# One-loop effect

1. Oct 21, 2005

### Kruger

I read in wikipedia that the Casimir-effect is a one-loop quantum electrodynamical effect. This seems to be wrong to me. I always thought that it is raised due to electromagnetic zero point oscillations. And this oscillations have nothing to do with virtual photons either virtual particles. So how must I understand this text in wikipedia?

2. Oct 21, 2005

### hellfire

I had a similar question a time ago. My understanding is now the following: You are right that, in general, the Casimir effect can be understood as a difference in the infinite integral in the Hamiltonian inside and outside the plates. This is the usual way to derive the Casimir effect in textbooks making use of a model of a non-interacting scalar field. However, if you go to more realistic scenarios you should consider the electromagnetic field as the main contributor to the Casimir force. This means, the Dirac field, the photon field and the interaction between both. As soon as there exist interacting fields you are able to talk about virtual particles and loop effects, because the ground state is not equal to |0> of the single fields anymore. I assume that one may describe then the Casimir effect in that terms. I think this is a very subtle topic and I might be wrong.

3. Oct 21, 2005

### dextercioby

Technically, the em. vacuum includes quantum corrections which means that various loop (electronic-positronic) corrections to the free photon propagator must be added. Fortunately, QED is renormalizable (at any loop order) and these corrections can be computed.

Daniel.

4. Oct 21, 2005

### Kruger

This is weird. Why are then only virtual photons considered. There are also virtual electrons and positrons in the vacuum and myon-anti-myon pair? And I mean a virtual photon has energy hw and these oscillations have energy hw/2.
Sorry Dexter, I was writing, while you was writing. But do we really get the same result? I mean are this vacuum bubbles in quantum electrodynamics the same thing as the oscillations with energy hw/2 in quantum field theory?

5. Oct 21, 2005

### vanesch

Staff Emeritus
It is technically difficult to claim that the Casimir effect is a one-loop effect, because in its expression, the (EM) coupling constant doesn't appear, which I presume it would - by definition - if it were the first order term in a series devellopment towards that coupling constant.
But my reasoning here might miss some subtility of which I'd like to know something more then.

6. Oct 21, 2005

### Kruger

I think, if we can explain the Casimir-effect by a one-loop effect, then there must not be a coupling constant present in the formula. This is because a coupling constant arrises when two particle interract with each other. And virtual particles, coming as vacuum fluctations, don't couple with a real particle.

Dexter, if you know more about this one-loop Casimir-effect, it would be instructive for me and I think others when you can make this point clear.

For me it is a difference if we derive the Casimir-formula witht hese scillations or with virtual particles. Because virtual particles have another origin.

7. Oct 22, 2005

### Kruger

Is this correct:

The zero point oscillations of the electromagnetic field are the same as virtual photons in quantum electrodynamics.

There are a lot of authors (also Milonny: the quantum vacuum) that say that we can explain the Casimir-effect as a pressure coming from virtual particles with momentum 1/2h(bar)k. But this momentum correspondands to the zero point oscillations with energy 1/2h(bar)w. So this should be the momentum of the zero point oscillations and not of virtual particles. Because virtual particles have momentum hk. That is why I ask if the above sentence is correct.

8. Oct 22, 2005

### CarlB

Julian Schwinger (co-recipient of Nobel prize for invention of QED) argued against the physical existence of the vacuum for the most of the rest of his life. His students have continued the fight, with the Casimir effect one of the major battlegrounds. Here's a relatively recent paper that shows other ways of calculating the effect:
http://arxiv.org/abs/hep-th/9811054

This is not a subject about which I know much. That it is a one-loop effect comes about from the presence of $$\alpha^4$$ in the formula, I guess.

One of Feynmann's books referenced the concept that a theoretical physicist should know a lot of different ways of showing the same thing.

Carl

9. Oct 23, 2005

### Kruger

Thanks Carl. But where do you see any a^4 in the formula of the Casimir-force?

10. Oct 24, 2005

### hellfire

As far as I know the problem is that the vacuum energy is actually equal to the integral of all "zero point oscillations" only for free fields. In case of interacting fields this is not true anymore, because there are virtual particles creating and annihilating which, I assume, must contribute to the energy of the ground state. I assume that if one considers the ground state of the electromagnetic field only with one-loop corrections, one may be able to calculate the formula of the Casimir force. I guess it is not necessary that the coupling constant appears to say that it is a one-loop effect. Note that h appears, which might be enough. But, as I said before, this is beyond my knowledge and I never have done these calculations.

11. Oct 24, 2005

### Kruger

Is this correct:

The zero point oscillations of the electromagnetic field are the same as virtual photons in quantum electrodynamics.

There are a lot of authors (also Milonny: the quantum vacuum) that say that we can explain the Casimir-effect as a pressure coming from virtual particles with momentum 1/2h(bar)k. But this momentum correspondands to the zero point oscillations with energy 1/2h(bar)w. So this should be the momentum of the zero point oscillations and not of virtual particles. Because virtual particles have momentum hk. That is why I ask if the above sentence is correct.

In my oppinion we must, if we can calculate the Casimir-force in two ways, with virtual particles and with zero point oscillations and get the same result and these things are not the same, we must in order to get the whole Casimir-force make this: Fcas=Fvirtualparticlepressure+Fzeropointoscillation.

If not, then virtual photons in QED must be the same thing as zero point oscillations in QFT.

12. Oct 25, 2005

### Blackforest

I am not a specialist of this question but it looks very interesting. Intuitively, I would defend this point of view: EM fields owns fluctuations. These fluctuations could be short life polarisations ... (see my homepage)

13. Oct 25, 2005

### hellfire

I would say this is right.