Virtual Photons as Force Carriers

[kcodon]

Hi everyone,

I have some questions on the nature of virtual photons. Firstly, are they actually particles (in the sense that a normal photon is a particle), or are they just mathematical consequences...and if so, what is their nature?

Secondly I have some questions still unanswered from a post a little while ago on virtual photons...

If anyone could answer this question of Sneills I'd be greatly interested in the answer...

I have a question, if a photon is exchanged between the two electrons as in the diagram, why is it that the e- emitting the photon recoils only when it is close to another e-? Or, why is it that it emits a photon only when it come close to another e-? This doesn't make sense to me as according to this theory an e- should be continuously emitting photons regardless of how close it is to another e- and therefor be continuously loosing energy. Do we observe a decrease in energy for the e- which emitted the photon and a gain in the absorbing e-?

So either Sneills point about how does the electron know when to emit a virtual photon, or the electron is continuously emitting virtual photons, but yes, how does the electron "know" which virtual photons to recoil from, as if it were continually emitting them in every direction the net force would be zero. This indicates some other form of field that tells the virtual photons either where to point or the like, to describe the change in motion...again back to a field. Also how does one photon "know" the speed to be emitted so as to cause an equal change in momentum for both electrons? I don't like using the term "know" but there is little else for it.

And is it, or will it ever be possible to measure an EM force to the degree so as to see if it is quantised to the degree predicted by QM i.e. virtual photons? Millikan got some fairly accurate measurement going on, and that was a wee while ago...

Thanks,

Kcodon

 

[Meir Achuz]

Hi everyone,

I have some questions on the nature of virtual photons. Firstly, are they actually particles (in the sense that a normal photon is a particle), or are they just mathematical consequences...and if so, what is their nature?

Thanks,

Kcodon

Virtual photons are mathematical constructs of using perturbation theory in momentum space.

 

[Meir Achuz]

"I have a question, if a photon is exchanged between the two electrons as in the diagram, why is it that the e- emitting the photon recoils only when it is close to another e-?"
Since the exchange takes place in momentum space, "close to" has no menaing.
The 1/r of Coulomb's law comes from a Fourier transform from momentum space to coordinate space.

 

[kcodon]

Thanks Meir Achuz,

However Sneill didn't mean the closeness in that sense...moreso in that how does the electron know to emit the virtual photon only at that point in time, in that exact direction, to cause the electrons to repel. Sorry I am not familiar with the maths, but if it turns out that the virtual photons are simply mathematical creations so to speak, then would not any attempt to explain how they actually carry force be completely pointless and irrelevant? Does quantum mechanics actually state this is how non contact forces transfer the force? There just seems so many inconsistencies in thinking of virtual photons actually as real and carrying force, that it would make more sense just to think again in terms of an EM field...the virtual photon is just a mathematical quantization of this field.

Kcodon

 

[ZapperZ]

Your question really has more to do with the issue of QFT, rather than "virtual photon" itself. Note that such field-theoretic method is common in almost all of physics, even in the description of the semiconductors that you are using in your modern electronics.

Now, if all QFT does is describe the brief "exchange" of a virtual particle between two entities, then it would have been utterly useless and unimpressive. We already know how to calculate such "forces" using ordinary methods. What QFT does well is treat the perturbative, higher-order interactions that are either not present, or not apparent in our older techniques. What happens, let's say, that on the way from one entity to the other, this virtual particle actually produces 2 virtual particles on its way there, and then one of them interacts with the final intended entity? One can imagine several "higher-order" scenarios, and each one of them will produce a different "channel" and different coupling strengths. This is where QFT truly becomes a tour de force method in not only calculating the results, but also in "visualizing" the interactions.

This technique is highly necessary in condensed matter physics. One of the best text to illustrate its usefulness is Mattuck's "Guide to Feynman Diagram in Many-Body Problems". One can clearly see how such techniques can reduce the intractable one many-body problem into many one-body problem. I would even say that a number of phenomena, such as the Kondo effect, would have been difficult to understand without such field theoretic method (the Kondo effect is, I think, THE first example of an "asymptotic freedom" that was discovered, long before this was done for elementary particle/nuclear physics).

