# Spin vs orbital angular momenta in QFT

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.

• Why is "that not QFT?" Angular momentum, both spin and orbital, is strongly related to rotational symmetry. (Paging E. Noether. Will E. Noether please pick up a white house phone.) You still have symmetry in QFT. You still get the same reps of the rotation group. – puppetsock Jul 26 '19 at 18:45
• @puppetsock no I am not saying it's not QFT per se; rather, I do not see the distinction in standard texts. In standard QM, it is very clear that spin and orbital angular momenta are two different operators and hence furnish different quantum numbers. – Everiana Jul 30 '19 at 0:50

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.

• Thanks I will look it up. But does that mean that the notion orbital angular momentum does not directly descent from Poincare group (while spin does)? Good quantum numbers are associated to operators which commute with the Hamiltonian, but it's not obvious to me they come from representation theory since stuff like $T_{\mu\nu}$ is not an element of the Poincare group (just as Casimir operators are not). – Everiana Jul 30 '19 at 0:54
• @Everiana $T^{\mu\nu}$ is not an element of the Poincare group, but it's only a 4-current density. The total energy-momentum is $P^\nu = \int T^{0\nu} \, d^3x$ and is the generator of translations. The details are all in Weinberg Ch. 7. – Robin Ekman Jul 31 '19 at 14:41

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.

• Yes I get the part that physically we should still make sense of orbital angular momentum, and as some of the answers above (yours included) show, one can construct conserved quantity that describes angular momentum using $T^{\mu\nu}$. This seems to tell me that orbital angular momentum does not automatically arise from Poincare group, since $T^{\mu\nu}$ is neither associated to the group or the algebra (just as Casimir operator is not an element of the algebra), and I am wondering if this is the case. – Everiana Jul 30 '19 at 0:58
• @Everiana, $T^{\mu\nu}$ has a lot to do with the Poincare group and is not a Casimir. The generators of translations in the Poincare group are $P^\mu = \int d^3 x T^{\mu 0}$. In any case the fact that you can construct an explicit angular momentum operator in terms of fields was a side point. My my main point is that if you want to consider angular momentum from a representation theory perspective, it is the momentum label that tells you everything about the orbital angular momentum. You left it out of the states $|j,s\rangle$ in your formulation of the question, but it is essential. – octonion Jul 30 '19 at 4:30

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.

• Your comment is very interesting. You are essentially saying that I need to go beyond the 1-particle Hilbert space for orbital angular momentum, which makes sense if I think about hydrogen atom. However, I could do the same for two particles of spin-1/2 (e.g. positronium) and do the tensor product. It doesn't seem to tell me (via the Clebsch-Gordan decomposition) that it is orbital-spin mixing instead of spin-spin mixing. – Everiana Jul 30 '19 at 1:01
• In that case you have both. Total spin will be the Clebsch-Gordon addition of spins and orbital angular momentum. – Miguel Correia Apr 28 at 1:16