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Why don't virtual particles cause decoherence? 
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#37
Feb1913, 02:11 AM

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



#38
Feb1913, 04:09 AM

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#39
Feb1913, 04:26 AM

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#40
Feb1913, 04:33 AM

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#41
Feb1913, 04:37 AM

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



#42
Feb1913, 11:29 AM

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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 eletronfield. 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 photonsin 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 nonperturbatively as well. If for finite "t", the vacuum state is transformed to nonzero photon number, then i would call these particles real, even though they disappear for t→∞. 


#43
Feb1913, 12:42 PM

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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.



#44
Feb1913, 01:32 PM

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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) 


#45
Feb1913, 04:32 PM

P: 381

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 nonperturbatively, i bet that we will see excitations appearing that die out when t→∞. This is my intuition ofcourse, and is based on [itex]\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, [/itex] (1) where [itex]\hat U\left( t \right)[/itex] is the evolution operator, and [itex]\left {vac} \right\rangle [/itex] 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 [itex]\left\langle n \right\hat U\left( {t \to \infty } \right)\left {vac} \right\rangle = 0\,,\,\,\,\,\forall n \ne vac [/itex], meaning that only the vacuum survives. My question is: Are these [itex]\left n \right\rangle [/itex] 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 nonzero number. (n = could be the occupation number of momentum eigenstates, whatever the basis, doesn't matter) 


#46
Feb1913, 04:48 PM

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P: 5,366

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 ... 


#47
Feb1913, 04:51 PM

P: 381

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.



#48
Feb1913, 06:16 PM

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The n> are not virtual particles but Fock states with welldefined 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)



#49
Feb1913, 06:21 PM

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P: 1,677

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. 


#50
Feb2013, 03:28 AM

P: 381

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 


#51
Feb2013, 08:55 AM

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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. 


#52
Feb2013, 12:44 PM

P: 719

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 onshell...still I'm wondering if there is a connection. After all, these are two approaches for the same physical situation. 


#53
Feb2013, 04:40 PM

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#54
Feb2113, 05:53 AM

P: 381

[itex]\left\langle n \right\hat U\left( t \right)\left {vac} \right\rangle [/itex] is the amplitude of having n photons, [itex]\left n \right\rangle [/itex], so in this case we have an amplitude associated with a quantum state, [itex]\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 }. [/itex] 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/onshell particles!). A quantum object is described by a quantum state, period. If for any reason my understanding is mistaken, please let me know. 


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