Considering the Lagrangian of two scalar fields in $d=4$: $$\mathcal{L}=\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2+\frac{1}{2}(\partial\chi)^2-\frac{1}{2}M^2\chi^2-g\phi^2\chi$$

Let's assume that $M>2m$. The decay rate is given by $$\int d\Gamma=\frac{1}{2M}\int\frac{d^3p_1d^3p_2}{(2\pi)^6}\frac{1}{4E_{p_1}4E_{p_2}}(2\pi)^4\delta^{(4)}(p_\chi-p_1-p_2)|M(\chi(0)\rightarrow\phi(p_1)\phi(p_2)|^2$$ Now after some algebra we get to first order (with coupling $g$): $$\Gamma=\frac{g^2}{M}\int\frac{d^3p_1}{(2\pi)^3}\frac{1}{4E_{p_1}^2}(2\pi)\delta(M-2E_{p_1})$$ We can integrate the delta and we get (with $E^2=m^2+p^2$) $$\frac{g^2}{M}\int\frac{d^2p_1}{(2\pi)^2}\frac{1}{M^2}$$ Now I would just carry out the remaining integral and get something of the form $$\Gamma=\frac{g^2}{16\pi M}\left(1-\frac{4m^2}{M^2}\right)$$ But in the solutions (see A Complete Solution to Problems in “An Introduction to Quantum Field Theory” by Peskin and Schroeder, p.25 f.) they get $$\Gamma=\frac{g^2}{8\pi M}\left(1-\frac{4m^2}{M^2}\right)^{1/2}$$ Where does the square root come from? Is there anything to consider while integrating?

  • $\begingroup$ A very detailed derivation (for arbitrary dimension, using also the right-hand side of the optical theorem) of the result you quote is in inspirehep.net/record/264517?ln=en, see eqn (9). (Your result follows from taking $\lambda=2g$ and $d=4$.) $\endgroup$ Commented Jul 2, 2019 at 9:56

1 Answer 1


If you look in the frame of the decaying particle (you have written $\chi(0)$), then its four-momentum is $(M,0)$, while for the decay products it is $(M/2,\vec{k})$ and $(M/2,-\vec{k})$. The square root comes from $|\vec{k}|=((M/2)^2 - m^2)^{1/2} = M/2(1 - 4m^2/M^2)^{1/2}$.


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