 Solution

For a separated weighted density, write $P_n=e^{\gamma t}p(\sigma )$. Its eigenvalue equation isSet $p=\psi \exp [-{\sigma }^2/(2\kappa \tau )]$. Differentiation eliminates the first derivative of $\psi$, giving the tilted Ornstein-Uhlenbeck oscillator transformationComplete the square and introduce the dimensionless variable $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$. The equation becomesComparing with the quantum harmonic oscillator gives $2E=1+\kappa n^2{\tau }^3-2\gamma \tau$. Its energies $E_m=m+1/2$ therefore correspond to ${\gamma }_m=\kappa n^2{\tau }^2/2-m/\tau$. The dominant nonnegative separated mode is the ground state, which is nodeless. ConsequentlyFor an independent check, $\log [\tilde{B}(t)/B_0]={\int }_0^t\tilde{\sigma }(s)ds$ is a centered Gaussian random variable with varianceThe exponential moment of a Gaussian linear functional gives $\mathbb{E}[B^n]=B_0^n\exp [n^{2}V(t)/2]$, whose long-time growth rate is the same ${\gamma }_n$. This exact Ornstein-Uhlenbeck multiplicative amplification formula also distinguishes the finite-time transient from the asymptotic exponential law. A general transient can contain sign-changing excited eigenfunctions in its expansion while the total weighted density remains nonnegative.