## Why don't virtual particles cause decoherence?

 Quote by JK423 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?

 Quote by JK423 $\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|} \hat U\left( t \right)\left| {vac} \right\rangle \,\,\,\left| n \right\rangle,$
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.

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 Quote by kith 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.

 Quote by kith 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.

 Quote by kith 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 gives the probability for state 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.

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.

 Quote by JK423 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.

 Quote by kith 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..

Mentor
 Quote by JK423 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.

 Quote by mfb 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.

Mentor
 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?

 Quote by mfb 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?

Mentor
 Quote by JK423 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?

 Quote by mfb 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.

 Quote by mfb 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.

 Quote by mfb 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.

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see #8
 Quote by Demystifier 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.

Mentor
 Quote by JK423 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.

 Recognitions: Science Advisor 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 wont 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).

First of all, thank you for the analysis.
 Quote by Haelfix 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.

 Quote by Haelfix 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 wont 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.

 Recognitions: Science Advisor 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.
 Recognitions: Science Advisor "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 lets 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).
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