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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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  • $\begingroup$ You can have wavefunctions in momentum space for matter particles as well. $f(\nu)$ is (with appropriate caveats) a legitimate wavefunction. Can you provide references to places where it is interpreted directly as an amplitude for a spatial probability density? $\endgroup$ – Emilio Pisanty Mar 21 '14 at 13:27
  • $\begingroup$ For example here on page 3: arxiv.org/abs/0911.5139 . It's not directly an interpretation as an amplitude for a spatial probability density but a direct connection between the frequency distribution and the spatial shape of the photon. Mybe I should modify my question: Why is the frequency distribution of a photon connected to its spatial shape? $\endgroup$ – thyme Mar 24 '14 at 8:53
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There is no position operator for photons, so photons do not have a spatial probability density. Associated with a photon (in a laser beam, say) one has only a probability density of hitting any given surface crossing the beam at a particular point of the surface.

See Chapter B2: Photons and Electrons (and the entries ''Particle positions and the position operator'' and ''Localization and position operators'' of Chapter B1: The Poincare group) of my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html

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  • $\begingroup$ Ok, but when I have the frequency distribution $f(\nu)$ of a single photon, how can I interpret its fourier transformation? As the spatial shape of the photon? $\endgroup$ – thyme Mar 24 '14 at 9:01
  • $\begingroup$ @thyme: The Fourier transform is a function of time and describes the oscillations in time. $\endgroup$ – Arnold Neumaier Mar 24 '14 at 10:18
  • $\begingroup$ This is not clear to me. What do you mean with "oscillations in time" and "describes"? If we talked about a classical electromagnetic field I would agree, but for a single photon I don't know the meaning... $\endgroup$ – thyme Mar 24 '14 at 10:57
  • $\begingroup$ @thyme: Independent of the application, the Fourier transform of a function in frequency space is always a function in time. For photons, the fourier transform of $f(\nu)$ is meaningless, as the time oscillations are extremely rapid, while the observation process is slow. E.g., our eyes observe the frequency distribution $f(\nu)$ itself, not its Fourier transform. $\endgroup$ – Arnold Neumaier Mar 24 '14 at 12:10
  • $\begingroup$ In any case, the Fourier transform of f(ν) can never be interpreted as something in space. For a spatial interpretation you'd need to Fourier transform the direction-dependent density f(nu,p) with respect to momentum, but because of the transversal nature of photons, this gives something easily interpretable only along planes perpendicular to the momentum p. $\endgroup$ – Arnold Neumaier Mar 24 '14 at 12:13

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