2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jahan Claes Jun 26 '19 at 2:54
  • $\begingroup$ The reference to saddle-points reminds me in part of the method of steepest descent if that helps? $\endgroup$ – CR Drost Jun 26 '19 at 3:20
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I agree, I was thinking of $A$ constant. I edited now. $\endgroup$ – MannyC Jun 26 '19 at 15:42
2
$\begingroup$

Gaussian integration is a particularly simple case of the WKB expansion, cf. e.g. this Phys.SE post. Of course, the caveat is that the saddle point may be complex-valued. In other words, in the 1D case, the saddle point may lie in the complex plane, and one has to show that one can close the integration contour between the real axis and the line of steepest descent through the saddle point. Some of these issues are addressed in e.g. this & this related posts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.