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

[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>.

 

[geoduck]

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

So there can be scattering.

 

[Dickfore]

no, there is no scattering.

 

[geoduck]

no, there is no scattering.

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, m_λ> where P is the total momentum eigenvalue of all particles, and m_λ satisfies p²+m_λ²=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,m_λ>. So this set of eigenvectors is smaller than what's needed to describe your Fock space.

 

[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?

 

[geoduck]

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?

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

 

[Dickfore]

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

 

[Chopin]

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

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. :)