Why don't virtual particles cause decoherence?

[tom.stoer]

But, why on Earth would anyone give "reality" to propagators, and call them particles? It doesn't make sense

Agreed!

But see #36 - #38; this is due to the way QFT is presented in lectures, most textbooks and popular books; it's due to decades of perturbative calculations; it's due to a very limited focus on QFT w/o being aware of the limitations and their implications; if you only have a hammer, you tend to see every problem as a nail.

 

[kith]

Anyway, my point is to see what happens to the field's state during the interaction.

That's quite interesting. Does somebody know some literature which talks about this?

\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|} \hat U\left( t \right)\left| {vac} \right\rangle \,\,\,\left| n \right\rangle, <br />

Is it really a coincidence that something like a propagator appears in this expression? I'm not very familiar with QFT but in non-relativistic QM, the propagator <x|U(t,t')|x'> gives the probability for state <x| at time t, if we startet in state |x'> at time t'. In your expression, we start with state |vac> at t'=0 and get the probability for the (intermediate) state |n> -which contains n photons- at (an intermediate) time t.

Can this somehow be related to the propagators in Feynman diagrams? An obvious difference would be that the particles in your expression need to be on-shell...still I'm wondering if there is a connection. After all, these are two approaches for the same physical situation.

 

[tom.stoer]

Can this somehow be related to the propagators in Feynman diagrams? An obvious difference would be that the particles in your expression need to be on-shell...still I'm wondering if there is a connection. After all, these are two approaches for the same physical situation.

You get the propagators by splitting the time evolution U(t2,t1) into infinitesjmal steps U(t1+dt,t1) and by expanding U in dt. I think you can find this derivation of Feynman path integrals and their perturbative calculations in any advanced QM textbook.

 

[JK423]

That's quite interesting. Does somebody know some literature which talks about this?

It's very interesting, but this problem for finite interaction time t is unsolvable for (unknown to me) mathematical reasons, as tom.stoer has pointed in a previous post i think.

Is it really a coincidence that something like a propagator appears in this expression? I'm not very familiar with QFT but in non-relativistic QM, the propagator <x|U(t,t')|x'> gives the probability for state <x| at time t, if we startet in state |x'> at time t'. In your expression, we start with state |vac> at t'=0 and get the probability for the (intermediate) state |n> -which contains n photons- at (an intermediate) time t.

Can this somehow be related to the propagators in Feynman diagrams? An obvious difference would be that the particles in your expression need to be on-shell...still I'm wondering if there is a connection. After all, these are two approaches for the same physical situation.

Yes, these things are similar and Feynman diagrams are derived using this quantity. Note however that the quantity
\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle
is the amplitude of having n photons, \left| n \right\rangle, so in this case we have an amplitude associated with a quantum state,
\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle \left| n \right\rangle }.
In Feynman diagrams, the amplitudes that we draw are not associated with any quantum states, hence, to my current understanding, virtual "particles" (corresponding to these drawings) are not even quantum objects (not even talking about off/on-shell particles!). A quantum object is described by a quantum state, period.
If for any reason my understanding is mistaken, please let me know.

 

[kith]

Thanks tom.stoer and JK423 for pointing me in the right direction about the propagators. I'd like to add a comment to the real vs. virtual issue.

Why would you say that they would be "internal lines" inside the Feynman diagrams? No they wouldn't, quantum states are quantum states, and real particles are described by such even when interactions are present, and you can measure them!

Whenever we detect a "real" photon, we can at least in principle track down the emission process for it. So we have a charged particle emitting a photon and a charged particle in the detector absorbing a photon. Formally, we can write down a scattering Feynman diagram between the emitting particle and the absorbing one - which makes the "real" photon a virtual one. I think this is why some people claim that every photon is virtual. Technical arguments aside, the question would be a question of ontology. If every measurement involving photons can be described by virtual photons why should we postulate the existence of "real" photons? If we don't, a quantum state is only a calculational tool and not a state of an actual element of reality.

I'm not sure if what I'm writing is correct. At least, Griffith's claims in his "Introduction to elementary particles" that we get the same answer if we (1) calculate the Feynman diagram I mentioned above or if we (2) calculate the Feynman diagrams for emission of a real photon followed by an absorption of a real photon.

 

[JK423]

Whenever we detect a "real" photon, we can at least in principle track down the emission process for it. So we have a charged particle emitting a photon and a charged particle in the detector absorbing a photon. Formally, we can write down a scattering Feynman diagram between the emitting particle and the absorbing one - which makes the "real" photon a virtual one. I think this is why some people claim that every photon is virtual. Technical arguments aside, the question would be a question of ontology. If every measurement involving photons can be described by virtual photons why should we postulate the existence of "real" photons? If we don't, a quantum state is only a calculational tool and not a state of an actual element of reality.

I'm not sure if what I'm writing is correct. At least, Griffith's claims in his "Introduction to elementary particles" that we get the same answer if we (1) calculate the Feynman diagram I mentioned above or if we (2) calculate the Feynman diagrams for emission of a real photon followed by an absorption of a real photon.

Thank you for the explanation, it seems plausible to me that this is what most people have in mind when say "real" particles are actually "virtual". However, as far as i can see, there is a huge difference between
1) virtual particles
2) real particles, being emitted and absorbed, hence described by an amplitude as well.
When you derive "virtual particles", you take the amplitude \left\langle {vac} \right|\hat U\left( {{t_0}} \right)\left| {vac} \right\rangle for a specific time instant (usually being t₀→∞), then you time-slice the evolution operator and form various sub-propagators inside this original one and draw them as diagrams/particles proapagating. However, these sub-propagators do not correspond to any quantum state, as the amplitude \left\langle {vac} \right|\hat U\left( {{t_0}} \right)\left| {vac} \right\rangle corresponds to the state \left| {vac} \right\rangle. Hence, virtual particles are not described by a quantum state at any time instant and by this we can safely conclude that they are not even quantum objects.
On the other hand, a real particle is described by a quantum state between the time of emission and absorption, so by definition this has nothing to do with internal lines inside Feynman diagrams..

 

[mfb]

Edited:
mfb you argued, that, since interactions are always present even "real" particles are "virtual". I really cannot understand this argument! Real particles are described by quantum states; if interactions are present then simply the number of real particles will be in superposition of different values, but still these are quantum states! Why would you say that this has any relevance to the propagators and "virtual particles", i.e. "particles" which are not described by quantum states (hence not particles!)? Why would you say that they would be "internal lines" inside the Feynman diagrams? No they wouldn't, quantum states are quantum states, and real particles are described by such even when interactions are present, and you can measure them! While internal lines are only pictures you draw when you perturbatively expand propagators..
Please elaborate on this because i am more confused than before

I am a bit confused how to interpret your question, so maybe this does not directly answer your question:

Let's consider Z-bosons produced at the LHC (I think we all agree that we have particles produced in the collisions, right?). Are they real? Their mass spectrum has a broad peak with long tails, so they are not on-shell - a property usually assigned to virtual particles.
If we go to longer-living particles, we get all particles in our everyday life. They can be slightly off-shell as well, or have other "unusual" properties - the deviations from ideal, non-interacting particles are just too small to notice it.
If we go in the other direction: W-bosons in weak decays (apart from top-decays) are off-shell as well. Are they real? The usual answer is no. But where is the difference between those W-bosons and electrons in a vacuum tube? It is just the timescale of their existence.

