Phys 624 Questions

Phys 624 - Quantum Field Theory
University of Maryland, College Park
Fall 2016, Professor: Ted Jacobson

Questions

10. Has the assumption that the vacuum is nondegenerate been used in anything we've done so far, i.e. the LSZ reduction, spectral representation, or expression for vev of time-ordered products of Heisenberg fields in terms of the free vevs of interaction picture fields?

9. In hw3, for the single QM oscillator we can't invoke the Reeh-Schlieder theorem to argue that O(t)|Ω> = 0 implies O(t)=0. Is there another way to argue that nevertheless O(t)=0 in this case?

8. Is normal ordering a linear operation on operators? Answer: It seems that we must define the normal ordering of a c-number to be zero, contrary to what Timo says under (2.121). Consider aa* = a*a + 1. If we apply normal ordering operation to both sides, and assume linearity, we get a*a = a*a + N(1), so it must be that N(1)=0. I don't know if this consistenly defines a linear operator on operators...but in any case, it seems that in stating Wick's theorem, we should not include the totally contracted terms inside the normal ordering symbol.

7. Concerning Weigand (2.107), the expression of interacting vevs of time-ordered Heisenberg fields in terms of free vevs of interaction picture fields and the evolution operator, it was asked: is this formula always valid? Answer: It requires that there is nonzero overlap of the free and interacting vacua. I think that is the only caveat, but I'm not certain. It would be nice to have an example with zero overlap of the free and interacting vacua, to see how to deal with that. My guess is that this would happen when there is a phase transition. Can there be an example in quantum mechanics? Finite dimensional? Infinite dimensional?

6. Suppose you have a vacuum energy and pressure corresponding to a Lorentz invariant energy-momentum tensor, i.e. pressure = - energy density. Then if you integrate the energy density over all space you get an (infinite) energy, and if you integrate the 3-momentum density you get zero, so the energy-momentum 4-vector looks like P = (E,0). This 4-vector is not invariant under Lorentz transformations, but the energy-momentum tensor *is* invariant. What gives?! Answer: This total 4-momentum P is defined relative to a particular spatial slice of spacetime. This is different from the 4-momentum of a particle, which is defined independent of any slice choice (although it's components depend on the reference frame). When you make a Lorentz transformation on the total 4-momentum of the spacetime, you should also transform this slice to a new slice, so you're not only transforming between the components of P in two different bases. When you do that, for the case of a vacuum energy-momentum tensor proportional to the metric, you'll find that the transformation of the components is cancelled by the effect of changing the slice.

5. If the momentum space Schrodinger equation has a constant potential in it, then the position space potential should be a delta function. This sounds weird. Isn't a constant in the Hamiltonian just the same constant in any representation? Answer: Yes. To transform the potential operator between the position and momentum representations, you don't just Fourier transform the function. Say it's constant in momentum space: <p|V|p'> = c δ(p-p'). Since this is proportional to the unit operator in one basis, it is in any other one as well: <x|V|x'> = <x|p><p|V|p'><p'|x'> = c <x|x'> = c δ(x-x').

4. Is it necessary to normal order the charge operator in order to show that it just counts the number of particles minus the number of antiparticles? Answer: Yes. As far as I can see, there is an ambiguity here: A constant 4-vector is an identically conserved current, which can always be added to another conserved current. Some criterion is needed to select which, if any, is the preferred current. In the case of charge, it seems to me at present that we can just choose to add the appropriate constant so that the charge of the vacuum is defined to be zero.

3. Does the LSZ reduction formula contain the spectral (Kallen-Lehmann) representation information? Answer: Only the poles.

2. The Feynman propagator doesn't vanish at spacelike separation. Why doesn't this mean the physics violates causality? Answer: It doesn't imply failure of observables to commute at spacelike separations. Second Answer: The propagator characterizes the correlation between fluctuations in the vacuum, and those fluctuations can be correlated without violating causality. Third Answer: If we could prepare and measure an infinitely localized particle state, even in a limiting sense, as we can in nonrelativistic quantum mechanics, then the spacelike nonvanishing propagator would indeed imply we could violate causality. But in QFT, we can't prepare and measure at a point. If we tried, we'd create many particles initially in the preparation, and in the detection we wouldn't know whether we were detecting a particle that started at the initial point, or one that we created while trying to detect it.

1. Can anyon commutation relations satisfy the causality condition of commuting at spacelike separation? Answer: I doubt it. There seems to be something very nonlocal about anyons. See the Supplements link for references showing there can be no free relativistic thery of anyons, and also a reference showing there can if they are charged.