Can the Schrodinger equation satisfy Laplace's equation?

[bb1414]

The time-dependent Schrodinger equation is given by:

##-\frac{\hslash^{2}}{2m}\triangledown^{2}\psi+V\psi=i\hslash\frac{\partial }{\partial t}\psi##​

Obviously, there is a laplacian in the kinetic energy operator. So, I was wondering if the equation was rearranged as

##-\frac{\hslash^{2}}{2m}\triangledown^{2}\psi=i\hslash\frac{\partial }{\partial t}\psi-V\psi##​

then does there exist a wave function ## \psi## that satisfies Laplace's equation

##\triangledown^{2}\psi=0##

​

so that

##\triangledown^{2}\psi=i\hslash\frac{\partial }{\partial t}\psi-V\psi=0##
​

If so, can the solution then be a set of spherical harmonics, which is commonly found when dealing with Laplace's equation in other areas?

 

[blue_leaf77]

My memory about the solution of Laplace's equation is a bit hazy but after checking wikipedia about spherical harmonics, the general solution of this equation takes the form of
$$
\psi(r,\theta,\phi) = \sum_{l=0} \sum_{m=-l}^l c_{lm} r^l Y_{lm}(\theta,\phi)
$$
which is clearly not normalizable and hence cannot serve as a square integrable solution required to be an element of Hilbert space.

 

[Demystifier]

An attempt to solve the equation by a Fourier transform would lead to ##{\bf k}^2=0##, implying ##k_x=k_y=k_z=0##. That would correspond to a constant function, which is also not square integrable.

 

[stevendaryl]

An attempt to solve the equation by a Fourier transform would lead to ##{\bf k}^2=0##, implying ##k_x=k_y=k_z=0##. That would correspond to a constant function, which is also not square integrable.

That's true for an infinite 2-D plane. However, if your "universe" consists of a half-infinite cylinder, parametrized by [itex]x,y[/itex] according to:

• [itex]0 \leq x \lt \infty[/itex]
• [itex]0 \leq y \leq L[/itex]
• The point [itex](x,y)[/itex] and the point [itex](x, y+L)[/itex] are identified.

Then there are normalizable solutions of the form:

[itex]\psi(x,\theta) = e^{\frac{2n\pi}{L} (-x \pm i y)}[/itex]