 

[JK423]

I am a bit confused how to interpret your question, so maybe this does not directly answer your question:

Let's consider Z-bosons produced at the LHC (I think we all agree that we have particles produced in the collisions, right?). Are they real? Their mass spectrum has a broad peak with long tails, so they are not on-shell - a property usually assigned to virtual particles.
If we go to longer-living particles, we get all particles in our everyday life. They can be slightly off-shell as well, or have other "unusual" properties - the deviations from ideal, non-interacting particles are just too small to notice it.
If we go in the other direction: W-bosons in weak decays (apart from top-decays) are off-shell as well. Are they real? The usual answer is no. But where is the difference between those W-bosons and electrons in a vacuum tube? It is just the timescale of their existence.

Thank you for taking the time to explain.
I think that the discussion about what virtual particles actually are, is irrelevant to the off/on shell issue. When we talk about off/on shell particles, we talk about quantum states that are in a superposition of energy eigenstates: Still we talk about quantum objects described by quantum states. So, whether a particle is off- or on-shell, it's still described by a quantum state at each given time instant, and that makes it real for its own sake. Generally an existing quantum object described by a quantum state at some time instant is real, i don't think that you disagree with that, right? I take this as a definition of real objects.
Now let's talk about virtual particles as defined by the internal lines of Feynman diagrams. If you want to talk about excitations of the electromagnetic field during the interactions of two electrons, then just evolve the vacuum of the E/M field with the appropriate evolution operator for finite t,
\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle \left| n \right\rangle }. (1)
During the interaction at time t, the states |n> are the excitations of the electromagnetic field described by quantum states, hence they are real quantum objects.

Question: Are these excitations {|n>} the virtual particles defined by the internal lines of Feynman diagrams?
Answer: No!

The latter virtual "particles" appear when we time-slice the amplitude \left\langle {vac} \right|\hat U\left( {t \to \infty } \right)\left| {vac} \right\rangle and get various sub-propagators. Note that this former amplitude corresponds (from (1) ) to the real quantum object \left| {vac} \right\rangle at the specific time instant t→∞.

Question: Do the aforementioned sub-propagators (virtual particles) correspond to any quantum state created during the interaction?
Answer: The only quantum states created during the interaction are the {|n>} in (1) with corresponding propagators (amplitudes) {\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle }, and these have nothing to do with these sub-propagators. Hence, you cannot ascribe a quantum state to these sub-propagators, so virtual particles are not even quantum objects! As you see, whether they are off/on-shell is irrelevant, since they are not even quantum states.. In other words, there is no instant in time -during this whole interaction- that these "virtual particles" popped out from the vacuum as quantum states disappearing in t→∞. The excitations that popped out from the vacuum during the interaction are the states \left| n \right\rangle in (1), but it's not them that appear in Feynman diagrams.. (and ofcourse they are not virtual since they actually existed at some time instant plus they are measureable in principle).

Conclusion: Virtual particles are neither "particles" nor quantum objects in general. They do not exist, since they are not described by a quantum state at any instant of time, hence they are just mathematical artifacts.

I hope that i made my point clear. Please tell me what you think.

 

[mfb]

we talk about quantum states that are in a superposition of energy eigenstates

Everything in our world is.

Generally an existing quantum object described by a quantum state at some time instant is real, i don't think that you disagree with that, right?

The longer this discussion goes on, the less I like the word "real".
Quantum states are always a property of the whole system. And I think this whole system is the universe. It can be described in terms of quantum states, and I agree that "the universe is real" is a reasonable definition.Again, I don't see any fundamental difference between a W boson in a weak decay and an electron in a vacuum tube. Do you consider them as real or not? In addition, I showed a continuous transition between both.
If you want to sort them in two different categories, where exactly do you draw the line between them? Is a W boson in a weak decay different from a Z boson decaying in 2 leptons with sqrt(s) = 20 GeV? Is this Z boson different from a Z boson with sqrt(s) = 80 GeV? sqrt(s)=90 GeV? Is this different from a short-living electron? And a long-living one?

 

[JK423]

Everything in our world is.The longer this discussion goes on, the less I like the word "real".
Quantum states are always a property of the whole system. And I think this whole system is the universe. It can be described in terms of quantum states, and I agree that "the universe is real" is a reasonable definition.Again, I don't see any fundamental difference between a W boson in a weak decay and an electron in a vacuum tube. Do you consider them as real or not? In addition, I showed a continuous transition between both.
If you want to sort them in two different categories, where exactly do you draw the line between them? Is a W boson in a weak decay different from a Z boson decaying in 2 leptons with sqrt(s) = 20 GeV? Is this Z boson different from a Z boson with sqrt(s) = 80 GeV? sqrt(s)=90 GeV? Is this different from a short-living electron? And a long-living one?

It's a good thing that we agree that things that exist are described by quantum states.

It's not about what "I" consider real, the definition of real is the same for all of us. Are all these particles that you mention described by a quantum state during their "living time" Δt? (whatever this Δt is!)
If the answer is yes, then they are real! (since we agree on the definition of real at least!)
Real particles leave traces on detectors (since they are quantum states that interact with the detector), virtual particles don't leave traces because they are not described by a quantum state and hence cannot (by definition) interact with anything.
A W boson in weak decay, when seen as the internal line of the lowest order relevant Feynman diagram, is a virtual one, hence you are never going to see its trajectory on any detector (in the case you had the resolution to do such a thing) since it's not described by a quantum state at any instant of time during the whole interaction process. A real W boson, on the other hand, whatever its lifetime, will leave its trajectory.

Do you agree on this?

 

[mfb]

It's a good thing that we agree that things that exist are described by quantum states.

I did not say this. But if all our universe follows the laws of quantum mechanics (and there is absolutely no measurement showing anything else)... sure.

It's not about what "I" consider real, the definition of real is the same for all of us.

I am not aware of any authority defining "real" for all physicists.

Are all these particles that you mention described by a quantum state during their "living time" Δt? (whatever this Δt is!)

I am not sure what exactly you mean by "described by a quantum state".

Real particles leave traces on detectors (since they are quantum states that interact with the detector), virtual particles don't leave traces because they are not described by a quantum state and hence cannot (by definition) interact with anything.

It is the very point of the concept of virtual particles that they interact. They leave a trace in the particles produced in the interaction - so you could consider those produced particles as detector.

A W boson in weak decay, when seen as the internal line of the lowest order relevant Feynman diagram, is a virtual one, hence you are never going to see its trajectory on any detector (in the case you had the resolution to do such a thing) since it's not described by a quantum state at any instant of time during the whole interaction process. A real W boson, on the other hand, whatever its lifetime, will leave its trajectory.

Do you agree on this?

Again, where is the border between both? At 1 MeV away from the mass? At 1 GeV? At 10 GeV?

 

[JK423]

I am not sure what exactly you mean by "described by a quantum state".

It's simple; do they have a wavefunction during the time of their existense? Or, in other words, are they an excitation of their corresponding field during the time of their existense? If your answer is positive then please read the rest of the post. If your answer is negative, virtual particles are mathematical artifacts and case is solved.

It is the very point of the concept of virtual particles that they interact. They leave a trace in the particles produced in the interaction - so you could consider those produced particles as detector.

Didn't you ever wonder why "virtual particles" cannot interact with anything else?

