Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 22, 2012 #1
    In field theory:


    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. jcsd
  3. Aug 22, 2012 #2
    oops, P is degenerate, so in general e^(-iHT) rotates |p1p2> in the subspace of eigenvalues p1+p2.

    So there can be scattering.
  4. Aug 22, 2012 #3
    no, there is no scattering.
  5. Aug 22, 2012 #4
    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.
  6. Aug 23, 2012 #5
    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)):


    Unless your commutator is not a poncaire commutator. Could you be more explicit?
  7. Aug 23, 2012 #6
    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
  8. Aug 23, 2012 #7
    Then there will be scattering provided that H contains an interaction term (containing a product of more than two creation/annihilation operators).
  9. Aug 23, 2012 #8
    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. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook