HUP in QFT and QM:virtual particles

In summary, the philosophers discussed Whitehead's book "Process and Reality" in a funny way. Feynman didn't understand what they were saying, and eventually he was invited to their seminar.
  • #1
haushofer
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Hi,

I have a question about reconciling two pictures of virtual particles and the Heisenberg Uncertainty Principle (HUP).

In QFT "virtual particles" show up in perturbative calculations. We try to calculate an amplitude in interacting theories, this can not be done in an exact way, so we use Taylor expansions, and in this expansion intermediate states show up which we call "virtual particles".

In non-rel. QM people often say that virtual particles can exist because of the uncertainty principle between energy and time, in which one interprets the "time" in the appropriate way (see e.g. Griffiths).

My question is: how to reconcile these two pictures? If we would find the mathematical tools to calculate amplitudes in interaction theories in an exact way analytically, what would happen to these "virtual particles"? On the one hand I would say they wouldn't show up in your calculations, just as e.g. all the intermediate steps in

[tex]
\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \ldots = 2
[/tex]

wouldn't show up; we know the answer is "2". But on the other hand, if their existence can be argued by the uncertainty principle, their existence should not depend on our ability to solve function integrals analytically in an exact way, right?

Are my analogies bad, or are the textbook statements not that accurate, or something else?
 
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  • #2
I don't think it's accurate to say that virtual particles can exist thanks to HUP. When this is stated in a text, it is misleading, I would say.
 
  • #3
haushofer said:
In QFT "virtual particles" show up in perturbative calculations. We try to calculate an amplitude in interacting theories, this can not be done in an exact way, so we use Taylor expansions, and in this expansion intermediate states show up which we call "virtual particles".
Yes.
In non-rel. QM people often say that virtual particles can exist because of the uncertainty principle between energy and time, in which one interprets the "time" in the appropriate way (see e.g. Griffiths).
IMHO, such "explanations" are rubbish. (I always become suspicious of physics authors who are subtly dismissive about mathematical rigor.) Is it only in Griffiths where you've seen such arguments, or are you thinking of other common textbooks also?

My question is: how to reconcile these two pictures? If we would find the mathematical tools to calculate amplitudes in interaction theories in an exact way analytically, what would happen to these "virtual particles"? On the one hand I would say they wouldn't show up in your calculations, just as e.g. all the intermediate steps in
[tex]
\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \ldots = 2
[/tex]
wouldn't show up; we know the answer is "2". But on the other hand, if their existence can be argued by the uncertainty principle, their existence should not depend on our ability to solve function integrals analytically in an exact way, right?
Right. Virtual particles are not real. :-)
Are my analogies bad, or are the textbook statements not that accurate,
The latter.
 
  • #4
haushofer said:
In non-rel. QM people often say that virtual particles can exist because of the uncertainty principle between energy and time, in which one interprets the "time" in the appropriate way (see e.g. Griffiths).
as strangerep says: it's rubbish
1) it's s difficult to state the time-energy HUP in non-rel. QM in a general sense; it's even more unclear how to formulate it in QFT
2) virtual particles /as defined in perturbation expansion) are not Hilbert space states but "integrals of propagators"; so you can't define Δt and ΔE (or Δx and Δp) for these expressions in the usual sense
3) somethimes people claim that virtual particles "borrow energy" or "violate energy conservation for some short time Δt"; this is nonsense as well b/c in the Feynman diagrams energy and momentum are conserved exactly at each vertex; what is violated is the mass-shell condition m² = E² - p²
4) last but not least it's b...sh.. to discuss the "existence" of virtual particles


