# Divergences in tree-level diagrams?

Consider the Feynman diagram in $$\phi^4$$ theory where there are three incoming momenta ($$p_1$$, $$p_2$$, and $$p_3$$), three outgoing momenta ($$q_1$$, $$q_2$$, and $$q_3$$), and one internal line so that this is a connected tree diagram. Using the momentum space Feynman rules this diagram can be computed up to symmetry factors as $$\frac{i (-i\lambda)^2}{(p_1 + p_2 + p_3)^2 -m^2 + i\epsilon} \tag{1}$$ where each factor of $$(-i\lambda)$$ comes from the two vertices and I have used momentum conservation to obtain the propagator $$\frac{i\delta^4(k - p_1 - p_2 - p _3)}{k^2-m^2 + i\epsilon}.$$

Is this correct? If so, how can we interpret (1) as a probability amplitude if the numerator is complex? More importantly, as we take (1) to be on-shell wouldn't this quantity diverge? Since we are not integrating in (1) we cannot use regularization and renormalization.

I think I am making a mistake somewhere. Any help would be very much appreciated.

• 1. As $p_1$, $p_2$, $p_3$ are on-shell, there is no singularity in (1). 2. The meaning of your second formula is unclear. There is no delta function in a momentum-space propagator. 3. Which complex amplitude? You can safely set $\epsilon=0$ in (1) as you have no loop integral where the $i \epsilon$ prescription tells you how to go around poles in the complex plane. But apart from that, why should a complex amplitude pose a problem? 4. Consulting a good text-book on QFT might be helpful. Commented Apr 2 at 5:59
• @Hyperon Thank you for your comments. I have been mostly following Peskin and Schroder. 1. Isn't $m^2 = (p_1 + p_2 + p_3)^2$ by the energy momentum relation which will cause (1) to be singular? 2. The delta function in my second formula is to enforce momentum conservation at the vertex. 3. The numerator in (1) has a factor of $i$ and so setting $\epsilon = 0$ would make (1) complex. I thought (1) corresponds to a probability and so should be in $[0, 1]$. Commented Apr 2 at 6:08
• (1) is a probability amplitude $A$. A probability amplitude is related to a probability $p$ via $p=|A|^2$. So there is no requirement that $A$ be real. It's actually a little more subtle since we're really dealing with probability densities, and there's no requirement that the probability density be less than or equal to 1. The requirement that the set of amplitudes $A$ yields a consistent probability distribution does lead to constraints on the form of $A$, but the constraints are more complicated than what you are supposing. Commented Apr 2 at 6:18
• @CBBAM 1. As you assume $p_{1,2,3}$ being the on-shell momenta of the inital state, you always have $(p_1+p_2+p_3)^2 \ge (3m)^2$ and no singularity in this case. However, the presence of a kinematic singularity would not hurt anyway. 2. If you work out the pertinent $6$-point function going on-shell afterwards by the LSZ-formalism, momentum conservation at the vertices comes out automatically, there is nothing to "enforce". 3. An amplitude $A$ is not a probability. $|A|^2$ is a probability. Commented Apr 2 at 6:21
• Scattering amplitudes are used to compute scattering cross sections, which aren't unitless probabilities but rather have units of area. The particles that come in and out of an idealized scattering process in QFT are infinite plane waves present everywhere, with infinite norm, and what one is really computing is not a bare "probability". Recall, Scattering rate = beam luminosity * cross section. Beam luminosity is what experimenters control, scattering rate is what experimenters measure, and cross section is the physics contained in the ratio of the two, and what feynman diagrams calculate. Commented Apr 2 at 7:00

1. More generally in perturbation theory, a connected $$n$$-point tree-diagram $$(2\pi)^d\delta^d(\sum_{j=1}^np_j)~{\cal M}(p_1,\ldots,p_n)$$ in momentum space $$(p_1,\ldots,p_n)~\in~(\mathbb{R}^d)^n$$ still has [besides a $$d$$-dimensional Dirac delta distribution enforcing total momentum conservation] resonance poles from denominators of propagators becoming zero. Depending on the whether the external momenta are incoming/outgoing, the poles may or may not belong to a physical region/channel. For generic external momenta $$(p_1,\ldots,p_n)$$ away from a measure-zero set, the tree-diagram is finite.
2. Observable quantities, such as, probabilities, scattering cross sections, decay rates, etc, are proportional to the dimensionful quantity $$|{\cal M}|^2$$ integrated over the pertinent Lorentz-invariant phase space (LIPS), cf. e.g. Ref. 1.