Consider the following facts:(adsbygoogle = window.adsbygoogle || []).push({});

1. For a particle with momentum ##k##, the two transverse polarization vectors ##\epsilon({\bf k}, \lambda_{1})## and ##\epsilon({\bf k}, \lambda_{1})## are purely spatial and orthogonal to ##\bf k##, that is,

##\epsilon^{0}({\bf k}, \lambda_{1}) = 0,##

##\epsilon({\bf k}, \lambda_{1})\cdot{k} = 0,##

##\epsilon^{0}({\bf k}, \lambda_{2}) = 0,##

##\epsilon({\bf k}, \lambda_{2})\cdot{k} = 0.##

2. The third, longitudinal, polarization vector ##\epsilon({\bf k}, \lambda_{3})##}, for a particle with momentum ##k##, is timelike positive, orthogonal to ##k## as well as the transverse polarization vectors, and has unit negative norm, that is,

##\epsilon^{0}({\bf k}, \lambda_{3}) > 0,##

##\epsilon({\bf k}, \lambda_{3})\cdot{k} = 0,##

##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{1}) = 0,##

##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{2}) = 0,##

##\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{3}) = -1.##

3. We can infer from the orthogonality of the polarization vectors that the longitudinal polarization vector of a particle points in the direction of momentum of the particle.

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Are these facts true for any gauge choice of the electromagnetic four-vector ##A^{\mu}##?

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# A Polarization of gauge bosons and gauge choice

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