# Does the path Integral contain virtual particles?

• I
So Feynman's path integral considers every possible path that a particle could take from start to end. In that process, there would be a path which contains a segment from, say, A to B at time t. But there could also be a path with a segment from B to A at that same time, t. If so, would this appear to be the antiparticle of the previous path segment? And would this represent the usual picture of a virtual particle pair created at some point and annihilating at a different point?

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ftr
Aren't you mixing up antiparticle with virtual particle?

Aren't you mixing up antiparticle with virtual particle?
I'm understanding that an antiparticle is the same as the particle only traveling backwards through time. And I'm assuming that traveling from B to A in time, t, is the reverse direction in time than traveling from A to B in time, t. And if both happen in the same time interval, that sounds to me like the usual description of virtual pair production. Please correct me if any of this is just wrong. Thanks.

PeterDonis
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I'm understanding that an antiparticle is the same as the particle only traveling backwards through time.
And with the charge conjugation and parity reversal operators also applied. In other words, antiparticles are the CPT conjugates of particles, not just the T conjugates.

if both happen in the same time interval, that sounds to me like the usual description of virtual pair production.
That's one way of looking at it, yes. More generally, the answer to your title question is yes: the Feynman diagrams that all have to be summed in the path integral will include diagrams in which virtual pairs are produced and annihilated. They will also include diagrams in which a virtual particle is exchanged between two other particles. Any diagram that is consistent with the Feynman rules for the theory in question will be included.

More generally, the answer to your title question is yes: the Feynman diagrams that all have to be summed in the path integral will include diagrams in which virtual pairs are produced and annihilated. They will also include diagrams in which a virtual particle is exchanged between two other particles. Any diagram that is consistent with the Feynman rules for the theory in question will be included.
Thank you. Are these "virtual particles" generic virtual particles, or do they somehow apply only to the propagation of the original particle? I ask because if these virtual particles are indistinguishable from any others, then I wonder if they are available for propagation of some third particle in the vicinity. I guess I'm asking if a single virtual particle can be used in the propagation of more than one particle. Correct me if I'm wrong, but it does seem that the same virtual process at a particular time can be arrived at by different paths that would also require the same virtual process at the same time. This sounds like that one virtual process at that time is being used by many paths and it does not care which path. Perhaps it could even be from paths for a different particles that are propagating nearby, I wonder. Thanks again.

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PeterDonis
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Are these "virtual particles" generic virtual particles, or do they somehow apply only to the propagation of the original particle?
I don't understand the distinction you are trying to draw here. Virtual particles are virtual particles.

I ask because if these virtual particles are indistinguishable from any others, then I wonder if they are available for propagation of some third particle in the vicinity.
I don't understand what you mean by this either.

This sounds like that one virtual process at that time is being used by many paths and it does not care which path
Or this. A "virtual process" is a Feynman diagram. One diagram is one diagram.

Here's a suggestion: instead of speaking in generalities, let's pick a particular quantum field theory--say QED, since that is pretty simple conceptually. Can you describe the question you want to ask in terms of QED Feynman diagrams?

Here's a suggestion: instead of speaking in generalities, let's pick a particular quantum field theory--say QED, since that is pretty simple conceptually. Can you describe the question you want to ask in terms of QED Feynman diagrams?
I'll give it a try. There are two situations. First, there is one particle propagating through space. The Feynman diagrams would have one input state and one output state, and there is a myriad of virtual particles exhibited in the calculation of it propagation. The second is that we then have two such single particles propagating though space. If the two particles are far apart, then we only consider the virtual particles associated for the propagation of first particle to be independent of the virtual particles associated for the second particle. But if the two propagating particles are brought close, then can the virtual particle associated with the propagation of the first particle actually be used in the propagation of the second? The Feynman diagram there has two inputs and two outputs. That usually requires every possible interaction between the two via virtual particles. But I wonder if this can also be said to be just the sharing of the virtual particles associated with one particle propagating with the virtual particles of the second particle propagating. Or would the second particle have its own virtual particles at a specific place and time that would never be drawn as a virtual particle at the same place and time in the diagram for the first particle?

PeterDonis
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First, there is one particle propagating through space. The Feynman diagrams would have one input state and one output state, and there is a myriad of virtual particles exhibited in the calculation of it propagation.
Ok so far.

