Monday, July 10, 2006

What's wrong with the Hamiltonian of QED?

I mentioned several times on this blog that the non-trivial action of the QED interaction V on the vacuum and one-particle states is deeply disturbing. What's wrong with it?

The interaction Hamiltonian V written in terms of particle creation and annihilation operators contains so-called tri-linear terms. An example of such a term is the product of three particle operators a*c*a, where I denoted 'a*' an electron creation operator, 'a' an electron annihilation operator, and 'c*' a photon creation operator. Let us focus just on this term and write the full Hamiltonian H in the truncated form (all momentum, spin, and polarization labels are omitted as well as numerical factors, integration signs, etc.)

H = H_0 + a*c*a + ... (1)

Having the full Hamiltonian H one can form the time evolution operator U = exp(iHt) and study the time dependence of states and observables. Let us consider the time evolution of a state a*|0> which at time t=0 had only one electron (here |0> denotes the vacuum state). We would expect (according to observations) that a single electron always remain in the one-electron state. Let us now find out whether this expectation is fulfilled with our Hamiltonian. We are going to find out how this state will look like after short time t. Let us take t small enough, so that the time evolution operator U can be approximated by linear terms in the series over t. Then

exp(iHt) a*|0> = exp(it(H_0 + a*c*a + ...)) a*|0>
= (1 + it H_0 + it a*c*a + ...) a*|0>
= a*|0> +it H_0 a*|0> + it a*c*a a*|0> + ... (2)

The first two terms in this expansion are harmless. They both correspond to states with one electron, as expected. However, the third state is different. After bringing the particle operators to the normal order and omitting, again, all numerical factors, this term will look like

a*c*|0> (3)

and this is now a state with one electron and one photon. So, the time evolution with the QED Hamiltonian (1) leads to the "emission" of photons by a single free electron. Such an effect has never been seen experimentally. The problem becomes even more severe if we take more terms in the expansion of the operator U and consider additional terms in the interaction V. It is easy to show that the vacuum state |0> will also become unstable with respect to the decay into multiparticle states containing photons and electron-positron pairs. This is also unphysical.

The cure for this desease is known for many years. It is often said that creation and annihilation operators, like a* and a, do not correspond to real physical particles. They describe creation and annihilation of fictitious "bare" particles. Real electrons are complex linear combinations of multiparticle "bare" states. Although, this has been known for decades, there were suprisingly few attempts to reformulate QED in terms of physical particles that can be observed. In view of the complete failure of QED to describe the time evolution, such reformulation of the theory seems to be a reasonable step. This step is exactly the essence of the "dressed article" program initiated in 1958 by the work of Greenberg and Schweber [1].

I would like to mention just one more thing. It appears that in a properly dressed theory all ultraviolet divergences disappear from both the S-matrix and the Hamiltonian. More about the "dressed particle" approach in later posts.

[1] O. W. Greenberg and S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8, (1958) 378.

11 Comments:

Anonymous Chris Oakley said...

Wait a minute ... not only can you have a trilinear electron-electron-photon term in the interaction Hamiltonian, but people do, and one can calculate radiative emission and absorption of photons by atoms this way. Energy conservation then prevents the emission of a photon by a free electron - see e.g. Dirac, Principles of QM, section 63. Problems of course start when one looks at second order, but that is a separate issue.

Tuesday, July 11, 2006 at 1:09:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris,

Energy conservation then prevents the emission of a photon by a free electron - see e.g. Dirac, Principles of QM, section 63


Unfortunately, I don't have Dirac's book. Could you please remind me in what context he made this statement. In interacting theory, the transition "electron -> electron + photon" is forbidden in the S-matrix, i.e., when the time evolution is integrated over infinite time interval. However, there is nothing preventing appearance of the terms like (3) on the right hand side of the time evolution formula (2) in my post.
I am going to write a separate post explaining why the photon emission by a free electron is allowed in the finite-time evolution in QED and forbidden in the S-matrix.

...one can calculate radiative emission and absorption of photons by atoms this way...

I am going to write a special post about bound states (=atoms) calculations in QED and in the "dressed particle" formalism. Stay tuned.

Tuesday, July 11, 2006 at 7:05:00 PM PDT  
Anonymous Chris Oakley said...

Hi Eugene,

It is not just Dirac's book ... every text book that treats radiation in the proper (i.e. second-quantized) way (and not just Einstein A and B coefficients) should have the same thing.

Wednesday, July 12, 2006 at 3:27:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris,

I am a bit confused. If you are talking about radiation of atoms, then I haven't posted about that at all. I promise to return to this discussion later.

If you are talking about the photon emission by a free electron, then we can have this discussion here. I think we both agree that this emission is not observed experimentally. However, you disagree with my statement that the Hamiltonian of QED (erroneously) predicts such an emission via formula (2) in my post. If I understand you correctly, you are saying that such an emission is forbidden by the energy conservation. I don't see how this follows from formula (2).

I would agree with you completely if you said that matrix elements of the S-matrix between states |1 electron > and |1 electron+1 photon > are exactly zero. But in my post I talked about the time evolution at finite times, not about the S-matrix.

Wednesday, July 12, 2006 at 4:39:00 PM PDT  
Anonymous Anonymous said...

Well, free electrons may not produce entire "photons", but they certainly do have an effect on the Maxwell field, because each electron has its own Coulomb potential well around it. This is made out of the A field, i.e. "photon-stuff", but is not a particle that can propagate (c.f. bound states of particles, which are non-perturbative). When another particle enters this field, its reaction to it can be described in terms of "virtual" photons. So the A field is affected, but not in such a way that you can get actual photons escaping to infinity from it.

Monday, July 30, 2007 at 4:44:00 AM PDT  
Blogger Eugene Stefanovich said...

Hi Anonymous,

yes, electrons have a Coulomb field around them. However, I don't think it is possible to represent this field as a bound state of photons. If we followed your suggestion, then even a simple non-relativistic two-electron scattering should have been treated as a multiparticle problem (2 electrons + infinite number photons). The hydrogen atom would not be represented as a 2-particle system. Your proposal would require a complete rewrite of quantum mechanics and QFT.

Eugene.

Monday, July 30, 2007 at 11:03:00 AM PDT  
Blogger Eugene Stefanovich said...

Adding to what I wrote before, I think that we have quite different philosophies. I tend to think that one electron is simply "one electron" and nothing else. The idea of the Coulomb field or potential comes into play when we have two or more electrons. Indeed, the only purpose of Coulomb potentials is to describe electron interactions. There should be at least two electrons for the interaction to take place. So, in my philosophy, Coulomb potentials are terms that are present in the Hamiltonian describing systems of several charged particles. Such terms are not present in the Hamiltonian describing one electron.

Monday, July 30, 2007 at 1:13:00 PM PDT  
Anonymous Anonymous said...

"even a simple non-relativistic two-electron scattering should have been treated as a multiparticle problem (2 electrons + infinite number photons)". Yes, now you understand basic QED.

"The hydrogen atom would not be represented as a 2-particle system." NO it isn't. The correction is called "Lamb shift" and it is what lead to the development of modern QED.

This is the problem with you high energy people. With nothing real to measure you end up understanding nothing about the real world.

Tuesday, June 1, 2010 at 1:41:00 AM PDT  
Blogger kof9595995 said...

“The cure for this desease is known for many years. It is often said that creation and annihilation operators, like a* and a, do not correspond to real physical particles. They describe creation and annihilation of fictitious "bare" particles.”
Does this "cure"(renormalization) give a mathematically precise proof to the electron decay problem?Or just a handwaving argument like "after renormalization, in principle we expect to see this problem solved...."

kof9595995

Thursday, September 1, 2011 at 7:38:00 AM PDT  
Blogger kof9595995 said...

“The cure for this desease is known for many years. It is often said that creation and annihilation operators, like a* and a, do not correspond to real physical particles.”
Does this cure(the renormalization) give a mathematically precise proof to the electron decay problem? Or just a hand-waving argument like "after the renormalization, we expect to see..."

Thursday, September 1, 2011 at 7:43:00 AM PDT  
Blogger Eugene Stefanovich said...

kof9595995,

No, renormalization does not provide a solution for the "electron decay" problem. If the original QED Hamiltonian had the form

H = a*a + c*c + a*c*a + ... (1)

then after renormalization it acquires a different form, comething like

H' = Aa*a + Bc*c + Ca*c*a + ... (1)

where A, B and C are infinite renormalization constants. In spite of these weird divergences, the renormalized Hamiltonian describes scattering amplitudes very accurately. However, the troubling term a*c*a is still present, which means that formulation in terms of physical (stable) electrons has not been achieved. You can read more about renormalization in chapter 9 of my book.

The true "cure" for the "electron decay" disease is in transition to the dressed particle picture.

Thursday, September 1, 2011 at 12:07:00 PM PDT  

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