# Spin part of the angular momentum from the Lagrangian

For fermions of spin $$1/2$$ the angular momentum has following form: $$\mathcal{J}_z = \int d^{3}x \ \psi^{\dagger} (x) \left[i(- x \partial_y + y \partial_x) + i\sigma^{xy} \right] \psi(x)$$ Here the first term is orbital part and the latter one is the spin part of angular momentum.

However, in the general prescription for derivation of the angular momentum: $$\mathcal{J}^{ij} = \int d^{3} x (x^i T^{0 j} - x^j T^{0 i}).$$ I cannot see, where the spin part can actually arise.

For example for the photon field, where: $$T^{\mu \nu} = F^{\mu}_{\alpha} F^{\mu \alpha} - \frac{1}{4} g^{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} .$$ This procedure seems to provide only the orbital part of the angular momentum, and no the spin. Or it is implicitly included in this expression?

For the spin $$1/2$$ field the term $$\sigma^{xy}$$ emerges, when one considers corrected energy tensor https://en.wikipedia.org/wiki/Belinfante%E2%80%93Rosenfeld_stress%E2%80%93energy_tensor. (formula 6 in https://arxiv.org/abs/1508.06349).

However, for the spin-1 theory this expression incorporates everything and is the most symmetric one.

I would strongly appreciate any help and comments!

The division of the total angular momentum into orbital and spin parts is unphysical, a bit like how the division of energy into potential energy and kinetic energy is unphysical. (Are static electric and magnetic fields purely potential energy while electromagnetic waves are purely kinetic? This is not a physically meaningful question, so we don't ask it.)

Only the total angular momentum is a physically meaningful quantity, and to define it in terms of the stress-energy tensor is to put the cart before the horse. The proper way to define the total angular momentum is by applying Noether's theorem to the group of rotations. In the case of a Dirac field, one indeed obtains the expression for $$\mathcal{J}_z$$ given by the OP.

The Belinfante–Rosenfeld correction term mentioned by the OP is defined in terms of the spin tensor which comes from Noether's theorem, so to define angular momentum in terms of the stress-energy tensor would be a circular definition. However, provided that this correction term is in fact included in the definition of the stress-energy tensor, the OP's expression will yield a sensible total angular momentum density tensor and will thus "include spin", whatever that means.

• Thanks for the answer, but there are a few points I would like to clarify. For the massless particles therer is no rest frame, but for the massive it seems like one make sense of the spin angular momentum and potential energy - like the potential energy is the energy associated with respect to some field, something that would be present in the particle's rest frame, and the spin is part present, when one moves to corotating frame. – spiridon_the_sun_rotator Nov 19 '20 at 7:10
• @spiridon_the_sun_rotator You can certainly define the orbital and spin parts of the angular momentum, but it does not mean they have any physical meaning, although in certain special cases they may be individually conserved. – Brian Bi Nov 19 '20 at 17:15

The symmetric EM tensor does contain the spin. There is a discussion by Michael Berry about this, and also some in Wikipedia here

Your observation is correct. The total angular momentum, integrated over volume, contains both contributions as pointed out in the other answer. However it takes the overall for of an orbital, position dependent, angular momentum while clearly the electromagnetic potential has internal AM as well.

My answer is written down in a peer reviewed, published paper, https://arxiv.org/abs/physics/0106078. Based on the so called Fermi lagrangian I derive a theory in which spin and angular momentum are separate observables.