Does this seem so natural to you and you think that "this is the concept of virtual particles"? When you argue about the reality of virtual particles, there is no "concept" behind them or any fundamental law of physics that prevents them from interacting with anything else! Think about it! What laws of physics prevents a wavefunction from interacting with other quantum systems? None! Then why virtual particles cannot interact with anything?
Perhaps, because they are not described by a quantum state after all and are just mathematical artifacts of perturbation theory?
Keep an open mind here and do not reject anything before you even think about it. When you argue that a virtual particle lives for time Δt, you imply that during this time it's described by a quantum state. This quantum state, if it exists, SHOULD be able to interact with the detector (or any other quantum system) during this Δt. But it can't. This says a lot by itself.

Again, where is the border between both? At 1 MeV away from the mass? At 1 GeV? At 10 GeV?

If you want a border line, this is not dermined by the deviation of mass. In plain words, real particles (whatever Δm and Δt!) are described by quantum states and hence leave traces on detectors, while virtual "particles" are not described by quantum states, hence, cannot (and do not) leave traces on detectors. That's the border line.

 

[tom.stoer]

see #8

The confusion stems from the unfortunate fact that physicists use two different DEFINITIONS of the word "virtual particle". According to one, it as any off-shell particle. According to another (more meaningful, in my opinion), it is any internal line in a Feynman diagram. The two definitions are not equivalent.

 

[mfb]

It's simple; do they have a wavefunction during the time of their existense? Or, in other words, are they an excitation of their corresponding field during the time of their existense? If your answer is positive then please read the rest of the post. If your answer is negative, virtual particles are mathematical artifacts and case is solved.

With my favorite interpretation of QM, the whole universe is a single wavefunction. But even with other interpretations, "the wavefunction of a particle" does not exist if the system consists of multiple, possibly interacting particles.

Didn't you ever wonder why "virtual particles" cannot interact with anything else?

What do you mean with "anything else"? Again, what do you mean with virtual particles? You always assume that you can separate all particles in two categories, and you don't tell me where exactly you see the boundary.

Keep an open mind here and do not reject anything before you even think about it.

Don't worry about that part.

When you argue that a virtual particle lives for time Δt, you imply that during this time it's described by a quantum state. This quantum state, if it exists, SHOULD be able to interact with the detector (or any other quantum system) during this Δt. But it can't. This says a lot by itself.

It can do it in the same way every other particle can.

If you want a border line, this is not dermined by the deviation of mass. In plain words, real particles (whatever Δm and Δt!) are described by quantum states and hence leave traces on detectors, while virtual "particles" are not described by quantum states, hence, cannot (and do not) leave traces on detectors. That's the border line.

No QFT-electron leaves a trace in the detector: It will interact and become a different electron afterwards. And, indeed, virtual particles share the same possible description.

@tom.stoer: I'm just considering off-shell particles at the moment, but as we discussed before, those definitions are not completely unrelated.

 

[Haelfix]

This is a classic discussion, and there isn't really a right or wrong answer here. Just levels of approximation, and what you are or are not prepared to take as a true statement (always like this with why questions), as Demystifier correctly points out, there is definitely semantics here.

Let me make an analogy to illustrate the problem. Suppose that on Alpha Centauri, a photon is emitted from an atom, and travels to earth. Scientists would draw a vertex with say an outgoing 'real' photon. This is then 'real' all the way up until it is absorbed by a cell in your eye. Indeed the photon is very nearly on shell, but not quite and thus becomes a virtual particle and an internal leg in a larger diagram. It is in this sense that Mfb is correct.
Now, mathematically, this is quite fictional. The problem is we have just drawn a tree level 'Feynman' diagram that strictly speaking, isn't properly a diagram in the first place. It's not just one term in a perturbative expansion, but the literal picture of what happens. It is a matter of definition, but we always assume in the mathematics that the particles are really irreducible representations of the Poincare group, where we choose prepared plane wave states off in the infinite past and infinite future and we have in mind some sort of LSZ procedure, where we interpret particle states as poles in the SMatrix.

But the particles we drew in the above diagram are nothing like that at all. Instead they are messy realworld objects that have no such ontology. Indeed your eyeball is not a perfectly massless Von Neuman machine off at future infinity. It's atoms will mix with the photon, and you won't have a well defined Fock space or number operator. So there won't even be a proper particle state to talk about, only a world state.

Still, in so far as it might be *useful* to picture the photon that you absorb as really being there, then you are allowed to make use of such an ontology, with the caveat that there are perfectly acceptable methods that make no use of perturbative methods at all and that you have to be careful ascribing reality to things that are strictly speaking mathematical fictions (for all the correct reasons that Tom pointed out).

 

[JK423]

First of all, thank you for the analysis.

Let me make an analogy to illustrate the problem. Suppose that on Alpha Centauri, a photon is emitted from an atom, and travels to earth. Scientists would draw a vertex with say an outgoing 'real' photon. This is then 'real' all the way up until it is absorbed by a cell in your eye. Indeed the photon is very nearly on shell, but not quite and thus becomes a virtual particle and an internal leg in a larger diagram. It is in this sense that Mfb is correct.
Now, mathematically, this is quite fictional. The problem is we have just drawn a tree level 'Feynman' diagram that strictly speaking, isn't properly a diagram in the first place. It's not just one term in a perturbative expansion, but the literal picture of what happens.

If the photon that travels from Alpha Centauri is real, then we can manipulate it; We can do unitary -non destructive- measurements on it at will.
Now take the two electrons that electromagnetically interact. In the first approximation, they "exchange a virtual photon". I argue that this virtual photon doesn't exist since we cannot manipulate it even in principle, it cannot interact with any quantum systems during the time of its existense.
So, in the first case with the real photon from Alpha Centauri, it may be absorbed in the end (and techically draw a diagram for it) but as you point out this is not a term in a perturbative expansion, and as i point out we can manipulate it during the time of its existense. I don't care if it's off/on shell or not, it doesn't matter! Can we manipulate the second virtual photon from the electron-electron scattering? No, we cannot design such an experiment even in principle with the quantum mechanical formalism, because virtual photons are not described by a quantum state at any instant of time, hence impossibe to interact with them!

So i really don't understand why it's a matter of semantics. I feel that the difference between real/virtual photons is perfectly clear and that has nothing to do with the off/on shell thing that mfb always points out.

Am i considering something wrong here?

Let me phrase my conclusion in the following (semantics-free) way:
Real particles are manipulable, they are described by quantum states. Non-real particles are not described by a quantum state at any instant of time, so they are not manipulable even in principle. (manipulable=general local non-destructive operations, e.g. unitary)
Do you all agree with that? I don't even mention the word "virtual" so we won't play with semantics.

It is a matter of definition, but we always assume in the mathematics that the particles are really irreducible representations of the Poincare group, where we choose prepared plane wave states off in the infinite past and infinite future and we have in mind some sort of LSZ procedure, where we interpret particle states as poles in the SMatrix.

But the particles we drew in the above diagram are nothing like that at all. Instead they are messy realworld objects that have no such ontology. Indeed your eyeball is not a perfectly massless Von Neuman machine off at future infinity. It's atoms will mix with the photon, and you won't have a well defined Fock space or number operator. So there won't even be a proper particle state to talk about, only a world state.

