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Let $\mathbf{q}$ be a complex vector of three elements defined as:

$$ \mathbf{q}:=\pmatrix{ E_x + iB_x\\ E_y + i B_y\\ E_z +i B_z }.\tag{1} $$

I define the function $f(\mathbf{q})$:

$$ \begin{align} f(\mathbf{q})&=\mathbf{q}^T\mathbf{q}=\pmatrix{ E_x + iB_x& E_y + i B_y& E_z +i B_z }\pmatrix{ E_x + iB_x\\ E_y + i B_y\\ E_z +i B_z }\\ &=E_x^2+E_y^2+E_z^2-B_x^2-B_y^2-B_z^2+2i(E_xB_x+E_yB_y+E_zB_z)\\ &=||\mathbf{E}||^2-||\mathbf{B}||^2 +2i\mathbf{E}\cdot\mathbf{B} \end{align} \tag{2} $$

where $\mathbf{E}:=(E_x, E_y,E_z)$ and $\mathbf{B}:=(B_x,B_y,B_z)$.

The equation produces the Lorentz invariant of electromagnetism.

What is the invariance group of $f(\mathbf{q})\to f(O\mathbf{q})$ under a linear transformation $O$?

$$ \begin{align} f(O\mathbf{q})&=(O\mathbf{q})^T(O\mathbf{q})\\ &=\mathbf{q}^TO^TO\mathbf{q}\\ &\implies O^TO=I \end{align}\tag{3} $$

Consequently, since $\dim_{\mathbb{C}} (\mathbf{q})$ is 3, we have $O(3,\mathbb{C})$.

I am a bit baffled as to why am I getting $O(3,\mathbb{C})$ here? I was expecting anything else; for instance Lorentz group $O(3,1;\mathbb{R})$; or even $U(1)$, the gauge group associated with electromagnetism in the literature. Why are the Lorentz invariants of electromagnetism not Lorentz invariant but $O(3,\mathbb{C})$ invariant - where is the mistake?

Qmechanic
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Anon21
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2 Answers2

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The Riemann-Silberstein vector is a construction that works because of a duality that exists between electric and magnetic fields in the absence of sources. It is not consistent with the full description of electrodynamics as found in QED for instance. In the latter the electric and magnetic fields are combined into the Faraday tensor which correctly transforms under SO(3,1). As such one finds that the electric field part transforms as a vector while the magnetic field part transforms as a pseudo-vector. So the Riemann-Silberstein vector combines a vector and a pseudo vector, which makes the question of its transformation properties challenging.

BTW, usually one would contract a complex vector with its Hermitian adjoint, which includes complex conjugation. That would then give the magnitude of the vector without an imaginary term. This magnitude is invariant under unitary transformation of the vector. So the associated symmetry group would be SU(3).

flippiefanus
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TL;DR: $SL(2,\mathbb{C})$ is the double cover of both $SO(3,\mathbb{C})$ and the restricted Lorentz group $SO^+(1,3;\mathbb{R})$, cf. e.g. this Math.SE post & and this Phys.SE posts, so $$ SO(3,\mathbb{C})~\cong~PSL(2,\mathbb{C})~\cong~SO^+(1,3;\mathbb{R}). \tag{A}$$ This Lie group isomorphism (A) makes OP's observation somewhat less surprising.

In more detail:

  1. The Faraday E&M tensor $F_{\mu\nu}$ decomposes into (anti)self-dual parts $F^{\pm}_{\mu\nu}$, which can rewritten as the two Riemann-Silberstein (RS) vectors $$\vec{F}_{\pm}~=~\vec{E}\pm i \vec{B}.\tag{B}$$

  2. In the Van der Waerden notation, the decomposition takes the form $$ F_{\alpha\dot{\alpha},\beta\dot{\beta}}~=~F_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}+\epsilon_{\alpha\beta}F_{\dot{\alpha}\dot{\beta}}, \tag{C}$$ i.e the RS vectors $\vec{F}_{\pm}$ can be identified with a (complex conjugate) symmetric second-rank spinor $F_{\alpha\beta}$ ($F_{\dot{\alpha}\dot{\beta}}$), respectively.

  3. In particular, OP's bilinear form $$ \underbrace{\vec{F}\cdot \vec{F}}_{\text{inv. under }O(3,\mathbb{C})}~\propto~\underbrace{\det(F_{\alpha\beta})}_{\text{inv. under }SL(2,\mathbb{C})} .\tag{D}$$

Qmechanic
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