Why is the wick rotated heat equation equivalent to Mellin transform followed by Fourier transformed heat equation?

46 Views Asked by At

I was studying this equation:

$$\frac{\partial^2}{\partial x^2}\Psi(t,x)=c(t,x)\frac{\partial}{\partial t}\Psi(t,x) \tag{1}$$

where $c(t,x)=-t/x.$

I found a particular solution to this equation, $\Psi(t,x).$ Next, I performed the substitution $x\mapsto -ix$ (this is the Wick rotation). The equation became:

$$i\frac{\partial^2}{\partial x^2}\Psi(t,x)=c(t,x)\frac{\partial}{\partial t}\Psi(t,x) \tag{2}$$

where $c(t,x)=t/x.$

I made a note that a particular solution is simply $\Psi(t,ix)$ or a complex version of the purely real equation.

I decided to take the Mellin transform of $(1)$ to obtain:

$$ \frac{\partial ^3\Psi(x,s)}{\partial x^3}= c(x,s)\frac{\partial \Psi(x,s)}{\partial s} \tag{3}$$

where $c(x,s)=s^2/x^2.$

And I used the Fourier transform to finally obtain:

$$i\frac{\partial^2}{\partial\rho^2}\Psi(\rho,s)= c(\rho,s) \frac{\partial}{\partial s}\Psi(\rho,s) \tag{4}$$

where $c(\rho,s)=s^2/\rho^3.$

Notice that $(4)$ and $(2)$ are essentially the same equation.

I was surprised and delighted to discover this. Is this to be expected?

Why exactly do the two processes result in the same equation? (Schrodinger wave equation for a free particle).