# Why don't virtual particles cause decoherence?

by Mukilab
Tags: decoherence, particles, virtual
 Sci Advisor P: 5,437 I think the problem is due to the way QFT lectures and textbooks are structured. 99% are perturbative methods.
P: 4,591
 Quote by tom.stoer 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.
Mentor
P: 11,837
 Quote by Demystifier 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 .
P: 4,591
 Quote by mfb 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.
 Mentor P: 11,837 I'm fine with "more real". A continuous spectrum from "very real" to "very unreal".
 P: 381 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→∞.
 Sci Advisor P: 5,437 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.
P: 5,437
 Quote by JK423 ... 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)
P: 381
 Quote by tom.stoer 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
$\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|} \hat U\left( t \right)\left| {vac} \right\rangle \,\,\,\left| n \right\rangle,$ (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)
P: 5,437
 Quote by JK423 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 ...
 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.
 Sci Advisor P: 5,437 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)
 Sci Advisor P: 1,685 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.
P: 381
 Quote by 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)
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
P: 5,437
 Quote by JK423 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.
P: 740
 Quote by JK423 Anyway, my point is to see what happens to the field's state during the interaction.

 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.
P: 5,437
 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.
P: 381
$\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle$
is the amplitude of having n photons, $\left| n \right\rangle$, so in this case we have an amplitude associated with a quantum state,
$\hat U\left( t \right)\left| {vac} \right\rangle = \sum\limits_n {\left\langle n \right|\hat U\left( t \right)\left| {vac} \right\rangle \left| n \right\rangle }.$