So is it real, or is it Memorex? Note that a "mathematical artifact" cannot have any influence on physical measurement. One certainly would not say that the higher order corrections to the electron gyromagnetic ratio obtained from QED is only a "mathematical construct" (whatever that means), especially when the experiment agrees so well with it. If one can, then Newton's Laws are "mathematical constructs" as well. One delves into this line of questioning at one's own risk, because after a while, one realizes that one doesn't have a proper way to define what is meant by "real" and "mathematical constructs", especially when combined with physical measurements and what have been verified.

Zz.

 

[blechman]

Let me take a stab at this:

(1) First, in regards to virtual photons being "real" - The way I always thought of it is that a virtual photon is an excitation of the electromagnetic field (therefore it is "real" in that sense) that has the "wrong" dispersion relation. Remember that the usual photon that you learn about before QFT comes along is described by the linear equation:

$$k=\omega/c$$

and this equation is now generalized to

$$k=f(\omega)$$

where f is some function (not even necessarily real-valued). The first thing to realize is that this change in the dispersion relation does not need QFT - it happens in classical E&M inside conductors, superconductors, crystals, etc. The idea behind QFT is that the vacuum is not the free space medium that we thought it was, so that photons can propagate with such a non-trivial dispersion relation. Virtual photons are a way of these "medium effects" to materialize in the context of perturbation theory. So in that sense, they are as "real" as the usual on-shell photon, although we cannot "see" them in the usual way (they won't register on our photo-diode) - but we can "see" them indirectly (through the Coulomb interaction, for example).

(2) About your question of which electron gets "kicked" by the other: that one is easier to answer. Remember that when you write down a Feynman diagram, you are doing a "sum over all paths" (a path integral) - so the answer is "Both!" - just like the particle going through the double-slit apparatus does not pass through slit A or slit B but it passes through "Both slits" - the two recoiling electrons are constantly exchanging photons back and forth, and they both recoil off of each other. When you actually write down the quantum amplitude for the scattering, you are implicitly summing over both scenarios where electron A recoiled off electron B's photon, and the other way around.

As far as knowing whether to attract or repel - that is simply because the electron photon vertex is signed, and thus the quantum amplitude knows whether the charges are like or different. As far as "how much momentum to transer?" That just follows from conservation of energy and momentum (remember that the photon has these things too!)

As far as "when do they know to exchange photons?" Again, this is quantum mechanics, so the answer is: they are always exchanging photons with a certain probability. So both electrons are constantly emitting (and absorbing) photons, thanks to Heisenberg's uncertainty principle. As they get "closer together" the probability that they exchange photons begins to mount, until (assuming they haven't scattered already) the probability effectively saturates at 1 and a photon-exchange is more-or-less inevitable.

***ASIDE***

It's good that you pointed this out, because this is the resolution to the IR divergence problem of QFT. One way to state this problem is that you compute that the probability of an electron to NOT emit a photon through bremsstrahlung as it loses energy goes to (MINUS) infinity - this is the famous "Sudakov double log". However, this paradox is resolved because as the probability for you NOT to brem off a photon **de**creases, you are more and more likely to brem off a photon! Once you do, then you start all over again. Mathematically, this means that the probability of emitting more and more photons sums up to an exponential, which is always between 0 and 1, and there is no paradox!

Of course, there are a lot of subtleties that I am not getting into (QFT loops, etc) but this is the idea. And it provides an answer to your question of "when does the virtual photon get exchanged?" Just remember the "Q" is QFT - it's all about probability.

 

[kcodon]

Thanks blechman and ZapperZ,

Unfortunately I know nothing of QFT, so I'm somewhat in the dark here, and I fear I won't appreciate your points until I understand QFT more so. If you could recommend a low level introduction to QFT, that would be greatly appreciated...maybe a book or online resource.

Actually a point you raised made me realize something...even if are constantly emitting virtual photons, the net force on the charge would be zero...it is the virtual photons from other charges that cause the force. Thats embarrassing = )

Anyway sorry then for the bother, though I hope it clears up some problems for others with a bit more understanding than I,

Kcodon

 

[blechman]

Hi kcodon.
I'm sorry if you found my response a little technical, and re-reading that ASIDE, I probably should have kept my mouth shut about much more advanced things like that. I guess when I get going, I can't stop!

