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So how is it possible to have a quantized outcome from a symmetric continuous event? Easily. So easily that I'll describe the easiest example to me. Which is to describe what happens when a Stern-Gerlach device interacts with a spin 1/2 particle. You could have a particle with any spin whatsoever, but no matter what single particle state you pick it ...

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I know that photons are quantized, they are not continuous. Photons are not quantised, nor are they continuous. They are the charge carriers of the electromagnetic field as arising in quantum field theory. An accelerated charge generates an electromagnetic field whose carriers are, in turn, the photons whose energy might be quantised. So how is it ...

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I believe it would be incorrect to term this as "photon-splitting". What is really happening is that a photon is being annihilated and subsequently two new photons are created by means of a non-linear optical process. Such a process that is routinely harnessed in quantum optics laboratories is spontaneous parametric downconversion (SPDC). The reason why ...

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The spin of the electromagnetic field tensor $F_{\mu\nu}$ is best understood by writing it as a spinor. A spin 1 field is a represented by a symmetric spinor $\xi^{AB}$ or by a dotted symmetric spinor $\eta_{\dot{A}\dot{B}}$. In order to get the field transforming correctly under parity, the electromagnetic field has to be a direct sum using the symmetric ...

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In QED, this superposition principle is still valid at least to first order perturbation theory. The deeper reason behind this superposition principle from the QED point of view, is, that photons do not interact with each other. They do not carry charges. They do not "see" each other. So they can be safely superposed without having an effect on each other. ...

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Consider for the shake of simplicity a free neutral scalar field $\phi$. Passing to the second quantization picture, it is a operator valued distribution $$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$ where $M$ is Minkowski spacetime and $\phi(f)$ is a densely defined symmetric operator on the Hilbert space F_+(\cal H) = \mathbb C \oplus \cal H ...

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My question is whether individual photons also carry orbital angular momentum? Yes. See https://en.m.wikipedia.org/wiki/Orbital_angular_momentum_of_light If yes, what are the values of orbital angular momentum in one-particle states? To quote the wikipedia page In particular, in a quantum theory, individual photons may have the following ...

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Yes, single photons can have orbital angular momentum. However, unlike spin, they are not required to have any. Just like in the classical case, the orbital momentum of single photons is determined by the shape of their EM mode- roughly speaking, the wavefront must have a helical aspect to it. In particular, this means that the eigenmodes of light in a 3D ...

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