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Does the field of a set of photons behave differently from a single photon.

e.g. Suppose I have a group of photons with their Electric fields $\mathbb{E}_n$ all aligned. So $$\mathbb{E}_{\text{tot}} = \mathbb{E}_1 + \mathbb{E}_2+ \text{...} + \mathbb{E}_n$$

and upon this field a single photon with parallel $\mathbb{E}_i$ field is incident. Does $\mathbb{E}_i$ scatter?

Or do I require such a large number of photons in $\mathbb{E}_{\text{tot}}$ that the field polarizes the underlying quantum vacuum to produce particle pair that scatter the $\mathbb{E}_i$ particle.

More simply put, can an electric field by itself scatter a photon?

Urb
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metzgeer
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    If you consider the QED lagrangian, there is no term involving the vector potential coupling to itself. That is, there is no vertex with two photons going in/coming out. – kηives Feb 08 '12 at 04:59
  • Related: http://physics.stackexchange.com/q/1819/2451 – Qmechanic Feb 08 '12 at 06:17
  • There is the possibility of 4th order photon-photon scattering (each photon is connected to an electron-positron pair), but of course you'd need very strong fields to have a chance of observing it. AFAIK it hasn't been seen yet. – twistor59 Feb 08 '12 at 08:05
  • @twistor59 Can you give a reference to this photon-photon scattering that examines the term by term expansion. Which I assume by knives comment is of the EM four vector? – metzgeer Feb 09 '12 at 02:15
  • I don't know of any book where anybody writes it out - it's just a straightforward application of the Feynman rules to the box diagram (top right) here. Because each vertex contributes a multiplicative alpha factor, the resultant amplitude is a bit on the low side! – twistor59 Feb 09 '12 at 07:49
  • A closely related question form a bit of time ago: http://physics.stackexchange.com/questions/1361 – Slaviks Mar 09 '12 at 11:02
  • Photons do not have an electric field. They have energy, momentum, spin. – my2cts Dec 21 '21 at 10:27

2 Answers2

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To be slightly pedantic, there isn't really such a thing as the field of a photon, if you think of photons as being things we can count.

The state $|p\rangle$ in the Hilbert space of the quantum Maxwell theory is a superposition of all possible states where the EM field has a definite value. This is because the field operator is a superposition of photon creation and annihilation operators. So if the EM field has a definite value, we aren't in an eigenstate of the photon number operator.

That said, there is such a thing as photon-photon scattering in particle physics. Two photons can scatter via intermediate virtual processes in the Standard Model. The effect isn't very strong though, so we don't see it at ordinary distance scales.

user1504
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  • Can you give a reference for this photon-photon scattering. I see this a lot, people referring to nth order scattering but I'm having trouble tracking down an actual detailed expansion. – metzgeer Feb 09 '12 at 02:16
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''More simply put, can an electric field by itself scatter a photon?''

Yes, though the dominant processes are higher order nonlinear effects and hence quite weak; still too weak to be detected experimentally. But this may change soon. Some representative papers:

However, there are experimental observations of virtual photon-photon interactions from electron-positron collisions producing hadrons; see, e.g., https://arxiv.org/abs/hep-ex/9906039.

Urb
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Arnold Neumaier
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    In the future please link to arXiv abstract pages rather than pdf files if possible, e.g., http://arxiv.org/abs/hep-ex/9906039 – Qmechanic Mar 18 '12 at 16:12
  • @Qmechanic: you may edit that if you like. (I usually choose the form that is quickest for me to write, as my time for discussion here is limited. So if I find the pdf before the abstract, I refer to that. Others can find the abstract in the same way as I'd have to find it.) – Arnold Neumaier Mar 18 '12 at 16:16
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    The link to the Springer paper seems dead now. – Urb Dec 21 '21 at 10:16