# Virtual Particles?

1. Jul 18, 2012

### JPBenowitz

So as I was taught in Modern Physics virtual particles are allowed to exist in virtue of "borrowing" energy from the vacuum as long as it is "re-paid" in a short enough time to satisfy the Heisenberg Uncertainty Principle. However after doing further research this doesn't seem to be the whole story. Can someone explain what is really happening during these "borrowing" and "re-paying" stages and the mechanism behind the "on-shell in and out states" and the "off shell intermediary state"?

2. Jul 18, 2012

### The_Duck

I think most people (at least on this forum) don't like to speak of virtual particles as "borrowing energy." Energy is always exactly conserved, even in QM. Rather, virtual particles violate the "mass shell" relation of special relativity:

E^2 = p^2 + m^2.

For instance a virtual particle can have an energy much less than its rest mass; an example is the virtual W boson in beta decay. A process that involves a virtual particle that is very far off the mass shell is "penalized" for this violation and has a smaller probability of occurring. This is why the weak interaction is so weak: low energy weak processes involve W and Z bosons far off their mass shells, and thus don't happen very quickly. Furthermore, I believe we can qualitatively think of virtual particles as typically only existing for a period of time approximately equal to 1/(delta E) where "delta E" is the amount of energy by which the virtual particle fails to satisfy the mass shell relation.

An "on shell" particle, by contrast, is one that does satisfy the special relativistic relation E^2 = p^2 + m^2. Since delta E = 0 here, on-shell particles can exist for arbitrarily long time periods. Thus, the persistent particles we see around us are all on-shell. "In states" are states of two particles that you have arranged to collide with each other. "Out states" consist of the resulting debris particles that remain after the collision has finished. In both in and out states all the particles must be on-shell, because these states represent collects of stable, persistent particles.

3. Jul 18, 2012

### dipstik

virtual particles are mathematical oddities used when computing probabilities of events occuring. because the probability of an event with a particular energy goes to infinity, they use a virtual particle with varying energy and sum over all those possibilities to get the answer out.

so for electron scattering you might intergrate over photons of varying energy, but it is just a virtual photon that mediates forces and balances energy (in theory... using a bit of interpretation and other possible misgivings). whats interesting, it that the integration involves using the complex plane (so you use residue theory), so you will see a bunch of literature talk about the complex plane when reffering to virtual particles etc.

but i admit i have forgotten much of my qed/qft

4. Jul 18, 2012

### JPBenowitz

So then where exactly does the borrowing and repaying of energy from the Heisenberg Uncertainty Principle come in? Is this simply a matter of interpretation?

5. Jul 18, 2012

### tom.stoer

It' simply wrong; no energy is borrowed

6. Jul 18, 2012

### JPBenowitz

Then why is widely taught in universities?

7. Jul 18, 2012

### JPBenowitz

Also, if energy is not borrowed then doesn't this necessitate the existence of "virtual particles" with negative energy?

8. Jul 18, 2012

### nonequilibrium

I've often read it say "in quantum mechanics energy is exactly conserved". What's the backing of this claim?

9. Jul 18, 2012

### Mark M

The following:

"In quantum mechanics, energy of a quantum system is described by a self-adjoint (Hermite) operator called Hamiltonian, which acts on the Hilbert space (or a space of wave functions ) of the system. If the Hamiltonian is a time independent operator, emergence probability of the measurement result does not change in time over the evolution of the system. Thus the expectation value of energy is also time independent. The local energy conservation in quantum field theory is ensured by the quantum Noether's theorem for energy-momentum tensor operator. Note that due to the lack of the (universal) time operator in quantum theory, the uncertainty relations for time and energy are not fundamental in contrast to the position momentum uncertainty principle, and merely holds in specific cases (See Uncertainty principle). Energy at each fixed time can be precisely measured in principle without any problem caused by the time energy uncertainty relations. Thus the conservation of energy in time is a well defined concept even in quantum mechanics."

From wiki, conservation of energy.

To say that particles 'borrow energy' is nonsense. The conservation of energy is respected by quantum mechanics.

10. Jul 18, 2012

### tom.stoer

I don't know. Unfortunately you can read it in some text books, too. It's wrong; even as an interpretation it's wrong

Each virtual particles comes with a continuum of energy and momenta; see below

Regarding virtual particles i.e. internal lines in Feynman diagrams: these diagrams are graphical rules how to write down and calculate matrix elements; the virtual particles are nothing else but the internal lines - but please be aware of the fact that there is not one single particle that is exchanged but rather a continuum (of four-momenta), each coming with its propagator; now one has to do two things: glue these lines together at the vertices and integrate over all exchanged four-momenta; the interesting point is that each vertex comes with a delta function which guarantuess energy-momentum conservation at each vertex!

So E²-p² = m² (with rest mass m) does not hold along the lines, but energy-momentum-conservation holds at each vertex.

