Our Schrödinger’s equation: ∂ψ/∂t = i∂2ψ/∂x2
and parallel classical heat equation: ∂ψ/∂t = ∂2ψ/∂x2
Initial conditions for t=0: ψ = e–x2

We postulate a general solution of the form
ψ = pe–(px)2
where p is a complex function (to be determined) of t. p = t–1/2/2 solves the heat equation numerically!
(except t = 1/4 when ψ = e–x2)

We know already that p(0) = 1 and that it is not always real.
For reference d((px)2)/dt = 2p'px2.
For reference d(e–(px)2))/dt = –2p'px2e–(px)2.
For reference d((px)2)/dx = 2p2x.
For reference d(e–(px)2))/dx = –2p2xe–(px)2.
Starting from ∂ψ/∂t = i∂2ψ/∂x2 and p' = dp/dt
∂ψ/∂t = p'e–(px)2–2p2p'x2 e–(px)2 = p'(1 – 2(px)2)e–(px)2.

∂2ψ/∂x2 = ∂/∂x(∂/∂x(pe–(px)2)) = ∂/∂x(–p2p2xe–(px)2) = –∂/∂x(2p3xe–(px)2)
= –2p3∂/∂x(xe–(px)2) = –2p3(1 – x2p2x)e–(px)2 = –2p3(1 – 2(xp)2)e–(px)2

Canceling e–(px)2 and (1 – 2(xp)2) we get:
p' = –2p3
for the heat equation and
p' = –i2p3
for Schrödinger.
These serve as an ordinary differential equation in p of t. p(t) = (1+4it)–1/2 solves p' = –i2p3.

σ = standard deviation of the distribution ψ2.

This Concrete track found several errors in doing the differentiation. It would have been faster to write a symbolic differentiatior!