Anyway, there are several great books for beginners. Griffiths has a great intro aimed at undergrads about particle physics; I hear a new edition of the text is coming out next year. Tony Zee has a book, "QFT in a Nutshell" that does a really good job explaining things to the beginners (and also to more advanced people, for that matter). I would recommend you start there. Another slightly older book with more emphasis on particle physics is Mandl+Shaw, called "The Plumber's Guide to QFT" by my old experimental particle physics prof (that's not the title - just a nickname!). When the time comes, the more advanced and quantitative texts by Peskin+Schroeder or Itzykson+Zuber are a must.

None of these books will mean anything to you if you do not have a strong grounding in quantum mechanics (Sakurai or Shankar) and E&M (Jackson, or at least Griffiths in its entirety).

Have fun!

 

[Count Iblis]

As Meir pointed out, the virtual particles are just a name for certain lines you draw in Feynman diagrams. You want to compute some function f(g) where g is a coupling constant. You do a perturbative expansion f(g) = a + b g + c g^2 + ... and the Feynman rules tell you how to obtain the terms in this series.

Now, besides the obvious objections of calling virtual particles "real", the fact that the perturbative expansion does not converge is another objection. When you measure f(g), you cannot say that it is built up of the individual terms and that these terms have a well defined physical meaning if the entire sum diverges. The late terms in the series will become largere and larger and, as 't Hooft has shown, the series is not Borel resummable.

In some cases it is possible to extract non-perturbative effects, i.e. effects that are zero to all orders in perturbation theory that e.g. behave as Exp(-1/g^2). Such terms are invisible in perturbation theory. So, they cannot possibly be interpreted in terms of virtual particles.

 

[kcodon]

Hi kcodon.
I'm sorry if you found my response a little technical, and re-reading that ASIDE, I probably should have kept my mouth shut about much more advanced things like that. I guess when I get going, I can't stop!

Don't worry about it...I'm sure I'd get into it to if I knew how, and it was probably a bit silly to post something like this without the knowledge to back it. So thanks too Count Iblis, however totally over my head

Thanks also blechman for those ideas on books to read...I look forward to reading them...however maybe some more QM books and University first!

Thanks again,

Kcodon

 

[pellman]

kcdodon, you don't have to know much about QFT to understand some important concepts here.

First, EM forces are mediated by photons, period. Not necessarily "virtual" photons.

Photons are simply quanta of the electromagnetic field. You mustn't think of them as point particles. The EM field is build up of waves and quantum theory (and 100 years of experimental data) tells us that the smallest unit of energy by which waves of frequency $$\nu$$ may be built is $$\hbar\nu$$, which we call a photon of frequency $$\nu$$ .

So consider two charged pith balls brought close together. You release them at rest and then they accelerate under the Coulomb force. So now they have more energy and the EM field has less energy. That is, they absorbed some photons. "Real" photons.

"Virtual" photons come in when we start calculating certain processes at the quantum level. In calculating the transition from an initial to a final state of a system there are certain terms which must be included representing the emission/absorption of photons which individually would violate conservation of energy. So we call them virtual. Whether this is merely math or that conservation laws are actually broken in a unobservable fashion is something of a metaphysical question.

It is similar to quantum tunneling, if you are familiar with that. A particle is on one side of some small region within which the potential is higher than the particle's energy. So the particle won't be observed to be in that region. But there is a non-zero probability of finding it on the other side of the region. It had to "virtually" pass through that region--"virtually" breaking conservation of energy--in the same way that an EM interaction involves "virtual" photons.

Todd

 

[blechman]

So consider two charged pith balls brought close together. You release them at rest and then they accelerate under the Coulomb force. So now they have more energy and the EM field has less energy. That is, they absorbed some photons. "Real" photons.

Except what you find is that the "'Real' photons" you are referring to are virtual photons! They have spacelike 4-momentum. In NRQED, they are known as "potential photons" since they generate the Coulomb potential.