In the Graduate College dining room at Princeton everybody used to sit with his
own group. I [Feynman] sat with the physicists, but after a bit I thought: It would be nice to see what
the rest of the world is doing, so I'll sit for a week or two in each of the other groups.
When I sat with the philosophers I listened to them discuss very seriously a book
called Process and Reality by Whitehead. They were using words in a funny way, and I
couldn't quite understand what they were saying. Now I didn't want to interrupt them in
their own conversation and keep asking them to explain something, and on the few
occasions that I did, they'd try to explain it to me, but I still didn't get it. Finally they
invited me to come to their seminar.
They had a seminar that was like a class. It had been meeting once a week to
discuss a new chapter out of Process and Reality some guy would give a report on it
and then there would be a discussion. I went to this seminar promising myself to keep my
mouth shut, reminding myself that I didn't know anything about the subject, and I was
going there just to watch.
What happened there was typical so typical that it was unbelievable, but true.
First of all, I sat there without saying anything, which is almost unbelievable, but also
true. A student gave a report on the chapter to be studied that week. In it Whitehead kept
using the words "essential object" in a particular technical way that presumably he had
defined, but that I didn't understand.
After some discussion as to what "essential object" meant, the professor leading
the seminar said something meant to clarify things and drew something that looked like
lightning bolts on the blackboard. "Mr. Feynman," he said, "would you say an electron is
an 'essential object'?"
Well, now I was in trouble. I admitted that I hadn't read the book, so I had no idea
of what Whitehead meant by the phrase; I had only come to watch. "But," I said, "I'll try
to answer the professor's question if you will first answer a question from me,
so I can have a better idea of what 'essential object' means. Is a brick an essential object?"
What I had intended to do was to find out whether they thought theoretical
constructs were essential objects. The electron is a theory that we use; it is so useful in
understanding the way nature works that we can almost call it real. I wanted to make the
idea of a theory clear by analogy. In the case of the brick, my next question was going to
be, "What about the inside of the brick?" and I would then point out that no one has
ever seen the inside of a brick. Every time you break the brick, you only see the surface.
That the brick has an inside is a simple theory which helps us understand things better.
The theory of electrons is analogous. So I began by asking, "Is a brick an essential
object?"
Then the answers came out. One man stood up and said, "A brick as an
individual, specific brick. That is what Whitehead means by an essential object."
Another man said, "No, it isn't the individual brick that is an essential object; it's
the general character that all bricks have in common their
'brickness' that is the essential object."
Another guy got up and said, "No, it's not in the bricks themselves. 'Essential
object' means the idea in the mind that you get when you think of bricks."
Another guy got up, and another, and I tell you I have never heard such ingenious
different ways of looking at a brick before. And, just like it should in all stories about
philosophers, it ended up in complete chaos. In all their previous discussions they hadn't
even asked themselves whether such a simple object as a brick, much less an electron, is
an "essential object."


We should send Griffiths et al. to those seminars 24 hours a day and 7 days a week in order to prevent them writing books!
 
  • #5
@strangerep
" Virtual particles are not real."
OTOH, Feynman said (in his lectures on gravtitation) that all particles we ever observe are "virtual", because when they are measured, they correspond to internal lines of Feynman diagrams.
But of course, in a sense, particles themselves are not real but just a convenient way of looking at fields.
 
  • #6
strangerep said:
Yes.
IMHO, such "explanations" are rubbish. (I always become suspicious of physics authors who are subtly dismissive about mathematical rigor.) Is it only in Griffiths where you've seen such arguments, or are you thinking of other common textbooks also?
Plenty of textbooks, like those of Beiser, but also popular literature. That's why it's hard for me to believe that all those people are sloppy, and there is no rigorous justification for it.
 
  • #7
tom.stoer said:
as strangerep says: it's rubbish
1) it's s difficult to state the time-energy HUP in non-rel. QM in a general sense; it's even more unclear how to formulate it in QFT
2) virtual particles /as defined in perturbation expansion) are not Hilbert space states but "integrals of propagators"; so you can't define Δt and ΔE (or Δx and Δp) for these expressions in the usual sense
3) somethimes people claim that virtual particles "borrow energy" or "violate energy conservation for some short time Δt"; this is nonsense as well b/c in the Feynman diagrams energy and momentum are conserved exactly at each vertex; what is violated is the mass-shell condition m² = E² - p²
4) last but not least it's b...sh.. to discuss the "existence" of virtual particles: ...
Yes, this is also how I understand it; I've even once wrote a more or less popular article about it, and saw that the editor added the "Δt and ΔE" HUP to the text which I tried to circumvent. As I said, I find it rather amazing that so many people quote this explanation for virtual particles, and it made me doubt my own understanding of QM and QFT.

In a topic about virtual particles I onced asked the question what happens with our notion of virtual particles and "our vacuum filled with virtual particles" if we were able to solve our amplitudes in an exact way, without perturbation theory. You would then agree, looking at your post, that the whole notion of virtual particles would disappear, and that they solely are artifacts of doing pertubation theory, right?

