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Virtual particles

  1. May 4, 2006 #1
    According to what I have read, virtual particles are those which are intermediate photons in interactions, which cannot be directly observed. If my understanding is correct that would mean that they consist of (in low order perturbation theory), (among other things) u and t channel photons. But electron electron scattering processes are described by such diagrams. That leads me to believe that all electrostatic coulomb force is mediated by virtual photons. Am i therefore to conclude that coulomb force over observable distance is mediated by photons of sufficiently low frequency so as to be unobservabe?

    Help would be super.

    Regards,
    Gordon.
     
  2. jcsd
  3. May 4, 2006 #2
    A virtual photon is an internal photon line, i.e. a propogator.

    One measures the presence of a Coloumb potential by the results of the interaction between the source and some charged body. The Coloumb force is the result of a virtual interaction. Two positively charged particles repel by exchanging a virtual photon. I can't trick the system and jump between the two of them and exclaim "hoozah there's a real photon there, look I just detected it", because you'd have just detected it by interacting with the source, assuming you were charged and if you weren't you wouldn't interact with it (the coupling between you and the EM force would be zero, you'd just be a free field).

    Non-virtual photons are external photon lines in Feynman diagrams, and are just radiated or absorbed photons.

    I'm probably not reading your post right, but are you saying that virtual photons only occur at low orders in perturbation theory? If so, contributions from virtual interactions are less prevalent in the low orders of perturbation theory than in the higher orders.
     
    Last edited: May 4, 2006
  4. May 4, 2006 #3

    Physics Monkey

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    Hi freeman,

    I just wanted to add to Perturbation's good answer and toss in some other interesting tidbits. It is important to realize that while one often draws a distinction between real and virtual photons, even so called real photons are detected by being absorbed by an atom or a crystal or whatever. That means even real photons are internal lines in some larger system that includes the atoms or whatever is doing the detection. So you can think of real photons as photons that are close to being on the mass shell, but you know that even real photons can't actually have definite energy. If they did have definite energy then they would be an energy eigenstate and hence could never disappear.
     
    Last edited: May 4, 2006
  5. May 5, 2006 #4
    Hello,

    Thanks for the help.

    Just to clarify, what you have said is that all photons, are intermediate states between charged particles and therefore the distinction between real and virtual particles is based on their proximity to the mass shell. All photons real and virtual can be detected by charged objects.

    On the basis of this it seems like a very ad hoc distinction to me. Is it really necessary, or is it just there to explain away the temporary violation of conservation of energy that is often given as a justification in more hand wavy explanations?

    Many thanks,
    Gordon.
     
  6. May 5, 2006 #5
    I didn't mean to say that, it's just that I'm only thinking in terms of lowest order diagrams, which I would hope are the leading order in behaviour anyway.

    Regards,
    Gordon.
     
  7. May 5, 2006 #6

    nrqed

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    Yes, roughly speaking you may think of the photons exchanged in th eu and t channels, at lowest order of perturbation theory (I say roughly because virtual photons are just a gimmick to visualize things...one should reallythink of excitations of the field and a nonperturbative treatment would not involve an expansion in exchanges of virtual photons. So I see them more as a trick related to PT than something physical).

    In any case, you should not think of virtual photons as being photons of small "undetectable" frequencies...There is no such interpretation. A virtual photon may have any frequency. The point is rather that virtual photons are off-shell, i.e. they don't obey the usual dispersion relation associated with real photons (in other words, the square of their four-momentum is not zero, as one expects for a massless particle).
     
  8. May 5, 2006 #7
    Thanks.

    I don't really understand the on/off-shell business. I understand what it means,

    [tex]p^{\mu}p_{\mu}=m^2[/tex]

    but where does that condition come from, and why doesn't it apply to intermediate particles.

    Many thanks,
    Gordon.
     
  9. May 5, 2006 #8
    A physical particle is said to be on mass-shell when the norm of its four momentum is that of the square of its observable mass. Off or near mass-shell is when [itex]p^2[/itex] doesn't equal or is almost the square of the rest mass.

    Propogators do not generally correspond to physical particles, that's usually the job of external lines, they are merely the transition amplitudes between fixed field configurations, the fixed config's being the end points in space-time, this propogation is not confined merely to those configurations with [itex]p^2=m^2[/itex]. The form of a two point function for a vector gauge boson is of the form

    [tex]\int\frac{dq^4}{(2\pi )^4}\frac{i\delta^{ab}g_{\mu\nu}}{q^2}[/tex] (Feynman gauge)

    (a and b label the gauge group indices and [itex]\mu,\nu[/itex] the space-time indices.)

