If we solve a crossbreed of Schr\"odinger and Heat

\begin{equation}

u_t =au_{xx}

\tag{1}

\end{equation}

with $a\in \mathbb{C}$ with the positive real part (which is well-posed in the direction of positive $t$) we get a formula with kernel $\frac{1}{\sqrt {4\pi a t}} e^{-x^2/4at}$. This square root is positive for real $a>0$.

Let us look, what happens if $a=bi+\varepsilon$, $\varepsilon \to +0$ and $b>0$. We get

$\frac{1}{\sqrt {4\pi (bi+\varepsilon) t}} e^{-x^2/4(bi+\varepsilon)t}$. But if $\varepsilon \to +0$ then this equation (1) becomes Schr\"odinger equation

\begin{equation}

u_t =ibu_{xx}

\tag{2}

\end{equation}

exponent factor tends to $e^{-x^2/4b i t}=e^{ix^2/4b t}$, but since $bi +\varepsilon \to bi$ from the right, then its argument tends to $\pi/2$ from below, and argument of the square root tends to ${\pi/4}$, so we get

\begin{equation}

\frac{1}{\sqrt {4\pi bt}} e^{-i\pi/4}e^{ix^2/4b t}

\tag{3}

\end{equation}

Recall that $b>0$, $t>0$.

The same is true for $b<0, t<0$. Indeed, changing $t\mapsto -t$, $b\mapsto -b$ preserves equation (2).

So (3) holds for $bt>0$. But taking complex conjugation and $b\to -b$ also preserves (2). This operation with (3) brings

\begin{equation}

\frac{1}{\sqrt {4\pi |bt|}} e^{i\pi/4}e^{ix^2/4b t}.

\tag{4}

\end{equation}

So, correct formula (I suspect I skipped on the lecture) for convolution kernel is

\begin{equation}

\frac{1}{\sqrt {4\pi |bt|}} e^{\mp i\pi/4}e^{ix^2/4b t}\qquad \text{for} \ \ bt \gtrless 0.

\tag{5}

\end{equation}