Spin vs orbital angular momenta in QFT

Question

I couldn't find past answers that quite match what I wanted, so I will try to ask in slightly different manner. In QM, we have total angular momentum operator $\vec{J}$ (I dropped the hat for convenience) \begin{align} \vec{J} = \vec{L} +\vec{S} \end{align} where $\vec{L}$ is for orbital angular momentum and $\vec{S}$ is for spin angular momentum. For hydrogen atom, this (not addition, but rather $\vec{L}^2$ and $L_z$) gives us two of three quantum numbers, thus we often label the state for fixed energy as $\left|l,m\right>$ (for some reason $m$ is used in standard texts instead of, say, $s$). I think I understand how they work in QM, or at least the fact that $\vec{J},\vec{L},\vec{S}$ all share the same commutation relations (i.e. share the same Lie algebra structure). The addition comes from when one has stuffs like spin-orbit coupling and the like.

In QFT, stuff starts to get a bit/lot confusing. First, angular momentum operator is really used to label states in the irreducible unitary representation of the Poincare group, which is written as $\left|p,\sigma\right>$ (or $\Psi_{p,s}$ following Weinberg); it is an operator in the sense that it appears as a representation of rotation element of the group. Wigner's classification then gives us that we also label states with two quantum numbers: we sometimes also write this as $\left|m,j\right>$, where $m$ is the rest mass and $j$ is the spin. Nowhere in any of these constructions do I see the conventional orbital angular momentum (in the text by Schwartz Chapter 11, for example, he made a comment when there is no angular momentum, so he got the "easy" case of $\vec{J}=\vec{S}$.) In standard QM, we clearly distinguish two versions of angular momenta and their eigenvalues.

Question: does orbital angular momentum make sense in QFT and how does it arise if it does? Should I think of $\vec L$ as furnishing some kind of tensor product representation so that I should be labelling states with $\left|m,j,s\right>$ or something? Another possibility is that one simply does not work with addition of angular momenta in QFT, especially in free theory, but I would like to have an explanation on why this should or should not be the case. I may have understood something really basic but I can't fish out exactly what.

Note: I believe the usual thing about spin-orbit coupling should not work, because that's QM (think about hydrogen atom) and strictly speaking QFT, unless we do something more (2-particle states?). In those cases, the operator $\vec{L}$ even comes from $\vec{r}\times\vec{p}$ which does not appear naturally in QFT (what's $\vec{r}$ in QFT?). I don't think I should go and define $\vec{L}:=\vec{r}\times\vec{\hat{\pi}}$, where $\hat\pi$ is the conjugate momentum of the field.

Answer 1

Since you mention Weinberg, have a look at section 7.4, especially (7.4.10). Under a homogeneous Lorentz transformation generated by the anti-symmetric tensor $\omega^{\mu\nu}$, a field transforms like $$\Psi^\ell \mapsto \Psi^\ell + (\mathcal{J}_{\mu\nu})^\ell{}_m \omega^{\mu\nu} \Psi^m$$ i.e. with the representative of $\omega^{\mu\nu}$ in whatever representation $\Psi$ belongs to. Then according to Noether's theorem $$ 0 = \partial^\kappa \left[\frac{i}{2} \frac{\partial \mathcal{L}}{\partial (\partial^\kappa \Psi^\ell)} (\mathcal{J}_{\mu\nu})^\ell{}_m \Psi^m \right] - \frac{1}{2}(T_{\mu\nu} - T_{\nu\mu}) \tag{7.4.10}$$ where $T_{\mu\nu}$ is the Noether current of translations, i.e., the canonical stress-energy tensor.

Now, $T_{\mu\nu}$ is itself conserved, $\partial^\mu T_{\mu\nu} = 0$, so $$\partial^\kappa (x_\mu T_{\kappa\nu} - x_\nu T_{\kappa \mu}) = T_{\mu\nu} - T_{\nu\mu}$$ and we can write (7.4.10) as $$ 0 = \partial^\kappa \left[ \frac{i}{2} \frac{\partial \mathcal{L}}{\partial (\partial^\kappa \Psi^\ell)} (\mathcal{J}_{\mu\nu})^\ell{}_m \Psi^m - \frac{1}{2}(x_\mu T_{\kappa\nu} - x_\nu T_{\kappa \mu}) \right ] $$ and this is quite clearly expressing that spin plus orbital angular momentum is conserved.

Read further in Weinberg 7.4 and see also the second section of the Wikipedia page on the Belinfante-Rosenfeld tensor.

Answer 2

Orbital angular momentum has to do with position. It is the 'ordinary' angular momentum $r\times p$ we learned about in Physics 101 whereas spin is an extra internal angular momentum that doesn't have to do with position. Now there is one extra label you forgot on your Wigner one particle state basis. That is the momentum $p$.

When we rotate the state $p$ rotates too, so the angular momentum operator should have a part related to the momentum operator. If we have a wavepacket superposition this extra piece will act like the 'ordinary' orbital angular momentum part.

This is accounted for automatically if you find the angular momentum operator in terms of fields via the Lagrangian. The angular momentum density will look something like $x^\lambda T^{\mu\nu}-x^\mu T^{\lambda\nu}$ where $T$ is the energy momentum tensor, plus an aditional 'spin' part if the fundamental field(s) in your Lagrangian have non-trivial rotation properties.

Answer 3

If you think about Wigner's classification, then one-particle states are irreducible representations of the Poincare group, as you said. To label these representations you need the Casimir invariants of Poincare - the particle's mass and the little group Casimir, which for massive particles corresponds to the symmetries in is its rest frame, the SO(3) rotation group. That's how you get spin and that's all the angular momentum you have in one-particle states. Orbital angular momentum requires the concept of distance, which for 1-particle states does not exist.

However, if you consider two-particle states you can just take it as a tensor product between 2 one-particle states with 2 different spin labels. OR, you can take it as a single state with definite energy-momentum and total angular momentum label. This angular momentum is just the Clebsch-Gordon sum of both spins and the orbital angular momentum between the particles. How you move from one description to the other is by going to the center of mass of the two particles. Here you have to specify the direction of the relative linear momentum on the 2-sphere, this two-angle dependence can be expanded into spherical harmonics, eigenfunctions of angular momentum.

As a concrete example, consider two spin-0 bosons. This 2-particle state depends on 6 variables, 2x linear momentum vectors. If you go to CM frame this state depends on the total linear momentum and the relative momentum. The absolute value of the relative momentum can be put into the total Energy, and you're only missing the orientation of the relative momentum. The 2 angle dependence can be traded for (l,m) by a spherical harmonic decomposition. If the particles have internal spin then you can sum these spins using Clebsch-Gordon coefficients, to get the usual total angular momentum 'j'. You can then check that these 'j' states are indeed eigeinstates of the 2-particle representation of the rotation generators of the Poincare group.