# Calculation of one-loop diagram in $\phi^4$ theory

In Folland's book Quantum Field Theory, page 207, he gives the value of the amputated one-loop $$\phi^4$$ diagram as $$I(p) = \frac{(-i\lambda)^2}{2} \int \frac{-i}{-q^2 + m^2 - i\epsilon} \cdot \frac{-i}{-(q+p)^2 + m^2 - i\epsilon} \frac{d^4q}{(2\pi)^4}. \tag{1}$$ To evaluate this integral, he first Wick rotates by substituting: $$q^0 \rightarrow iq^0 \\ p^0 \rightarrow ip^0$$ which he claims gives $$\frac{(-i\lambda)^2}{2} \int \frac{-i}{(|q|^2 + m^2)(|q+p|^2 +m^2)} \frac{d^4q}{(2\pi)^4}. \tag{2}$$ He then goes on to introduce Feynman parameters and carries on with the computation. I am confused on the numerators of (1) and (2). After carrying out the multiplication of the two fractions in (1), shouldn't the numerator of (2) be $$-1$$? Also, how is it justified to drop the terms containing $$\epsilon$$?

I am confused on the numerators of (1) and (2). After carrying out the multiplication of the two fractions in (1), shouldn't the numerator of (2) be $$-1$$?

Don't forget that you have a $$dq_0$$ in your measure $$d^4q = dq_0 d^3q$$.

So you pick up another factor of $$i$$ when you make the $$q_0\to i q_0$$ substitution (change of variables), since $$dq_0 \to i dq_0$$.

Also, how is it justified to drop the terms containing $$\epsilon$$?

It is justified to set $$\epsilon$$ to zero because the rest of the denominator can no longer be zero. For example, there is no way for $$|q|^2 + m^2$$ to be zero since both terms are non-negative, and $$m^2$$ is positive (since $$m\neq 0$$).

The $$\epsilon$$ is there to make the integration meaningful when the denominator can be zero, and it is understood that we want $$\epsilon$$ to go to zero. So, after the Wick rotation we can just explicitly set $$\epsilon$$ to zero, since that is the same as the limit as it goes to zero.

• Thank you! I completely overlooked how the measure transforms under the Wick rotation. Commented Mar 8, 2023 at 2:54