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QFT Comments and discussions

  1. Sep 13, 2003 #1


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    Jeff mentioned he thought a thread on this might be good, and I decided to start it here so the board wouldn't be entirely blue sky.

    I'll start with a question raised by Patrick van Esch on s.p.r. In the Peskin and Schroeder textbook, they develop the expansion of the wave φ in the free theory into normal modes with the ladder operators a and a+. So far so good.

    Then when they come to discuss interacting field theory, they exploit that same expansion, ladder operators and all, for their development. But they haven't redefined their expansion technology in terms of the new "interacting" φ, so where do they get off doing this?

    All comments welcome but references to Haag's theorem should be tied to the issue at hand.
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  3. Sep 13, 2003 #2
    Good question, do the equations represent exact or approximate solutions? I would guess the latter.
  4. Sep 13, 2003 #3


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    Well it's not to the equations yet. It's a question of how to represent things. Ultimately the work will be done in pertubative mode, which is an approximation, but very accurate in the case of QED. But that is way down the road from this problem.
  5. Sep 14, 2003 #4


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    I don't have peskin and schroeder, but I'm pretty sure I know what's going on.

    Consider an interacting system with hamiltonian H(Q,P) in which Q and P are a canonical position operator and it's conjugate momentum, both in the heisenberg picture, which means that given their values Q(x,0) and P(x,0) at time t = 0, they're time evolution is governed by the full hamiltonian H as

    Q(x,t) = eiH(Q,P)tQ(x,0)e-iH(Q,P)t
    P(x,t) = eiH(Q,P)tP(x,0)e-iH(Q,P)t.

    In order to define the S-matrix perturbatively in the usual way as an expansion in powers of an interaction V requires we separate the hamiltonian into "free" and interacting parts as

    H = H0 + V.

    We then require - and this is the point - the "zeroth order" part of the theory to be the free theory. Thus we must write H(Q,P) in terms of free particle operators q and p whose time evolution is governed by the free hamiltonian H0(q,p) alone as

    q(x,t) = eiH0(q,p)tq(x,0)e-iH0(q,p)t
    p(x,t) = eiH0(q,p)tp(x,0)e-iH0(q,p)t.

    The conjugate pair (q,p) are said to be in the interaction picture. We then have

    Q(x,t) = eiVtq(x,t)e-iVt
    P(x,t) = eiVtp(x,t)e-iVt

    and in particular

    Q(x,0) = q(x,0)
    P(x,0) = p(x,0).

    So since H(Q,P) is time-independent (though H0 and V separately are usually not), it may be written at time t = 0 by replacing heisenberg with interaction picture operators. Doing so gives H0 = H0(q,p) which is the free hamiltonian and is trivially in the interaction picture since

    H0(q,p) = eiH0(q,p)tH0(q,p)e-iH0(q,p)t.

    On the other hand, V is not invariant under time translations generated by H0 and so in the interaction picture it becomes

    V(t) = eiH0(q,p)tVe-iH0(q,p)t.

    Then using the free particle commutation relations satisfied by (q,p) and the schrodinger equations based on H0 giving their time-evolution, V(q,p) may be expressed as sums of linear combinations of the free particle annihilation and creation operators.

    Haag is irrelevant here: His theorem says representations of the commutation relations will be inequivalent for scalar interactions differing only in the value of the coupling constant giving their strength.
    Last edited: Sep 14, 2003
  6. Sep 14, 2003 #5


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    Thank you. This is clearly the answer, and I now recall from my studies in P&S that they do this same development a little later. The problem Patrick was led to find was caused by their habit of postponing the rigor until after they have shown the application. Good pedagogy perhaps, but a trap for "inquiring minds".

    Another possible factor is that Patrick and I, who studied the material together in an online group, learned the material but perhaps didn't internalize it to such a recoverable level, as would a grad student in that first year total immersion.
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