Momentum space amplitudes for phi-four theory

Momentum space amplitudes for phi-four theory

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4 Responses to Momentum space amplitudes for phi-four theory

  1. AryaDatta says:

    In the last reply to my comment above, by “space-time volume integral term” I mean to say the measure of the integration which is $d^4z$ here.


  2. AryaDatta says:

    For Example 3, the given vacuum diagram appears at the first order in the expansion of the S-matrix amplitude, and it appears along with another connected vacuum diagram that involves no interactions- just a particle propagating freely. This has been shown in Fig. 19.4(a) of L&B. This other connected part appears as a delta function and it is multiplied by the vacuum part given in the problem (equation 19.11). There are 3 such scenarios possible, and when divided by the 4! term in the denominator, we get back the symmetry factor of 8. When we consider both the connected and vacuum diagrams in calculating the amplitude, we have the desired delta function term.


    • AryaDatta says:

      In equation 7 (Example 5), the propagator term corresponding to internal momentum $k_1$ is to be squared, as it occurs twice in the diagram.


      • AryaDatta says:

        The expression of problem 19.4(a) (Equation 19.25) fits into all of this as well. The 2nd term in Eqn. 19.10 leaves just the space-time volume integral term alone, as the contraction of phi(z) with itself gives the Feynman propagator with an argument of 0, thus, the exponent term in it becomes 1, and independent of the integration variable.


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