Homework 8 – due Friday, Nov 17th – 8A,8B,8C

(8A) Field redefinitions and scattering Consider a seemingly complicated theory with the basic complex scalar field B (note that |x|2 = x∗x): g L = |(∂ B)(1 + gB)|2 − m2|B + B2|2 µ 2 1. Divide the Lagrangian for B into a free part and interaction terms. Compute the elastic N + N¯ → N + N¯ scattering amplitude in the tree level approximation, imagining that you follow the LSZ formula. Tree level means that there are no loops and therefore you don’t need any counterterms. However, you will need a g2 vertex diagram as well as a diagram with two g cubic vertices.

Also, you will need derivative interactions. With derivative interactions, it is not true that L = −H, and also, there are problems with commuting the time-derivatives through the normal-ordering symbol. Believe me that these two problems cancel and can be collectively ignored, as will be clear from Feynman’s path integral formulation later. Assume L = −H and be sloppy about permuting the T , ∂µ symbols. 2. Now pretend that someone brings you a secret message that says: g ψ = B + B2 2 Using this field redefinition, guess what is the Lagrangian for ψ – it is very simple! – and verify it by an actual calculation. What is the scattering amplitude for ψ quanta? This question should be supersimple. Use this simple answer as a hint to fix any combinatorial errors in the calculation with B if you have errors.

3. If you do the same calculation in the B variables or the ψ variables, will the off-shell Green’s functions of B and those of ψ be equal (yes/no)? Should they be (yes/no)? Should the on-shell amplitudes be equal (yes/no)?

Comment: An advantage of the LSZ formalism is that it doesn’t matter whether you assign a field like B or a field like ψ to the particle. As long as the vacuum-to-one-particle matrix element is properly normalized, you will always obtain the same S-matrix as the results above should have convinced you.

(8B) Nucleon self-energy On page 248 of Coleman’s notes – the file Coleman 215-442, page “34 of 227” – there is a calculation of Π0(k2), the renormalized one-loop meson self-energy operator. The result is written as an integral over one Feynman parameter and Coleman shows that it is analytic except for a cut running from 4m2 to ∞. In the same theory, compute the renormalized nucleon self-energy Σ0(k2) using similar methods, again up to the order g2 but with somewhat different Feynman diagrams. Show that it is analytic except for a cut going to infinity from a certain point whose position you should also determine.

1 (8C) Quartic coupling Consider the complex scalar field with a quartic interaction:

λ 2 L = |∂ ψ0|2 − m2|ψ0|2 − |ψ0|2 + L µ 4   counterterms Note that the mass m and λ in the main part of the Lagrangian are the physical parameters, and ψ0 is properly normalized to create nice one-particle states. The whole Lagrangian including the 2 counterterms will have coefficients of the |∂µψ| kinetic term normalized to one, while the mass term 2 2 2 02 will be −m0|ψ| , written in terms of the “unphysically” normalized fields ψ = Z3ψ . Is the quartic vertex properly normalized to simplify the Feynman rules? Incidentally, we define this particular λ to be 4m2 λ =Γ0 at s = t = u = 3 which is fine because the four-point function depends on the Mandelstam variables only (because of Lorentz invariance of the theory), although a priori it could depend on the four four-momenta arbitrarily. Note that the sum s + t + u is what it should be. Also, realize that Γ0 is the stripped version of the four-point function:

i π 4δ(4) p . . . p 0 p ,p ,p ,p X(Diagrams with 4 ext. legs and 1PI image in the blob) = − (2 ) ( 1+ + 4)Γ ( 1 2 3 4) Note that two “charge” arrows (drawn directly on the lines) are incoming and two arrows are outgoing. But all momenta (with arrows drawn away from the propagators) are treated as incoming here. Sorry that I didn’t draw the picture with the disk and four external legs on the left-hand side, it’s too time-consuming and I’m not experienced in texing these figures. You will get it. Task: We have decided that the vacuum energy counterterms will always be ignored because they never influence the S-matrix. There are still three counterterms left: for the wavefunction renormalization; for the mass; and for the coupling. Compute them at the one-loop level.

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