It is similar to quantum tunneling, if you are familiar with that. A particle is on one side of some small region within which the potential is higher than the particle's energy. So the particle won't be observed to be in that region. But there is a non-zero probability of finding it on the other side of the region. It had to "virtually" pass through that region--"virtually" breaking conservation of energy--in the same way that an EM interaction involves "virtual" photons.

Tunneling: that's actually a much better way of saying what I was trying to say before. Cheers!

 

[kcodon]

Hmmm I think I see what you are saying Todd...about virtual photons that is. Again I don't have the knowledge to understand "calculating certain processes", and maybe this would help, however I think I have at least a clearer understanding now than I did before

Yes, I also think my question was in the metaphysical sense, which unfortunately makes any chance of a clear, unrefuted answer go down the gurgler doesn't it = )

Thanks a heap though,

Kcodon

 

[pellman]

Except what you find is that the "'Real' photons" you are referring to are virtual photons! They have spacelike 4-momentum. In NRQED, they are known as "potential photons" since they generate the Coulomb potential.

That's interesting.

And it just goes to show how intuitively misleading the labels "real" and "virtual" are.

 

[country boy]

Except what you find is that the "'Real' photons" you are referring to are virtual photons! ...

Has anyone mentioned that all photons are virtual, in the sense that every photon is emitted and absorbed. It's the approximation of going to very long times between emission and absorption that permits you to treat photons as "real."

 

[lightarrow]

As far as knowing whether to attract or repel - that is simply because the electron photon vertex is signed, and thus the quantum amplitude knows whether the charges are like or different. As far as "how much momentum to transer?" That just follows from conservation of energy and momentum (remember that the photon has these things too!)

Sorry if I ask you only now:
It's not clear to me how virtual photons are exchanged between two oppositely charged particles (e.g. electron and positron); maybe every particle sends (more) virtual photons in the opposite direction with respect to the other particle, or what?

 

[pellman]

Has anyone mentioned that all photons are virtual, in the sense that every photon is emitted and absorbed. It's the approximation of going to very long times between emission and absorption that permits you to treat photons as "real."

This is true. But what makes a photon "virtual" is that its presence violates a conservation law.

 

[blechman]

This is true. But what makes a photon "virtual" is that its presence violates a conservation law.

No, that's not true. QED doesn't violate conservation laws. The thing that makes a photon "virtual" (or any other particle as well) is that it is not "on-shell"; or that it does not obey the dispersion relation $k=\omega/c$. But it still conserves energy and momentum.

lightarrow: I do not understand your question. What do you mean, "opposite direction"? Can you rephrase it?

 

[Count Iblis]

Causality is not violated in any physical process. But if you assign physical meaning to indivudual Feynman diagram, causality can be violated. E.g., in case of Compton scattering, there are two diagrams. In one diagram the electron absorbs the photon, becomes a virtual electron and then emits a photon. In the other diagram, when the photon is still on its way to the electron, the electron decides to emit a photon ahead of time, becoming a virtual electron, and then it absorbs the photon to become a real electron again. That's like me buying a new house because I know that I will win the lottery next week.

So, the sum over the two diagrams gives a physical meaningful result, while one of the diagrams cannot be interpreted as a physical process. This proves that virtual photons do not have a well defined physical meaning. It's just writing the effect you want to calculate as a sum of terms in a handy way.

 

[Meir Achuz]

the electron decides to emit a photon ahead of time,

But there is no time in a Feynman diagram. The variables are the four k^\mu.

 

[pellman]

No, that's not true. QED doesn't violate conservation laws. The thing that makes a photon "virtual" (or any other particle as well) is that it is not "on-shell"; or that it does not obey the dispersion relation $k=\omega/c$. But it still conserves energy and momentum.

Ok, dropping all pretense now that I understand this, what does it mean?

If energy is not, $$E^2 = p^2 +m^2$$, then what is it? How do you define it? That is, if energy is conserved and also $$E^2 \ne p^2 +m^2$$, then what is the "energy" that is conserved. I'm not arguing; I read some more and realize that what you are saying is standard. I just don't know what it means.