The second is that we then have two such single particles propagating though space. If the two particles are far apart, then we only consider the virtual particles associated for the propagation of first particle to be independent of the virtual particles associated for the second particle.
Not really, but it's not worth delving into this since you're really interested in the case where there is interaction; see below.

if the two propagating particles are brought close, then can the virtual particle associated with the propagation of the first particle actually be used in the propagation of the second?
You're confusing yourself by mixing up Feynman diagrams. Unfortunately I don't have a handy way to draw such diagrams here, but these are simple enough that they can be described reasonably well in words.

So we have an "in" state consisting of two electrons (remember we're talking QED here), and an "out" state consisting of two electrons. The simplest diagram for this case is just two straight electron lines, with nothing else present. (Note that there are actually two cases of this diagram, one with the lines just going straight up and one with them swapping places; but we won't go into that here.) Note that there are no vertexes in this diagram, just lines, and the lines are all external lines, so there are no virtual particles at all in this diagram.

The next simplest diagrams are those with two vertexes. (Btw, you might want to ask yourself why there are no diagrams with a single vertex for this case. Think about what a single vertex would correspond to.) There are three diagrams with two vertexes:

(1) Both vertexes are on electron line number 1. This corresponds to electron number 1 emitting a virtual photon and then absorbing the same virtual photon later on.

(2) Both vertexes are on electron line number 2. This corresponds to electron number 2 emitting and then absorbing a photon.

(3) One vertex is on each electron line. This corresponds to electrons 1 and 2 "exchanging" a photon. (The word "exchanging" is used because, as you will see if you think about it, there is no way to specify once and for all which one is "emitting" and which one is "absorbing" the photon, since the vertex on each line could be anywhere.)

Now: try to reformulate your question in terms of the above diagrams. Can you do it? I think you will find that you can't: that the question you were trying to ask can't even be asked in terms of the above diagrams. Which means the question isn't really well-defined.

Now: try to reformulate your question in terms of the above diagrams. Can you do it? I think you will find that you can't: that the question you were trying to ask can't even be asked in terms of the above diagrams. Which means the question isn't really well-defined.
You might be right. I'd have to think about it awhile. But I have other questions. From your previous answer in post #4 we have that virtual particle pairs are included in the Feynman path integral. Are these virtual pairs of the same nature as those found in the cosmological constant? Is there a zero point energy associated with them? Are they in fact the same virtual pairs of normal space being used if a particle happens to transverse that space? Thanks again.

PeterDonis
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From your previous answer in post #4 we have that virtual particle pairs are included in the Feynman path integral.
More precisely: diagrams that include virtual particle pairs are included in the path integral. But there will be many different diagrams that include virtual particle pairs; it won't be just one.

Are these virtual pairs of the same nature as those found in the cosmological constant?
What does "of the same nature" mean? We are talking about a general framework of quantum field theory. That framework can be applied to many different physical situations. One such situation might be explaining the cosmological constant, but so far nobody has been able to do this successfully: every attempt to construct a QFT model ends up predicting a value for the cosmological constant that is about 120 orders of magnitude larger than the one we actually observe.

Is there a zero point energy associated with them? Are they in fact the same virtual pairs of normal space being used if a particle happens to transverse that space?
Same response as above to these questions.

The general error you appear to be making here is a common one: thinking of "virtual particles" as actual objects, with actual properties, that you can ask about and compare with other properties of other actual objects. They're not. They're just mathematical objects that occur in one particular type of model--a perturbative QFT model. Thinking of them as "real" is only going to confuse you.

You might want to read Arnold Neumaier's Insights article on this topic:

https://www.physicsforums.com/insights/misconceptions-virtual-particles/

bhobba
PeterDonis
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Moderator's note: I have adjusted the level of this thread to "I" since that seems more appropriate for the discussion taking place in the thread.

vanhees71
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So Feynman's path integral considers every possible path that a particle could take from start to end. In that process, there would be a path which contains a segment from, say, A to B at time t. But there could also be a path with a segment from B to A at that same time, t. If so, would this appear to be the antiparticle of the previous path segment? And would this represent the usual picture of a virtual particle pair created at some point and annihilating at a different point?
In relativistic QT the "1st-quantization approach" is misleading for precisely the reason that leads to the introduction of antiparticles, i.e., there's always the possibility that particles are created or destroyed in scattering process. The adequate formulation is thus "2nd quantization", aka quantum field theory, which easily lets you describe such creation-annihilation processes (of course always in accordance with the fundamental conservation laws for energy, momentum, angular momentum and the various charges in the Standard Model).

