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When reading about quantum mechanics, I've stumbled upon a description of the force between two electrons as a result of virtual particles being "exchanged" by the two electrons.

So far I've understood that the virtual particles are a "prediction" of the uncertainty principle, and that their lifespan is inversely proportional with their mass.

Is an electron (or real particles in general) somehow stimulating the creation of virtual particles? If so, why?

If these guys pop in and out of existence, then won't they interfere with real particles? If so, then I guess the average effect on a particle is zero, as the creation is completely random along with the momenta of the particles. Is this correct?

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A bit of advice, when you start discussing physics at such a fundamental level as this, "why?" questions become truly futile. Virtual particle exchange is how we describe the fundamental interactions in QFT. If we could explain why particles interact in such ways in terms of more fundamental interactions, then that would just pass the buck as to what kind of interactions are the fundamental ones. I believe there's a great Feynman clip addressing this point somewhere on the net. –  David H Sep 3 '13 at 8:50
    
The question is fine. How many times do we hear academic physicists, Professors asking "why?" questions?! Clearly, no one is making a ludicrous anthropomorphism with an electron. In your example of fundamental interactions. Salaam might have asked; why is hypercharge quantised? Ah, because SM gauge groups are embedded in a non-Abelian simple group :) –  innisfree Sep 3 '13 at 9:01
    
I agree that the word "why" can ultimately be rendered meaningless in physics, but I use it here as a short form of "How is this explained, or imagined to be so?". –  lejon Sep 3 '13 at 9:24
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@DavidH youtube.com/watch?v=36GT2zI8lVA –  Chris White Sep 3 '13 at 21:57
    
@ChrisWhite Ah yes, that's exactly the clip I was talking about. I could listen to that man talk for hours. –  David H Sep 5 '13 at 1:59
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2 Answers 2

First of all, virtual particles are indeed a consequence of the uncertainty principle – without any quotation marks. Virtual particles are those that don't satisfy the correct dispersion relation $$ E = \sqrt{m^2 c^4 +p^2 c^2}$$ because they have a different value of energy by $\Delta E$. For such a "wrong" value of energy, they have to borrow (or lend) $\Delta E$ from the rest of the Universe. This is possible for a limited amount of time $\Delta t$ as long as the "negated" time-energy uncertainty relationship $$\Delta t \cdot \Delta E \leq \hbar / 2$$ is obeyed. One simply can't measure energy $E$ during too short an interval $\Delta t$ more accurately than with the error $\Delta E$ given by the formula above which makes it possible to borrow/lend this much energy for such a short time.

Pretty much by definition, virtual particles are effects that look like a temporary existence of a real particle which is bounded in time by the inequality above. The more virtual the particle is – the greater the deviation of the energy $\Delta E$ is – the shorter is the timescale over which the virtual particles may operate. In the limit $\Delta E\to 0$, the virtual particles become "real" which means that they may also be observed. For a nonzero value, they can't be observed and they're just "intermediate effects in between the measurements" that modify the behavior of other particles. Most explicitly, virtual particles appear as propagators (internal lines) of a Feynman diagram.

The electron is not necessarily "simulating" anyone, whatever "simulating" was supposed to mean. Instead, the electron may "emit" a virtual particle such as a photon. The emission of a real photon is impossible by the energy/momentum conservation: in the initial electron's rest frame, the energy is just $m_e c^2$ but it would get increased both by the extra kinetic energy of the final moving electron and by the positive photon's energy, thus violating the energy conservation law. But the electron may emit a virtual photon for which the energy conservation law is effectively violated (or the photon has a different energy, perhaps negative one, than it should have) which is OK for the time $\Delta t$ described above. As long as the photon disappears before this $\Delta t$ deadline arrives – it is absorbed by another charged particle, everything is fine and this intermediate history contributes to the probability amplitudes. That's why charged particles influence each other due to electromagnetism; this is how the virtual photons operate.