You might say, that well, the actual relation is $$(E-A^0(x))^2 = (p^k-A^k)^2 +m^2$$ in the presence of a potential, but this is true of macro particles as well, so that can't be the meaning of "virtual", can it?

Maybe I will just have to wait until I come to this part in the QFT book I'm reading.

 

[lightarrow]

lightarrow: I do not understand your question. What do you mean, "opposite direction"? Can you rephrase it?

In the case of two equally charged particles I can understand repulsion as given by the emission of a virtual photon from a particle1 towards particle2: because of momentum conservation, particle1 withdraw from particle2; the same with particle2.
What happens instead in the case of 2 oppositely charged particles, that is in the case of attraction?

 

[jtbell]

Perhaps this will help:

http://www.physics.adelaide.edu.au/~dkoks/Faq/Quantum/virtual_particles.html, from the Usenet Physics FAQ. Among other things, it discusses attractive forces.

 

[blechman]

several replies at once

So, the sum over the two diagrams gives a physical meaningful result, while one of the diagrams cannot be interpreted as a physical process. This proves that virtual photons do not have a well defined physical meaning. It's just writing the effect you want to calculate as a sum of terms in a handy way.

This is a little misleading, Count. Technically, I suppose you are not wrong, but remember that Feynman diagrams are not "what is REALLY happening" - they are shorthand notations for rather complicated integrals.

The way I prefer to think about it is that there is always a cloud of (virtual) photons surrounding an electron. Sometimes they make contact with other electrons (mediating the Coulomb force), sometimes they don't. That's all - no violation of causality here!

And as you say, a particular F.D. does not have any physical meaning - it violates causality, gauge invariance, etc. But all these violations are canceled by other diagrams, so the physical process is okay.

If energy is not, $$E^2 = p^2 +m^2$$, then what is it? How do you define it?

It's the "m" in your equation that I take issue with. Look: a field has energy E and momentum p. These quantities are components of a four-vector (q) that is conserved: that is, each component of the four-vector is conserved. Now if the particle is "on-shell", that means that the square of the four-momentum is a fixed constant that we call "mass":

$$q^2=E^2-p^2\equiv m^2$$

However, if the particle is "off-shell" or "virtual", then $q^2\neq m^2$. There's no problem with this. Everything is still conserved, but the virtual particle doesn't have a "mass" that is the usual mass. In fact, it could even have a "tachyonic" (imaginary) mass - this is what happens in a t-channel F.D. - prove it yourself if you don't believe me! Furthermore, the quantity $q^2$ is still a Lorentz scalar (again, easy to prove!) so it is meaningful to talk about that quantity as a "physical" quantity, although it's a little misleading to think of it as a "mass" - it's just a Lorentz-invariant quantity that describes the kinematics of the process.

In the case of two equally charged particles I can understand repulsion as given by the emission of a virtual photon from a particle1 towards particle2: because of momentum conservation, particle1 withdraw from particle2; the same with particle2.
What happens instead in the case of 2 oppositely charged particles, that is in the case of attraction?

Virtual photons are not ball-bearings! In particular, they can carry "negative energy" (see my comment above), so that's where the attraction channel comes from.

 

[blechman]

But there is no time in a Feynman diagram. The variables are the four k^\mu.

Wha?? Time is one axis of a F.D., I'm not sure what you mean.

 

[quantumfireball]

Virtual particles are decribed by propogators
where the propogator is <01 phi_-(x)*phi_+(x)10>
where phi(x) is the operator field associated with the particle

 

[reilly]

Meir Achuz has it right: virtual particles,( photons or electrons, or...) are artifacts of perturbation theory -- my words, not his. They provide a concept that helps, many of us at least, to think intuitively about QFT interactions and dynamics. In other words, they are simply metaphorical. Some insight into the subject can be obtained by looking at quantum transitions over a finite period of time.
Regards,
Reilly Atkinson

Electrons do not have a clue about when to emit or absorb a photon; the two process tend to happen (sorta')ndependently( as Feynman diagrams show. And all one can say is about probabilities of radiation emitted or absorbed -- that's a fundamental concept in QM. (Much the same thing can be said about classical radiation phenomena.)