Still, in so far as it might be *useful* to picture the photon that you absorb as really being there, then you are allowed to make use of such an ontology, with the caveat that there are perfectly acceptable methods that make no use of perturbative methods at all and that you have to be careful ascribing reality to things that are strictly speaking mathematical fictions (for all the correct reasons that Tom pointed out).

You lost me here. I don't think that the virtual particle concept has anything to do with the measurement problem of quantum mechanics.. Moreover real-life photons are semi-localized in space and are not plane waves, just like all the particles. This is due to decoherence i think. Why does that have anything to do with the "reality" of the particle? Why is its "reality" at stake? I don't understand.

 

[tom.stoer]

I think it makes sense to come back to the original question, whether virtual particles do cause decoherence. The answer "no" and the following explanation was always based on a very precise, formal definition, namely "virtual particle = internal line in a Feynman diagram = propagator". So this result is still correct.

A discussion mixing "interpretation of QM and measurement" and "ontology of QFT" is certainly interesting but will unfortunately not lead to a reasonable result.

Instead one should ask "what caused decoherence?" and one will find out that the cause "separation of the world into system - pointer - environement and tracing out environment d.o.f." is quite robust and does not depend on any interpretation of real or virtual particles; it's a mathematical result. So I still think that real and virtual particles are rather irrelevant for decoherence.

 

[Haelfix]

"Real particles are manipulable, they are described by quantum states. Can we manipulate the second virtual photon from the electron-electron scattering? No, we cannot design such an experiment even in principle with the quantum mechanical formalism, because virtual photons are not described by a quantum state at any instant of time, hence impossibe to interact with them"

Well again it depends on if you want to define the problem away (and it is perfectly consistent to do just that) =) If you don't, you will find that it is actually ambiguous. You can define an experiment, it just won't necessarily tell you what you are looking for.

So what does it mean to be "manipulable" in an operational manner. It means that the particle can interact or scatter in some way.

So let's say we are looking at a weak process, something with say W exchange. A student asks, I want to know if virtual particles are really there or not. How would we find out?
Well, the only way you could find out, is by sending a probe in, with the right wavelength to actually resolve the very short times that the virtual W exists for. So draw the Feynman diagram, and add a very high energy photon that interacts with the internal W. Lo and behold you will indeed get a W particle out. So yes, the student says, it really is there! Virtual particles exist afterall!

But look again and manipulate the diagram, you can always reinterpret the diagram in another way. For instance, that the high energy photon actually decays *into* a new W, rather than kicking out something that was already there. This is a very typical complementarity of description. In order to see something you have to do to the system the very thing that you were trying to find out.

Now, the very nearly on-shell photon arriving from Alpha Centauri, is absolutely NO different in this regard, except that you don't need much energy at all to resolve it (read you don't need anything remotely like an ideal measuring device).

So again, when observing real-world particles, there is always some level of idealization involved (indeed they really are NOT particle states at all in the Fock space sense of the word).

 

[Haelfix]

Instead one should ask "what caused decoherence?" and one will find out that the cause "separation of the world into system - pointer - environement and tracing out environment d.o.f." is quite robust and does not depend on any interpretation of real or virtual particles; it's a mathematical result. So I still think that real and virtual particles are rather irrelevant for decoherence.

Absolutely, that much is assured.

 

[JK423]

So what does it mean to be "manipulable" in an operational manner. It means that the particle can interact or scatter in some way.

So let's say we are looking at a weak process, something with say W exchange. A student asks, I want to know if virtual particles are really there or not. How would we find out?
Well, the only way you could find out, is by sending a probe in, with the right wavelength to actually resolve the very short times that the virtual W exists for. So draw the Feynman diagram, and add a very high energy photon that interacts with the internal W. Lo and behold you will indeed get a W particle out. So yes, the student says, it really is there! Virtual particles exist afterall!

Two choices:

1) Let me known if i got this wrong, but you propose that an interaction between the probe-photon and the virtual W (internal line) is possible (see bold). First of all, can you write down (mathematically) this interaction (just use an arbitrary unitary operator, not a specific H_int? I would like to see the state space you will use for the internal W and how this connects with the propagator that defines the internal W in the first place. Also this statement is in sharp contrast to what tom.stoer says, so you can't agree with him at the same time. If you let the internal lines -in diagrams of perturbation theory- acquire a state, during the time Δt of their supposed existence, then they are able to cause decoherence as well!

2) In the case where what you propose has nothing to do with direct interaction between the probe and the internal W, and you mean something else, how would that establish the internal exchanged W as real, when the particular internal W that we are talking about has nothing to do with the interaction that produced the real W? I don't understand your argument at all. If the probe interacted with something else and a real particle W was emitted, the latter has nothing to do with the "internal-line' Ws.

So, what's your position?

But look again and manipulate the diagram, you can always reinterpret the diagram in another way. For instance, that the high energy photon actually decays *into* a new W, rather than kicking out something that was already there.

What does this have anything to do with the "reality" of the propagators that define the internal lines in Feynman diagrams? See (2) above.

So again, when observing real-world particles, there is always some level of idealization involved (indeed they really are NOT particle states at all in the Fock space sense of the word).

Real-world particles are not particle (Fock) states? Why would you say that!? Then what are they? I thought quantum mechanics was the most elementary description we had. If they are not states, really, what are they?
EDIT: I just noticed the "Fock space sense" thing, you aren't saying that they are not described by a state. If you'd like please elaborate a little on what you mean with "Fock space sense".

 

[Haelfix]

"If you let the internal lines -in diagrams of perturbation theory- acquire a state, during the time Δt of their supposed existence, then they are able to cause decoherence as well!"

First off.. Strictly speaking, particles don't *cause* decoherence. The environment does! You can never say too exactly what causes decoherence, you can only trace out the degrees of freedom in the joint density matrix and write down a characteristic time scale.

Second of all, in the above example it is now the REAL outgoing W that acquires a particle state (where the W is measured off at asymptotic infinity) WWbar just increased by one! The problem is in the interpretation. You can't tell whether a real high energy photon shimmied a virtual W and made it real, or alternatively a real high energy photon decayed into a real W. In either case, we are indeed talking about a real W end state.

But this is exactly the same thing that happens with the Alpha Centauri photon. Where say the virtual photon shimmies an atomic electron in an eye cell and then scatters off and becomes a real onshell particle. In both cases, you have 'detected' something. We just ascribe to the photon more reality, b/c it is so much longer lived.

"Real-world particles are not particle (Fock) states? Why would you say that!? Then what are they? I thought quantum mechanics was the most elementary description we had. If they are not states, really, what are they?"

In quantum field theory, mathematically, there is no such thing as a Fock space in the case of an interacting system. Much less an interacting system that has time dependance. This is a troubling statement when you first see it, but it is in fact true (it is called Haags theorem). Strictly speaking, we can only really do quantum mechanics with free particles.
That doesn't stop us from formally proceeding with such a construct anyway, but just be aware in the back of your head that there is always an approximation that is being taken when you apply mathematics to the realworld. And in this case, the approximation involves timescales.

 

[JK423]

"If you let the internal lines -in diagrams of perturbation theory- acquire a state, during the time Δt of their supposed existence, then they are able to cause decoherence as well!"

First off.. Strictly speaking, particles don't *cause* decoherence. The environment does! You can never say too exactly what causes decoherence, you can only trace out the degrees of freedom in the joint density matrix and write down a characteristic time scale.

