Why don't virtual particles cause decoherence?

[Maui]

Imagine I develop a new mathematical formalism, that is a good enough mathematical approximation to, say, Newtonian Mechanics, based on a given mathematical Serie.

Imagine I call each element of the Serie, "a little green dwarf", because I like it.

Would you say that those "little green dwarfs" are "real" or "physical"?

If i see a multitude of little green dwarfs all around me for a life time, i might be inclined to believe they are real and exist. The mathematics wouldn't work if it had no resemblance to reality. Why would it work otherwise? Just a happy coincidence?

Would you say that gravity exists because of the actions of those "little green dwarfs"

If little green dwarfs are the curvature of spacetime, yes.

 

[tom.stoer]

I guess we should come back to the question

I was recently told virtual particles don't cause decoherence. Why not?

My first answer was

Decohence is due to factorizing the full Hilbert space H in H_system, H_pointer and H_environment and then "tracing out" the environment degrees of freedom. The remaining "subsystem" can be described by an "effective density matrix" which is nearly diagonal in the pointer basis, so it seems as if it collapsed to the a pointer state with some classical probability.

Virtual particles are artifacts of perturbation theory, i.e they are not present in the full theory w/o using this approximation. Using virtual particles does not introduce the above mentioned factorization of H. And last but not least virtual particles are not states in any Hilbert space H_system, H_pointer or H_environment , but they are "integrals over propagators".

The discussion over the last couple of days did not change anything; the first answer is still correct.

Let me summarize some additional ideas

[one] can formulate QFT [and non-rel. QM] non-perturbatively w/o Feynman diagrams;

... there is no need to introduce perturbation theory and propagators when studying density operators.

... attributes of internal lines are gauge dependent whereas our observations aren't.

But all this is not directly relevant for the original question b/c virtual particles are completely irrelevant in the context of decoherence: they are not present in the full theory; they do neither introduce the above mentioned factorization of H nor the partial trace; and they are not states in any Hilbert space H_system, H_pointer or H_environment .

Last but not least: nobody would assume that any approximation like a Taylor series (or green dwarfs) do introduce additional effects which are not already present in the full theory w/o the approximation (w/o green dwarfs). So if the theory w/o virtual partices green dwarfs) already contains decoherence (gravity) it would be silly to say that decoherence (gravity) is due to virtual particles (green dwarfs). This changes if the theory cannot be formulated w/o virtual particles (w/o green dwarfs), or if the formulation is conceptally simpler (in the sense of Ockham's razor) using virtual particles (green dwarfs).

I am not an expert regarding green dwarfs, but I know that perturbation theory is incomplete and misses relevant non-perturbative effects. So I can't see any reason to rely on the interpretation of partially unphysical artifacts due to an incomplete approximation instead of using the full theory.

 

[Demystifier]

Where is the difference? An internal line in a Feynman diagram is not exactly on-shell, and particles not exactly on-shell are internal lines in Feynman diagrams.
Some particles are just more off-shell than others.

There are internal lines which are exactly on-shell.
There are external lines which are slightly off-shell (see e.g. http://en.wikipedia.org/wiki/Scharnhorst_effect ).

 

[TrickyDicky]

I guess we should come back to the question

Hi Tom, I tend to sympathize with the way you present the answer to this FAQ, but I have a doubt about the "virtual particles are just artifacts of perturbation theory" issue, you often use the example of QCD where this methodology is not necessary, but if we consider only QED ("the jewel of the physics crown") for a moment it does seem that perturbation is needed to obtain reasonable results, so in that sense at least for QED VP seem like something you can't get rid of so easily.

 

[tom.stoer]

Hi Tom, I tend to sympathize with the way you present the answer to this FAQ, but I have a doubt about the "virtual particles are just artifacts of perturbation theory" issue, you often use the example of QCD where this methodology is not necessary, but if we consider only QED ("the jewel of the physics crown") for a moment it does seem that perturbation is needed to obtain reasonable results, so in that sense at least for QED VP seem like something you can't get rid of so easily.

Neither QED nor QCD require perturbation theory or virtual particles. In QED non-perturbative aspects are either discused using rel. QM + radiative corrections a la Lamb shift, or they are not relevant at all (due to small alpha 1/137 and abelian gauge symmetry). So virtual particles are common standard and mostly sufficient in QED, but not required. One can quantize QED non-perturbatively w/o using 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.

 

[Demystifier]

... so perturbation theory does not make sense to arbitrary high order; its radius of convergence is zero.

Yes, and this important fact is usually ignored by those who try to ascribe some reality to virtual particles. Insisting on incorporating this fact into a realistic interpretation of virtual particles would be like saying that virtual particles are real only when their number is sufficiently small (say less than 137 in the case of QED). Which, of course, would not make sense.

 

[tom.stoer]

I think the problem is due to the way QFT lectures and textbooks are structured. 99% are perturbative methods.

 

[Demystifier]

I think the problem is due to the way QFT lectures and textbooks are structured. 99% are perturbative methods.

Yes. Another problem is due to the way popular-science books on quantum physics are written. They talk about virtual particles as of very vivid objects jumping around and sending messages between real particles, making them (real particles) know about each other. Once you get such a vivid picture, later it is very difficult to abandon it.

 

[mfb]

Yes, and this important fact is usually ignored by those who try to ascribe some reality to virtual particles.

I think you ascribe too much reality to real particles .

 

[Demystifier]

I think you ascribe too much reality to real particles .