The path-integral formulation is just one way to formulate quantum theory. In the case of relativistic quantum field theory it's applied to the "2nd quantization" formalism, and thus the integrals are not over trajectories in phase space as in the "1st quantization" formalism but over field configurations. Then the quibbles with particles and antiparticles resolve themselves. Note, however, that path integrals are far from trivial and not even rigorously defined from a mathematical point of view. A very good book on QFT with a "path-integral first" approach is

D. Bailin and A. Love, Introduction to Gauge Field Theory, Adam Hilger, Bristol and Boston, 1986.

How far you can come with the 1st-quantization method and path integrals, see the "bible on path integrals", which you can download for free directly from the author (although it's also published as a textbook to a very fair prize):

http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pi.pdf

https://smile.amazon.de/Integrals-Q...60874&sr=8-1&keywords=Kleinert+Path+Integrals

bhobba
For example, as I recall, it used to be said that when an electron jumps from a high orbit to a lower orbit, this is done by means of a virtual photon causing the electron to fall from its unstable high orbit to a lower one. IIRC, this virtual photon did not belong to any path integral calculation but was simply part of the background fluctuations inherent in space. Is this much correct? Or is this an outdated explanation? Thanks.

vanhees71
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Nothing jumps in quantum theory. The unitary time evolution is continuous.

If you take into account quantum field theory, i.e., the quantization of the electromagnetic field, and you have an atom in an excited state there is a certain transition-probability rate for this atom to emit a photon and going into a lower-energy state. This is called spontaneous emission and is indeed due to the quantum fluctuations of the electromagnetic field due to the charged particles (mostly the electrons) in the atom.

bhobba
This is called spontaneous emission and is indeed due to the quantum fluctuations of the electromagnetic field due to the charged particles (mostly the electrons) in the atom.
OK, thank you. So would these quantum fluctuations not exists apart from the electron in the region? Or are they part of the background independent of whether there are charged particles in the region?

vanhees71
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This is a mute question, because you can probe the electromagnetic response of empty space only by putting some charges in. The vacuum itself is stable, i.e., if there's nothing according to QED nothing will spontaneously pop up although many popular-science writers try to convince you from the opposite :-(.

bhobba
This is a mute question, because you can probe the electromagnetic response of empty space only by putting some charges in. The vacuum itself is stable, i.e., if there's nothing according to QED nothing will spontaneously pop up although many popular-science writers try to convince you from the opposite :-(.
This only brings up the question: is there a vacuum energy, zero point energy, dark energy, cosmological constant, quantum fluctuations, apart from any stable charge in the region?

bhobba
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For example, as I recall, it used to be said that when an electron jumps from a high orbit to a lower orbit, this is done by means of a virtual photon causing the electron to fall from its unstable high orbit to a lower one. IIRC, this virtual photon did not belong to any path integral calculation but was simply part of the background fluctuations inherent in space. Is this much correct? Or is this an outdated explanation? Thanks.
The phenomena of spontaneous emission you are referring to is not explained that way.

Here is the correct explanation:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

The summary is the reason it will, unpredictably, but with a predictable probability (see Fermi's Golden Rule), emit a photon and be found in a lower energy 'orbit' is that the solutions to the Schrodinger equation are stationary states - which means an electron in that state remains in that state forever. But the model is too simplistic - it really is not in a stationary state - the electrons field is part of the universal quantum EM field that permeates all of space and that was not taken into account. So you model it as a 'perturbation' (ie a small change to the stationary state) and you get the Golden Rule mentioned before:
http://staff.ustc.edu.cn/~yuanzs/teaching/Fermi-Golden-Rule-No-II.pdf

That is why it eventually emits a photon - its just the normal evolution of a non-stationary system. But note - it's modeled using perturbation theory.

