What is the interpretation of the Chern-Simons electromagnetic spin density?

Question

Hans de Vries (who happens to be a no-longer-active physics.SE user) has an online book (referenced below) in which ch. 6 is a presentation of an object he calls the Chern-Simons current, electromagnetic spin density, or Chern-Simons electromagnetic spin:

$ C^a = \frac{\epsilon_0}{2} \epsilon^{abcd} A_b F_{cd} = \epsilon_0 \epsilon^{abcd} A_b \partial_c A_d $ .

He has a long and detailed presentation of this thing, including graphs and examples. Unfortunately I'm not having much luck extracting from this what he claims is the interpretation of it, whether his interpretation is standard, and whether it has a classical interpretation. He references Mandel and Wolf (which I don't have access to), but what they apparently present is a different expression, $\epsilon_0 \textbf{E}\times\textbf{A}$, and refer to it as the intrinsic angular momentum of the electromagnetic field. De Vries says that $C^a$ is the natural way of making this tensorial. It seems hard to check whether what he's saying is standard, since he says, "The derivations (which I had to do myself since somehow one can't find these anywhere) and many details can be found in my paper..." (linking to a paper that duplicates the material in the book).

The expression is manifestly classical, so I don't really see how it's to be interpreted as the intrinsic or spin contribution to the field's angular momentum. The classical/quantum interpretation is also obscured because factors of $\hbar$ start appearing at de Vries' eq. 6.6, but these equations are supposed to be justified somewhere later on.

It seems odd to me that this is written as a product of the four-potential and a derivative of the four-potential. This makes it not manifestly gauge-invariant. If I was going to write down a density of angular momentum for the electromagnetic field, I would start from the stress-energy tensor, which is a product of $F$ with $F$, and therefore independent of gauge.

It's not at all obvious to me what one would even mean by a spin density for the electromagnetic field. I guess for a classical fluid of electromagnetic radiation in equilibrium (e.g., the kind of environment we had during the early universe), I would define a comoving frame and look at the amount of angular momentum $L^{ab}=r^ap^b$ in a small volume element. But that clearly isn't going to work for the electromagnetic field in general, since you can't define a comoving frame for, e.g., an electromagnetic plane wave.

It does make sense that the expression is manifestly translation-invariant, since if there is some sensible way to split the angular momentum into spin and orbital parts, only the orbital part should depend on one's choice of axis.

De Vries, Understanding Relativistic Quantum Field Theory, http://www.physics-quest.org/ , ch. 6

Answer 1

The electromagnetic field angular momentum has the following decomposition:

$\mathbf{J} = \frac{1}{4\pi c}\int d^3x \mathbf{x}\times(\mathbf{E}\times \mathbf{B} )= \frac{1}{4\pi c}\int d^3x\left [ \mathbf{E}\times \mathbf{A} + \sum_{j=1}^3 E_j(\mathbf{\nabla}\times\mathbf{x})A_j \right]$

This expression appears for example in Jackson: Classical electrodynamics (second edition) in problem: 7.19. The second term can be interpreted as the orbital angular momentum for the following reasons: 1) It is an weighted average of the angular momentum operator: $(\mathbf{\nabla}\times\mathbf{x})$, 2) Its density identically vanishes for a plane wave.

The first term's density, on the other hand, is not dependent on the position in space. Moreover, in the temporal gauge, where the canonical Poisson brackets are:

$\left \{ E_i(\mathbf{x}), A_j(\mathbf{y}) \right \} = -i \hbar \delta^3(\mathbf{x} - \mathbf{y})\delta_{ij}$

The first term $ \frac{1}{4\pi c}\int d^3x \mathbf{x}\times(\mathbf{E}\times \mathbf{B} )$, satisfies the angular momentum commutation relations (in the Poisson bracket level). Moreover at the quantum level, this term has an eigenvalue of $+1$ on left polarized photons and $-1$ on right polarized photons. Thus it can be interpreted as the helicity operator.

The only difficulty in the above decomposition is that it is not gauge invariant (Only the components, because the total angular momentum is gauge invariant). This is a general property of relativistic systems.

The total angular momentum is the Noether charge of the rotation group $SO(3)$. Thus a covariantization of the "spin density", would be the Noether charge of the Lorentz group $SO(3, 1)$.

