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I How to understand the derivation for this process in QFT?

  1. Apr 30, 2017 #1
    I'm reading the book "Quantum Field Theory and the Standard Model" by Matthew Schwartz and I'm finding it quite hard to understand one derivation he does. It is actually short - two pages - so I find it instructive to post the pages here:
    The point is that the author is doing this derivation without any of the tools that would be needed. He doesn't use the LSZ formula, nor Wick's theorem, nor perturbation theory, nor Feynman Diagrams, absolutely nothing. Also, he doesn't use spinors, nor Dirac fields in order to deal with this.

    Actually he discusses this prior to introducing all these things. Now, how he does this derivation is quite confusing for me.

    I mean, he first talks about a propagator that would be [itex]1/k^2[/itex]. Now I know that the Fourier transform of the classical Klein Gordon propagator actually is [itex]1/k^2[/itex], but I can't understand where this enters the discussion here, nor how this [itex]k[/itex] he associates to a propagator is the total four-momentum.

    I also don't understand this discussion that leads to the formula for [itex]\mathcal{M}[/itex]. Up to this point the only thing the author has told about [itex]\mathcal{M}[/itex] is that it is related to the S-matrix by [tex]\langle f | S- \mathbf{1}|i\rangle = i(2\pi)^4 \delta^4\left(\sum p_i^\mu - \sum p_f^\mu \right) \langle f | \mathcal{M} | i\rangle[/tex] being [itex] |i\rangle, |f\rangle[/itex] respectively the initial and final states.

    Only in the next chapter he derives the LSZ formula that tells how to compute [itex]S[/itex] and hence [itex]\mathcal{M}[/itex] in terms of correlation functions, and only in the chapter after that he derives Wick's theorem to finaly compute this perturbatively.

    So what is really going on here? How to understand this derivation the author presents? What is the point with this [itex]1/k^2[/itex] propagator and why it relates to the center of mass energy? How this all leads to the formula for [itex]\mathcal{M}[/itex]?
  2. jcsd
  3. May 1, 2017 #2
    Actually after thinking a little bit more about it, I believe I started to get the idea. The initial and final states are both two-particle states of definite momentum. My guess is that the author is saying that spin can be introduced here as another degree of freedom of these two particle states, included by a tensor product.

    In other words, without spin we would have the initial state [itex] |p_1 p_2\rangle = |p_1\rangle \otimes_S |p_2\rangle[/itex] being [itex]\otimes_S[/itex] the symmetric tensor product. The author then considers that for each particle the one-particle state is actually [itex] |p_1; s_1\rangle = |p_1\rangle \otimes |s_1\rangle[/itex]. In that setting we end up with the conclusion that

    [tex] \langle f | S | i \rangle = \langle p_3 p_4 ; s_3 s_4 | S | p_1 p_2 ; s_1 s_2 \rangle [/tex]

    and considering that spin states and orbital states evolve independently, since [itex]S[/itex] is just the time evolution operator acting between assymptotic states we have

    [tex] \langle f | S | i \rangle = \langle p_3 p_4 | S_A | p_1 p_2 \rangle \langle s_3 s_4 | S_B | s_1 s_2 \rangle [/tex]

    now the one in the orbital states is related to the LSZ formula. There indeed all these cancelations involving the propagator occur and that would be just a dimensionless number as the author states. We are thus left with just the spin part. Finaly since the initial and final states are different, and since [itex]S = \mathbf{1}+iT[/itex], the identity part goes away, and we are left exactly with the conclusion that the matrix element is determined entirely by spin.

    Is that the correct reasoning? The only last thing to understand is: why the S-matrix spin part is [itex]\langle s_3 s_4 | S_B | s_1 s_2 \rangle[/itex] becomes a sum over projections onto the photon's spin states?
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