11. Jul 18, 2012

### JPBenowitz

I think I understand what you are saying. This sums it up real well:

"The most common use of the propagator is in calculating probability amplitudes for particle interactions using Feynman diagrams. These calculations are usually carried out in momentum space. In general, the amplitude gets a factor of the propagator for every internal line, that is, every line that does not represent an incoming or outgoing particle in the initial or final state. It will also get a factor proportional to, and similar in form to, an interaction term in the theory's Lagrangian for every internal vertex where lines meet. These prescriptions are known as Feynman rules.

Internal lines correspond to virtual particles. Since the propagator does not vanish for combinations of energy and momentum disallowed by the classical equations of motion, we say that the virtual particles are allowed to be off shell. In fact, since the propagator is obtained by inverting the wave equation, in general it will have singularities on shell.
The energy carried by the particle in the propagator can even be negative. This can be interpreted simply as the case in which, instead of a particle going one way, its antiparticle is going the other way, and therefore carrying an opposing flow of positive energy. The propagator encompasses both possibilities. It does mean that one has to be careful about minus signs for the case of fermions, whose propagators are not even functions in the energy and momentum (see below).

Virtual particles conserve energy and momentum. However, since they can be off shell, wherever the diagram contains a closed loop, the energies and momenta of the virtual particles participating in the loop will be partly unconstrained, since a change in a quantity for one particle in the loop can be balanced by an equal and opposite change in another. Therefore, every loop in a Feynman diagram requires an integral over a continuum of possible energies and momenta. In general, these integrals of products of propagators can diverge, a situation that must be handled by the process of renormalization.
" From Wikipedia

12. Jul 19, 2012

### tom.stoer

it does

13. Jul 19, 2012

### haushofer

Virtual particles are remnants of the fact that one is doing perturbation theory. If you would be able to solve e.g. in QED for the amplitude of an interaction analytically, you wouldn't need to do perturbation theory. And no "virtual particle" would show up, only in- and outstates.

14. Jul 24, 2012

### TrickyDicky

what I don't understand is the insistence on saying virtual particles are just a mathematical construct from working with perturbative methods while at the same time not acknowledging that they are a key element of the physical explanation of QFT and the SM independently of the calculation method.
I mean what about the 5 gauge bosons mediating interactions, they are virtual particles regardless the calculation method, they are considered physical, and are included in the SM tables of particles (where they are called the force carriers so no confusion with their field excitation counterparts is possible) even if they are actually the quantum fields of QFT, I can see calling them VP can lead to confusion since as commented above they are actually part of a momentum-energy continuum and maybe we should be always calling them quantum fields instead but as I said the high energy physics tendency to call anything a particle should in any case be the cause of the confusing terminology.
The bottom line is that most people understands that when talking about VP, what is actually meant is quantum fields, and quantum fields are a fundamental part of QFT and the SM that obviously should not be dependent on the particular calculation method, they are physical in as much as QFT can be considered a physical theory.

15. Jul 25, 2012

### tom.stoer

They are not independent; they exist only in the context of perturbation theory; there e.g. no such virtual particles in lattice gauge QCD.

Yes!

Yes!

The problem with the virtual particles is that they are presented such that many people think that they are in some sense the defining entities of QFT. It's like saying that numbers like +1,+1,+1,+1, ... define the concept of holomorphic functions simply b/c you know that 1/(1-z) = 1+z+z²+z³+... There are functions like exp(-1/z²) where these numbers do no longer exist (b/c there is no such series for z=0) but of course the function as a whole still does exist.

If you look at non-perturbative QCD you will find out that the whole concept of virtual particles is of rather limited importance (they are used to probe this regime, but they fail to define it).

Mathematically they are something totally different than "real particles". Real particle 'are' Hilbert space states, whereas virtual particles are integrals of propagators. The whole mathematical concept is different. Assume you have a theory dealing with functions f(z) and you call these object f(z) "functions". Now assume that you identify expansions like 1/(1-z) = 1+z+z²+z³+... and for some reason you call (+1,+1,+1,...) a function. That would be strange. Now think about an object like (+1,+1,+1,...) w/o any context of a "real function" like f(z) and assume that you call it a "virtual function". Then tell a friend that you have a theory of functions where the basic ingredient is a "virtual function" like (+1,+1,+1,...). I guess you agree that somewhere there was a wrong turn.

In addition there is this statement that virtual particles are allowed to borrow energy from the vacuum and therefore violate energy-mometnum conservatioin for a short time. Of course this is nonsense but you can still read this b...s... in many places. Even Steven Hawking explains his famous Hawking radiation in terms of virtual particles, but if you look at the math he used there are no such virtual particles; there is something totally different.

16. Jul 25, 2012

### vanhees71

I never understood what the expression "virtual particle" is good for. As this vague idea is used in the literature, I'd call it a propagator, and this is a rather abstract mathematical object in the formalism of QFT. It's defined as a two-point correlation function of the corresponding quantum fields. So indeed, virtual particles are simply representants of certain aspects of quantum fields.

