Quantum Mechanics Notes — Dale Matthews

Laplacian in Spherical Coordinates

Here we derive an expression for the Laplacian in spherical coordinates using only the chain rule.

The definition of spherical coordinates (from Stewart Calculus):

\[x=\rho \sin(\phi) \cos(\theta)\]

\[y=\rho \sin(\phi) \sin(\theta)\]

\[z=\rho \cos(\phi)\]

and we have:

\[\rho^2=x^2+y^2+z^2\]

from these;

\[\rho = \sqrt{x^2+y^2+z^2}\]

\[\phi=\arccos(z/\rho)\]

\[\theta = atan2(y,x)\]

The arctangent function returns angles in range \((-\pi/2,\pi/2)\) whereas atan2 returns angles in the range \((-\pi,\pi]\) (depends on definition). atan2 is defined piecewise in terms of the arctangent function.

We want to find the Laplacian of a function \(u\) expressed in terms of spherical coordinates.

In Cartesian coordinates the Laplacian of \(u\) is given by:

\[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\]

By the chain rule we can write:

\[\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \rho} \frac{\partial \rho}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x}+\frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}\]

Similarly we can compute the other first derivatives of \(u\) wrt \(y\) and \(z\) which have the same form.

First compute \(\rho_x,\rho_y,\rho_z\)

\[\rho=\sqrt{x^2+y^2+z^2}=(x^2+y^2+z^2)^{1/2}\]

\[\rho_x=\frac{1}{2}(x^2+y^2+z^2)^{-1/2} \times 2x= x / \sqrt{x^2+y^2+z^2}=x/\rho\]

Similarly for \(\rho_y\) and \(\rho_z\) we have:

\[\rho_y=\frac{1}{2}(x^2+y^2+z^2)^{-1/2} \times 2y= y / \sqrt{x^2+y^2+z^2}=y/\rho\]

\[\rho_z=\frac{1}{2}(x^2+y^2+z^2)^{-1/2} \times 2z= z / \sqrt{x^2+y^2+z^2}=z/\rho\]

Now compute \(\theta_x,\theta_y,\theta_z\)

\[\theta=atan2(y,x)\]

\[\tan(\theta)=y/x\]

this is valid away from the \(x=y=0\) axis.

\[d \tan(\theta)=\sec^2(\theta) d\theta\]

By the quotient rule:

\[\sec^2(\theta) d\theta = d (y/x)=\frac{x dy - y dx}{x^2}\]

\[\sec^2(\theta)=1+\tan^2(\theta)=1+(y/x)^2 = \frac{x^2+y^2}{x^2}\]

\[\frac{(x^2+y^2)}{x^2} d\theta =\frac{x dy - y dx}{x^2}\]

\[d\theta = \frac{x dy - y dx}{x^2+y^2}= -\frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2} dy\]

But we know from calculus that the differential \(d\theta\) is given by:

\[d\theta = \frac{\partial \theta}{\partial x} dx + \frac{\partial \theta}{\partial y} dy+\frac{\partial \theta}{\partial z} dz\]

so we can identify:

\[\frac{\partial \theta}{\partial x} = -\frac{y}{x^2+y^2}\]

\[\frac{\partial \theta}{\partial y} = \frac{x}{x^2+y^2}\]

\[\frac{\partial \theta}{\partial z}=0\]

Let’s write these in terms of the spherical coordinates. From the definition of spherical coordinates;

\[x=\rho \sin(\phi) \cos(\theta)\]

\[y=\rho \sin(\phi) \sin(\theta)\]

\[z=\rho \cos(\phi)\]

\[x^2+y^2= \rho^2 \sin^2(\phi) \cos^2(\theta)+\rho^2 \sin^2(\phi) \sin^2(\theta)=\rho^2 \sin^2(\phi)\]

therefore

\[\frac{\partial \theta}{\partial x} =-\frac{y}{x^2+y^2}= -\frac{\rho \sin(\phi) \sin(\theta)}{\rho^2 \sin^2(\phi)}=-\frac{\sin(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{\rho \sin(\phi) \cos(\theta)}{\rho^2 \sin^2(\phi)}=\frac{\cos(\theta)}{\rho \sin(\phi)}\]

and we found earlier

\[\frac{\partial \theta}{\partial z}=0\]

Now we want to find \(\phi_x,\phi_y,\phi_z\).

