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Are Virtual Photons Responsible For Real Photons?

  1. May 16, 2005 #1
    1) What is the nature of the electric field that comes from a single charge? Do virtual particles create this field?

    2) What is the nature of the coulombic interaction between charges? (I know that like charges repel and opposite charges attract. But what is the nature of the repellence and the attraction?) If virtual particles are responsible for this field, how do they create this electric field?

    3) How many fields comprise the universe? What are those fields? Are virtual particles assumed to cause these fields? How do virtual particles create these fields?

    4) What came first, light or the photon? (Genesis of the Christian Bible tells us that light came first. What does quantum mechanics tell us?)

    5) Any thought as to how the first photon originated?

    6) How do DC and AC electric fields differ from electric fields that generate light?

    7) Do you know of any websites, books or journal articles that I can read to gain an in depth understanding of these concepts?
  2. jcsd
  3. May 16, 2005 #2
    Please introduce yourself to this fine forum

    Hi Kris,
    IMO - (Its My Opinion): that James C Maxwell and his Classical collegues discovered the Alternating Electric field several decades before the advent of quantum physics. Obviously, the question you ask was ask by Einstein and has not been adequately answered for a whole century. With QM, the two electrons of each atomic orbit move forever in the manner of a dog chasing its tail; I'd call that Direct Current. Of course I have difficulty understanding the virtual "standing wave" modeling of atomic orbits. Cheers, Jim

    P.S. Your profile is completely empty so that the fine people in this forum cannot respond usefully to you without some knowledge of whether you are a pre-teen or College Student or whatever. Don't you agree?
  4. May 16, 2005 #3


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    For #5,why don't u read Planck's original 14th Dec.1900-paper...?

  5. May 16, 2005 #4
    In the domain of classical electrostatics, the field is given by an inverse square law, and the theory has no regard for virtual photons, or even photons. My knowledge of QED/QFT is not good enough for me to give a clear mathematical explanation of how electric fields are related to virtual particles.

    The Coulombic interaction is given by
    which in combination with Newton's second law will describe motion in the classical regime. Again, virtual particles are an aspect of QED.

    This is more questions relating to QFT. I could answer some of these, but feel it's best left to someone else who could provide a better explanation.

    A photon is a quantum of the electromagnetic field; oscillations of the electromagnetic field are light. Anyway, in who's reference frame are you asking this question? :wink:

    I would imagine some interaction in the very first few moments of the universe (at least until after the electromagnetic force became "separated" from the others) released a photon of some kind.

    Electric fields do not generate light (well they may do, but in an indirect way) -- light is a self sustaining oscillation of electric and magnetic fields, something which you seem to have overlooked.

    If your mathematics is up to scratch (i.e. integral and differential calculus, geometry) then I would recommend W.J.Duffin, "Electricity & Magnetism" 4th ed.
  6. May 16, 2005 #5


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    The quantum field view on electrostatic fields is indeed given by virtual particles. This is a rather complicated issue in fact. Although the "electrostatic effect" in a collision can easily be handled to first order, true bound states due to electrostatic binding are a much more difficult (although not impossible) problem in QFT.

    Again, there's an easy, and a difficult answer to this question. The easy one is the electrostatic interaction observed during collisions. What you can then find out is that to first order, the exchange of a single virtual photon, the way it is described in QED, gives you exactly the same final state as the semiclassical scattering to a Coulomb potential. That's the easy answer.

    That's given by the model you take. The standard model of elementary particles has as many fields as it has particle types. I'd rather say that it are the fields that describe virtual particles. See, the question is a bit analogous to: "how do electrostatic forces create potentials" in a classical context.

    I think that Genesis is a particularly bad physics book :-)

    What came first: acceleration or force (in a Newtonian context) ?

    By a light-chicken ?

    Surprise: they are the same fields ! Only, light are AC fields of rather high frequency: namely around 400 THz. Power line AC is 50 Hz in Europe and 60 Hz in the US.

    Go easy. Learn things step by step. Learn first classical physics. Then learn quantum mechanics. Then learn quantum field theory. Don't do it the other way around.

  7. May 16, 2005 #6
    If i set for example in vacuum

    [tex]\vec{E}(x,t)=\vec{e}_z\int E_0(k)e^{ik(x-ct)}dk [/tex]
    [tex]\vec{B}(x,t)=-\vec{e}_y\int ik'E_0(k')e^{ik'(x-ct)}dk'[/tex]

    which satisfies (i don't know maybe i'm mistaken), Maxwell's equ. in vacuum....

    The question is :

    how many photons with which energy are there (as a function of the envelope of the wavepacket, the frequency, eventually of x ant t) ?

    Is that anyhow an allowed question, since this field does not seem to be quantized ?
  8. May 16, 2005 #7


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    With each classical field solution, there corresponds a quantum state of the (QED) electromagnetic field which is its "best description", namely the corresponding coherent state.
    A coherent state is an eigenstate of the annihilation operator of the corresponding mode, with as an eigenvalue the complex amplitude of the field.

    You know that corresponding to each plane, harmonic wave (+ polarization direction) corresponds a wave vector k and a polarization direction epsilon ; well, there corresponds also an annihilation (and a creation) operator a(k,epsilon) to it. This is the "photon of momentum k and polarization epsilon" annihilation operator.

