Thursday 4 August 2016

quantum optics - Wave function of a photon?


consider a single photon. Since it is not possible to create a photon with a certain frequency it can be characterized by a normalized frequency distribution $f(\nu)$ that is peaked around some mean frequency.


Now I sometimes hear or read that the fourier transformation of $f(\nu)$ is considered as the wave function of the photon (interpreted as the probability density in space). This is especially done in quantum optics. But I don't understand that.


The reason is the following: In classical optics it's completely clear to consider the wave vector and the position as conjugate variables. Also in standard textbook QM this is clear due to the commutator relation of the Position and Momentum Operator (dealing with massive particles). But for a single photon, described by a creation operator, I can not find a reason to interpret the fourier transformation of $f(\nu)$ as the spatial probability density of the photon.




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