Angular momentum in the Maxwell field theory/Chern-Simons theory?

Question

I'm trying to calculate the angular momentum in the chern simons theory. But equivalently, I was trying a calculation of angular momentum in the Maxwell field theory, which will hopefully be insightful in getting the answer for CS theory.

$\mathcal{L} \propto f^{\mu\nu}f_{\mu\nu} $

The generalised angular momentum is defined as:

$p^i = \frac{\partial \mathcal{L}}{\partial (\partial_0 a_i)} = E^i$

Now, the angular momentum would be defined as:

$L^i = \epsilon^{ijk}a_j p_k$

However, the angular momentum in a Maxwell theory is given by $L^i \propto x \times (E\times B)$, which is nowhere near the answer I get for the angular momentum I've defined. Am I wrong in the definition of the angular momentum? There should be a B field present to get the correct answer, which is nowhere to be found in my answer.

I would also appreciate the calculation for a pure CS theory, if there is any fundamental difference in the steps for either theory. The answer for the CS theory: $\mathcal{L} = \frac{k}{4\pi}ada$

$L = \frac{-k}{4\pi} \int x^i \epsilon^{ij} (a_j f + f a_j)$

Which is completely against any intuition I have. This is given in Carl Turner's notes (https://arxiv.org/abs/1905.12656) in eq 3.67.

Edit: pasting a clarification from the comments, I'm not sure if my definition for angular momentum is correct (I haven't seen any mention of generalised angular momentum in my quick research). I'm literally just basing it out of a naive generalisation of angular momentum in classical mechanics. Hence my question.

Answer 1

In this answer I work in natural units, and deal mostly with the Maxwell Lagrangian density \begin{equation} \mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu\nu} . \end{equation}

Canonical momentum

The canonical momentum conjugate to the field $A_\mu$ is \begin{equation} \Pi^\mu = \frac{\partial\mathcal{L}}{\partial (\partial_0a_\mu)} = (0,E^i). \end{equation} However this is not the same as the (physical) linear momentum of the electromagnetic field.

Linear momentum

The electromagnetic stress-energy tensor is \begin{equation} T_{\mu\nu} = \frac{1}{4\pi} \left(F_{\mu\rho}F_{\mu}^{\rho} - \frac{1}{4}\eta_{\mu\nu} F_{\rho\sigma}F^{\rho\sigma}\right) . \end{equation}

Its form can be derived as a functional derivative of the action with respect to the metric. This gives a straightforward way to compute it, and allows to quickly discuss Chern-Simons theories (below). Alternatively, more intuition can be gained by a bottom-up construction of the corresponding continuity equations (see e.g. Poynting's theorem).

The linear momentum density of the electromagnetic field is given by the components \begin{equation} p_i = \frac{1}{4\pi} F_{0j}F_{i}^j \ . \end{equation}

which in 3d can be expressed (involving the Poynting vector) \begin{equation} \vec{p} = \frac{1}{4\pi} \vec{E}\times \vec{B}. \end{equation}

Angular momentum

The angular momentum density is defined in general as the antisymmetric tensor $\ell_{ij} = x_i p_j$. In three dimensions this corresponds to the (pseudo-)vector obtained from the cross product \begin{equation} \vec{\ell} = \frac{1}{4\pi} \vec{x}\times (\vec{E}\times\vec{B}). \end{equation}

Chern-Simons

The Chern-Simons action can be expressed in terms of differential forms, \begin{equation} S_{CS} = \frac{k}{4\pi} \int \mathrm{Tr} (A \wedge \mathrm{d}A + \frac{2}{3} A \wedge A \wedge A) \end{equation} Therefore it does not depend on the metric, and therefore the stress-energy tensor vanishes. In other words, since $S_{CS}$ is a topological term, it does not contribute to the energy, momentum or angular momentum.

It does have canonical angular momentum, i.e. the conserved angular momentum operator of (3.67) in Turner's notes. As pointed out in section 3.5.1 (p.40), once the gauge is fixed you have two fields $a_1,a_2$, which are canonically conjugate to each other. The canonical angular momentum should be derived from there. Note that (3.67) is essentially the integral of (the 2d version of) $\vec{x} \times \vec{p}$.