# Coupled Spin operators

1. Apr 21, 2014

### cooev769

Trying to get my head around this one. Given that you can have a proton and an electron in a hydrogen atom for example, and they can create a singlet or triplet configuration, with spin 1 and spin 0 respectively. The total spin operator can be derived as:

S^2 = (Se + Sp)^2 = Se^2 + Sp^2 + 2Se.Sp

Where the dot is the dot product, but then what you get is a bunch of Sx,Sz,Sy acting on the tensor products to provide eigenvalues and the eigenvectors out. The math is all well and good, but the problem I have is that the spin operators Sz,Sy and Sx don't commute, so how come you can just apply the operators to the respective spins to get out an answer, seems odd to me. Any help would be appreciated.

Cheers.

2. Apr 21, 2014

### Jilang

What, are you saying that the components of the spin of the proton don't commute with those of the electron?

3. Apr 21, 2014

### cooev769

No. I'm saying that the component of the electron don't commute with each other yet they're evaluated in the dot production order to solve for the S^2 operator. But this seems daft to me given they do not commute.

4. Apr 21, 2014

### strangerep

Write out the dot product. Do you get any terms that are affected by noncommutativity?

5. Apr 21, 2014

### cooev769

I understand what you're saying strange rep, but by that logic wouldn't that mean I can just write out the total spin operator as:

S = Sz + Sy + Sx

And apply this to any state to get the total spin. It's effectively the same thing.

6. Apr 22, 2014

### strangerep

If each of your terms on the rhs are vectors, then S is also a vector. If that's what you meant, then lets write them properly, in bold, i.e.,

S = Sz + Sy + Sx.

Now draw a diagram with 3 orthogonal axes in 3D space and draw a vector from the origin to the point (1,1,1), i.e., x=1, y=1, z=1.

The vector S corresponds to spin-projection along that direction. This is quite different from the spin magnitude, i.e., the length of the vector.

The direction of a vector is not invariant under rotations, but the length is.

Hopefully you can now see that it's not the same thing?

7. Apr 22, 2014

### cooev769

8. Apr 22, 2014

### cooev769

Ah I get it now, thank you.