Maybe. But even if real particles are less real than I think, I am quite confident that at least real particles are more real than the virtual ones.

 

[mfb]

I'm fine with "more real". A continuous spectrum from "very real" to "very unreal".

 

[JK423]

This thread is utterly confusing.. In order to investigate what virtual particles actually are, i would like to propose the following gedanken procedure. "Gedanken" because i don't think it's solvable, but it's very intuitive.
Since we care about what happens to the E/M field, let's compute the reduced density matrix of the E/M field at some arbitrary time t during the interaction of two electrons! What would we see?

Mathematically:
So, we start with the following initial states: two electrons in momentum eigenstates, and the E/M field in the vacuum state. We evolve this state via the (electromagnetic) interaction Hamiltonian, which couples the two electrons and the E/M field, and we evolve it for some finite time "t". We compute the reduced density matrix of the E/M field, by integrating the degrees of freedom of the eletron-field. Suppose that this interaction does not create any photons at the end of the day (i.e. for t→∞).

Question: What would we see for finite t? Would the vacuum state of the electromagnetic field transform to superpositions (or mixtures) of some number of photons? Since the final state of the field, for t→∞, is going to be a vacuum state, are these photons-in the aforementioned supeposition- what we call "virtual particles"? Can all this be, actually, calculated? Note that i have made no reference to perturbation theory, suppose that we could do the calculations non-perturbatively as well. If for finite "t", the vacuum state is transformed to non-zero photon number, then i would call these particles real, even though they disappear for t→∞.

 

[tom.stoer]

I would not say that 'real particles' are real, but I would start an 'ontological' interpretation on QFT based on Hilbert space states and their properties, not based on Feynman diagrams.

 

[tom.stoer]

... let's compute the reduced density matrix of the E/M field at some arbitrary time t during the interaction of two electrons!
... two electrons in momentum eigenstates, and the E/M field in the vacuum state.
... we compute the reduced density matrix of the E/M field, by integrating the degrees of freedom of the electromagnetic field.

So you want treat the el.-mag. field as environment to be integrated out?

The starting point with two electrons plus el.-mag. field in vacuum is strange; we should at least add the static Coulomb field (nondyn. d.o.f in Coulomb gauge)

 

[JK423]

So you want treat the el.-mag. field as environment to be integrated out?

The starting point with two electrons plus el.-mag. field in vacuum is strange; we should at least add the static Coulomb field (nondyn. d.o.f in Coulomb gauge)

You have incorrectly changed the quote of my post; i propose to integrate out the electron field, not the E/M field, since we care about the electromagnetic field's reduced state.

Anyway, my point is to see what happens to the field's state during the interaction. I've seen you tom.stoer arguing that virtual particles are just propagators in some integrals, not states. Well, ofcourse they are, because in some sense you integrate out the E/M field and are left with the propagators. But if we try to follow the time evolution of the E/M field's state during the interaction, even non-perturbatively, i bet that we will see excitations appearing that die out when t→∞. This is my intuition ofcourse, and is based on
\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 /> (1)
where \hat U\left( t \right) is the evolution operator, and \left| {vac} \right\rangle the E/M field's vacuum, while i have neglected the states of the electrons. In the case where no "real photons" are produced at the end of the interaction, it's
\left\langle n \right|\hat U\left( {t \to \infty } \right)\left| {vac} \right\rangle = 0\,,\,\,\,\,\forall n \ne vac, meaning that only the vacuum survives.

My question is:
Are these \left| n \right\rangle in (1) what we call virtual particles?

If the answer is positive then virtual particles are quite real to me, because if these equations are correct, a hypothetical measurement of the occupation number n during the interaction (i.e. for finite t) will reveil a non-zero number. (n = could be the occupation number of momentum eigenstates, whatever the basis, doesn't matter)

 

[tom.stoer]

You have incorrectly changed the quote of my post; i propose to integrate out the electron field, not the E/M field ...

Sorry for that, inserting, editing, backspace, "corrections", ... my fault!

But now I am even more confused b/c in QM it's the environment = the d.o.f. which are traced out which 'cause decoherence'. So if you want to see how photons create decoherence you have to trace them out.

In addition I do not understand your formulas; you seem to introduce basis states |n>, but as I said what we call virtual particles are not states. And you try to say something regarding decoherence, but you don't use a density operator ...

 

[JK423]

Well my post had nothing to do with decoherence, perhaps the "reduced d.m. formalism" created such an impression. What i care about is to see what happens to the state of the E/M field as i have pointed out in the previous posts.

 

[tom.stoer]

The |n> are not virtual particles but Fock states with well-defined momentum, spin etc., nothing I would call a virtual particle. Particles described by Fock states which can be counted by the number operator are not virtual particles. There is no number operator for virtual particles (propagators)

 

[Haelfix]

The proper way to talk about 'measurement' here, is to introduce Von Neuman measuring devices together with the appropriate kernel and response functions.

This further muddles the interpretation of what a 'real' particle is, since it invariably mixes with the measuring device and you don't have a pure plane wave state off at asymptotic infinity.

As a general rule, decoherence doesn't tell you what happens during a measurement. It only tells you what happens if you forget about some details of the system (similar to how entropy is often described). If you try to be specific about what exactly 'causes' this or that, then you enter a world of pain.

 

[JK423]

The |n> are not virtual particles but Fock states with well-defined momentum, spin etc., nothing I would call a virtual particle. Particles described by Fock states which can be counted by the number operator are not virtual particles. There is no number operator for virtual particles (propagators)

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

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

 

[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?