    Where q is the four-momentum of the gauge boson. Obviously if the norm of q were identically zero this amplitude would be rather poorly defined and totally useless. If you work out propogation amplitudes with just on-shell particles one gets violations of causality, which is one of the reasons physics made the progression from relativistic quantum mechanics to quantum field theory.

    Similarly with fermions

    [tex]\int\frac{dq^4}{(2\pi )^4}\frac{i}{\gamma^{\mu}q_{\mu}-m}[/tex]

    This whole idea of being "on/near mass-shell" is important in LSZ scattering theory, where isolated poles or branch cuts in amplitudes correspond to physical particle states.

    Thanks for adding that bit, Physics Monkey, I was going to put it, but I thought I'd leave it implicit, don't know why.
     
    Last edited: May 6, 2006
  10. May 5, 2006 #9

    nrqed

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    I know that this is true but it's one of those things that I keep hearing about and never see explicitly worked out. Do you have a reference (online or not) where this is explicitly worked out? I would find this instructive.
     
  11. May 6, 2006 #10
    Thanks for your help all.

    I suppose my problem now boils down to this. In Feynman diagrams we have internal lines and external lines. We call external lines real particles with on shell momentum, and internal lines virtual particles with no such on-shell constraint. Now if we have a one vertex diagram in QED, we have one external photon. This has definite on-shell momentum. However, if we were to detect this particle we could only do so by observing its scattering from a charged particle, in which case the whole process can be described by a two vertex diagram that is essentially electron electron scattering. Now the photon is an intermediate line. What happens now? What is going on here? I think there is something fundamental I am missing.

    Regards,
    Gordon.
     
  12. May 6, 2006 #11

    nrqed

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    This is exactly the issue that PhysicsMonkey was discussing in post number 3.
     
  13. May 6, 2006 #12

    selfAdjoint

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    I think that the Feynmann diagram in this case would have the real photon coming in as one external line and the outgoing scattered photon as another, with similar lines for the charged particle. The interaction is represented by an undirected line (frequently shown as wavy) indicating mediation by virtual states. Of course these lines, and the two vertices, are used to construct the integral for the amplitude.The essential thing for an external line in the momentum representation is that there be a real momentum, and this the photon of the example has, both before and after scattering. But I live for correction!:biggrin:
     
  14. May 6, 2006 #13
    Work out the usual propogation amplitude from quantum mechanics for propogation from one position eigenstate to another in a time t, i.e.

    [tex]\langle \vec{x}|e^{-iHt}|\vec{x}_0\rangle[/tex]

    With [itex]H=E=\sqrt{|\vec{p}|^2+m^2}[/itex] and with [itex]\vec{x}-\vec{x}_0\neq 0[/itex], the above amplitude should vanish, otherwise a quantum particle could traverse a space-like distance in an arbitrary time t, but it doesn't, though it is quite small. QFT resolves this problem by making matter an operator valued distribution, and now the problem of causality becomes getting two operators at space-like separation to commute, which they do with the introduction of antiparticles.

    If you want me to show you a quick run down of showing it, give me a shout.

    Incidently, does anyone know if there's a latex string for the Feynman slash notation?
     
    Last edited: May 7, 2006
  15. May 7, 2006 #14
    So from above statement, assume the matter operator is the deuteron [NP], then what is the mathematics to show how this would commute with introduction of the antiparticle, anti-helium-3 [P^N^P^], where P = matter proton, P^ antimatter proton, N = matter neutron, N^ = antimatter neutron ?
     
  16. May 7, 2006 #15

    nrqed

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    Yes I would like to see how you would present this. I would appreciate it. This is one of those physics things that I always took for granted in order to move on and do QFT but now it is bothering me. It is also realted to what one really means by [itex] \phi(x) [/itex] applied on the vacuum. One would expect that to check causality, one would create a particle at position x, destroy a particle at position y and verify that the overlap between the two states vanishes for spacelike intervals. But it does not (this expression is simply the propagator D(x,y)).

    At this point, people mumble something about ''well, this is not what we should really look at, we should really look at whether a measurement at x may affect a measurement at y and this is given by the *commutator* between the two fields, and *this* vanishes for spacelike intervals'' (that's the ''derivation'' of P&S). I have to admit that I don't follow this argument (and I wonder how an EPR situation would be treated in QFT! This is one annoying thing about QFT...It is more general than NR quantum mechanics and yet, almost none of the applications of NQRM is revisited with QFT. Normally when teaching a new technique, the logical thing to do is to show how it reproduces previous techniques in some limits...This is what everyine does in introducing SR and GR (relating the results to Newtonian and Galilean expressions) but this not done very much when introducing QM (i.e. showing the connection with classical physics) and not at all with QFT (showing how to reproduce NRQM).

    I'd like to see your approach.

