The heat equation
$$u(t,x)= \frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right)$$
is presented as a Gaussian distribution changing with time from a point source at time zero at the origin (Dirac). It makes sense that the variance of the Gaussian increases with time, and it is $\sigma^2 = kt:$
$$\mathcal N(0,\sigma^2)= \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right).$$
Well, almost! Then there is that pesky $4$ turned into a $2$ in the Gaussian.
I understand that as long as it fulfills the heat equation, $u_t=k\nabla^2u,$ we are good. And it does:
$$\frac{\partial}{\partial t}u(t,x)=k\frac{\partial^2}{\partial x^2}u(t,x)=k\frac{1}{8\sqrt{\pi k t^5}}\exp \left(-\frac{x^2}{4kt}\right)\left(x^2-2kt \right)$$
But why is the denominator doubled in the heat equation? There surely is a physical way to just see it...
Note that for a normal random variable satisfying $$ dX_t = \sigma dW_t $$
Its PDE is, $$ \frac{\partial f(t,x)}{\partial t} + \frac{1}{2}\sigma^2 \frac{\partial^2 f(t,x)}{\partial x^2} = 0 $$
Compared with the PDE for the heat equation,
$$\frac{\partial u(t,x)}{\partial t}=k\frac{\partial^2 u(t,x)}{\partial x^2}$$
So, the correspondence is $2k$ vs. $\sigma^2$.