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

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

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

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

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 $O$?

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

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


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

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    $\begingroup$ Fundamentally it's because $\mathfrak{so}(3, 1)$ (when complexified, since you're using complex variables) is just two copies of $\mathfrak{su}(2)$, which is the Lie algebra of $SO(3)$. This trick only works in $3+1$ dimensions. $\endgroup$ – knzhou Feb 8 at 0:25
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    $\begingroup$ @knzhou For $\mathfrak{so}(3,1)$, I need my vector $\mathbf{q}$ to have 4 components. Right now I have three: $E_x+iB_x,E_y+iB_y, E_z+iB_z$. What would the 4th component be? $\endgroup$ – Alexandre H. Tremblay Feb 8 at 0:34
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    $\begingroup$ This is the Riemann-Silberstein vector from 1907. $\endgroup$ – G. Smith Feb 8 at 0:36
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    $\begingroup$ @AlexandreH.Tremblay The point is that the EM field strength is a 6-dimensional representation of $\mathfrak{so}(3, 1)$ (not every representation of the Lorentz group has to be 4-dimensional), but this is in turn a combination of two 3-dimensional representations of $\mathfrak{su}(2)$. This works only because of the connection between $\mathfrak{so}(d-1, 1)$ and $\mathfrak{su}(2)$ when $d = 4$. $\endgroup$ – knzhou Feb 8 at 0:41
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    $\begingroup$ OP note that your group is not $O(3,R)$ but $O(3,C)$ this matters. @knzhou if you would write up an answer explaining the subtleties regarding complexification and algebra/group distinctions I think that might clarify this? $\endgroup$ – jacob1729 Feb 8 at 0:57
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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).

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