The rotation of a photon around an axis with orientation given by the vector n⃗ is the action representated by the operator $ U = e^{ i \theta \vec{n}. \vec{J} } $. But to get a clearer understanding of what that would mean in a specific case... $$ $$ Suppose $ \vec{n} $ were parallel to $ \vec{J} $ and that and the photon's direction of propagation is along $z$ and that at time $t = t_0$, the photon's electric field $ \vec{E} $ were pointing up along $y$ direction. $$ $$ Suppose now at $t_0$, the photon is 'rotated' about $ \vec{n} $ by an angle $ \Pi $, (180 degrees) what now are the directions of $ \vec{E} $ and $ \vec{J} $ ?
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$\begingroup$ Please use mathjax for equation rendering. The link has a detailed tutorial on how. Thanks. $\endgroup$– joseph hCommented Feb 6, 2023 at 4:26
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$\begingroup$ "the photon's electric field" The photon does not have an electric field, you are confusing it with the classical electromagnetic wave, which emerges from a large number of photons. this question and answers are relevant physics.stackexchange.com/questions/90646/… $\endgroup$– anna vCommented Feb 6, 2023 at 19:14
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$\begingroup$ @anna if a photon doesn't have an associated electric field / electric vector, how can 2 photons have an electric field, how can 2,000,000,000 photons? where does the 'bunch' of photons get this electric field when none of the photons comprising that 'bunch' has any electric field? $\endgroup$– peteCommented Feb 6, 2023 at 21:53
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$\begingroup$ @anna A photon does have an electric field associated with it, otherwise how can a single photon knock out a single electron via the photoelectric effect, and a single photon make a dot on a screen in a double-slit nterference experiment! $\endgroup$– peteCommented Feb 6, 2023 at 22:05
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1$\begingroup$ Bit lost. Isn't the $\vec J$ in $e^{i\theta\vec n\cdot\vec J}$ a vector of angular momentum operators? Does it make sense to talk about $\vec n$ being parallel to $\vec J$? And anna v has a point. The $\vec E$ field of a photon is no longer a vector field. It is a field of vector-valued quantum observables. For a single photon, $\vec E$ has such high uncertainty that "$\vec E$ pointing up along the $y$ direction" no longer makes sense. You need many photons to build up a coherent ("classical looking") state (in which the uncertainty in $\vec E$ becomes small) $\endgroup$– HTNWCommented Feb 7, 2023 at 4:41
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