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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 that in classical optics it's completely clear to consider the wave vector and the position as conjugate variables. Also in standard QM textbook 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.

AlexH
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thyme
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  • 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? – Emilio Pisanty Mar 21 '14 at 13:27
  • For example here on page 3: http://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? – thyme Mar 24 '14 at 08:53

2 Answers2

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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://arnold-neumaier.at/physfaq/physics-faq.html

Community
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Arnold Neumaier
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  • 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? – thyme Mar 24 '14 at 09:01
  • @thyme: The Fourier transform is a function of time and describes the oscillations in time. – Arnold Neumaier Mar 24 '14 at 10:18
  • 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... – thyme Mar 24 '14 at 10:57
  • @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. – Arnold Neumaier Mar 24 '14 at 12:10
  • 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. – Arnold Neumaier Mar 24 '14 at 12:13
  • Suppose photons are gamma quants. Can't I put Geiger counter in any place and register the spatial probability density, which you said doesn't exist? – Dims Feb 20 '23 at 18:55
  • @Dims: You get a coarse-grained spatial probability density that depends on the way you observe it and cannot be refined arbitrarily. That the spatial probability density exist s(as a mathematically well-defined object) means that there is a measurement independent density that gives a well-defined probability at any resolution. – Arnold Neumaier Feb 21 '23 at 10:47
  • @ArnoldNeumaier can you measure probabily down to an arbitrary resolution for electron? I don't think so. And I don't think practical inability to measure probablility density with infinite resolution means anything. When we are discussing two-slit experiment, we are measuring probability with screens and probes with finite resolution and just approximate, say, interference picture with smooth curves... – Dims Feb 21 '23 at 13:30
  • @Dims: Theoretical physics is about models defined by precise concepts. Theoretical consequences are exact in the model; approximations are involved in (i) numerical calculations given the model (except for exactly solvable models), and (ii) the comparison with experiment. Statements of existence in theoretical physics always refer to a particular model. The standard model of a photon is the mass 0, helicity 1 description, where a spatial probability density does not exist. But there are many inequivalent coarse-grained versions that match experimental spatial probability densities. – Arnold Neumaier Feb 21 '23 at 15:04
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One can certainly define the wave function a photon in the representation of second quantization: $$|\mathbf{k},\lambda\rangle = b_{\mathbf{k}\lambda}^\dagger|0\rangle,$$ where $\lambda$ is the photon polarization state, and $|0\rangle$ is the photon vacuum.

In fact, one can make an even stronger claim: for massive particles the first quantization means describing them by a wave equation (Schrodinger's, Dirac, etc.), whereas the second quantization is the description by the filling numbers. For photons the Maxwell equations are already wave equations, and their description by the filling numbers is their first quantization.

Roger V.
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