# Annihilation of virtual e-e+ pair

1. Jun 17, 2005

### Kruger

I read a book and there it is written something as following:

"By the lifetime of a virtual pair (e-e+) (that is given by the HUP) a virtual e- can remove with the distance s from the e+ that is given by: s=c*d(t) (d(t) from HUP)."

My question: How can they annihilate if they are away at a distance s? I mean haven't they to collide to annihilate?

2. Jun 19, 2005

### Kruger

doesn't somebody know that?

3. Jun 19, 2005

### marlon

Virtual particles have a definite momentum value, the HUP then says their uncertainty on position is infinite. Basically virtual particles are everywhere...you cannot make statements concerning their path or their position at some time t. No, they don't have to be together to annihilate. Don't think of this annihilation as a collision like real on mass shell particles, though ofcourse you can look at it like that. All i am saying is that this is not required because of the infinite uncertainty on the position. Suppose two virtual particles are very far apart, i can also say they are very close to each other. Why ? Well, because suppose you are able to measure their position once, you will get a totally different value, the second time you measure their position (let us assume they are standing still). This is the manifestation of the HUP.

Besides, your formula implies that they move at the speed of light, but then again virtual particles are off mass shell, so they can exceed the speed of light.

Really what book did you read this in and within what context ?

regards
marlon

4. Jun 19, 2005

### Kruger

@marlon: I'm reading P.W. Milloni's book (The quantum vacuum and an introduction to QED). I'm now at the part where he wrote something about the Dirac equation and Dirac's hole theory.

The second thing that confused me is that he says we can look on zero point energy in two ways. One way is that zero point energy is a energy that can be withoug any photons and one is that zero point energy is made up by virtual photons (of course 0-point energy of the em field).

But what is the common interpretation of 0-point-energy of em field

and can these virtual (zero point energy) photons make pair creation (if momentum is conserved)?

Last edited: Jun 19, 2005
5. Jun 19, 2005

### marlon

What they mean by the first vision is just that the ZPE corresponds to physical phenomena (like the strong interaction or EM-radiation, ...) in their lowest possible energy-state which is ofcourse non-zero. Since a photon (just like any other particle in QFT) is a manifestation of the transition of one energy level to another, you do not need them because the energy value of the EM-radiaiton is constant and non-zero. Just compare this with the surface of the sea. The waves correspond to particles but if the surface is flat, there are no particles to talk about.

The second vision with virtual particles is just the fact that during a period of time the energy conservation can be violated and transition from energy level to another is possible : this is the creation of a virtual particle. Ofcourse, in that short period the energy has to be 'restored' to the original zero-point value and thus the virtual particle must be annihilated shortely after the creation. Just think of the sea as going up and down in that period delta t. Hence this is the virtual particle-sea picture that was invented by Dirac.
This creation can be one virtual particle that dies shortely after. the pair creation only is necessary when you are working with charged virtual particles and charge conservation needs to be respected. ame goes for spin conservation and so on...the conservation laws are the actual referee that govern the game that is going on. Momentum is always conserved by virtual particles. Just look at the Feynmann-diagram-rules and this also implies that virtual particles are everywhere in space which has a marvellous calculatory advantage in both QED and QCD when calculating propagators.

Study this matter well because a lot of people have difficulties understanding this. Good Luck

regards
marlon

6. Jun 19, 2005

### Kruger

I try it. But sometimes I've about one hour to understand the sense behind an equation.

Last edited: Jun 20, 2005
7. Jun 20, 2005

### marlon

don't worry, i know perfectly what you mean :)

marlon, hang on in there...you will get by

8. Jul 8, 2005

### Kruger

In virtual pair creation where a virtual photon created from vacuum creates an electron-positron-pair the electron and positron must have an opposite momentun.
Why? I mean momentum conservation is not there: p(photon)=p(e-)+p(e+)=0 while p(e-)=-p(e+).

9. Jul 10, 2005

### Kruger

Mhh, don't look at my last post. marlon, you say that only one virtual photon can be created from the vacuum with a well defined momentum. But where is momentum conservation? I mean there must be two virual photons created. else you have no momentum conservation.

Last edited: Jul 10, 2005
10. Jul 10, 2005

### marlon

Kruger,

First of all you need to be aware of the fact that the momentum conservation needs to be respected along the internal Feynman-loops and at the initial and final vertex points.

Besides, you impose the condition for momentum conservation at a vertex point, so the number of external incident lines at one vertex point determines the number of internal lines, starting from that vertex point. The number of vertices equals the order of perturbationtheory you are working in.