    Well, to get the corresponding quantum state of a classical plane wave with wavevector k, polarization direction epsilon, and complex amplitude Ec (giving you the amplitude and the phase of the classical wave), you have to find the eigenstate of a(k,epsilon) with eigenvalue Ec (one can prove that such a state exists).

    So there is a state (in Fock space) psi, such that:

    a(k,epsilon) |psi> = Ec |psi> (1)

    That psi is then called a coherent state, and is the quantum description of the classical plane wave with given wave vector, polarisation, amplitude and phase.

    In fact, to be truely the description of a classical plane wave, one should also have that a(k',eps') |psi> = 0 (2).

    Note that a(k,epsilon) only acts on the subspace in Fock space of the k,epsilon photons. So there are in fact zillions of coherent states that correspond to equation (1): you can add as many OTHER photons (other k or epsilon) to the state psi as you want, a(k,epsilon) only acts on the part in the (k, epsilon) subspace. That's why we needed to eliminate all those others by equation (2).

    So if you have a general classical EM field with complex amplitude Ec(k,eps) as a function of k, for the two possible directions of eps, then one should generalize (1) and (2) to one common equation set:

    a(k,eps) |psi> = Ec(k,eps) |psi> for all k and eps (3).

    I am not quite sure, but I think that (3) specifies the quantum state |psi> uniquely in Fock space. This psi is then the coherent state corresponding to the classical field with complex amplitude Ec(k,eps).

    (I write this from memory, and I extrapolated a bit, I hope I'm not making a fundamental error here - feel free to correct me).

    It will turn out that |psi> contains ALL photon number states.

    Note that a "pure photon number state" is not a quantum state of the EM field that has a classical correspondence.

  9. May 16, 2005 #8
    I have a questionair.

    what is the difference from a virtual photon and a photon?

    -- "The dumbest question is the one you dont ask."
  10. May 16, 2005 #9
    The problem is that I don't know the answer so I cannot correct anyhow.

    The thing I saw was that the Fock space can be described in the "photon numer basis" [tex] |n_{k1}>....|n_{kn}>[/tex]...a photon number state [tex]|n_k>[/tex] is an eigenstate of the energy operator of the corresponding mode, and it's energy is [tex]\hbar\omega_k(n_k+.5)[/tex].

    So I thought we could just ask the following : The energy density of the classical field is : [tex]En(x,t)=\frac{1}{2}\left(\epsilon_0E(x,t)^2+\mu_0B(x,t)^2\right) [/tex].

    So that now one could just compute [tex]En(x,t)=<x|\bigotimes_{k=-\infty}^\infty|n_k>[/tex]

    Where the tensor product is run over continuous values....

    The problem is that then you have quantum mechanically a continuous infinity of coordinates : [tex]x_{k,n}[/tex] : the coordinate of [tex]n[/tex]th photon in mode k......Does this apply to the times [tex]t_{k,n}[/tex] too ?? I don't know, because

    [tex]<x_k|n_k>=N_{n_k}e^{-x_{k,n_k}^2/2}H_{n_k}(x_{k,n_k})e^{in_k kct_{k,n_k}} [/tex]

    So that [tex]<x_{k_1,n_k_1}...x_{k_n,n_k_n}|\bigotimes_{k=k_1}^{k_n}\sum_{n=1}^{n(k)}|n>=\int_{-\infty}^\infty dk\sum_{n=1}^{n(k)} N_{n}e^{-x_{k,n}^2/2}H_{n}(x_{k,n})e^{inkct_{k,n}}[/tex]

    Where [tex] n(k) : \mathbb{R}->\mathbb{N}[/tex] is the number of photons having momentum k.

    What do I do with the many space-time coordinates [tex]x_{k,n},t_{k,n}[/tex] in order to recover the classical space-time (x,t) ???

    So how could the manyparticle view hence with many coordinates of QM be reconciliate with the space-time dependent (x,t) only classical view....and are the times for every particle (photon in mode k) needed ??
  11. May 16, 2005 #10
    I don't remember about virtual photons very well, I think of 2 different things :

    1) Take a charged point particle moving, this create a EM field...discover (somehow) (for example Fourier transform) to which quantum state of the EM this corresponds (spectral decomposition)...those photons are the virtual photons corresponding to the charged particle.

    2) Take just a charge...then this create an electric field....if I remember well, you can modelize the electric field with scalarly polarized photons, those photons are virtual and are responsible for the Coulomb force.....

    I'm really not sure about what I say here those are just vague reminders of small passages read somewher.....I'll look in the litterature....
  12. May 16, 2005 #11


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    I'd say: a virtual photon is one that doesn't appear in the initial state, nor in the final state, but is used as an intermediate step when calculating the transition amplitude.

  13. May 22, 2005 #12
    A virtual photon doesn't obey the usual energy / momentum relations. For example, in electron electron scattering, the exchanged photon's momentum can be shown to be non-zero, and its energy to be zero. This violates the relativistic relation [itex]E^2=m^2+p^2[/itex] (in natural units, c = hbar = 1).

    See if you can show this - all you need is the formula given above, and the fact that photons are massless.

    Also remember that virtual particles can, loosly speaking, only violate these laws within timeframes given by the Heisenberg uncertainty principle.
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