\phi_k {k^2 + m^2 \over 2} 
\phi_{-k} \right]. \nonumber \\
=&& {\cal N} \exp \left[ -\beta \int_{-\infty}^{\infty} d^D x ~~{1 \over 2}  
\left( \nabla \phi(x) \right)^2 + {m^2 \over 2} \phi^2(x) \right]. \nonumber\end{aligned}$$

To compute the eigenfunctions corresponding to non-zero eigenvalues we need to solve the eigenvalue problem $$\begin{aligned}
\left[ {1 \over 2 \beta} {\delta^2 \over \delta { \phi_k^{R,I}}^2}  
- (k^2 +m^2) \phi_k^{R,I}  {\delta \over \delta \phi_k^{R,I}} 
+E_N \right] F_N = 0\end{aligned}$$ where we wrote $P_k = P_0 F_N$. This equation has a familiar solution. To see this explicitly we perform a change of variables to obtain $$\begin{aligned}
\left[ {\delta^2 \over \delta X }  
- 2 X  {\delta \over \delta X} + {2 E_N  \over k^2+m^2} \right] F_N = 0
\label{Hermite}\end{aligned}$$ where $X=\sqrt{\beta(k^2 +m^2)} \phi_k^{R,I}$. Eq. (\[Hermite\]) has the solution $$\begin{aligned}
&& E_N = N (k^2 +m^2)  \\
&& F_N = H_N (X) = H_N(\sqrt{\beta(k^2 +m^2)} \phi_k^{R,I})\end{aligned}$$ where $H_N$ is the Hermite polynomial of order $N$.

From the properties of the Hermite polynomials we know that these solutions are orthogonal under the measure $e^{-X^2}$, which is just the k-part of the canonical distribution. The set of functions $\{H_N(X) e^{-X^2/2} \}$ thus constitutes an orthogonal basis. It can additionally be shown 