Yes, your environment in this case is the sea of virtual particles (=internal lines). If you say that these are described by a quantum state, then they can cause decoherence to other quantum systems. So, please state your position precisely,
During the time Δt of the supposed existence of an internal line (in a Feynman diagram of perturbation theory), is it described by a quantum state or not?
It's a yes/no answer.. We don't have to complicate things.

Second of all, in the above example it is now the REAL outgoing W that acquires a particle state (where the W is measured off at asymptotic infinity) WWbar just increased by one! The problem is in the interpretation. You can't tell whether a real high energy photon shimmied a virtual W and made it real, or alternatively a real high energy photon decayed into a real W. In either case, we are indeed talking about a real W end state.

But this is exactly the same thing that happens with the Alpha Centauri photon. Where say the virtual photon shimmies an atomic electron in an eye cell and then scatters off and becomes a real onshell particle. In both cases, you have 'detected' something. We just ascribe to the photon more reality, b/c it is so much longer lived.

Here you are saying that the real W acquires a quantum state, implying that the initial internal line did not (your position is not clear at all). If the internal W is not described by a quantum state then how did your high energy photon interact with it? I keep asking you the same question all the time (just like i did with mfb-except that he didn't answer because he didn't understand the question), and i hope that you will answer in your next post, since it's the main crucial point (in my opinion) in this whole discussion:

Are the internal lines (in Feynman diagrams of perturbation theory) described by a quantum state during the time Δt of their supposed existence? Yes or No?

--- If your answer is Yes there are many implications; you come in conflict with tom.stoer first thing, you cannot agree with him, it's inconsistent. Internal lines can cause decoherence to other quantum systems. Moreover, this makes internal lines measureable, which is wrong. If that was the case then internal lines would be real as hell! Nothing virtual about them.

--- If your answer is No , then a system that is not described by a quantum state is not a quantum system by definition, hence internal lines will simply be a mathematical artifact. Hence we come to Demystifier's initial conclusion in an earlier post #4, that virtual particles (defined as internal lines in Feynman diagrams of perturbation theory) simply do not exist at all.

My answer is No. What is yours?

EDIT: Ofcourse the correct answer is not a matter of taste. The 'Yes' answer is mathematically inconsistent.

"Real-world particles are not particle (Fock) states? Why would you say that!? Then what are they? I thought quantum mechanics was the most elementary description we had. If they are not states, really, what are they?"

In quantum field theory, mathematically, there is no such thing as a Fock space in the case of an interacting system. Much less an interacting system that has time dependance. This is a troubling statement when you first see it, but it is in fact true (it is called Haags theorem). Strictly speaking, we can only really do quantum mechanics with free particles.
That doesn't stop us from formally proceeding with such a construct anyway, but just be aware in the back of your head that there is always an approximation that is being taken when you apply mathematics to the realworld. And in this case, the approximation involves timescales.

Very interesting! (and peculiar!) I'll search on this, thanks!

 

[TrickyDicky]

JK423, I'd like to know if you consider neutrinos, quarks and gluons as real particles according to what you are stating?

 

[JK423]

@TrickyDicky
A particle species can be both real and in an internal line as a mathematical artifact. I cannot tell you that ALL quarks are real! If you do perturbation theory and start drawing internal lines of quarks then these are fake.. If you want to be exact, start with a quantum state of quarks and gluons (which is real by definition) and take the time evolution; this will give you at each time 't' all the excitations of the quark and gluon field. These excitations are the real ones, even if they are not "external lines" but are in a bound state. Mathematically this seems like the most difficult task to do, to calculate the exact amplitudes of this time evolution, but that's irrelevant.

EDIT: I should express myself more clearly. All particles are real. It's the interpretation of perturbation theory that is not correct.

 

[rkastner]

'Virtual particles' have a clear physical meaning in my relativistic development of John Cramer's Transactional Interpretation (Possibilist TI or 'PTI'). Virtual particles can be understood as possible 'offer waves', in the sense that the amplitude for them to be emitted is equal to the coupling amplitude for the interaction under study.

Taking the coupling amplitudes as amplitudes for generation of both offer and confirmation waves provides for a clear account of measurement and the micro/macro 'cut'. This is similar to the idea of decoherence but it is more direct and does not suffer from the serious weakness of the improper mixture (see below). It is not primarily based on arguments about 'many degrees of freedom' (although that does play a role), but on taking into account the relativistic domain which is habitually ignored in discussions of the measurement problem.

See Chapter 6 of my new book, http://www.cambridge.org/us/knowledge/discountpromotion/?site_locale=en_US&code=L2TIQM

and this paper: http://arxiv.org/abs/1204.5227

In TI it's easy to see why virtual particles don't cause the effects usually attributed to 'decoherence' -- they are not really offer waves and they do not result in confirmations.
However, you don't need the usual 'decoherence' arguments (which don't really solve the measurement problem anyway, since decoherence just leads to an improper mixture that can't be interpreted as epistemic probabilities). Genuine collapse occurs in TI due to the presence of confirmations from absorbers, which dictate the measurement basis and provide a truly epistemic ('proper') mixed state.

 

[rkastner]

This is a classic discussion, and there isn't really a right or wrong answer here. Just levels of approximation, and what you are or are not prepared to take as a true statement (always like this with why questions), as Demystifier correctly points out, there is definitely semantics here.
...
Still, in so far as it might be *useful* to picture the photon that you absorb as really being there, then you are allowed to make use of such an ontology, with the caveat that there are perfectly acceptable methods that make no use of perturbative methods at all and that you have to be careful ascribing reality to things that are strictly speaking mathematical fictions (for all the correct reasons that Tom pointed out).

We can clear up the semantic problem by having a better definition of 'real' and 'virtual' photons. And there is one. Davies began grappling with this issue in his QED treatment of the Wheeler-Feynman theory (Davies 1971, 1972, email me if you want refs) but I don't think he quite solved it. The problem is solved unambiguously in the transactional picture as follows: virtual particles are unconfirmed, nascent offer waves; whereas real photons are confirmed offer waves resulting in actualized transactions. Virtual photons do not transmit real energy, while real photons do. But in TI, a 'real photon' is just an actualized transaction.
More quantitatively, in terms of the Davies theory, a virtual photon is just the time-symmetric propagator while a real photon corresponds to the pole in the Feynman propagator. (Davies considers the difficulty of a 'real' photon being an internal line -- but never quite solves it. It is solved by defining the real photon as an actualized transaction.)

As Davies noted, it is misleading to try to define the real vs virtual distinction in terms of 'off-shell' vs 'on-shell', since any emitted and detected photon has a finite lifetime. But the photon discussed in the example above is certainly a real photon since it transferred real energy to your eyeball. However virtual photons need not be regarded as mere 'fictions' -- yes, they are sub-empirical, but that doesn't mean they don't exist. People sometimes want to throw them out as mere artifacts of a purely mathematical process (perturbation theory), but arguably they play a genuine physical role, for example, in the Kondo effect. The mistake is to equate 'real' with 'empirical' i.e. to say if something is not detected or does not transfer real energy it cannot be 'real'. This is just a metaphysical presupposition. Remember that Ernst Mach thought atoms were fictitious, and he turned out to be wrong.

I argue in my book that the fundamental message of quantum theory is that there is a level of physical possibility beneath physical actuality. That is, as Heisenberg said, "Standing in the middle between the idea of an event and the actual event, a strange kind of physical reality just in the middle between possibility and reality.” He wasn't just speaking figuratively here. He was onto something.

 

[JK423]

@rkastner
Whether internal lines in Feynman diagrams of perturbation theory (=virtual particles) are described by a quantum state or not, during the time Δt of their supposed existence, is irrelevant to any interpretation of quantum mechanics.

 

[rkastner]

@rkastner
Whether internal lines in Feynman diagrams of perturbation theory (=virtual particles) are described by a quantum state or not, during the time Δt of their supposed existence, is irrelevant to any interpretation of quantum mechanics.

Why?

 

[JK423]

Why?

Interpretations are about interpreting the quantum formalism. The question whether something is described by a quantum state or not, is irrelevant to how this quantum state is interpreted. Am i wrong?

 

[rkastner]

Interpretations are about interpreting the quantum formalism. The question whether something is described by a quantum state or not, is irrelevant to how this quantum state is interpreted. Am i wrong?

Propagators are certainly part of the quantum formalism at the relativistic level. They describe how the quantum field is propagated. If one wants to argue that virtual particles (i.e. propagators) don't exist, then one is essentially saying that quantum fields don't exist -- yet they are the very basis of quantum field theory. It makes no sense to say that only excited states of the field exist but the fields themselves do not. If they don't exist, then there is nothing there to get excited.

And if one is trying to understand the physical meaning of quantum states (which is what theory interpretation is about), then it's relevant to consider whether a given entity is or is not described by a quantum state, isn't it?

 

[JK423]

Propagators are certainly part of the quantum formalism at the relativistic level. They describe how the quantum field is propagated. If one wants to argue that virtual particles (i.e. propagators) don't exist, then one is essentially saying that quantum fields don't exist -- yet they are the very basis of quantum field theory. It makes no sense to say that only excited states of the field exist but the fields themselves do not. If they don't exist, then there is nothing there to get excited.

And if one is trying to understand the physical meaning of quantum states (which is what theory interpretation is about), then it's relevant to consider whether a given entity is or is not described by a quantum state, isn't it?

I am not talking about propagators! A propagator is defined only when you have two events, that take place e.g. at times t₁ and t₂. What i am talking about is what happens inbetween these two times. A real excitation has a quantum state during that time. Does a virtual particle (defined as an internal line in Feynman diagrams of perturbation theory) acquires a quantum state during this time of its supposed existence?

EDIT:
Are you implying that there (quantum?) entities that are not described by a quantum state? The only quantum entities are quantum states! I am not aware of any other entity. Can you give me an example?

 

[tom.stoer]

Propagators are certainly part of the quantum formalism at the relativistic level.

yes, for a limited range of applicability.

If one wants to argue that virtual particles (i.e. propagators) don't exist, then one is essentially saying that quantum fields don't exist ...

in which sense do propagators exist? and in which sense do quantum fields exist?

It makes no sense to say that only excited states of the field exist but the fields themselves do not. /QUOTE]nobody is saying that.

suppose you have N radioactive atoms with decay constant k; in which sense do the atoms exist? and in which sense do the numbers (-k)^n / n! exist? do you understand why I think that you reasoning is not very plausible?

 

[Haelfix]

Yes, your environment in this case is the sea of virtual particles (=internal lines). If you say that these are described by a quantum state, then they can cause decoherence to other quantum systems. So, please state your position precisely,
During the time Δt of the supposed existence of an internal line (in a Feynman diagram of perturbation theory), is it described by a quantum state or not?
It's a yes/no answer.. We don't have to complicate things.

It is precisely the fact that you don't know the details of the environment (including what its quantum state is) that allows you to organize your Hilbert space in such a way as to allow for decoherence. If you did know what was in the environment, you would be forced to treat the problem differently. I keep harping on this point, b/c it makes the distinction between real and virtual completely irrelevant. At the level of detail that would allow you to discern the difference between the two (say by probing the system), would in fact also necessarily impy that you no longer have an 'environment'.

Anyway, once more. As to your other point. We agree. Virtual particles do not have a state. They do not even have energy-time uncertainty relationships full stop, period. At the level of the mathematics, this is completely unambigous. The problem is the model we use of the real world is not isomorphic to the above model, it is an approximation. The above model describes an idealized world where you measure particles off at infinity (and it makes certain assumptions about clustering, the asymptotic boundary conditions must be specified, and strictly speaking is only valid for nonabelian gauge fields, and really requires a lattice cutoff as well as perfect physically impossible detectors etc etc).

We'd like to use the same mathematics to describe particle interactions in a laboratory, which assuredly does not satisfy the above conditions. And indeed this works. However the price you pay for this, is that the distinction between real and virtual can be made arbitrarily small. So while it's true that the virtual particles don't have a state in this picture, NEITHER DO THE REAL 'realworld' particles.

 

[JK423]

Virtual particles do not have a state. They do not even have energy-time uncertainty relationships full stop, period. At the level of the mathematics, this is completely unambigous.

It's good that we agree on this.

The problem is the model we use of the real world is not isomorphic to the above model, it is an approximation. The above model describes an idealized world where you measure particles off at infinity (and it makes certain assumptions about clustering, the asymptotic boundary conditions must be specified, and strictly speaking is only valid for nonabelian gauge fields, and really requires a lattice cutoff as well as perfect physically impossible detectors etc etc).

We'd like to use the same mathematics to describe particle interactions in a laboratory, which assuredly does not satisfy the above conditions. And indeed this works. However the price you pay for this, is that the distinction between real and virtual can be made arbitrarily small. So while it's true that the virtual particles don't have a state in this picture, NEITHER DO THE REAL 'realworld' particles.

I agree that these mathematical issues are present, i read about Haag's theorem and its implications that you told me in an earlier post. But no one i think is completely certain as to what this theorem actually means since QFT works perfectly well in practice. But is this thing relevant to our discussion? All these problems have to do with the particular QFT model that we are using! Noone said that quantum theory is wrong due to these problems. For example, if string theory turns out to be correct then all these problems would dissappear. But string theory is based on quantum mechanics!

My point is the following:
According to quantum mechanics, there are only quantum systems and quantum systems are described by quantum states. There is nothing else! If something is not described by a quantum state then it does not exist(!) since it's not a quantum system, thus it can't be anything else. And that is independent of any model of quantum mechanics that you will use.

Do you agree?Edit: The fact that virtual particles (defined as internal lines in Feynman diagrams of perturbation series (sigh..)) do not acquire a quantum state has nothing to do with the mathematical difficulties that you mention. That's why i think that these mathematical difficulties are irrelevant to our discussion about the "reality" of virtual particles.

 

[rkastner]

"According to quantum mechanics, there are only quantum systems and quantum systems are described by quantum states. There is nothing else! If something is not described by a quantum state then it does not exist(!) since it's not a quantum system, thus it can't be anything else. And that is independent of any model of quantum mechanics that you will use.
Do you agree?"

These are all metaphysical presuppositions stated dogmatically. You're presupposing that quantum fields do not exist (and therefore don't propagate -- since you deny that propagators have physical content). So the burden is on you to explain how there can be excitations of something that doesn't exist (quantum states being excitations of quantum fields).

And Tom, yes I think people are denying the existence of fields when they deny the existence of the vacuum expectation value of the field (which is what a propagator is).

For further clarification of my proposed ontology of fields in the context of TI, please see Chapter 6 of my book or this paper: http://arxiv.org/abs/1204.5227

 

[JK423]

"According to quantum mechanics, there are only quantum systems and quantum systems are described by quantum states. There is nothing else! If something is not described by a quantum state then it does not exist(!) since it's not a quantum system, thus it can't be anything else. And that is independent of any model of quantum mechanics that you will use.
Do you agree?"

These are all metaphysical presuppositions stated dogmatically. You're presupposing that quantum fields do not exist (and therefore don't propagate -- since you deny that propagators have physical content). So the burden is on you to explain how there can be excitations of something that doesn't exist (quantum states being excitations of quantum fields).

First of all, what i said in the post you quoted is absolutely correct and standard. If you know any other systems other than quantum systems, and you know what physics describes these other systems, write a paper and publish it.

As for the explanation that you seek:
There is always an "excitation" present, the vacuum state |vac> of the field. This state can interact with other excitations and exchange energy so that other excitations can be formed {|n>}. All these things are described by a quantum state. Virtual particles are not. Period.

 

[rkastner]

There is always an "excitation" present, the vacuum state |vac> of the field. This state can interact with other excitations and exchange energy so that other excitations can be formed {|n>}. All these things are described by a quantum state. Virtual particles are not. Period.

Yes, there is energy in the vacuum state. What is it that has the energy? This actually supports my point: it is the field that has the energy. If you deny the existence of the field then there is nothing to which the real energy can be attributed.

Remember also that Feynman described the coupling constant for QED as the amplitude for a fermion to emit a virtual photon. There is a lot of really interesting physics going on in the area to which you want to summarily deny eligibility in the realm of physical systems.

 

[JK423]

Yes, there is energy in the vacuum state. What is it that has the energy? This actually supports my point: it is the field that has the energy. If you deny the existence of the field then there is nothing to which the real energy can be attributed.

Remember also that Feynman described the coupling constant for QED as the amplitude for a fermion to emit a virtual photon. There is a lot of really interesting physics going on in the area to which you want to summarily deny eligibility in the realm of physical systems.

First of all, please quote the post where i said that

(1) "fields do not exist".

I didn't say such a thing. What I said was :

(2) "Virtual particles (defined as internal lines in Feynman diagrams of perturbation theory) are not described by a quantum state, hence they are not quantum systems, thus they do not exist".

If you think that these two statements, (1) and (2), are equivalent, then it's you that have a great misunderstanding of the theory, not me. The vacuum state, is a quantum state, hence it's a quantum system. This state always exists, hence the field always exists. The virtual particles (defined as internal lines in Feynman diagrams of perturbation theory) are NOT described by the vacuum state, they are not described by ANY quantum state! Hence they are not quantum systems! If they are not quantum systems, then they cannot be anything else since only quantum systems exist, thus virtual particles (defined as internal lines in Feynman diagrams of perturbation theory) do not exist! Why is it so difficult to accept such a thing?

 

[tom.stoer]

And Tom, yes I think people are denying the existence of fields when they deny the existence of the vacuum expectation value of the field (which is what a propagator is).

Translated to my example you are saying that I deny the existence of the radioactive atoms b/c I deny the existence of the coefficients of a Taylor series. The atoms (and their mathematical) reprentations do exist on a different ontological level; there is no reason why the coefficients of the Taylor series which arise in an approximation shall be 'real' in the same way as the atoms.

A quantum state (like a free particle Fock state) is not identical with reality, but it represents something which has attributes observed in the real world (momentum, spin, physical polarization) whereas a (gauge boson) propagator has no such attributes (no 4-momentum, no spin, partial unphysical polarization, it may be a ghost d.o.f., it is gauge dependent whereas the attributes aren't, ...). Therefore it does not represent reality (in the sense we can define it when we want to have agreement with observations).

In your paper

For further clarification of my proposed ontology of fields in the context of TI, please see Chapter 6 of my book or this paper: http://arxiv.org/abs/1204.5227

I cannot see how you address and resolve gauge issues, especially in non-abelian gauge theories. In QCD there is no single phenomenon which can be understood using only perturbation theory. Therefore any interpretation focussing strictly on perturbation theory misses all key insights for QCD (it's like trying to understand logarithms, cuts and Riemann sheets using Taylor series; it's undefined or wrong)

Please refer to my post #35

Neither QED nor QCD require perturbation theory or virtual particles ... In QCD nearly everything requires non-perturbative methods (even in DIS - using perturbation theory - one probes non-perturbative structure functions)

There is one problem, namely that QED is ill-defined in the UV (Landau pole), in contrast to QCD which is UV complete.

Anyway, most perturbation series (QED, QCD, phi^4 theory, ...) are ill-defined and divergent, so perturbation theory does not make sense to arbitrary high order; its radius of convergence is zero.

I don't see how you can derive a correct interpretation (of a formalism) based on an undefined formalism producing incorrect results or no results at all.

 

[TrickyDicky]

I always understood "virtual particles"-internal Feynman propagator diagram lines- as the "dress" of those physical particles that don't have physical reality as bare particles(bare particles too only have mathematical existence actually). In that sense they exist as mathematical entities that are necessary to give a complete picture of certain physical particles, and the discussion about their existence is quite meaningless. Both the bare particle and its "dress" are necessary to describe what we observe as physical particles (for those that need it, quarks for instance don't follow this scheme but then again they are not directly observable, only indirectly thru the traces left by other particles).
What is misleading and really problematic in view of the repeated discussions is the name "virtual particles", because it seems to imply the autonomous existence of certain particles that for sure don't exist. But then again the same thing could be said about pointlike bare particles, they need to be dressed to count as real or existing physically.

 

[JK423]

@TrickyDicky
A bare electron always interact with the vacuum state of the electromagnetic field, no one says otherwise. This interaction gives rise to the so called "dressing" and renormalization etc. But the vacuum state is not virtual particles..
The latter are not described by the vacuum state or any other state, so it's wrong to think that around the electron there are virtual particles -being continuously created and annihilated- that "dress it".
So we shouldn't play with words here. The physical entity in this case is the vacuum state, so if you want to interpret something just interpret what the vacuum state is, that's fine with me. Sure thing is, that it has nothing to do with the "reality" of internal lines in Feynman diagrams of perturbation theory, since the latter are only mathematical artifacts.
"Dressing" exists irrespectively of whether you do perturbation theory or not.

 

[TrickyDicky]

@TrickyDicky
A bare electron always interact with the vacuum state of the electromagnetic field, no one says otherwise. This interaction gives rise to the so called "dressing" and renormalization etc. But the vacuum state is not virtual particles..
The latter are not described by the vacuum state or any other state, so it's wrong to think that around the electron there are virtual particles -being continuously created and annihilated- that "dress it".

Perturbative vacuum state(vanishing VEV) or the one associated to zero-point energy and that has measurable effects(Casimir effect) is what I usually identify with what is misnamed as "virtual particles". Do you know any other meaning for "virtual particles"?

So we shouldn't play with words here. Sure thing is, that it has nothing to do with the "reality" of internal lines in Feynman diagrams of perturbation theory, since the latter are only mathematical artifacts.

I get the impression that most of this thread is just that. You seem to be interested in imposing a clear cut semantic separation between existence-non existence and virtual-real, without realizing that taking that extreme position all particles must be considered as idealized mathematical artifacts.

