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Suppose we have a functional integral of the form $$ \int D\psi \exp\left( -S \right) $$ over a real field $\psi$ with an action $$ S=\int_r L(r,\psi(r))=\int_r \frac{1}{2}\psi(r)^2+i\psi(r) j(r). $$ Now we could perform a saddle point analysis which would mean we search for the field configuration $\psi(r)$ which solves the Euler Lagrange equation $$ \frac{\delta S}{\delta\psi(r)}=\frac{\partial L}{\partial\psi}=0. $$ In this case this is the field configuration $$ \psi(r)=-ij(r). $$ Now my question: How can a real field $\psi(r)$ have an imaginary saddle point? I think it has something todo with the fact that using the Gaussian integral $$ \int_{-\infty}^{\infty}dx \exp(-ax^2)=\sqrt{\frac{2\pi}{a}}, $$ a shift of integration into the complex plane $x\to x-ib/a$ to prove the relation $$ \int_{-\infty}^{\infty}dx \exp(-ax^2+ibx)=\sqrt{\frac{2\pi}{a}}\exp(-b^2/a) $$ is actually a variable transformation that makes the integration boundaries complex.

user203417
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    Well, the given action $S$ is complex. Is the question why $S$ is complex? If so consider to include some context. – Qmechanic Jun 15 '23 at 09:38
  • The action is complex because it comes from a Hubbard Stratonovic decoupling of a term $\exp(-1/2 \int_r jAj)=\int D\psi \exp(\int_r 1/2 \psi A^{-1} \psi +\int_r i \psi j)$ where the $i$ on the right side in front of the $\psi j$ is due to the minus signs on the left in front of the $\int_r jAj$. – user203417 Jun 15 '23 at 10:17
  • My confusion comes from the fact that the field $\psi$ is real and the saddle point apparently is not in the area of definition of $\psi$. – user203417 Jun 15 '23 at 10:25
  • Right. So a real-valued classical description is absent, but the quantum theory is nevertheless well-defined. That happens. – Qmechanic Jun 16 '23 at 08:22

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