Recall

\[\phi=\arccos(z/\rho)\]

Introduce the variable \(v=z/\rho=z \rho^{-1}\) so \(\phi = \arccos(v)\)

Using the standard formula from calculus we have:

\[d\phi =-\frac{1}{\sqrt{1-v^2}} dv\]

Now calculate \(dv\) by calculating \(v_x,v_y,v_z\).

Use \(\frac{\partial \rho}{\partial x} = x/\rho\) which we calculated earlier.

\[\frac{\partial v}{\partial x}= \frac{\partial}{\partial x} [z \rho^{-1}]=-z\rho^{-2}\frac{\partial \rho}{\partial x}=-\frac{z}{\rho^2} \cdot \frac{x}{\rho} = -\frac{zx}{\rho^3}\]

By symmetry \(v_y\) has the same form as \(v_x\)

\[\frac{\partial v}{\partial y} = -\frac{zy}{\rho^3}\]

Calculate \(v_z\)

\[\frac{\partial v}{\partial z}=\frac{\partial}{\partial z} (z \rho^{-1})=\rho^{-1}+z(-\rho^{-2})\frac{\partial \rho}{\partial z}=\frac{1}{\rho}-\frac{z}{\rho^2 }\frac{\partial \rho}{\partial z}= \frac{1}{\rho}-\frac{z}{\rho^2 }\frac{z}{\rho}=\frac{1}{\rho} - \frac{z^2}{\rho^3} = \frac{\rho^2-z^2}{\rho^3}\]

therefore

\[\frac{\partial v}{\partial z}=\frac{\rho^2-z^2}{\rho^3}=\frac{x^2+y^2}{\rho^3}\]

For \(\sqrt{1-v^2}\) we have;

\[\sqrt{1-v^2}=\sqrt{1-z^2/\rho^2}=\sqrt{\frac{\rho^2-z^2}{\rho^2}}=\frac{\sqrt{x^2+y^2}}{\rho}\]

Recall we are trying to find \(d\phi\) by calculating \(dv\). So far what we have calculated is;

\[\frac{\partial v}{\partial y} = -\frac{zy}{\rho^3}\]

\[\frac{\partial v}{\partial x} = -\frac{zx}{\rho^3}\]

\[\frac{\partial v}{\partial z}=\frac{x^2+y^2}{\rho^3}\]

\[\sqrt{1-v^2}=\frac{\sqrt{x^2+y^2}}{\rho}\]

and

\[\begin{aligned} d\phi & = -\frac{1}{\sqrt{1-v^2}}dv=-\frac{1}{\sqrt{1-v^2}}(v_x dx + v_y dy + v_z dz) \\ & = \left (-\frac{\rho}{\sqrt{x^2+y^2}} \right ) \left ( \frac{-zx}{\rho^3} \right ) dx+\left (-\frac{\rho}{\sqrt{x^2+y^2}} \right ) \left ( \frac{-zy}{\rho^3} \right )dy+\left (-\frac{\rho}{\sqrt{x^2+y^2}} \right ) \left ( \frac{x^2+y^2}{\rho^3} \right ) dz \\ &= \frac{zx}{\rho^2 \sqrt{x^2+y^2}} dx+\frac{zy}{\rho^2 \sqrt{x^2+y^2}} dy - \frac{\sqrt{x^2+y^2}}{\rho^2} dz \end{aligned}\]

But we know from calculus;

\[d\phi = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz\]

so we can identify;

\[\frac{\partial \phi}{\partial x} = \frac{zx}{\rho^2 \sqrt{x^2+y^2}}\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}\]

Let’s convert these results to spherical coordinates.