    Bets regards

    Patrick
     
  17. May 7, 2006 #16
    In basic QM, operators commute at space-like separation. So any particle state shouldn't be able to affect another if the two are space-like separated, i.e. they commute. The two terms in the commutator can be interpreted as the propogation of two particles simultaneously, one particle from x to y and an antiparticle from y to x.

    [tex]\left[\phi (x), \phi (y)\right]=\langle 0|\left[\phi (x), \phi (y)\right]|0\rangle =\langle 0|\phi (x)\phi (y) |0\rangle-\langle 0|\phi (y)\phi (x) |0\rangle[/tex]

    The two amplitudes in the above then cancel if [itex](x-y)^2<0[/itex].

    Relativistic QM wise,

    [tex]\langle \vec{x}|e^{iHt}|\vec{x_0}\rangle[/tex]
    [tex]=\langle \vec{x}|e^{-it\sqrt{|\vec{p}|^2+m^2}}|\vec{x}_0\rangle[/tex]
    [tex]=\int\frac{d^3p}{(2\pi )^3}e^{-it\sqrt{|\vec{p}|^2+m^2}}e^{i\vec{p}\cdot(\vec{x}-\vec{x}_0)}[/tex]

    You can do stationary phase on this and then take [itex]x^2>>t^2[/itex], which is well outside the light-cone, and get a kind of order of magnitude guesstimate for the integrand. Or you can switch to spherical coordinates in momentum space and get

    [tex]\frac{1}{2\pi^2|\vec{x}-\vec{x}_0|}\int^{\infty}_0 p\sin (p|\vec{x}-\vec{x}_0|)e^{-it\sqrt{p^2+m^2}}dp[/tex]

    and use Bessel functions. However, I'm not very adept with Bessel functions. Being self taught I skipped over them as I've never found differential equations particularly interesting. I'll probably go over series solutions and what not later and have a crack at it.
     
    Last edited: May 8, 2006
  18. May 7, 2006 #17

    nrqed

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    you meant QM or QFT here??

    I think that I am a bit confused by the term ''affect''. How would that be defined, precisely? (for example, how would EPR fit in this terminology...)

    And what type of measurement are we talnking about here? Why must one introduce commutators to discuss measurements. what about the question ''if a particle is detected at this point over here at time t_0, what is the probability that it will be observed at that point over there at time t''...what would that calculation look like and why can't we focus on this?

    Thanks for your feedback
     
  19. May 7, 2006 #18
    QM has space-like operators commuting, but it would extend into QFT, since we're also bothered about operators here.

    What I mean is that some measurement made at some point x cannot interfere, convolute etc. a measurement made at some point y space-like separated from x. I.e. if I find that some measurement performed at x gives an observable A, then the same procedure performed at y to give an observable B is not affected at all by that made at x if the two are space-like separated. Hence, the operator giving observable A at x commutes with that giving B at y. A simple such attempt at a "measurement" would be of [itex]\phi[/itex]. Any proper observables will be constructed from [itex]\phi[/itex] and would then have to commute at space-like separation if the field does. Since causality is really about information, what we're actually bothered about is observation, not particle propogation.

    The question "if a particle is detected at this point over here at time t_0, what is the probability that it will be observed at that point over there at time t" is just particle propogation. You take states at x and y, compute their inner product and square the modulus. Observation and measurement is concerned with physical observables, i.e. operators and their eigenvalues.

    Or at least that's how I understand it. Maybe someone more qualified could better me.
     
    Last edited: May 7, 2006
  20. May 7, 2006 #19

    nrqed

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    It makes much sense and I do appreciate your point of view (I am not trying to be difficult, I am just using this thread to think aloud and get some feedback).

    I see what you are saying. Measuring something at x and at y in one order or the other should not make any difference for spacelike separations. I would like to understand better the connection between applying the field operator on the vacuum and the process of measurement. What is the connection between the two?

    And on ething that bothers me is why th approach is so different than in NRQM. There, we show that there is violation of SR by looking at the probability of observing a particle at (x,t_0) and then at (y,t_0). Why does one then change completely the approach in QFT...

    But the key point is probably: what does one really mean by the expectation value of phi(x) phi(y) between vacuum states....In words, what does this quantity represent?

    Thanks for your input!!
     
  21. May 7, 2006 #20

    nrqed

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    But if a particle can propagate faster than light then causality is surely violated. It seems to me that proving that matter cannot propagate faster than light should be easier than proving that no information can...And yet this issue seems completely avoided. Worse yet, if one sees the expectation value of two fields at two different spacetime points to mean what it seems to mean, particles *would* have a nonzero probability of propagating faster than light. So I don't know what the expectation value of two fields mean...
     
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