Does [H,P]=0 imply no scattering in field theory?

1. Aug 22, 2012

geoduck

In field theory:

[H,P]=0

So shouldn't that mean there can be no scattering? If you have momentum state |p1p2> then it is an eigenvalue of P with eigenvalue p1+p2. But it should be a simultaneous eigenstate of H too, so

e^(-iHt)|p1p2>= |p1p2> up to a phase factor.

But in general, |p1p2> can go into |p3p4> so long as p1+p2=p3+p4. However, the above argument seems to say that |p1p2> must stay at |p1p2>.

2. Aug 22, 2012

geoduck

oops, P is degenerate, so in general e^(-iHT) rotates |p1p2> in the subspace of eigenvalues p1+p2.

So there can be scattering.

3. Aug 22, 2012

Dickfore

no, there is no scattering.

4. Aug 22, 2012

geoduck

My notation was bad. By P, I meant the total momentum operator, summed over all particles. That means P is degenerate, so it's not true that H (the total Hamiltonian) and P share eigenvectors.

You can however choose a basis where they share eigenvectors. The basis is written like:

|P, mλ> where P is the total momentum eigenvalue of all particles, and mλ satisfies p2+mλ2=E2, where E is the total energy of all the particles.

I do find this disturbing though, because the above set of eigenvectors seems to be smaller than the full set needed to describe the theory. Take the total momentum P=0, with two particles. Then they can have equal and opposite momenta in the x,y, or z directions (or any linear combination), but these states are all described by |0,mλ>. So this set of eigenvectors is smaller than what's needed to describe your Fock space.

5. Aug 23, 2012

jarod765

Is that commutator exactly the same as the poncaire group commutator. The reason you get scattering is because the amplitude for a scattering process is described by the S matrix (e^(-i S)):

http://en.wikipedia.org/wiki/S-matrix

Unless your commutator is not a poncaire commutator. Could you be more explicit?

6. Aug 23, 2012

geoduck

Yes, H generates time translations, and P generates space translations, so H and P are the Poincare generators.

The S-matrix should be equal to e^(-iHt) when t→ ∞ , so an eigenvector of H at t=-∞ should remain an eigenvectors of H at t=∞. But a mystery is what is this eigenvector? A momentum state such as |p1p2> is not an eigenvector of H.

My main motivation for this is to try to figure out Peskin and Schroeder's page 226 (1995 edition) where in the second equation, a <k1k2| Schrodinger state centered at <p1p2| somehow became a <p1p2,t=∞| Heisenberg state. <p1p2| and <p1p2,t=∞| are very different. In fact <p1p2,t=∞| can be anything, say <p3p4|, depending on the Hamiltonian, so long as p1+p2=p3+p4

7. Aug 23, 2012

Dickfore

Then there will be scattering provided that H contains an interaction term (containing a product of more than two creation/annihilation operators).

8. Aug 23, 2012

Chopin

I'm still a bit hazy on this as well (asymptotic multi-particle states seem to be one of the hairier things to get one's head around in QFT), but I believe the answer is that you isolate them from the non-interacting Fock states using the same sort of procedure that we used to get the real vacuum state from the non-interacting vacuum. Namely, send t to infinity, and all of the other states will run away (P&S page 86, starting after 4.26).

A similar procedure allows you to get one-particle states, and then they assume that since you're making localized wavepackets, you ought to be able to apply the operator multiple times in non-interacting regions of space to build up a multi-particle state. At any finite t, this procedure generates something that isn't an eigenstate of the Hamiltonian, but as t goes to infinity, it becomes one.

I think. :)