\[x=\rho \sin(\phi) \cos(\theta)\]

\[y=\rho \sin(\phi) \sin(\theta)\]

\[z=\rho \cos(\phi)\]

\[x^2+y^2=\rho^2 \sin^2(\phi)\]

\[\sqrt{x^2+y^2}=\rho \sin(\phi)\]

\[\frac{\partial \phi}{\partial x}=\frac{zx}{\rho^2 \sqrt{x^2+y^2}}=\frac{(\rho \cos(\phi))(\rho \sin(\phi) \cos(\theta))}{\rho^2 \cdot \rho \sin(\phi)}=\frac{\cos(\phi)\cos(\theta)}{\rho }\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}} =\frac{(\rho \cos(\phi))(\rho \sin(\phi) \sin(\theta))}{\rho^2 \cdot \rho \sin (\phi)}=\frac{\cos(\phi)\sin(\theta)}{\rho}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= -\frac{\rho \sin(\phi)}{\rho^2} = - \frac{\sin(\phi)}{\rho}\]

To summarise the results we have so far for the derivatives of the spherical variables we have;

\[\frac{\partial \rho}{\partial x} = \frac{x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho}\]

\[\frac{\partial \rho}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}=\frac{y}{\rho}\]

\[\frac{\partial \rho}{\partial z}=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\rho}\]

\[\frac{\partial \theta}{\partial x} =-\frac{y}{x^2+y^2}=-\frac{\sin(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial z}=0\]

\[\frac{\partial \phi}{\partial x}=\frac{zx}{\rho^2 \sqrt{x^2+y^2}}=\frac{\cos(\phi)\cos(\theta)}{\rho }\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}} =\frac{\cos(\phi)\sin(\theta)}{\rho}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= - \frac{\sin(\phi)}{\rho}\]

Now we find an expression for the Laplacian into which we will plug in these results. We also need to find the second derivatives

\[\rho_{xx},\rho_{yy},\rho_{zz},\theta_{xx},\theta_{yy},\theta_{zz},\phi_{xx},\phi_{yy},\phi_{zz}\]

But first I will find an expression for the Laplacian so we see why we need these second derivatives.

The Laplacian is;

\[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\]

So to find an expression for the Laplacian we have to find an expression for the second derivatives of \(u\).

We have for the first derivative \(u_x\);

\[\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \rho}\frac{\partial \rho}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}\]

taking the derivative again and using the product rule;

\[\frac{\partial^2 u}{\partial x^2}=\left [\frac{\partial}{\partial x} \left (\frac{\partial u}{\partial \rho} \right ) \right] \frac{\partial \rho}{\partial x} + \frac{\partial u}{\partial \rho} \frac{\partial^2 \rho}{\partial x^2} + \left [ \frac{\partial}{\partial x} \left (\frac{\partial u}{\partial \theta} \right ) \right ]\frac{\partial \theta}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial x^2} + \left [ \frac{\partial}{\partial x} \left (\frac{\partial u}{\partial \phi} \right ) \right ] \frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \phi} \frac{\partial^2 \phi}{\partial x^2}\]

Now we find each of the terms in brackets \([]\) using the chain rule.

From the chain rule we have;

\[\frac{\partial}{\partial x}=\frac{\partial}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial}{\partial \phi}\frac{\partial \phi}{\partial x}\]

so for the first bracketed term

\[\frac{\partial}{\partial x} \left ( \frac{\partial u}{\partial \rho} \right ) = \frac{\partial^2 u}{\partial \rho^2} \frac{\partial \rho}{\partial x}+\frac{\partial^2 u}{\partial \theta \partial \rho} \frac{\partial \theta}{\partial x} + \frac{\partial^2 u}{\partial \phi \partial \rho}\frac{\partial \phi}{\partial x}= u_{\rho \rho} \rho_x + u_{\theta \rho} \theta_x + u_{\phi \rho} \phi_x\]

for the second bracketed term;

\[\frac{\partial}{\partial x} \left ( \frac{\partial u}{\partial \theta} \right ) = \frac{\partial^2 u}{\partial \rho \partial \theta} \frac{\partial \rho}{\partial x} + \frac{\partial^2 u}{\partial \theta^2} \frac{\partial \theta}{\partial x} + \frac{\partial^2 u}{\partial \phi \partial \theta}\frac{\partial \phi}{\partial x}= u_{\rho \theta} \rho_x + u_{\theta \theta} \theta_x + u_{\phi \theta} \phi_x\]