"Dressing" exists irrespectively of whether you do perturbation theory or not.

I made clear I was not including the non-perturbative vacuum of quarks for instance.

 

[JK423]

Perturbative vacuum state or the one associated to zero-point energy and that has measurable effects(Casimir effect) is what I usually identify with what is misnamed as "virtual particles". Do you know any other meaning for "virtual particles"?

You can call the artifacts of perturbation theory however you want, that's fine with me. But the fact that you give them a fancy name doesn't make them real. The only relevant physical system is the vacuum state, and this has nothing to do with virtual particles since the latter are not described by a quantum state.

I get the impression that most of this thread is just that. You seem to be interested in imposing a clear cut semantic separation between existence-non existence and virtual-real, without realizing that taking that extreme position all particles must be considered as idealized mathematical artifacts.

Your impression is correct, because i am a very annoyed with all these misconceptions that are taught even in Universities and fool so many students, with "virtual particles" popping out from the vacuum etc. THERE IS a clear cut in this case, things that are described by states exist (well not always, depends on the model, see ghosts) and things that are not described by quantum states do not exist (in any model). This is such a clear definition of existence, which allows you to throw the notion of virtual particles as "something that exists" out of the window forever. They are only mathematical artifacts.
This is not an extreme position. If you think that it is, please elaborate so that i can understand what you have in mind.

 

[TrickyDicky]

You can call the artifacts of perturbation theory however you want, that's fine with me. But the fact that you give them a fancy name doesn't make them real. The only relevant physical system is the vacuum state, and this has nothing to do with virtual particles since the latter are not described by a quantum state.

If you only consider virtual particles as the pop-sci nonsense portraits, then you are right, otherwise you should know that Feynman internal lines have something to do with the perturbative vacuum state. Besides you are permanently using "quantum state" as synonym of "real" but that by itself is a matter of interpretation in quantum theory that opens a can of worms I'm not going to even try to address.

Your impression is correct, because i am a very annoyed with all these misconceptions that are taught even in Universities and fool so many students, with "virtual particles" popping out from the vacuum etc. THERE IS a clear cut in this case, things that are described by states exist (well not always, depends on the model, see ghosts) and things that are not described by quantum states do not exist (in any model). This is such a clear definition of existence, which allows you to throw the notion of virtual particles as "something that exists" out of the window forever. They are only mathematical artifacts.
This is not an extreme position. If you think that it is, please elaborate so that i can understand what you have in mind.

No serious physicists thinks about internal lines in Feynman diagrams as "particles popping in and out" , that's pop-sci stuff, I don't think people like Bill-k, mfb or Healfix take that nonsense seriously, you seem to be fighting a straw man.

 

[JK423]

I am not going to get inside people's heads to know what they are thinking. I made a simple question all this time about whether virtual particles are described by quantum states. The answer is "no", and for me that's everything i need to know about "virtual particles". Tom.stoer, Demystifier and Healfix agree on this answer.

Now, whether you can or cannot understand the significance of this fact is another matter. Saying that virtual particles are not described by quantum states during their "supposed existence", is such a great statement, that allows you to see these things as purely mathematical artifacts and stop considering ANY ontological significance that they may have in the real world. As Demystifier said, 1 Apple=+2 Apples + (-1) Apple, doesn't make the +2 and -1 Apples real.

I learned lots of things from this thread to be honest. I hadn't realized all these things before. Thanks PhysicsForums! ;)

 

[TrickyDicky]

I made a simple question all this time about whether virtual particles are described by quantum states. The answer is "no", and for me that's everything i need to know about "virtual particles".
Now, whether you can or cannot understand the significance of this fact is another matter

Fine, sci-popping in and out particles are not described by quantum states (in fact they are only described by pop-sci writers and some confused professors because they are just a silly picture for the math), that was agreed by me a few posts ago.

Saying that virtual particles are not described by quantum states during their "supposed existence", is such a great statement, that allows you to see these things as purely mathematical artifacts and stop considering ANY ontological significance that they may have in the real world.

If the ontology is what had you worried, rest assured "virtual particles" as autonomous entities have no ontological significance whatsoever. Note however that in general physicists are more interested in mathematical models that reflect as accurately as possible the measurement of observables, and ontological consideration are quite secondary.

I learned lots of things from this thread to be honest. I hadn't realized all these things before. Thanks PhysicsForums! ;)

Great.

 

[JK423]

Fine, sci-popping in and out particles are not described by quantum states (in fact they are only described by pop-sci writers and some confused professors because they are just a silly picture for the math), that was agreed by me a few posts ago.If the ontology is what had you worried, rest assured "virtual particles" as autonomous entities have no ontological significance whatsoever. Note however that in general physicists are more interested in mathematical models that reflect as accurately as possible the measurement of observables, and ontological consideration are quite secondary.

I am happy that we agree (and stopped playing with words!)!

I wonder, however, why in QFT textbooks the authors never (to my knowledge) warn the reader about the interpretation of perturbation theory and virtual particles, and talk about them like they are "really there" doing their stuff.

Example from Peskin (p. 13):
Even when there is not enough energy for pair creation, multiparticle states appear, for example, as intermediate states in second-order perturbation theory. We can think of such states as existing only for a very short time, according to the uncertainty principle ΔΕΔt=h. As we go to higher orders in perturbation theory, arbitrarily many such "virtual" particles can be created.

TrickyDicky you still think that i am fighting a straw man? Peskin completely confuses the reader from the first page. He says that "quantum states" are appearing that satisfy the energy-time uncertainty principle, when we said that this is not the case.

This thing is a crime to science and i am not exaggerating. Most of the PhD students (on experimental particle physics) that i have talked to about this issue, believe that "virtual particles are actually exchanged down there, real time". That's not their fault, it's scientific community's fault. Feynman, unwillingly, created a huge frustration to the future generation of students with his drawings..
And by the way it's not a coincidence that it's mostly the experimentalists (and not theorists) that are confused about virtual particles. They see diagrams with particles being exchanged for so many years, and at the same time most of them don't have the time to carefully study QFT and see for themselves what these things really are, so i cannot blame them.

 

[tom.stoer]

The 'explantation' from Peskin is unacceptable.

 

[TrickyDicky]

Example from Peskin (p. 13):
Even when there is not enough energy for pair creation, multiparticle states appear, for example, as intermediate states in second-order perturbation theory. We can think of such states as existing only for a very short time, according to the uncertainty principle ΔΕΔt=h. As we go to higher orders in perturbation theory, arbitrarily many such "virtual" particles can be created.

TrickyDicky you still think that i am fighting a straw man? Peskin completely confuses the reader from the first page. He says that "quantum states" are appearing that satisfy the energy-time uncertainty principle, when we said that this is not the case.

I wouldn't give so much significance to that introductory paragraph, when he writes "multiparticle states" he is simply justifying the necessity of dealing with more than a single particle in relativistic QM. The example is admittedly not very fortunate.
It is true that it might be misleading, but I don't know many textbooks on complex mathematical or physical matters that are not completely misleading at one point or another. Although it shouldn't be used as an excuse let's agree that writing/teaching is hard.

 

[tom.stoer]

I don't think that Feynman ever indicated how to interpret his drawings ontologically (I guess he would have hated this word) Feynman diagrams have been invented for bookkeeping.