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EM wave function & photon state vector.

  1. May 19, 2012 #1
    According to this review: http://lanl.arxiv.org/pdf/quant-ph/0508202v1.pdf
    A classical EM plane wavefunction is a wavefunction(in Hilbert space) of a single photon with definite momentum(c.f section 1.4) , although a naive probabilistic interpretation is not applicable. However, what I've learnt in some other sources(e.g. Sakurai's Advanced QM, chap 2) is that, the classical EM field is obtained by taking the expectation value of the field operator.Then according to sakurai, the classical E or B field of a single photon state with definite momentum p is given by [tex]\langle p|\hat{E}(or\ \hat{B})|p\rangle[/tex], which is 0 in the whole space. This seems to contradict the first view, but both views make equally good sense to me by their own reasonings, so how do I reconcile them?
     
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  3. May 19, 2012 #2

    Jano L.

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    These can hardly be reconciled. They do not agree on what the word "photon" is supposed to mean. Sakurai uses the word in the standard sense: as a Fock state in which only one oscillator is excited to the first excited state. Birula does not define the term photon explicitly, but his " wave function " is very far away from describing this Fock state. In the standard theory, this " wave function " is much closer to what is known as a coherent state.

    Do not worry about this. There never was a consensus on what the word photon is supposed to mean. The word is not essential for the quantum theory. The most common use is that which Sakurai uses, but then it is not a point-like nor extended particle and surely is not something that the theory would be based on.
     
  4. May 20, 2012 #3
    But taking Sakurai's view has its own confusing issues: for Dirac field the ket and classical field(or Dirac wavefunction) is related by [itex]\psi(x)=\langle 0|\hat{\psi}(x)|p\rangle[/itex] not expectation values(c.f.sakurai chap 3-10; weinberg chap 14.1), and it seems strange to me to give different prescriptions of relating quantum states and classical field configurations.
    (I also posted the same question here: http://physics.stackexchange.com/qu...nction-photon-wavefunction#comment63847_28616, there's been some discussion and may be of help)
     
  5. May 20, 2012 #4

    Jano L.

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    If I understand you correctly, you wonder the definition of the electric field in QTR is

    [tex]
    \mathbf E:= \langle \psi |\hat{\mathbf E} | \psi\rangle,
    [/tex]

    but the Dirac wave function is recovered in QFT by

    [tex]
    \psi = \langle 0 |\hat \psi |\mathbf p\rangle.
    [/tex]


    I do not understand quantum theory of field, but from what I know I would say this difference is because the things on the left-hand sides are different things that are incomparable in every possible respect. The electric field is a classical field with real value and is measurable. The wave function is a quantum mathematical function, which is complex in general and not measurable. If it was defined by an expectation value of an hermitian operator, it would be real and could not be complex exponential.

    I think it is good to think of these things, particularly because it shows that the relations between the concepts of the various partial theories are not as uniform and systematic as a naive overview would suggest. Perhaps both above relations are unsatisfactory and can be modified.
     
  6. May 20, 2012 #5
    I guess my confusion comes from the following content of quite a few textbooks:
    (1)Relativistic wave equations are understood as field equations(EM,Dirac etc.)
    (2)c-number solutions of field equations are understood as classical fields(this is usually mentioned for EM field, but I presume this is also the case for Dirac since they are both field equations)
    (3)In addition, the relation between classical EM field and a quantum state is given by the expectation value, and textbooks never discuss the meaning of [itex]0|Eˆ(x)|p[/itex]; On the other hand, for Dirac field the c-number quantity [itex]0|ψˆ(x)|p[/itex] is often discussed but never the expectation value.
    I understand your argument on Hermitean operator and observable, but I just don't feel it's a strong enough reason for such a sharp distinction about EM and Dirac field.
     
  7. May 21, 2012 #6

    Jano L.

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    I think looking at this from the historical perspective might help. You have to drop the false belief that the Dirac wave function is a classical field. It was never introduced that way. It was meant to be a Schroedinger-like wave function, and it is a complex valued function.

    There is no such thing as Dirac field in classical electromagnetic theory. There is only electric and magnetic field and electric charge, which is not a wave function.

    So i think the confusion is most probably created by unfortunate terminology.

    What is called classical Dirac field, should be called electron wave function.

    Perhaps you struggle with this because you want to derive classical theory of electromagnetic field from the quantum theory of field. However, this was never accomplished. There are various fatal problems. See for example papers by Mario Valente, especially

    http://ehu.es/ojs/index.php/THEORIA/article/download/754/901.
     
  8. May 21, 2012 #7

    DrDu

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    Somehow the quantum-classical transition always involves consideration of coherent states.
    Think e.g. of a highly excited hydrogen atom (a Rydberg atom). In the classical limit, the electron should spiral around the nucleus, but you have to take coherent states with small uncertainty of both position and momentum to approximate this quantum mechanically.
     
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