Concerning the last question, yes, virtual particles may interfere with the real ones. For example, if we study processes in an external electric field create by many coherent long-wavelength photons, there will still be Feynman diagrams with virtual photons in them. The amplitudes from these diagrams have to be added to the amplitudes with the real classical electric field, and only the result (sum) is squared in absolute value. That's what we mean by interference.

And yes, the effects of virtual particles on a isolated electron are equally likely in all directions and in this sense they "average out". An electron state with a sharply defined 3-momentum still remains an energy eigenstate and moves along a straight line. However, due to the constant emission and reabsorption of some virtual particles, the real electron-like energy eigenstate has a "cloud" of virtual photons around it. The symmetries of the theory such as the gauge symmetry and the Lorentz symmetry aren't broken by the virtual photons. After all, the virtual photons result from the theory whose Lagrangian does respect these symmetries and no anomaly breaks them.

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If the virtual photon can be absorbed before $\Delta t$ deadline arrives, shouldn't this mean that such interaction is possible only in very small space scales because of finite speed of light? How can electrons interact on larger scales then? Or maybe virtual photon can have arbitrarily small energy which would let it live for long enough to travel very far? –  Ruslan Sep 3 '13 at 11:45
    
Thanks for an elaborate answer. What exactly triggers the emission of a virtual photon from an electron? Since the force between two electrons is detectable, the electrons must emit more virtual photons than empty space for the force to be measured. –  lejon Sep 3 '13 at 12:26
    
@Nick: I have approved your suggested edit, bhut onnly because of the \geq vs. \leq. –  Dimensio1n0 Sep 3 '13 at 12:56
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I don't know if I agree with this answer--I don't really see how you can justify virtual photons as a "real" entity at all--they are constructions permitted by the fact that you happen to be able to write the interaction hamiltonian in terms of vertices and propogators. But they're an artifact of the perturbation theory--if you could solve the exact scattering problem, you woulnd't have a linear combination of Feynman diagrams, you'd just have a nonperturbative solution of the scattering problem (at least, i haven't seen a proof that the perturbation series converges to an exact solution). –  Jerry Schirmer Sep 3 '13 at 20:07
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Dear @Jerry, according to quantum mechanics, nothing - not even "real particles" - is real in the classical sense so your objection is completely immaterial in the context of QM. You say that virtual particles are artifacts of perturbation theory - propagators in perturbative Feynman diagrams - and you are right. But real particles of "any particular types" are also artifacts of perturbation theory. They're created by creation operators that only have the simple commutator with the Hamiltonian in the free-theory limit. –  Luboš Motl Sep 5 '13 at 15:08
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I have encountered this question elsewhere on the web, but I can't seem to remember. My answer assumes that you are referring to the virtual particles that mediate the forces.

The idea here is to understand the S-Matrix and scattering amplitudes. QFT is the study of the birth and death of particles between 2 observations. Every QFT scattering interaction has an 'in' and 'out' state, where the states are defined in the Hartree-Fock scheme. What we observe are the in and out states only.

I will give an example of the scattering of an electron in an electrical field. So the in state (eigenstate we created) consists of an electron in a certain state of momentum and spin and a photon (from the applied field) in a certain momentum and helicity.

We leave the system undisturbed and after a long time we observe it again. We now have the out state. There is a probability of getting all kinds of configurations of the electron and a photon (constrained only by the preservation of momentum, energy, charge etc.).

We want to find out the probability for obtaining all kinds of out states. We use series expansions to calculate this amplitude. Feynman and co-workers came up with tools to keep track of the terms of this series. We represent terms of each term of a series (where oen out state corresponds to a whole series) using Feynman diagrams. To facilitate the easy calculation, we use virtual particles in these diagrams.

As QFT deals with discontinuous observations, there are no actual photons being exchanged, just before and after states. To make the math easy we imagine that virtual particles were exchanged between the in and out states.

I recommend any standard QFT (Antony Zee for example) book to get a clear picture of how we use virtual particles as a tool. I hope this answer is satisfactory.

EDIT: I have taken the time interval to be infinite as I mentioned earlier. It is never really the case in real life, but it allows energy to be conserved in my scattering process.

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