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In principle, we could describe all physics without EM fields (or photons), as they are mainly a useful tool to describe "action at distance" (which does not mean instantaneous) between charged particle. In some sense, I could always integrate out photons, and describe only electrons and not have any changes in the observations (as detecting photons is done by observing how the motion of charged particles changes). With this picture, all photons are "virtual" (in the QFT sense), and we could expect them to always be off-shell.

My question is : why do we expect that photons that are "really emitted" (in a sense that should be made clearer, but that might mean "travelling very far before being absorbed") are always on-shell (i.e. having $E=p$)? Is it because the propagator of an off-shell photon decays very fast and therefore these photons can not interact with long distance charges ?

We could imagine that two very distant electrons (say, in two different galaxies) "scatter" each other (what we usually call "seeing a distant star") with off-shell photons. Why is it not so ? Is it just because the probability of this event is very small ?

Qmechanic
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Adam
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2 Answers2

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A previous question has been signaled by Chris White, and I think that the answer of Arnold Neumaier is great. Now, let us add some hints relatively to your question.

In principle, we could describe all physics without EM fields (or photons), as they are mainly a useful tool to describe "action at distance" (which does not mean instantaneous) between charged particle. In some sense, I could always integrate out photons, and describe only electrons and not have any changes in the observations (as detecting photons is done by observing how the motion of charged particles changes). With this picture, all photons are "virtual" (in the QFT sense), and we could expect them to always be off-shell.

No, this is only true if there were only photonic internal lines, and only electrons external lines. So, in QED, we would practically restrict to tree diagrams. With loop diagrams , we would have to consider internal electrons lines ("virtual electrons"). So, your view should be "correct" only if you would consider a mixed of classical electron field and quantum photon field. But if you want to consider a unified theory of quantum fields (like QED), it is not correct.

Is it because the propagator of an off-shell photon decays very fast and therefore these photons can not interact with long distance charges ?

You cannot mix momentum space and position space. Choose position space. If we look at the propagator $D(x)$, it is a function of $x^2 = \vec x^2-x_0^2$, $D(x)=D(x^2)$. So, the propagator, or the amplitude, does not decrease automatically because the spatial distance $|\vec x|$ is increasing. It depends on $x^2$. Of course, if $x_0=0$, the propagator is decreasing with $|\vec x|$ (in $\frac{1}{|\vec x|^2}$). This is true, also, that one may calculate the interaction energy between $2$ (fixed, eternal) electrons ($j^0_i(x) = \delta(\vec x- \vec x_i$)), for instance, and it turns that this interaction energy is in $\frac{1}{|\vec x|}$ (see Zee, Quantum field in a nutshell, Chapter I.4)

We could imagine that two very distant electrons (say, in two different galaxies) "scatter" each other (what we usually call "seeing a distant star") with off-shell photons. Why is it not so ?

But it is the case.

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Trimok
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  • First, one comment : your paragraph "No,..." is not true. The action of the EM field is quadratic, so I can integrate them out exactly. Then, I never have to talk about photons at all. You use a perturbative picture, where one treats electrons and photons on the same footing (thus photons and electrons loops). It is the most convenient way to do a calculation, but not the only one. – Adam Nov 01 '13 at 13:31
  • Second : because detecting photons is always seen as a scattering of electrons, why should the photons emitted by a distant star be on-shell, i.e. "real" photons ? If not the distance, then what ? – Adam Nov 01 '13 at 13:33
  • @Adam : "First" : OK, I agree, but you have a new formalism with a supplementary lagrangian term $\iint dx dy J^\mu(x) D_{\mu\nu}(x_y)J^\nu(y)$, and so, you have to define the new rules with this formalism – Trimok Nov 01 '13 at 17:07
  • @Adam : Second : Yes. On-shell is an abstraction. You may consider all photons in the universe as "virtual photons" exchanged by matter particles (This was the point of view of Feynmann). Heinsenberg inequalities give you the limit, if you are on shell, you will have : $\Delta \vec p = \vec 0$, but we know that this means $\Delta \vec x = \vec \infty$. So, there is no strictly " on -shell particle". Practically, I think that we have to compare $\Delta p^i$ and $p^i$.So, if $\Delta p^i$ is neglectible relatively to $p^i$ (with $p^2=0$), we may say that the particle is quasi "on-shell" – Trimok Nov 01 '13 at 17:25
  • I see. Is there a way to show in a more rigorous way that photons that are absorbed long after being emitted are always "quasi on-shell". Otherwise, why would be compute S-matrix elements with on-shell photons, if they can be off-shell anyway... – Adam Nov 01 '13 at 17:28
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    @Adam : For a Feynman diagram With no external (on-shell) photons, the transition amplitudes are calculated using only propagators/Green-functions for photons (and electrons). And these quantities (propagators/Green-functions) do not represent on-shell particles. You may see these as field perturbations or better field correlations. – Trimok Nov 01 '13 at 17:39
  • Yes, sure, I know that. What I meant was : When we compute a scattering amplitude for a photon (say, $e+\gamma\to e+\gamma$), we assume that the emitted photon is on-shell. But in the end, this photon will be absorbed (to be detected), so in principle it should not have to be on-shell. So why are we assuming it to be on-shell ? – Adam Nov 01 '13 at 19:09
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    @Adam : For each Feynman diagram, the external particles are assumed to be on-shell. In fact, the whole story is transitions amplitudes between the vacuum and the ... vacuum. so, it is a gigantic graph, and, a Feynman diagram is only a little piece of this gigantic graph, when we assume that external particles are on-shell, and we compare with experiment. – Trimok Nov 01 '13 at 19:26
  • Yes. I know. But we're getting to my point ;-)

    Why should we assume that the photons are on shell, if in the end they will be, one day, detected (i.e. absorbed) by an electron. Your answer is : "they are on shell because we put them on shell", but my question is "why should we do that, since they are in fact absorbed at some point, so we can keep them off shell"

     – Adam Nov 01 '13 at 19:54
  • @Adam : In a experiment, we are not interested in the correlations between the source of the incoming particle and the "sink" of these incoming particles, and idem for the outcoming particles, so we did not take in account these propagators (this is the LSZ reduction formula). We suppose that these real or quasi-real particles exist, and we measure the result of their interaction. – Trimok Nov 01 '13 at 20:05
  • I can cast my question otherwise : why does it work ? Why is it that only on-shell photons are important ? Why do a star only emit on-shell photons ? – Adam Nov 01 '13 at 20:12
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Our current mathematical models of everything depend on fields and interactions between them. These models have been verified for the microcosm, i.e. where the values of h_bar are, relative with the values of the variables, significant. In the quantum mechanical regime.

Since our view of fields is such that everything can interact with everything as long as there exists a coupling constant and only the probabilities of interaction play a role, you could say that everything after the bing bang is virtual, off mass shell.

In truth though it is the value of h_bar that defines what is measurable as probable or not. In the dimensions of the world we live in it has little meaning to postulate mathematical models where the events they predict with have a very small and unmeasurable probability, or very small and unmeasurable value, as in the measurement of mass.

More so as we have in these limits the classical physics models which also work beautifully outside the microcosm dimensions.

And in the end extending the mathematical models to unmeasurable unverifiable variable regions can tell us nothing new, it just adds complexity.

anna v
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