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Quantum Electrodynamics

  1. Oct 6, 2015 #1


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    What is the photon's equivalence of the electron's wavefunction? Can I measure qualities of the photon that will collapse into an eigenstate? What properties of the photon aren't fixed?
  2. jcsd
  3. Oct 6, 2015 #2
    The photon is inherently relativistic and only properly described by quantum field theory. With that said, the QM analogy to the single photon's wave function is the complex electromagnetic field, E + iB. Google "photon wave function" and you will get quite a few references. And to draw the QM analogies further, neither position, momentum, or polarization are necessarily fixed.
    Last edited: Oct 6, 2015
  4. Oct 6, 2015 #3


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    The photon follows Maxwell's equation to form a wave according to the wave formula with velocity c. But this is fundamentally different from an electron's wave function that obeys the Schrodinger equation (first partial derivative in time opposed to second...) and that describes the probalistic state of the electron.
    Where exactly is the uncertainty described in photons? How do I "measure" something and what "collapses" with respect to what "operator"?
  5. Oct 6, 2015 #4
    Essentially yes, so you should now go on to study quantum field theory, which is the proper theory for photons.
  6. Oct 7, 2015 #5
    I would say not. The photon is not an electromagnetic wave, even if it has some properties of the em wave.

  7. Oct 8, 2015 #6


    Staff: Mentor

    Photons can't be described by ordinary QM because position is not an observable for photons (that position is an observable is a basic assumption of QM). It can only be described by QFT where the concepts you are alluding to above are much more nebulous eg in QFT even particle number isn't certain. Before moving onto QFT it would be wise to be reasonably conversant with ordinary QM. However once you are some good books are starting to appear that will allow you to learn QFT eg

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