Schrodinger equation on the complex disk

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SUMMARY

The discussion centers on the application of the Schrödinger equation within the context of the complex disk, specifically the region defined by ##\mathbb{D}=\{z\in\mathbb{C}: |z|<1\}##. Participants highlight that the wave function ##\psi(z)## cannot be analytically defined in this space, as functions like sine or cosine lead to non-normalizable exponential behaviors when extended to the imaginary axis. The conversation also touches on the implications of using the real 2D disk ##\mathbb{D}_{\mathbb{R}}=\{(x,y)\in\mathbb{R}^{2}: x^2+y^2<1\}##, suggesting that energy eigenfunctions in this scenario can be expressed in terms of Bessel functions, specifically ##\psi(z) \propto J_l(k\sqrt{x^2 + y^2})##.

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Ssnow
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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
 
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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.
 
hilbert2 said:
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
 
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.
 
Ok thanks! I will think on.
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
 

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