- #1

CAF123

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## Homework Statement

A self-interacting real scalar field ##\psi(x)## is described by the Lagrangian density ##\mathcal L = \mathcal L_o + \mathcal L_I = \frac{1}{2} (\partial_{\mu}\psi)(\partial^{\mu}\psi) − \frac{1}{2}m^2\psi^2 − \frac{g}{4!}\psi^4 ## where g is a real coupling constant, and describes the interactions of spinless bosons, B. Write down the S-matrix expansion and identify the lowest-order normal-ordered term that gives rise to the two-body scattering process B(p1)+B(p2) → B(p3)+B(p4). Draw the Feynman diagram corresponding to this process and show that the Feynman amplitude is M = −ig.

Repeat the exercise but now for the Lagrangian ##\mathcal L = \mathcal L_o + \mathcal L_I = \frac{1}{2} (\partial_{\mu}\psi)(\partial^{\mu}\psi) − \frac{1}{2}m^2\psi^2− \frac{\lambda}{3!}\psi^3##.

(Answer: $$\mathcal M = −i\lambda^2\left( \frac{1}{ s − m^2} + \frac{1}{ t − m^2} + \frac{1}{u − m^2}\right))$$

## Homework Equations

Generic S matrix expansion element $$S^{(n)} = \frac{(-i)^n}{n!} \int d^4 x_1 \dots \int d^4 x_n T(\mathcal H_I(x_1) \dots \mathcal H_I(x_n))$$

## The Attempt at a Solution

Here, ##\mathcal L_I = -\frac{g}{4!}\psi^4 \Rightarrow \mathcal H_I = \frac{g}{4!}\psi^4## since there are no derivative terms. Then the lowest order term in the expansion would be $$S^{(1)} = -i \int d^4 x_1 T(\mathcal H(x_1)) = -i \int d^4 x_1 T(:\frac{g}{4!}\psi_1(x_1)\psi_2(x_1)\psi_3(x_1)\psi_4(x_1):)$$ I think it is correct and that the Feynman diagram is just a single vertex with four lines coming out of it (two incoming and two outgoing), so no propagators involved. This also makes the matrix element depend on one power of g so the answer makes sense but I am not sure how to get to ##\mathcal M## from the above expression for ##S^{(1)}##. It seems like I need to take to take the matrix element between initial and final states: ##|i\rangle = |B_1(p_1), B_2(p_2)\rangle## and ##|f\rangle = |B_3(p_3),B_4(p_4)\rangle## and deduce ##\langle i|S^{(1)}|f\rangle##

Thanks!