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Why photon wave function does not exist?

  1. Dec 18, 2012 #1
    I hear some reasons that photon exist in 4D space-time, wave function not and so on.

    But, an electron can be described with de Broglie waving and we can use wave function to describe electron. Frequency of the wave function is the same as energy/\hbar and k of wave function is the same as that of de Broglie k. But why this is not appropriate for the photon. Maybe because photon does not have rest energy?

    What is connection between wave function and de Broglie waving is described in
    http://iopscience.iop.org/0031-9120/43/4/013
     
  2. jcsd
  3. Dec 18, 2012 #2

    vanhees71

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    Well, that's a very subtle question. For the photon the facts are simple to state: As a massless particle with spin 1 there doesn't exist a position operator. That's nicely summarized in Arnold Neumaiers theoretical-physics FAQ

    http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

    For massive particles you can always switch to a non-relativistic description, valid for "slow" particles, which for bound states also includes the assumption of binding energies small compared to the masses of the involved particles.

    In this case often a physically sensible description in terms of single-particle wave functions is possible. This also holds true considering relativistic corrections in such cases, e.g., using the Dirac equation to describe an electron moving in a fixed Coulomb potential of a heavy nucleus.

    Generally, in the fully relativistic realm a single-particle interpretation in terms of wave functions as is possible in non-relativistic Schrödinger wave mechanics is not possible. It even leads to contradictions with causality for free particles. That's why Dirac has been forced to his "hole-theory interpretation", leading him (after some quibbles) to the prediction of the existence of antiparticles (particularly the positron as the electron's antiparticle). However, hole theory, i.e., the interpretation of the vacuum states as the one, where the single-free-particle states with negative frequency are occupied, and the holes in this "Dirac see" as antiparticles is in fact a many-body reinterpretation, which is better stated from the very beginning as quantum-field theory, which describes situations, where the particle number needs not be conserved. For interacting particles it's even highly non-trivial to define an observable, describing a "particle number" at all.

    That's why a good lecture on relativistic quantum theory is taught as quantum field theory right from the beginning. In my opinion, it's even a much better didactical approach in non-relativistic quantum theory, i.e., in the introductory lectures, not to use the historical way and introduce quantum theory as "wave mechanics". One has to do so to a certain extent to heuristically justify the (in the first encounter) very abstract formulation in terms of states (Hilbert-space rays/statistical operators) and the representation of observables with self-adjoint operators, but this is the best way to present quantum theory as early as possible. Also the predominance of solving bound-state problems (often in terms of wave mechanics) in the introductory QT lectures can mislead students to think about these as the only physically meaningful solutions, thereby missing the more general formulation of dynamics.

    All this is of course not possible for the treatment of modern physics in highschools. Here one must stay qualitative to a large extent, and the cited article has to be taken with a very large grain of salt. It's for sure not correct, but it's also not really wrong. I admit, it's a dilemma, how to teach quantum mechanics at the high-school level, because most mathematical tools necessary to solve the Schrödinger equation in these bound-state problems are not available. However, one should be careful not to teach wrong or "semi-right" things.

    The worst case has been my own high-school experience (Germany, Abitur 1990), where for the largest time we learned about the Bohr model of the hydrogen atom with it's electron orbits around the nucleus, which is simply plain wrong and has to be corrected afterwards. This time would have been much better spent, introducing modern quantum theory right away, and be it only in a qualitative (but correct!) way. Fortunately we have had a good teacher, who gave also an introduction to the Schrödinger equation afterwards and telling us, what's wrong with "old quantum theory", but many high-school students don't have good teachers and then they only get this Bohr-orbit stuff. At the university you then have a hard time to forget again these wron pictures!
     
  4. Dec 18, 2012 #3
    Do you have a link to this article where we don't have to pay $33?:redface:
     
  5. Dec 18, 2012 #4

    DrDu

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    I am not quite sure about the meaning of the statement either:
    Even accepting that there is no position operator for photons, I do not quite see why this precludes writing down a wavefunction although maybe only in momentum space.
    In QFT for a free photon field you have a Fock space with a vacuum vector |0> and hence also all kind of n-particle vectors which can be obtained by acting with various ##a_{k\sigma}^+## on it. I don't see any reason why not to call these vectors "momentum space wavefunctions".
     
    Last edited: Dec 18, 2012
  6. Dec 18, 2012 #5
    there is an argument by dirac about integral spin particles which states that momentum representation is sufficient for integral spin particles such as photon and there is no coordinate representation of these particles.
    He was talking about dirac eqn or I should say his own eqn.
     
  7. Dec 18, 2012 #6

    Demystifier

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    You are right, it is fully justified to call it momentum-space wave function. You can also take the Fourier transform of it and get something like a position-space "wave function", but physically it has a somewhat different interpretation than wave function in non-relativistic QM.
     
  8. Dec 18, 2012 #7
    What do you mean by the word wavefunction? I guess you can always go to the classical limit, where you have classical fields for photons.
     
  9. Dec 19, 2012 #8

    vanhees71

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    Usually the wave function of a particle is the position representation of a pure quantum state. Of course for particles with non-vanishing spin this wave function has also a spin (or polarization) index. The physical interpretation of such wave functions is clear in the non-relativistic case via Born's rule: The modules squared of the wave function is the probability density to find a particle at the point given by the spatial coordinates with the spin given by the spin index (usually meaning that the spin-z component is [itex]\sigma[/itex] with [itex]\sigma \in \{-s,-s+1,\ldots,s \}[/itex].

    This interpretation fails in the relativistic case. For photons, i.e., massless particles with spin 1, you cannot even define such a wave function formally since there is no position operator for photons. Thus the notion of a wave function for photons is not even defineable!
     
  10. Dec 19, 2012 #9

    DrDu

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    Yes, but this argument does not exclude the possibility to define a wavefunction in momentum representation?
     
  11. Dec 20, 2012 #10
    that is what is necessarily dirac's argument I have provided.
     
  12. Dec 27, 2012 #11
    Photons can be detected at a specific location (for instance on a CCD pixel or rods/cones of a retina). Is that detected position an eigenvalue of some operator? That operator would be a position operator, but at the same time it is said that there is no position operator for photon and that position of a photon is not an observable.

    Do I understand something wrong?
     
  13. Dec 28, 2012 #12

    Demystifier

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    In modern literature, measurements which can be described in terms of collapse into an eigenstate of an operator are referred to as projective measurements. There are also more general measurements described in terms of positive operator valued measures (POVM), which are not projective measurements. Measurement of position of the photon is an example of such a more general measurement.

    See e.g.
    http://arxiv.org/abs/1007.0460
     
  14. Dec 29, 2012 #13

    DrDu

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    Seemingly not, as the photon is destroyed in the process of measurement.
     
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