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