It's called "particles" since the poles of the propagator represent dispersion laws $\omega(\vec{k})$ if they are located not too far from the real axis of the complex energy plane, i.e., if the propagator has a sharp peak around the position of the real part of the pole.

In this sense condensed-matter physicists use the term "quasiparticles", which however can mean something completely different than is suggested by the terminus "particle" to the lay man. Take phonons: They are typical examples for quasiparticles in solid-state physics, but they describe (perturbatively) the interaction of lattice vibration of the crystal with the electrons within the body. Thus the quasi particles rather represent certain aspects of collective motions (lattice vibrations, sound waves) of the body as a whole and nothing like a little point-like lump of matter, I'd understand as a "particle" in the usual sense.

17. Jul 25, 2012

### mikeph

I guess I was another person who was told about these virtual particles borrowing energy. I never understood what that meant. From reading this thread I guess it was one of those "white lies" you are told when you don't have all the prerequisite knowledge to understand what is really going on.

Only this time it's not chemistry at age 13, but quantum field theory at age 21...

18. Jul 25, 2012

### Lapidus

Virtual particles always conserve energy, but violate the mass shell condition.

When you combine the uncertainty relations from quantum physics with the mass shell condition from relativity you end up having virtual particles. They are a central concept of relativistic quantum theory and not tied to any calculation procedure, though they arise in perturbation theory explicitly.

They physical explain interactions that respect quantum physics and relativity. Thus they explain the fundamnetal interactions as obeserved in Nature. Strictly speaking, every particle that takes part in some interaction, that has been emitted or absorbed, is to some degree virtual. The distinction between virtual and real particle is an artificial one.

You can make any virtual particle observable by waiting long enough after it has interacted, or by supplying it with enough energy to make live long enough, so that it becomes an 'almost on-mass shell' particle.

19. Jul 25, 2012

### A. Neumaier

They are a key element in _visualizing_ processes in QFT that defy any simple explanation.
This is why they are important in teaching.

But it is a (widespread) mistake to attach any physical reality to them. They exist only in perturbation theoretical approaches - and they exist in this sense even for systems of _classical_ anharmonic oscillators! See http://physics.stackexchange.com/questions/4349/are-w-z-bosons-virtual-or-not/22064#22064

Last edited: Jul 25, 2012
20. Jul 25, 2012

### tom.stoer

I don't agree.

A real particle is a Hilbert or Fock space state, whereas a virtual particle is a propagator.

In order to do that you don't need any virtual particle; look at a non-perturbative formulation of QTF e.g. using a Hamiltonian. There are interaction terms but no virtual particles, only field operators and Hilbert space states.

21. Jul 30, 2012

### eudo

Is it not true that every real particle is an internal line in a larger Feynman diagram? That photon emitted from a star a million light years away interacts with an electron in your eye. Or that electron in your experiment isn't really a free electron; it interacts with a component of your detector...

22. Jul 30, 2012

### tom.stoer

You are hitting a weak point!

You are right, once you include the measurement device you don't have any external lines, only internal ones. But in the QFT formalism you don't do that. You always assume that detected particles are out-states i.e. real particles. Therefore the measurement device is never described in the QFT formalism

23. Jul 30, 2012

### eudo

Someone once said that it was funny that "real" magic is the kind that doesn't really exist, while the kind of magic that does exist is fake.

Perhaps we have something like that here. "Real" particles are the mathematical constructs that we can't really detect, while the particles that we really detect are the virtual ones :)

24. Jul 30, 2012

### Ilmrak

I agree that in QFT you don't describe measurement device.

Nevertheless in QFT the distinction between real and virtual particle is ill-defined in my opinion.

Consider for example a $W$ boson. Even if we usually think about it as a real particle, it continuously interact with the vacuum fields which renormalise its mass giving to it a finite life or equivalently a finite width in mass distribution (or a imaginary part in the propagator's pole). So the $W$ isn't exactly on the mass shell even when it's "real". We can say the same for every unstable particle.
So are $W$ "real" when they have a mass within $1 \sigma$ from the real part of the renormalized mass? $2 \sigma$? It's quite arbitrary...

Moreover I think the usual concept of particle is a convenient tool "only" in the perturbative theory. In an exact solution of a (strongly) interacting QFT the fields configuration which corresponds to the solution of the non interacting theory aren't so fundamental.

The only definition of particle I think could be "fundamental" is an irreducible representation of the theory symmetry group.

Ilm

25. Jul 30, 2012

### tom.stoer

Ilmrak, I tend to agree to most what you are saying; we had this discussion many toimes in many posts and I always said that the notion of "virtual particles" is reasonable in perturbation theory, but not fundamentally. I believe that the notion "particle" is reasonable in the sense you are indicating. A particle is a Hilbert space state in a irred. rep. of certain symmetry groups including both internal symmetries and Poincare group.