BTW perturbation theory is where all this stuff about virtual particles etc comes from - its tied up, as had been mentioned many times to you, with the pictorial representation of a Dyson series:
https://en.wikipedia.org/wiki/Dyson_series

Note - it is, as the article says, a perturbation series. The way perturbation theory works is you assume to start with a simple solution like the stationary state mentioned before. Then you assume its not really that at all, but the difference between it and what it really is, is small. Perturbation theory can then be used to approximate the correct solution. Then one uses that approximation to get a better one and so it goes - you get a series of better approximations:
http://www.cims.nyu.edu/~eve2/reg_pert.pdf

Thanks
Bill

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vanhees71
bhobba
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This only brings up the question: is there a vacuum energy, zero point energy, dark energy, cosmological constant, quantum fluctuations, apart from any stable charge in the region?
In many other threads you have been told the answer to a lot of those - and the reasons. I will list them:

1. There is no vacuum energy - a process called normal ordering is used to ensure it is zero:
2. The same with zero point energy
3. Dark energy - still under investigation.
4. Why is the cosmological constant the value it is - still under investigation.
5. Quantum Fluctuations - Dr Neumaier's Insights article already mentioned had a sequel examining exactly that - its a myth::
https://www.physicsforums.com/insights/vacuum-fluctuation-myth/

Now can we please have you stop asking the same things over and over - they have been done to death.

Thanks
Bill

vanhees71
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This only brings up the question: is there a vacuum energy, zero point energy, dark energy, cosmological constant, quantum fluctuations, apart from any stable charge in the region?
In special relativity there's no way to make physical sense of an absolute energy level. Thus the ground-state energy can be freely chosen to take any value you want, but no physics changes. Thus within the Standard Model of particle physics there's nothing observable related to the absolute choice of the zero-point (or vacuum) energy. Quantum fluctuations are always related to the presence of matter and/or radiation (both are described on one footing in terms of quantum fields). There are some very nice Insights articles by Arnold Neumaier:

https://www.physicsforums.com/insights/misconceptions-virtual-particles/
https://www.physicsforums.com/insights/physics-virtual-particles/
https://www.physicsforums.com/insights/vacuum-fluctuation-myth/
https://www.physicsforums.com/insights/vacuum-fluctuations-experimental-practice/

Within General Relativity the picture changes a bit, but here we have only incomplete concepts since the quantization of the gravitational interaction is not yet satisfactorially worked out. So we have to restrict ourselves to the (semi-)classical picture. There the energy-momentum-stress tensor is the source of the gravitational field or, within General Relativity (GR) synonymous the curvature of spacetime, and thus the absolute value of the energy density in the universe has observable consequences. Also the fundamental principles of GR (the strong equivalence principle aka general covariance) admits a free parameter, the socalled cosmological constant, which you can reinterprete as another special form of energy-momentum tensor. Today from measurements of the cosmic microwave background fluctuations as well as the distance-redshift relation (aka Hubble relation) from measurements of supernovae indicates that about 70% of the energy-momentum content of the universe is due to the cosmological constant or equivalently Dark Energy, and the consequence is that the Hubble expansion today is accelerated. For a very good review, see the Nobel announcements and the Nobel lectures,

https://www.nobelprize.org/nobel_prizes/physics/laureates/2011/

Click on the popular as well the advanced information. The Nobel lectures you find by clicking on the recipients of the prize (both as text and as video).

bhobba
Now can we please have you stop asking the same things over and over - they have been done to death.
My problem is that we seem to be integrating virtual processes to come up with real effects but then denying any kind of independent ontological status of the virtual process that we are integrating. How can nothing real have real effects? I'm sorry, but that's going to remain a problem for me until I get a satisfactory answer. And I'm not sure that resorting to the language of quantum fields is going to help because fields are made up of infinitesimal little bits that have effects.

PeterDonis
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we seem to be integrating virtual processes to come up with real effects
No, we are integrating individual terms in a mathematical model to come up with predictions for the results of measurements. The fact that the predictions turn out to be correct does not mean every single term or symbol that appears in the mathematical model has to correspond to something "real".

that's going to remain a problem for me until I get a satisfactory answer
You have already been given the correct answer (multiple times, as @bhobba points out). If you can't accept it, that's something you are just going to have to work out for yourself.