$\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\Re}{\operatorname{Re}}$
Consider Schrödinger equation \begin{gather} u_t=i u_{xx}\qquad -\infty< t< \infty \label{eq-3.C.1} \end{gather} which we get from diffusion equation formally setting $k=i$. We can apply the same arguments arriving t \begin{gather} u(x,t)= \frac{1}{\sqrt{4\pi i t}} e^{ix^2/4t} \label{eq-3.C.2} \end{gather} but we need to figure out the "sign" of the square root. Also we need to extend $t>0$ for diffusion equation for all $t$ for Schrödinger equation.
Consider $k\in \mathbb{C}\colon \Re(k)>0$. Then the diffusion equation still has sense for $t>0$ and the square root is uniquely determined so for $k=i=e^{i\pi/2}$ we get instead a correct value of $\sqrt{4\pi i t}$, namely, $e^{i\pi/4}\sqrt{4\pi t}$.
Now, as $t<0$ we should consider $t>0$ and take a complex–conjugate, thus arrivng to \begin{gather} u(x,t)= \frac{e^{\mp \frac{\pi i}{4}}}{\sqrt{4\pi |t|}} e^{ix^2/4t} \qquad \text{as} \ \pm t>0 \label{eq-3.C.3} \end{gather} as a solution to Schrödinger equation (\ref{eq-3.C.1}) with initial condition \begin{gather} u|_{t=0}=\delta (x). \label{eq-3.C.4} \end{gather}