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Mathematical method to get Laplacian

The Laplacian in Cartesian coordinates is:

<math>\bigtriangledown^2=\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}</math>

We start the derivation of the Laplacian in spherical coordinates with the kinetic energy operator:

<math>\ <KE> = <\Psi|\frac{p^2}{2m}|\Psi> = \frac{1}{2m}*<\Psi|{p^2}|\Psi> </math>

For the moment we can disregard <math>\ \frac{1}{2m} </math> because it is a multiplicative constant.

In 1-D:

<math>\ <\Psi|p^2|\Psi> = <p\Psi|p\Psi> = \int (\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})^*(\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})dx </math>

In 3-D:

<math> <KE> </math> is proportional to <math> \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV </math>

Where:

<math> \bigtriangledown=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}+\frac{\partial\Psi}{\partial z}</math>

Legendre Polynomials

<math>\Theta(\theta):\alpha,\beta = m^2</math>

where <math>\alpha</math> is separation of the r-dependent part and <math>\beta</math> is the separation of the <math>\phi</math> dependent part.

Then the Differential equation form of <math>\Theta(\theta)= P_l^m (cos\theta)</math> replacing <math>z=cos\theta</math>

is: <math>(1-z^2)\frac{d^2U}{dz^2} - 2z \frac{dU}{dz} + \alpha U = 0</math>

Then we can take

<math>U(z)= \sum_{n=0}^\infty a_n z^n </math>

We can then take <math>\xi=\frac{1-z}{2}</math> where <math>-1 < z < 1 </math> and therefore <math>0 < z < 1</math> Then <math>U(\xi)</math> is also a differential equation where

<math>U(\xi)=\sum_{n=0}^\infty a_n \xi^2</math>

From this we get

<math> const. U'' + const. U' + const. U= const</math> for all <math>\xi</math>

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