# Can pair production happen in vacuum?

1. Apr 19, 2005

### skpang82

Hi.

Can pair production happen in vacuum?

I had this question for a test. And the answer is: no, it cannot because it would violate the conservation of momentum, since you can't have a momentum zero photon.

But i seem to recall that pair production happen in space (?). Due to Heisenberg's uncertainty relation, there is a constant energy flux, which allows for pair production. The pairs are produced, and they annihilate each other again within a certain time frame. And if it happens near a black hole, one of the pair may pass beyond the event horizon of the black hole, thus making it appear as if the black hole is radiating. I think i got this info quite some time ago when i was reading on Stephen Hawking. Not sure if i remember it correctly.

2. Apr 19, 2005

### marlon

The answer is YES. Indeed conservation of both energy and momentum will be violated but during some short period of time, that is actually allowed by the HUP. Ofcourse we are talking about virtual particles here...Check out :
https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=2 [Broken]

Scroll down to the 'what are virtual particles' entry

marlon

Last edited by a moderator: May 2, 2017
3. Apr 19, 2005

### dextercioby

Why are you treating differently "space" and "vacuum"...?In QFT,"vacuum" has another description,at least in the interacting theories...QED is a good example.

Free field theories have the same notion of vacuum just like the "space" in SR and Class.Field Theories.The fields are quantized,yet the "vacuum state" is seen as the state of zero particles and defined through

$$\hat{a}|0\rangle=|0\rangle$$

,where $$\hat{a}$$ is a generic notation for an annihilation operator...

In interacting field theories,the notion of vacuum is different.It can accomodate particles.

Daniel.

4. Apr 19, 2005

### ZapperZ

Staff Emeritus
You are confusing between "pair production" and "virtual particles".

Pair production produces a particle-antiparticle pair from high energy photons. These particles are REAL, they can interact, and they don't need to die before the rest of the universe detect them. Their lifetimes are not limited by the HUP. And yes, they need to obey both conservation of momentum and energy.

Virtual particles, on the other hand, can appear out of nowhere. When this happens, they temporarily violate conservation of energy (or energy+mass, to be exact), but not conservation of momentum (they exist only via an interaction). However, their lifetimes and other properties are confined within what are allowed by the HUP.

Do not confuse the two...

Zz.

5. Apr 19, 2005

### EL

Must say I'm a little sceptic when it comes to speaking about virtual particles as if they truely exist (which I'm not sure if you are doing or not). If something only can exist virtually, how can we say it is there?

6. Apr 19, 2005

### marlon

I suggest you look up the Casimir force or Casimir effect...Try Wikipedia.

ps : a general remark : i have been reading that only energy conservation is violated when virtual particles arise. This is only partially correct : also momentum conservation can be violated during the existence (ie the lifetime) of virtual particles. A particle can emit a virtual particle that is more heavy then the actual emitting particle. In calculating the propagator's amplitude one integrates over all possible internal momenta. Ofcourse eventually momentum conservation must be respected between the initial and final state. In interactions, the virtual particles will carry a definite momentum-value (so they are everywhere because of the HUP). One value will respect the momentum conservation law, however other values are also present. Besides, in QFT one can prove that the amplitude of the propagator lowers if virtual particles are more and more in violation with energy conservation laws.Besides it is also wrong to claim that virtual particles exist ONLY because of some interaction. This is certainly not enough. Other then that , they exist because of the non-zero-groundstate value (non-zero zero point energy) and MAINLY because of the HUP. However, we need to make sure that we are not talking about semantics here. It is only when you compare the very initial state with the very final state that both energy and momentum MUST be conserved...

regards
marlon

Last edited: Apr 19, 2005
7. Apr 19, 2005

### EL

Dear Marlon. I'm well aware of the Casimir effect. What I wanted to discuss was mearly the semantics.
I mean, virtual particles are not what we usually call particles, but only abstract interpretations of mathematical expressions arising in QFT.
Therefore, if someone asks wheter som "particles" are created, it sounds a little strange to answer: YES, "virtual particles". But of course, as long as everyone knows what it ment by a virtual particle it is ok. However it may be confusing for anyone not so familiar with it.

8. Apr 19, 2005

### marlon

Ahh so you are not argiung that virtual particles are indeed REAL...

Sorry but then i misinterpreted your words...again apologies

regards
marlon

9. Apr 19, 2005

### ZapperZ

Staff Emeritus
Here's the deal...

If virtual particles picture is valid, then there are a number of consequences. One of them being higher order interactions such as those we obtain when we do perturbation expansion. We see this often via the self-energy interactions such as in QED.

Guess what? This is how QED can make the most accurate prediction of the electron gyromagnetic, something that classical field theory could not. Higher order corrections based on quantum field theory picture produce a more accurate value of a physical quantity that we measure.

Now we can argue back and forth on the philosophical definition if something is "real". However as an experimentalist, if I see a theoretical idea in which the consequences of that idea actually predicts with very high accuracy of a measured quantity, I'd say there's something to that theory. Such a description is better and more accurate. In the end, that's as best as we can say.

Zz.

10. Apr 20, 2005

### EL

Just to make myself more clear. I'm really quite familiar with QED, and I'm absolutely NOT questioning the theory.
What I'm questioning is the use of the term "virtual particles" as if they really are "particles". We just choose to call parts of the mathematical expressions arising in QED for "virtual particles" since it makes it easier for us to keep track on what the end results, i.e. what can be measured, will be.

11. Apr 20, 2005

### dextercioby

Okay,let us call them "virtual particles" and u call'em "internal lines"...

Daniel.

12. Apr 20, 2005

### EL

I call the "internal lines" for "virtual particles" all the time. It's not about that. Just wanted to cleraify for "skpang82", who is maybe not so familiar with the concept.
I know you guys are quite fast to debunk (and you do it really well!), but in this case you have misinterpreted my intentions.

Last edited: Apr 20, 2005
13. Apr 20, 2005

### ZapperZ

Staff Emeritus
But think of it this way - if we consider them as real particles, we would not have attached the word "virtual" to them. We'd just call them.... er..... PARTICLES!

The fact that we don't, and call it by an additional qualifier (virtual), implies that they have additional limitations to them. We can go on with this forever and include other "particles" such as phonons, magnons, spinons, holons, polaritons, etc. etc.... Some of these do not even have the world virtual attached to them, yet, we still loosely call them "particles" in the sense that we treat them the same way as any exchange particle in QFT.

Zz.

14. Apr 20, 2005

### EL

Yes, yes, yes, yes, yes. I agree, I understand.
You are reading something between my lines which was not ment to be there.

15. Apr 22, 2005

### Sterj

I read this thread and it seems to be interesting (not the part of pair creaton where all particles are real but the part with virtual particles).

I think at something: Consider a system that is in vacuum state. The system contains a nuclei and a electromagnetic field in ground state (actually one oscillation).

To get an electron (that is in a negative energy state) go in a positive energy state we need a ground state oscillation as follows (w):
w*h/(4pi)=2*m(electron)*c*c. So this oscillation needs many energy. If the oscillation "decides " to create an e-e+ pair it needs that energy.

right?