Wednesday, July 05, 2006

Can relativistic QFT predict time evolution?

From what I know about relativistic quantum field theory, the answer is "no" (if you disagree, please provide a reference that proves otherwise). The explanation is simple. In quantum mechanics, one needs to have a well-defined Hamiltonian in order to form the time evolution operator exp(iHt) and to calculate the time dependence of wave functions and observables. However, in relativistic QFT, the only sensible Hamiltonians (i.e., those that can be used to calculate the S-matrix via Feynman-Dyson formula) must contain infinite renormalization counterterms. So, they are not well-defined. They are actually not defined at all. Even if we forget about the infinities in QFT Hamiltonians (for example, we can introduce artificial momentum cutoffs), they are still not good, because they normally contain terms (like trilinear terms in QED) which have a non-trivial action on the vacuum and one-particle states. This is completely at odds with the observed stability of vacuum and single particles.

QFT is good at calculating one and only one thing - the S-matrix (or the S-operator). All wonderful experimental predictions of QED or Standard Model (scattering amplitudes, anomalous magnetic moments, Lamb shifts, etc.) are related to the S-matrix elements. The time evolution enters in the S-matrix in an integrated form (from infinite past to infinite future). It is just impossible to recover the detailed form of a function by knowing its definite integral. By having exact knowledge about the S-operator we can say very little about the underlying Hamiltonian. It has been shown [1] that there exists a huge class of scattering-equivalent Hamiltonians connected to each other by unitary transformations.

So, the great successes of relativistic renormalized QFT in calculations of the S-matrix and related observable quantities do not at all contradict the miserable performance of QFT when it comes to calculations of the time evolution. This fact went unnoticed for a long time for a simple reason. Currently there are no experimental techniques capable of measuring the detailed time evolution of micro-particles. Existing time-dependent experimental data are of such a low quality that simple non-relativistic Hamiltonians are quite capable to describe them, and fine relativistic and radiative corrections are just not needed.

There is a way to fix this problem, i.e., to have a finite well-defined Hamiltonian which is useful for both S-matrix and time evolution calculations in QFT. This way is called the dressed particle approach. More about that in another posting.

[1] H. Ekstein, "Equivalent Hamiltonians in scattering theory", Phys. Rev. 117 (1960), 519


Anonymous Peter said...

You write

"It has been shown in H. Ekstein, Phys. Rev. 117 (1960), 519 that there exists a huge class of scattering-equivalent Hamiltonians connected to each other by unitary transformations."

In view of the physical relevance of |psi|^2 (mentioned already by Schrödinger in 1926), I guess that most QM is unitary equivalent/invariant. Ekstein's result looks like a consequence of this fact.

What do you think?

Thursday, July 6, 2006 at 1:06:00 PM PDT  
Blogger Eugene Stefanovich said...

The answer depends on what you mean by most QM is unitary equivalent/invariant. The theory does not change at all if one performs any unitary transformation of both physical states and observables. This is equivalent to the change of basis and has no effect on the results of calculations. However, the effect of an unitary transformation may be rather non-trivial if this transformation is applied only to states or only to observables. In Ekstein's work it is demonstrated that the S-matrix of a quantum theory remains the same if one applies certain kinds of unitary transformations to the Hamiltonian (state vectors are not affected). Basically, this work says that there are many different Hamiltonians which yield exactly the same S-matrix. This is not a trivial statement, and it does not follow simply from the non-observability of the phase of the wave function.

Thursday, July 6, 2006 at 4:00:00 PM PDT  
Anonymous Peter said...

I see, thank you

Friday, July 7, 2006 at 9:58:00 AM PDT  

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