Schrodinger equation on the complex disk

[Ssnow]

Hi to all member of the Physics Forums. I have this question: it is possible consider the analogue of the Schrödinger equation on the plane with configuration space $(x,p)\in\mathbb{R}^4$ on the complex disk $\mathbb{D}=\{z\in\mathbb{C}: |z|<1\}$?
Ssnow

 

[hilbert2]

You can describe a point on 2D plane as a complex number, but the product of two such complex numbers doesn't have any clear interpretation in that case. The wave function $\psi (z)$ could not be required to be analytical in that situation, because a sine or cosine type function will behave like an increasing exponential function (not normalizable) if analytically continued to the imaginary axis.

 

[Ssnow]

You can describe a point on 2D plane as a complex number, but the product of two such complex numbers doesn't have any clear interpretation in that case. The wave function $\psi (z)$ could not be required to be analytical in that situation, because a sine or cosine type function will behave like an increasing exponential function (not normalizable) if analytically continued to the imaginary axis.

Ok this is clear because from the complex point of view $z$ cannot be treated as a real position vector and, as conseguence, the physical interpretation of $\psi(z)$ is not so clear ... but if we consider the real 2D disk $\mathbb{D}_{\mathbb{R}}=\{(x,y)\in\mathbb{R}^{2}: x^2+y^2<1\}$ ? Is the situation similar to the 2D plane case?
Ssnow

 

[hilbert2]

If you absorb the coordinates $x$ and $y$ in the same complex variable $z=x+iy$ and write the energy eigenfunctions for a particle confined in a circular 2D box, they should be proportional to

$\displaystyle\psi (z) \propto J_l (k\sqrt{x^2 + y^2 })$,

where $k$ is a constant that depends on the energy eigenvalue and $J_l$ is a Bessel function.

If you're able to show that a complex function $\psi (x+iy) = J_l (k\sqrt{x^2 + y^2})$ with $k,x,y\in\mathbb{R}$ doesn't comply with Cauchy-Riemann equations, then the problem would be solved.

 

[Ssnow]

Ok thanks! I will think on.
Ssnow

 

[Ssnow]

Another specification, the Schrödinger equation on the disk will be of radial form (it is correct?) with $r< 1$ I suppose with the eigenfunctions proportional to Bessel functions ...
Ssnow