Somehow people make the notion of vp's in popular literature much more romantic. I guess a popular explanation like "artifacts of the fact that our mathematical skills are not developed enough" is not flashy enough. Thinking about it, also regarding your topic on the definition of energy and entropy in GR elsewhere, we could almost start a topic in which all these popular notions in physics literature are being critized :P


In the Graduate College dining room at Princeton everybody used to sit with his
own group...

Yes, I love that story, and recently also quoted it here in some topic :D
 
  • #8
I think there is some justification for quoting the HUP in this context:
Think of the ground state of the Harmonic oscillator: I think it is permissible to say that the x-positon of the particle cannot be exactly zero because of the HUP.
If we accept this, then the fact that the probability of finding a non-zero amplitude for a field in QFT is also in some sense due to the HUP because this is (in each mode) equivalent to a H.O. The fact that this probability is smaller if k²+m² is larger fits into this picture, ath least intuitively: It is less probable to find an excitation of a higher-energy mode. (At least intuitively, this smaller probability might be thought to realte to a "shorter time", although this is nothing I would ever use to calculate something).

And with an interaction added to the theory, this probability amplitude may be described via the concept of virtual particles.

So I think there is some justificaton for this in a non-rigorous explanation, but in a good textbook you should add a bazillion of caveats.

And to finish with a question: Could one use the fall-off of a not-on-mass-shell propagator to make this more rigorous (the fall-off is faster the greater the violation of the mass-shell condition is)?
 
  • #9
That means that even if we would be able to solve for an interacting theory exactly, we still would find "virtual particles"?

I'm not familiar with these things, but aren't we able to solve some 1+1 dimensional interacting QFT's exactly? Does the notion of virtual particles appear in such theories?
 
  • #10
John Baez also seems to justify the relation between vp's and the HUP:

http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

We are really using the quantum-mechanical approximation method known as perturbation theory. In perturbation theory, systems can go through intermediate "virtual states" that normally have energies different from that of the initial and final states. This is because of another uncertainty principle, which relates time and energy.

In the pictured example, we consider an intermediate state with a virtual photon in it. It isn't classically possible for a charged particle to just emit a photon and remain unchanged (except for recoil) itself. The state with the photon in it has too much energy, assuming conservation of momentum. However, since the intermediate state lasts only a short time, the state's energy becomes uncertain, and it can actually have the same energy as the initial and final states. This allows the system to pass through this state with some probability without violating energy conservation.

Some descriptions of this phenomenon instead say that the energy of the system becomes uncertain for a short period of time, that energy is somehow "borrowed" for a brief interval. This is just another way of talking about the same mathematics. However, it obscures the fact that all this talk of virtual states is just an approximation to quantum mechanics, in which energy is conserved at all times. The way I've described it also corresponds to the usual way of talking about Feynman diagrams, in which energy is conserved, but virtual particles can carry amounts of energy not normally allowed by the laws of motion.
 
  • #11
QM people often say that virtual particles can exist because of the uncertainty principle between energy and time, in which one interprets the "time" in the appropriate way

That's not been the previous consensus in these forums[nor in the posts here so far]:

Here is How Zapper explains it in his blog...

...the HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. ... there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single measurement with arbitrary accuracy that is limited only by our technology. However, physics involves the ability to make a dynamical model that allows us to predict when and where things are going to occur in the future. While classical mechanics does not prohibit us from making as accurate of a prediction as we want, QM does!
The HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. The uncertainty theorem is about the statistical distribution of the results of future measurements. The theorem doesn't say anything about whether you can measure both at the same time. That is a separate issue.

[hence, no support for virtual particles.]On QFT: explanation from prior discussions in these forums:
There is not a definite line differentiating virtual particles from real particles — the equations of physics just describe particles (which includes both equally). The amplitude that a virtual particle exists interferes with the amplitude for its non-existence; whereas for a real particle the cases of existence and non-existence cease to be coherent with each other and do not interfere any more. In the quantum field theory view, "real particles" are viewed as being detectable excitations of underlying quantum fields. As such, virtual particles are also excitations of the underlying fields, but are detectable only as forces but not particles. They are "temporary" in the sense that they appear in calculations, but are not detected as single particles. Thus, in mathematical terms, they never appear as indices to the scattering matrix, which is to say, they never appear as the observable inputs and outputs of the physical process being modeled. In this sense, virtual particles are an artifact of perturbation theory, and do not appear in a non-perturbative treatment.

