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QED, the exchange of photons

  1. Nov 29, 2013 #1
    I might be asking a question who's answer is way beyond my level but ill give it a shot. In QED, an electric or magnetic field is described in terms of exchanged photons, right? But if a photon itself is made of electric and magnetic fields...... a paradox to my ill-informed brain. :mad:
     
  2. jcsd
  3. Nov 29, 2013 #2

    ShayanJ

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    You're just confusing the things you have read!
    Electromagnetic field is somehow a way of saying how many photons you can find on average in a region of space.
    And electromagnetic force between two charged particles is described as a process of photon exchange.
    Although I think there should be more to it which I don't know,so I hope people more familiar with QED will answer too.
     
  4. Nov 30, 2013 #3
    Particles are vibrations in quantum fields. The fundamental "stuff" is the quantum field.

    Imagine a vacuum, a complete empty space, maybe far out in space somewhere. This empty space is still filled with quantum fields. That is a quantum field for photons, one for the electrons, one for the quarks, one for the Higgs particles, one for each fundamental particle.

    In vacuum all these quantum fields are not excited, but due their quantum nature are not completly still either, they vibrate very little by themselves. But the strength of E-M field in vacuum on average is zero.

    When you interact with a quantum field, e.g. move two electric charged bodies in the vacuum, you excite a quantum field above its vacuum state. Depending on how long-lived these excitations/ vibrations are, you call them particles.

    So always, when you through interaction excite/ vibrate a quantum field, in vacuum or not, and these excitations exist long enough in space and time to be measured, you have particles.
     
    Last edited: Nov 30, 2013
  5. Dec 1, 2013 #4
    Hmm, that's very interesting. Is there any Wikipedia page I could read that would tell me more about the quantum field? Is that QED?

    So is the classical view of an electromagnetic wave incorrect? If a photon is modeled as a point particle in the standard model, then it isn't composed of anything, it's just a fluctuation in the quantum field. On the other hand, classical physics says it is composed of oscillating magnetic and electric fields. But magnetic and electric fields are the exchange of photons. So is the classical view of a photon, the oscillating electromagnetic fields, only correct if you look at "wave" of light? (much more than just one photon)
     
  6. Dec 1, 2013 #5
    You can read more about QED here.

    To describe electromagnetic field QED is using the electromagnetic four-potential Aμ. The classical electric and magnetic fields can be calculated from the four-potential Aμ: see for example here (page 332 onwards).

    Photon is an excitation of the quantized field. After the quantization the electromagnetic field is no longer classical and things get tricky: the field turns into operator which means it does not have a definite value at each point in space, but is rather in a quantum superposition of many configurations.

    I am not sure what you mean by "classical view of a photon", because photons are quantum objects and do not make sense classically.
     
  7. Dec 2, 2013 #6
    Maybe this is a better way to pose my question. Hertz discovered the electromagnetic wave with a loop of conductive material and a tiny gap in which there would be spark from the induced current by the EM wave passing through the loop. If this experiment was done, but the EM wave consisted of just one single photon, would it still have the same effects, inducing a current in the conductor? (Although very very very small). Or is it the interaction of many photons that creates the phenomena of oscillating E&M fields that are both perpendicular to the direction traveled (this is what I mean by "classical view")?
     
  8. Dec 2, 2013 #7

    Avodyne

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    Yes, a single photon would produce a current. In fact, in leading order in time-dependent perturbation theory, emission or absorption of a single photon is equivalent to a classical EM wave of a certain amplitude (which depends on the initial and final states of the stuff emitting/absorbing the photon). It is for this reason that classical EM can be a useful approximation even in situations involving just one photon.
     
  9. Dec 2, 2013 #8

    ShayanJ

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    Of course doing Hertz's experiment with such a weak field won't have the same results that Hertz got.The point is,classical electromagnetism can be derived by getting averages of QED equations on a very large number of photons and on long enough time scales and and distances.
    The energy flux([itex]\frac{J}{s m^2} [/itex])of an EM wave is represented by its Poynting vector [itex] \vec{S}=\vec{E}\times\vec{H} [/itex].
    Now lets calculate the energy flux of a number of photons moving in a vacuum in the direction [itex] \hat{k} [/itex].Their velocity is c and lets call their number density n and if we consider a monochromatic wave with frequency [itex] \omega [/itex],then the energy flux becomes [itex] \vec{S}=n\hbar\omega c \hat{k} [/itex]
    So we have: [itex] \vec{E}\times\vec{H}=n\hbar\omega c \hat{k} [/itex].
    Let's simplify the matter by assuming that [itex] \vec{E} [/itex] and [itex] \vec{H}[/itex] are perpendicular to each other so,using [itex] H=\frac{E}{\mu_0 c} [/itex] we may write [itex] E=c\sqrt{n\hbar\omega \mu_0} [/itex]
    I don't know,maybe some people don't like it,and it may seem not so useful,but I think it can help for getting some insight.
     
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