Now, spin is a quantum phenomena in the context that its quantization cannot be explained classically. But it is has seeds in the classical theory. As vey well known, massless relativistic particles correspond the the Wigner's little group $E(2)$. A more precise description is through the theory of coadjoint orbits, where the massless particles correspond to the orbit: $(SO(3,1) \rtimes R(4))/(E(2) \times U(1)) = \mathbb{R}^3 \times (\mathbb{R}^3 - 0)$ , please see arxiv:0912.218 by: Carinena, Gracia-Bondia, Lizzi, Marmob, and Vitale. The coadjoint orbit describes the physical degrees of freedom (after the reduction of all gauge freedom), in the case of a massless particle 3 position coordinates and 3 momentum coordinates except for the zero momentum which is not allowed for a massless particle. The coadjoint orbit has a symplectic structure and fully describes the classical dynamics of a free massless particle. There are only certain values of (the coefficient of the) symplectic structure for which the coadjoint orbit can be quantized. This coefficient turns out to be the helicity, and after its quantization, the particle acquires a helicity. This is the process of prequantization in our case. It is on the border between classical and quantum and some people consider it a part of the classical theory. Actually, we work a lot in this framework, for example in the calculation of a classical partition function of a spin system, we use the fact that spin is quantized but nothing else from quantum theory, thus we are working within prequantization.

Answer 2

The axial spin vector of the EM field

In the context of the Pointcaré algebra we can consider $C^\mu$ as the Pauli Lubanski vector of the electro magnetic field.

\begin{equation} \begin{array}{|c|c|c|c|c|c|} \hline &&&&& \\ & \mbox{transl.} & \mbox{Lorentz} & \mbox{rotation} & \mbox{boost} & \mbox{Pauli Lubanski vect.} \\ \hline &&&&& \\ \mbox{Poincaré:} & P^\mu & J^{\mu\nu} & J^i & K^i & W^\mu=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,J_{\alpha\beta}\,P_\nu \\ &&&&& \\ \hline &&&&& \\ \\ \mbox{e.m. field:} & A^\mu & F^{\mu\nu} & B^i & E^i & \,C^\mu\,=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}\,A_\nu \\ &&&&& \\ \hline \end{array} \end{equation}

It transforms like an Axial vector in the relativistic Lorentz gauge which is an absolute requirement for an expression to be considered the spin of the EM field.

$C^\mu$ contains the more familiar expression $\epsilon_o\vec{E}\times\vec{A}$ but this term alone is not a 4-vector and does not transform correct, not even in the Lorentz gauge. There is a very interesting correspondence of $C^\mu$ with the axial vector of the fermion field: \begin{equation} j_\circlearrowleft^\mu\ =\ \frac{iq_e}{2m}\ {\bar \psi}\gamma^\mu\gamma^5\psi \end{equation}

Which we can consider the Pauli Lubanski vector of the fermion field if we perform the Gordon decomposition on it. We get:

\begin{equation} \begin{array}{|c|c|} \hline & \\ & \mbox{Pauli Lubanski vector} \\ \hline & \\ \mbox{Poincaré:} & W^\mu=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,J_{\alpha\beta}\,P_\nu \\ \\ \\ \mbox{e.m. field:} & C^\mu\,=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}\,A_\nu \\ \\ \\ \psi-\mbox{field:} & j_\circlearrowleft^\mu~=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,M_{\alpha\beta}\,j_\nu \\ \\ \hline \end{array} \end{equation}

Where $M^{\mu\nu}$ is the Magnetization Tensor of the electron field defined by.

\begin{equation} M^{\mu\nu} ~=~ \left(\frac{\mu_e}{2mc}\right)~\mbox{ $\bar{\psi}\sigma^{\mu\nu}\psi$} ~=~ \left( \begin{array}{cccc} 0 & ~~P_x & ~~P_y & ~~P_z \\ -P_x & 0 & - M_z & ~~M_y \\ -P_y & ~~M_z & 0 & - M_x \\ -P_z & - M_y & ~~M_x & 0 \end{array} \right) \end{equation}

and the $j_\nu$ is given by the phase change rates $\partial^\nu\phi$ of the field $\psi$.

In http://physics-quest.org/Book_Chapter_EM2_ChernSimonsSpin.pdf I calculate $C^\mu$ for some elementary cases and show that it has the values we may expect from it.

Hans

Answer 3

I puzzled over this same question a couple of years ago when I read De Vries article on line. I'm only guessing but I think it basically amounts to (for a single electron e.g.):

1. convering the spinor to an equivalent vector field by combining it with the charge density; and,

2. considering the resulting vector field as the curl of a current density.

In other words, by combining the two spinor components of the wave function, you get a spin density at every point in space which actually has the quality of being a vector field because in addition to the direction (which is all you get with a spinor) you have a magnitude which comes from the local charge density. If you take the anti-curl of that vector field you get a current density.

That resulting current density is what De Vries calls Chern-Simon. I think.