and a reminder from relativity:

We're familiar with other cases where geometric circumstances create real (not virtual) particles e.g. Hawking radiation at BH horizon and Unruh radiation caused by acceleration or felt by an accelerated observer. So it seems that expansion of geometry itself, especially inflation, can produce matter. ...Quantum fluctuations in the inflationary vacuum become quanta [particles]
at super horizon scales...it seems that expansion of geometry itself, especially inflation, can produce matter...The evolution of quantum fluctuations is from their birth [at Planck Scale] in the inflationary vacuum and their subsequent journey out to superhorizon scales where they become real life perturbations...[particles at detection]
..., is perhaps my favorite calculation in physics.
 
  • #12
"there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single measurement with arbitrary accuracy that is limited only by our technology"
Probably I'm misunderstanding that, but that seems to contradict what I've read on QM (and it also seems to contradict all those Gedankenexperiments of Bohr and other people who are discussing the HUP). So I'd be glad if you could explain (or put a link here).
Edit: See for example this experiment:
http://en.wikipedia.org/wiki/Uncertainty_principle See einstein's slit
 
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  • #13
Sonderval:
Probably I'm misunderstanding that, but that seems to contradict what I've read on QM...

Likely you ARE understanding what I posted...

the best interpretation is the second quote from Zapper's blog [above]..

try these sources on for size...and make your own judgments: I suspect the issue to which you refer is captured in post 2 and 3 here:

Is Heisenberg Uncertainty a problem with our measuring techniques
https://www.physicsforums.com/showthread.php?t=538854John Baez has a mathematical perspective here:

http://www.cbloom.com/physics/heisenberg.html

[Seems to me he is discussing a lower bound involving standard deviations, an ensemble of measurements, and NOT a single the measurement of conjugate observables. If so, this seems to me consistent with Zappers blog comments I quoted previously.

This is the discussion that led me to my interpretation...

Loooong discussionhttps://www.physicsforums.com/showthread.php?t=516224

I have zero confidence that everyone accepts this interpretation.

[edit: a related discussion brought this interesting piece:

"particles *may* have well-defined positions at all times, or they may not ... the statistical interpretation does not require one condition or the other to be true."
 
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  • #14
In rereading my notes on HUP, I came across this explanation from PAllen
that I really like [unsure which thread this is from]:

PAllen: If you are measuring position and momentum of the 'same thing' at two different times, the measurements are necessarily timelike. The measurements occur at two times on the world line of the thing measured. This order will never change, no matter what the motion of the observer is. If, instead, they occur for the same time on the "thing's" world line, they are simultaneous for the purposes of the uncertainty principle.to measure a particle's momentum, we need to interact with it via a detector, which localizes the particle. So we actually do a position measurement (to arbitrary precision). Then we calculate the momentum, which requires that we know something else about the position of the particle at an earlier time (perhaps we passed it through a narrow slit). Both of those position measurements, and the measurement of the time interval, can be done to arbitrary precision, so we can calculate the momentum to arbitrary precision. From this you can see that in principle, there is no limitation on how precisely we can measure the momentum and position of a single particle.

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wave function, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum. The width of this measured momentum distribution for many particles is what is limited by the HUP. In other words, the HUP says that the product of the widths of your measured momentum probability distribution, and the position probability distribution associated with your initial wave function, can be no smaller than Planck's constant divided by 4 times pi.

[Reading Wikipedia on HUP appears to give a different impression than these descriptions.]
 
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  • #15
I don't see why a momentum measurement necessarily involves two position measurements, at least not of the particle concerned. I could have a particle bounce from a ball or wall and measure the momentum change of the ball or wall at leisure. (As in the Einstein slit experiment I quoted above).
 
  • #16
Sonderval said:
Feynman said (in his lectures on gravtitation) that all particles we ever observe are "virtual", because when they are measured, they correspond to internal lines of Feynman diagrams.
That's a bit misleading, imho. I suppose he's referring to how a measurement corresponds to establishing a correlation between a system's initial state and an apparatus' final state by having them interact. (Ballentine covers this reasonably well in his QM text.)
[...] particles themselves are not real but just a convenient way of looking at fields.
I, too, prefer to focus on thinking about fields.
 
