but I don't know what expansion/what other commutation relation I need to derive in order to show these commutators are conserved.

Thanks for any help!
SK

2. Jan 18, 2013

### G01

You're on the right track. Define the total angular momentum vector as you did, and the individual components in a similar way. You should then be able to explicitly find the commutators by working them out explicitly in terms of the single particle operators and the commutation relations given.

3. Jan 18, 2013

### Sekonda

Ahh good to know, will give it a try now and get back if I'm having trouble!

4. Jan 18, 2013

### Sekonda

I managed to attain this commutation relation:

$$[\hat{J}^{(1)},\hat{J}^{(2)}]=0$$

and

$$[\hat{J},\hat{J}^{(1/2)}]=0$$

but that's all, I think those are all I need?

5. Jan 18, 2013

### vela

Staff Emeritus
You are keeping in mind that $\hat{\vec{J}}$ is a vector, right? The commutation relations involve individual components of this vector and the (non-vector) operator $\hat{J}^2$.

6. Jan 18, 2013

### Sekonda

Though I seem to be having trouble showing

$$[\hat{J},\hat{J}_{z}]=0$$

7. Jan 18, 2013

### Sekonda

Ok so the operator

$$\hat{J}^{2}$$

is equal to what? The individual components squared and then summed? Or the sum of the individual components squared?

8. Jan 18, 2013

### G01

The J operators are vector operators:

$$\hat{J}^2=\hat{\vec{J}}\cdot\hat{\vec{J}}$$

What is the dot product of two vectors in terms of the component vectors?

9. Jan 18, 2013

### vela

Staff Emeritus
That's really something you should know or at least look up in your textbook before asking here.

10. Jan 18, 2013

### Sekonda

No worry I've managed to do it now, was getting confused over the J^2 operator...

Cheers guys!
SK