For example in first order perturbation theory you can have a vaccuum fluctuation at a vertex. The associated Feynman diagram can be seen as an infinity-sign standing up like an 8 (ie two loops that touch each other at the vertex-point). The vertex is at the crossing of the two loops. The two loops of the 8 correspond to two virtual particles, each with one internal momentum (i and j) and i+j=0

Ofcourse, you know that when calculating the amplitude of this interaction, you will integrate over the internal momenta, expressing the summation over virtual transition states in QM-perturbation theory. This implies that the value for i or j basically can be anything that you want because of this integration. So i suggest you try calculating this and check out the possibilities for yourself.

regards
marlon

Last edited: Jul 10, 2005
11. Jul 12, 2005

### Kruger

Ok, First consider a vacuum diagramm:

1. there are no external lines
2. At vertex a virtual photon gets away, of course with an well defined momentum p.
3. The photon "gets" into an electron positron pair. momentum of electron and positron is something like p/2.
4. electron-positron annihilates, of course into two virtual photons.
5. the virtual photons annihilate into one virtual photon.
6. virtual photon gets away at vertex.

Watch the Feynman diagram I painted .
Hope that should be right.

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12. Jul 12, 2005

### marlon

you have two external lines actually. These lines correspond to one on mass shell particle each, so they cannot be virtual. then you have indeed two internal lines that stem from one vertex point to the other. These internal lines correspond to two virtual particles that do not mutually interact. At each vertex you impose conservation of momentum and along the internal lines, energy conservation is not respected. the fact that you have two vertices means you are working in second order perturbationtheory (ie second order in the coupling constant, whatever that may be here).

Just google for Feynmann diagrams to find out more on these interactions or consult your university course if you have one, or zee's book or Weinberg, Peskin,... :)

Due to the present external lines, this sure as hell cannot be a vaccuum fluctuation. In this case, the virtual particles are force carriers.
marlon

Last edited: Jul 12, 2005
13. Jul 13, 2005

### Kruger

Ok, I got now the two books (Weinberg and Peskin) - the two introductions to QFT.
I think I understand it now, the things with the vacuum bubbles or diagrams.
A virtual photon must not have energy of 2mc^2 to create virtual electron positron pair, since these virtual particles are off mass shell and do not have to obey Einstein's energy relation.

Last edited: Jul 13, 2005
14. Jul 14, 2005

### Kruger

Lets make a thought experiment:

virtual photon created:
momentum p, energy E

it "splitts" in e-e+:
momentum(e-) k, energy Ee,
momentum(e+) k-p, energy Ep

While the energy relation is not sadisfied:
Ee^2 not equal to c^2k^2+m^2c^4

The mass has to be that of an electon, the momentum has to be conserved.
Now, isn't E=Ee+Ep in this process. Energy conservation only at "begin" (no energy, after virtual photon created) and then at end (no energy, virtual photon "reabsorbed from the vacuum")?

You see my point?

15. Jul 14, 2005

### marlon

you are correct but you keep on mixing up things. I mean, the process that you describe is NOT a vaccuum fluctuation. You are describing the interaction between real particles (ie the external lines) via virtual gauge bosons. the mistake you make is tha you say well i start from a virtual photon. This photon must be real if you wanna discribe such a process.

A vaccuum fluctuation is described by a CLOSED loop that starts and ends at the very same point, so this cannot be the case in your example.

So just associate real particles with the external lines and you are completely correct here. Your words on the momentum and energy 'conservation' considerations are correct, though.

marlon

16. Jul 22, 2005

### Antiphon

Marlon, are you saying that for a virtual particle $$\psi(x,t) = e^{ikx} f(t)$$ because they have a definite momentum?

If so , what might the $$f(t)$$ look like?

17. Jul 23, 2005

### marlon

You know that virtual particles are described in terms of propagators because they arise thanks to an interaction going on. This equation does not describe this. Look up the Feynman diagram rules for that matter. Besides you need to apply a summation over all internal momenta value, which corresponds to the summation over intermediate states in QM. You 'll also integrate over all position coordinates at each vertex, expressing the fact there is no preferred place in space time where the interaction is going on. This implies that the mediating virtual particle (or the virtual particle associated with a vacuum fluctuation) really is everywhere once the interaction is 'turned on', hence we have an infinite spread on position. Thanks to Heisenberg we know that the momentum is exact.

marlon