  • #17
Sonderval said:
I don't see why a momentum measurement necessarily involves two position measurements, at least not of the particle concerned. I could have a particle bounce from a ball or wall and measure the momentum change of the ball or wall at leisure. [...]
Er,... how do you "measure the momentum change of the ball" ?
 
  • #18
haushofer said:
[...] it's hard for me to believe that all those people are sloppy, and there is no rigorous justification for it.
I've yet to anyone seriously advocating that stuff who is also capable of explaining the notion of "unitarily inequivalent representations" and "Haag's theorem" in QFT properly. ;-)
 
  • #19
Sonderval said:
Could one use the fall-off of a not-on-mass-shell propagator to make this more rigorous [...] ?
The way to be rigorous is to solve the theory exactly rather than via perturbation theory. :-)
 
  • #20
haushofer said:
I'm not familiar with these things, but aren't we able to solve some 1+1 dimensional interacting QFT's exactly? Does the notion of virtual particles appear in such theories?
In the (exactly solvable) Lee model, one basically finds the correct Hilbert space for the interacting theory, rather than trying to shoehorn the interacting theory into a Hilbert space which is only appropriate for the free theory.
 
  • #21
There are solvable models in 1+1 dim., or at least models (Schwinger model, Gross-Neveu model, 1+1 dim. QCD) where other approximations but perturbation theory are reasonable (bosonization, large-N limit, ...). In all those cases one can find non-perturbative solutions and phenomena (axial anomaly, mass generation, chiral symmetry breaking and quark condensate, mesons ins Hartree-Fock approx., baryons as topological solitons, hadrons spectrum, ...) w/o ever talking about virtual particles etc.; of course on could apply perturbation theory but nobody was interested.

But QFT was dominated by perturbation theory for decades; even standard textbooks like Ryder introduce QFT as PI + perturbation theory (and add some small chapters regarding topological effects, mostly in classical (!) field theory, like monopoles and instantons) Experimental physics was dominated by accelerators and scattering experiments as well. It would be interesting to scan the nobel prize list for HEP, elementary particle physics and QFT and try to find one single example which is not mostly related to perturbation theory.

If you only have a hammer, you tend to see every problem as a nail.
 
  • #22
@strangerep
I can measure the momentum change of the ball by looking at its velocity (i.e., measuring its position at two time-points), or, if I know that it was at rest initially, by looking at its change in kinetic energy. You could have the ball hit a surface and measure the dent it makes.

Of course the HUP applies to the ball as well - I just wanted to say that you need not to measure the position of an object X at two points in time to measure its momentum, you can as well transfer the momentum to another object Y and then measure the momentum of that at you leisure. Sorry if this was unclear.

Yes of course, rigorously it would be better to solve the theory exactly - I was just asking whether one could use the falling off of the propagator to give quantitative meaning to the "energy uncertainty" by using p²-m² (with p=four vector) as a measure of the "energy violation" and the fall-off length of the propagator as a measure of the uncertainty in time. Intuitively, this seems plausible to me, but I never saw something like that.
 
  • #23
I forgot, here is the exact Feynman quote on virtual particles:
Of course any photon that has a physical effect may be considered as a virtual photon since it is not observed unless it interacts, so that observed photons never really have ω=± k. There are, however, no difficulties in passing to the limit; physically, we know of photons that come from the moon, or the sun, for which the fractional difference between ω and k is very, very small.
 
  • #24
Of course any photon that has a physical effect may be considered as a virtual photon ...

Isn't this self contradictory?...can you observe a physical effect without an interaction?
 
  • #25
@Naty
Why should a virtual photon have anythingto do with "no interaction"? The Coulomb interaction is also exclusively due to virtual (longitudinally/timelike) polarised virtual photons.
 
  • #26
Sonderval said:
The Coulomb interaction is also exclusively due to virtual (longitudinally/timelike) polarised virtual photons.
No; the Coulomb interaction in QED can be formulated w/o virtual photons at all, using Coulomb or Weyl-gauge; so virtual photons are not gauge invariant concepts and are therefore unphysical!
 
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  • #27
[tex]
\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \ldots = 2
[/tex]
This is wrong. The summation of this Dirichlet series was known as the Basel problem and Euler showed that the sum is [itex]\pi^2/6[/itex], which is also [itex]\zeta(2)[/itex], the Riemann zeta function.
 
