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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> (yk says) Rather, <math> \bigtriangledown\Psi=\begin{bmatrix} \frac{\partial\Psi}{\partial x}
\frac{\partial\Psi}{\partial y}
\frac{\partial\Psi}{\partial z}\end{bmatrix}</math>

and

<math> dV = r^2 sin\theta dr d\theta d\phi </math>

The above integral can also be written if we introduce a few tensors:

The first one, <math>g_{ij}=\begin{bmatrix} 1 & 0 & 0
0 & r^2 & 0
0 & 0 & r^2\sin\theta\end{bmatrix} </math> is a tensor, when it is bracketed by row and column vectors <math>\begin{bmatrix}dr & d\theta & d\phi\end{bmatrix}</math> and <math>\begin{bmatrix}dr
d\theta
d\phi\end{bmatrix}</math> it is form proper 3D distance, <math> ds^2 = dr^2 + r^2 d\theta^2 + (r\sin\theta)^2 d\phi^2</math>.

The 2nd tensor is similar to the first, but it will allow <math>\begin{bmatrix}\frac{d}{dr} & \frac{d}{d\theta} & \frac{d}{d\phi}\end{bmatrix}</math> and <math>\begin{bmatrix}\frac{d}{dr}
\frac{d}{d\theta}
\frac{d}{d\phi}\end{bmatrix}</math> to form a proper 3D dot product of two gradients, even though some of the differentiation do not have proper units (like <math>d\theta</math>). This 2nd tensor takes the form of <math>g^{ij}=\begin{bmatrix} 1 & 0 & 0
0 & \frac{1}{r^2} & 0
0 & 0 & \frac{1}{r^2\sin\theta}\end{bmatrix} </math>.

These two tensors are both expressed with <math>g</math>, but one has indeces as subscripts, and the other, superscripts, in the convention we are using. In fact, these two matrices are inverse of each other.

In addition, a scaler quantity, <math>g</math> is introduced, which is the determinant of the tensor <math>g_{ij}</math>. Using these notations, the expectation value for the kinetic energy in the spherical coordinates can be written as,

<math> \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi </math>

where

<math> g^{(\frac{1}{2})} = r^2 sin(\theta) </math>

<math> i,j = r,\theta,\phi </math>.

This expression can be equated to that for the normal Cartesian version of the same calculation, <math> \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) g^{\frac{1}{2}} dr d\theta d\phi</math>.

i.e. <math> \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV</math>.

Both sides of this equation can be modified (using integral by parts) to move the differentiation operator on the left to be operating on the “right” function. i.e.

<math> \int (\Psi)^* \bigtriangledown_i [g^{ij} g^{(\frac{1}{2})} (\bigtriangledown_j \Psi) dr d\theta d\phi = \int (\Psi)^*\bigtriangledown(\bigtriangledown \Psi) dV = \int (\Psi)^*(\bigtriangledown^2 \Psi) dV</math>.

At this point, if we replace <math>dV</math> with <math>g^{\frac{1}{2}} dr d\theta d\phi</math>, we get:

<math> \int (\Psi)^* \bigtriangledown_i [g^{ij} g^{(\frac{1}{2})} (\bigtriangledown_j \Psi) dr d\theta d\phi = \int (\Psi)^*(\bigtriangledown^2 \Psi) g^{\frac{1}{2}} dr d\theta d\phi</math>.

Since this equation between two integrals hold for any wave functions, the integrand must be equal, too. Further, we can regard the equation for the operator in front of <math>\Psi</math>. From this, we can claim that

<math> \bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j = \bigtriangledown^2 g^{\frac{1}{2}} </math> or

<math> \bigtriangledown^2 = \frac{1}{g^{\frac{1}{2}}}\bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j </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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