# Gaussian path integral is equivalent to saddle-point?

If we have a path integral involving many fields,

$$Z = \int \mathcal D \phi_1 \cdots \mathcal D \phi_n \exp(-S[\phi_1,\ldots, \phi_n]),$$

and $$\phi_n$$ occurs only quadratically-- i.e. the $$\mathcal D \phi_n$$ integral is Gaussian-- I've been told that integrating over $$\phi_n$$ is equivalent to solving for $$\phi_n$$'s equation of motion

$$\phi_n= f(\phi_1,\ldots, \phi_{n-1})$$

using Euler-Lagrange and plugging in. Up to normalization. Can one show in general why this is true?

• I'm not sure that this could be true. If a field appears in the action as a Gaussian like $\phi_n(\nabla^2+r(\phi_1,...,\phi_{n-1}))\phi_n$, then its E-L equation is just $(\nabla^2+r(\phi_1,...,\phi_{n-1}))\phi_n=0$, and so plugging in a solution for $\phi_n$ just causes all terms involving $\phi_n$ to disappear from the action. On the other hand, doing the Gaussian integral for $\phi_n$ gives you a factor of something like $\det(\nabla^2+r)^{-1}$, which is not the same as just disappearing entirely. – Jahan Claes Jun 26 '19 at 2:54
• The reference to saddle-points reminds me in part of the method of steepest descent if that helps? – CR Drost Jun 26 '19 at 3:20

The gaussian integral $$\int dx\,e^{-\frac12 a x^2 + bx + c} = \sqrt{\frac{2\pi}{a}}\, e^{c+b^2/(2a)}\,,$$ is similar to its path integral counterpart, which is $$\int \mathcal{D}\phi\,e^{-\frac12\phi \cdot A\cdot \phi + \phi\cdot b + C} \propto \exp\left(C + \frac{1}{2} \, b\cdot A^{-1}\cdot b\right)\,.$$ By the dot I mean $$a\cdot b \equiv \int a(x)\, b(x)$$, $$a\cdot B \cdot c \equiv \int a(x)\, B(x,y)\, c(y)$$. Moreover $$A^{-1}$$ satisfies $$\int A(x,y)\cdot A^{-1}(y,z) = \delta(x-z)\,.$$
The equations of motion for $$\phi$$ are $$-A\cdot \phi + b = 0\qquad \Longrightarrow\qquad\phi = A^{-1}\cdot b\,.$$ Replacing this on the action yields the same result $$-\frac12\phi \cdot A\cdot \phi + \phi\cdot b + C \quad\to\quad -\frac12\,b \cdot A^{-1} \cdot A\cdot A^{-1}\cdot b + b\cdot A^{-1}\cdot b + C = \frac12\,b\cdot A^{-1}\cdot b + C\,.$$ If the dot notation is confusing I suggest to expand everything in integrals. The operator $$A$$ usually is just $$(\square_x + m^2) \delta(x-y)$$ and $$A^{-1}$$ is $$G_F(x-y)$$, the Feynman propagator.
Edit: As a comment pointed out, this does not take into account the $$(\det A)^{-1/2}$$. If $$A$$ is a constant operator, this does not pose any problem in perturbative computations because we only need the partition function modulo overall factors.
On the other hand, if $$A$$ is a function of the remaining fields $$A(\phi_1,\ldots,\phi_{n-1})$$, it will not pass through the subsequent integrals. The way this is normally handled is by exponentiating it as $$(\det A)^{-1/2} = e^{- \frac12 \mathrm{Tr}\log A}\,,$$ (with a suitable regularization procedure) and this typically yields a non-local action
• I agree, I was thinking of $A$ constant. I edited now. – MannyC Jun 26 '19 at 15:42