# Particle physics J=L+S

1. May 18, 2014

### Matterwave

I've never really been...convinced...of the statement

$$\bf{J}=\bf{L}+\bf{S}$$

I've always just gone along with it, but I've never seen why this is "right". So I guess now's as good a time as any to ask.

Thinking about this from a "classical" perspective (which obviously is not correct, but perhaps I can at least show where my doubt comes from), if the L stands for the angular momentum of the particle with respect to the center of mass, and the S stands for the angular momentum of the particle "spinning" around (again, obviously not right), then the two should be measured from different coordinate origins (e.g. L measured from the proton in a Hydrogen nucleus if we are looking at the electron, and S is measured from the "center" of the electron). So, I can not motivate the correctness of this statement from naive classical analyses.

Looking at this mathematically (e.g. from the analysis in Ballentine chapter 7), we have that the $\bf{L}$ operators act on the physical space while the $\bf{S}$ operators act on the internal space.

Ballentine then says something along the lines of, the total rotation operator $e^{in_\alpha J_\alpha }$ must be in the form:

$$e^{in_\alpha J_\alpha }=e^{in_\alpha L_\alpha }e^{in_\alpha S_\alpha }$$

From which the statement $\bf{J}=\bf{L}+\bf{S}$ is true if the L's and S's commute. But I'm not convinced that the above formula is a simple product, and not a direct product. The two different operators operate in different spaces, and so, shouldn't it be a direct product? If I express my state function as a 2 component vector e.g. $\Psi_i (x,t), i=1,2$, for example, the rotation dealing with $\bf{L}=-i\hbar \bf{x}\times\nabla$ must be applied to each component individually, while the rotation dealing with S applies to my 2 component vector as a whole. The whole J=L+S thing doesn't make sense to me taken as an operator equation since L is a differential operator, and S is a matrix. What's the sum of a derivative and a matrix supposed to mean? Unless I am now constructing a 2x2 diagonal matrix for $\bf{L}$? I'm confused. =/

This has always bothered me.

2. May 18, 2014

### Ravi Mohan

The correct form of the equation is
$$\hat{\bf{J}} = \hat{\bf{L}}\otimes\hat{\bf{I}} + \hat{\bf{I}}\otimes\hat{\bf{S}}$$
where the $\hat{\bf{L}}$ and $\hat{\bf{S}}$ generate rotations in different subspaces.

3. May 19, 2014

### Matterwave

And so should there be a direct product in:

$$e^{i\theta_\alpha J_\alpha}=e^{i\theta_\alpha L_\alpha}\otimes e^{i\theta_\alpha S_\alpha}$$

?

4. May 19, 2014

### Ravi Mohan

Yes, mathematically (and intuitively), there should be a direct product.
I would suggest Modern Quantum Mechanics, J.J. Sakurai, section 3.7 for further reading.
Edit:
There is a little discrepancy in the exponential operators. Please check Sakurai.

5. May 19, 2014

### dextercioby

Well, something like this:

$$e^{i\theta_\alpha J_\alpha}=e^{i\theta_\alpha L_\alpha}\otimes \hat{1} + \hat{1}\otimes e^{i\theta_\alpha S_\alpha}$$

6. May 19, 2014

### Matterwave

7. May 20, 2014

### Ravi Mohan

The infinitesimal rotation operator that affects the Hilbert space (made from tensor product of position space and spin space) is
$$\hat{I}-\frac{i(\hat{\bf{L}}\otimes\hat{I}_2+\hat{I}_1\otimes\hat{\bf{S}}).\bf{n} d\theta}{\hbar} =\left(\hat{I}_1-\frac{i\hat{\bf{L}}.\bf{n} d\theta}{\hbar}\right)\otimes\left(\hat{I}_2-\frac{i\hat{\bf{S}}.\bf{n} d\theta}{\hbar}\right).$$

For a fine angle rotation, the equation takes the form
$$\exp\left(\frac{-i\hat{\bf{J}}.\bf{n} \theta}{\hbar}\right)=\exp\left(\frac{-i\hat{\bf{L}}.\bf{n} \theta}{\hbar}\right)\otimes\exp\left(\frac{-i\hat{\bf{S}}.\bf{n} \theta}{\hbar}\right).$$

8. May 20, 2014

### Matterwave

Uh...isn't that what I had in post #3?

9. May 20, 2014

### ChrisVer

well of course there is a direct product, since we are talking about operators on the exponentials...

10. May 20, 2014

### Ravi Mohan

Yes you were right (I am sorry for creating the confusion). It was that you wrote the rotation operator as $e^{in_\alpha J_\alpha}$ in the first post and then you changed it to $e^{i\theta_\alpha J_\alpha}$.

11. May 20, 2014

### Matterwave

Indeed, I changed notations around and did not explain them. I get your point.

But the direct product of the two exponentials look a little different than what Dexter wrote, or are they equivalent?

12. May 20, 2014

### Ravi Mohan

Yes. They are equivalent. The exponential operators, that dextercioby mentions, give superimposition of the seperate rotations in position and spin space. This is the same rotation which $\hat{\bf{J}}$ generates in whole of position $\otimes$ spin space.

Last edited: May 20, 2014