Tuesday, January 19, 2010

Instability of the QED vacuum at large fine structure constant

Today, in the "Mathematical Quantum Mechnics" lecture, I learned that the QED vacuum (or at least the quantum mechanical sector of it) is unstable when the fine structure constant gets too big.

To explain this, let's go back to a much simpler problem: Why is the hydrogen-like atom stable? Well, a simple answer is that you just solve it and find the spectrum to be bounded above -13.6Z^2{\rm eV}. But this answer does not extend to other problems that cannot be diagonalised analytically.

First of all, what is the problem we are considering? It's the potential energy of the electron which in natural (for atomic physics) units is V(r)=-\alpha Z/r. And this goes to negative infinity when r goes to 0. But quantum mechanics saves you. Roughly speaking (this argument can be made mathematically sound in terms of Hardy's inequality), if you essentially localise the electron in a ball of radius R and thus have the potential energy V\le-\alpha Z/R, Heisenberg's uncertainty implies the momentun is at least of the order 1/R and thus the kinetic energy is at least of the order +1/R^2. Thus, when R becomes small and you seem to approach the throat of the potential the positive kinetic energy wins and thus the Hamiltonian of the hydrogen atom is bounded from below. This is the non-relativistic story.

Close to the nucleus however, the momentum can be so big that you have to think relativistically. But then trouble starts as at large momenta the energy grows only linearly with momentum and thus the kinetic energy only scales like +1/R which is the same as the potential energy. Thus a more careful calculation is needed. The result of it is that it depends on \alpha Z which term eventually wins. Above a critical value (which happens to be of order one) the atom is unstable and one can gain an infinite amount of energy by lowering the electron into the nucleus and quantum mechanics is not going to help.

Luckily, nuclei with large enough Z do not exist in nature. Well, with the exception of neutron stars which are effectively large nuclei. And there it happens. All the electrons are sucked into the nuceus and fuse with the protons to neutrons. In fact, the finite size of the nucleon is what regulates this process as the 1/r nature of the Coulomb potential is smeared out in the nucleus. But such a highly charged atom would be only of the size of the nucleus (about some femto meters) rather than the size of typical atoms.

But now comes QED with the possibility of forming electron-positron pairs out of the vacuum. The danger I am talking about is the fact that they can form a relativistic, hydrogen like bound state. And both are (as far as we know) point like and thus there is no smearing out of the charge. It is only that \alpha Z\approx 1/137 in this case which luckily is less than one. If it would be bigger you could create this infinte amount of energy from the vacuum by pair creation and bringing them on-shell in their relative Coulomb throat. What a scary thought. Especially, since \alpha is probably only the vev of some scalar field which can take other values in other parts of the multiverse which would then disappear with a loud bang.

Some things come to my mind which in principle could help but which turn out to make things worse: \alpha is not a constant but it's running and QED has asymptotic slavery which means at short distances (which we are talking about) it gets bigger and makes things worse. Further, we are treating the electromagnetic field classically which of course is not correct. But my mathematical friends tell me that quantizing it also worsens things.

We know, QED has other problems like the Landau pole (a finite scale where \alpha goes to infinity due to quantum effects). But it seems to me that this is a different problem since it already appears at \alpha\approx 1.

Any ideas or comments?


antin said...

This instability is of course most clearly expressed in the imaginary part of the Euler Heisenberg Lagrangian for the photon field. Attemps to see the effect in collsions of e.g. two U nuclei have failed so far.

Alejandro Rivero said...

Also, I remember a short physreview paper, about one page, about the stability and/or convergence of coupling constant expansion in QED.

Very famous, but I can not remember the author.

Anonymous said...

A relevant review paper is

" Quantum mechanics, the stability of matter and quantum electrodynamics"
Elliott H. Lieb, math-ph/0401004.

Apparently the vacuum is indeed unstable for \alpha > 2.72.

James said...

Can I ask, if this is a problem for QED, why isn't it a problem for QCD, which does have alpha~1. Isn't this just the QCD quark condensate?

Robert said...

James, you are right. At least for part of the story.

Strictly speaking, what I discuss for most part of the post is not really QED but a (relativistic) theory of quantized electrons that interact via a classical electromagnetic interaction (except for the last part where I mention the running coupling). This is a better approximation for QED than for QCD.

For QCD, I guess, its asymptotic safety that saves the day (physically speaking): At long distances there is no 1/r potential (that was my starting point) and at short distances the interaction becomes weak and the problem goes away. The cross over should be around the QCD scale and with the condensate this is exctly what one finds.