  • #28
@tom
Could you send me a link or something explaining that?
And why do you say that you fix the gauge in a certain way to show this and then say that virtual photons are not gauge invariant? I'm confused.
 
  • #29
strangerep said:
Yes.
IMHO, such "explanations" are rubbish. (I always become suspicious of physics authors who are subtly dismissive about mathematical rigor.) Is it only in Griffiths where you've seen such arguments, or are you thinking of other common textbooks also?

To give Griffiths some credit, in his elementary particles book, he expresses his unease with uncertainty principle arguments,

"When you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."
 
  • #30
Wiki explains it this way, which seems ok:

In physics, a virtual particle is a mathematical conception that arises in quantum field theory. It is used in an analysis of an actual elementary subatomic process or reaction that links some physically concretely detectable input and output subatomic particles or quantum fields.

From discussions in these forums, I use the following concepts...until somebody else explains why they are incorrect or provides a better interpretation:

. In the quantum field theory view, "real particles" are viewed as being detectable excitations of underlying quantum fields

Local particle states correspond to the real objects observed by finite size detectors.

Tom Stoer of these forums has posted:
Particles appear in rare situations, namely when they are registered.

meaning, I think, otherwise and normally, they are fields.

Another view is to say that the concept of a "particle" is not invariant. Two observers will in general not agree on the number of particles they observe.

I'm not asserting these are correct...only that they seemed agreed upon previously. If professional full time physicsts can't agree on exactly what a particle is, I'm not going to get all uppity about one view or another.
 
  • #31
Dickfore said:
This is wrong. The summation of this Dirichlet series was known as the Basel problem and Euler showed that the sum is [itex]\pi^2/6[/itex], which is also [itex]\zeta(2)[/itex], the Riemann zeta function.

Ah, so then I'm misstaking it for some other series, but you still get the point i assume ;)
 
  • #32
tom.stoer said:
No; the Coulomb interaction in QED can be formulated w/o virtual photons at all, using Coulomb or Weyl-gauge; so virtual photons are not gauge invariant concepts and are therefore unphysical!
That sounds interesting, tom. Could you give some references?:)
 
  • #33
haushofer said:
Ah, so then I'm misstaking it for some other series, but you still get the point i assume ;)

No, I don't, since I only paid attention to this wrong equation and discarded the rest.
 
  • #34
Sonderval said:
@tom
Could you send me a link or something explaining that?
And why do you say that you fix the gauge in a certain way to show this and then say that virtual photons are not gauge invariant? I'm confused.

haushofer said:
That sounds interesting, tom. Could you give some references?:)

To find the Coulomb term w/o virtual photons simply use Coulomb gauge; have a look at the (outline of a) derivation here https://www.physicsforums.com/showthread.php?t=553666

Deriving the Feynman rules in the path integral formalism requires gauge fixing. But different gauge choices produce different sets of Feynman rules, so the propagators and vertices depend on the gauge. Only physical matrix elements are gauge invariant. The most prominent example is the comparison of a non-abelian gauge theory like QCD formulated a) in Lorentz gauge (as an example) and b) in axial gauge (as an example). In the Lorentz gauge the gauge fixing procedure "creates a new species of particles", so-called Fadeev-Popov ghosts which have their own propagators and vertices. These ghosts are absent in physical gauges like the axial gauge. So the on the level of Feynman diagrams one has different diagrams, different terms, even different particles (!) but on the level of matrix elements both gauges are of course equivalent (this is expressed in the Slavnov-Taylor identities expressing gauge invariance in the QFT formalism). The ghosts do never appear as external particles, only as internal lines i.e. "virtual particels". But that means that the particle content as defined by the Feynman diagrams of the theory differs between these gauges, whereas the particle content on the level of the physical Hilbert space (which requires studying BRST symmetry in the ghost case) is identical.

For a reference any QFT book discussing non-abelian gauge theory will do. Try Srednidzki b/c it's free for download. Or try Ryder as an introduction.
 
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  • #35
Deriving the Feynman rules in the path integral formalism requires gauge fixing.
and I was thinking that feynman rules are derived first,then the gauge is fixed (which is supposed to be the key advantage of path integral formalism)
 

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