Chirality of the Electromagnetic Field Tensor

I have learned that chirality is a concept, that appears for $(A,B)$ representations of the Lorentz group, where $A\neq B$.

An example would be a Dirac spinor, corresponding to the representation $(\tfrac{1}{2},0)\oplus(0,\tfrac{1}{2})$, where we can identify left- and right-chiral components.

Wikipedia lists the electromagnetic field strength tensor $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ as transforming under the $(1,0)\oplus(0,1)$ representation of the Lorentz group.

Supposing my first sentence is true, where can I see chirality in the electromagnetic field strength tensor?

On can define the dual of a field strength tensor by $$F^*_{\mu\nu}= \frac 12 \epsilon_{\mu\nu\alpha\beta} F_{\alpha\beta}$$ and one can impose the condition that $F^*_{\mu\nu} =F_{\mu\nu}$ (self dual) or $F^*_{\mu\nu}=-F_{\mu\nu}$. These conditions are preserved by Lorentz tranformations and correspend to the $(1,0)$ and $(0,1)$ representations.

See the table in the Wikipedia article on the "Representation theory of the Lorentz group"

It is possible to write Maxwells' "curl" equations so that they look like a spin one version of the the Weyl equation for chiral fermions: $$i\hbar \frac{\partial \Psi_\pm}{\partial t}= \pm c(\Sigma\cdot {\bf P}) \Psi_\pm$$ where $\Psi_{\pm}$ are three component Riemann-Silberstein vectors $\Psi_\pm\equiv {\bf E}\pm i{\bf B}c$, ${\bf P}= -i\hbar \nabla$ and the spin-one matrices are $[\Sigma_i]_{jk}==-i\epsilon_{ijk}$. I've always assumed, but have not explicitly checked, that the $\Psi_\pm$ are the two chiral components described by the self-dual and anti-self-dual conditions.

Indeed under electromagnetic duality (and with $c=1$) we have $({\bf E,B})\mapsto ({\bf B,-E})$ so ${\bf E}+i{\bf B}\mapsto ({\bf B}-i{\bf E})=-i({\bf E}+i{\bf B})$ and ${\bf E}-i{\bf B}\mapsto i({\bf E}-i{\bf B})$. Note that $(F^*)^*=-F$, so that in Minkowski signature the eigenvalues of the duality transformation are $\pm i$. This means that real-valued EM fields cannot be self-dual in Lorenzian signature. They can be in Euclidean signature where the Lorentz group is replaced by ${\rm SO}(4)$.

• Amazing. I wonder if it is possible to move in the opposite direction and write Dirac equation as curl-like operators on linear combinations of the Dirac field – lurscher Sep 20 '18 at 15:11
• Where does the definition of the spin-1 matrices come from? – Stephan Sep 25 '18 at 8:16
• @stephan. They are the usual spin-1 matrices, but in the $x,y,z$ vector basis rather than the $|j,m\rangle$ basis. – mike stone Sep 25 '18 at 12:25
• @mikestone. So they are connected to the generators of SO(3)? Or am I mixing something up in my head? I‘ve got another question: Where is the connection between the Riemann-Silberstein vectors and the (anti-)self duality of the em field tensor? And shouldn‘t there be an i in these conditions such that $F**=-F$? – Stephan Sep 25 '18 at 12:53
• @stephan. I was a bit vague in my use of (anti)-self dual in my answer. We have $\star \circ \star=1$ in 4d Euclidean space but $\star \circ \star=-1$ in 4d Minkowski signature. The RS vectors are the Minkowski $\pm i$ eigenvectors of $\star$ in Minkowski, so my understanding is that they are the next-best-thing to self-dual and anti-self dual. And yes you can regard the $\Sigma_i$ as ${\rm SO}(3)$ generators and at the same time the $J=1$, ${\rm SU}(2)$ generators. The $J=1$ rep of ${\rm SU}(2)$ lifts to a rep of ${\rm SO}(3)$. – mike stone Sep 25 '18 at 17:59

There's a great existing answer, I just thought I'd check where the "rotation" comes from.

As you know, the electromagnetic field tensor decomposes under $SO(3)$ into two vectors, $\mathbf{E}$ and $\mathbf{B}$, which are preserved under rotation. In fact, any linear combination of $\mathbf{E}$ and $\mathbf{B}$ are preserved under rotations. Now if we add in the boosts, the specific combinations that are preserved under both rotations and boosts are $$\mathbf{E} = \pm i \mathbf{B}.$$ These correspond to the $(1, 0)$ and $(0, 1)$ irreps; they are called self-dual and anti-self-dual fields.

Here we're working with complex-valued electromagnetic fields, i.e. we have $$\mathbf{E} = \mathbf{E}_0 e^{ik \cdot x}, \quad \mathbf{B} = \pm i \mathbf{E}_0 e^{ik\cdot x}.$$ To get representative real-valued solutions, we may take the real part. For a wave propagating along $\hat{\mathbf{z}}$, guessing $\mathbf{E}_0 \propto (1, \pm i, 0)^T$, we find the self-dual and anti-self-dual fields correspond to light waves with clockwise and counterclockwise circular polarization, a clear manifestation of chirality. You can't boost or rotate a clockwise polarized wave into anything but a clockwise polarized wave.