For the third bracketed term;

\[\frac{\partial}{\partial x} \left ( \frac{\partial u}{\partial \phi} \right ) = \frac{\partial^2 u}{\partial \rho \partial \phi}\frac{\partial \rho}{\partial x} + \frac{\partial^2 u}{\partial \theta \partial \phi}\frac{\partial \theta}{\partial x} + \frac{\partial^2 u}{\partial \phi^2}\frac{\partial \phi}{\partial x} = u_{\rho \phi} \rho_x + u_{\theta \phi} \theta_x + u_{\phi \phi} \phi_x\]

collecting all these results together for the bracketed terms and plugging into the original equation for \(u_{xx}\) we get;

\[\begin{aligned} \frac{\partial^2 u}{\partial x^2} &= \left [ u_{\rho \rho} \rho_x + u_{\theta \rho} \theta_x + u_{\phi \rho} \phi_x \right] \frac{\partial \rho}{\partial x} + \frac{\partial u}{\partial \rho} \frac{\partial^2 \rho}{\partial x^2}\\ &+ \left [ u_{\rho \theta} \rho_x + u_{\theta \theta} \theta_x + u_{\phi \theta} \phi_x \right ]\frac{\partial \theta}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial x^2} \\ &+ \left [ u_{\rho \phi} \rho_x + u_{\theta \phi} \theta_x + u_{\phi \phi} \phi_x \right ] \frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \phi} \frac{\partial^2 \phi}{\partial x^2} \\ &= \left [ u_{\rho \rho} \rho_x + u_{\theta \rho} \theta_x + u_{\phi \rho} \phi_x \right] \rho_x + u_{\rho} \rho_{xx}\\ &+ \left [ u_{\rho \theta} \rho_x + u_{\theta \theta} \theta_x + u_{\phi \theta} \phi_x \right ] \theta_x + u_{\theta} \theta_{xx} \\ &+ \left [ u_{\rho \phi} \rho_x + u_{\theta \phi} \theta_x + u_{\phi \phi} \phi_x \right ] \phi_x + u_{\phi} \phi_{xx} \\ &= u_{\rho \rho} \rho_x^2 + u_{\theta \rho} \theta_x\rho_x + u_{\phi \rho} \phi_x \rho_x + u_{\rho} \rho_{xx}\\ &+ u_{\rho \theta} \rho_x \theta_x + u_{\theta \theta} \theta_x^2 + u_{\phi \theta} \phi_x \theta_x + u_{\theta} \theta_{xx} \\ &+ u_{\rho \phi} \rho_x \phi_x + u_{\theta \phi} \theta_x \phi_x + u_{\phi \phi} \phi_x^2 + u_{\phi} \phi_{xx} \\ &= u_{\rho \rho} \rho_x^2 + u_{\theta \theta} \theta_x^2 + u_{\phi \phi} \phi_x^2 \\ &+ u_{\rho} \rho_{xx} + u_{\theta} \theta_{xx} + u_{\phi} \phi_{xx} \\ &+ 2 u_{\rho \theta} \rho_x \theta_x \\ &+ 2 u_{\rho \phi} \rho_x \phi_x \\ &+ 2 u_{\theta \phi} \theta_x \phi_x \end{aligned}\]

therefore we have the following result

\[\begin{aligned} \frac{\partial^2 u}{\partial x^2} &= u_{\rho \rho} \rho_x^2 + u_{\theta \theta} \theta_x^2 + u_{\phi \phi} \phi_x^2 \\ &+ u_{\rho} \rho_{xx} + u_{\theta} \theta_{xx} + u_{\phi} \phi_{xx} \\ &+ 2 u_{\rho \theta} \rho_x \theta_x \\ &+ 2 u_{\rho \phi} \rho_x \phi_x \\ &+ 2 u_{\theta \phi} \theta_x \phi_x \\ \end{aligned}\]

Notice we did not actually plug in the derivatives for \(x\) anywhere - we just found the form of the expression and did so in a very general way - so the same form of expression holds for \(u_{yy}\) and \(u_{zz}\) with \(x\) replaced by \(y\) or \(z\). Therefore we can combine these into a single expression for the Laplacian which we will plug the derivatives into.

\[\begin{aligned} \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} &= u_{\rho \rho}( \rho_x^2 + \rho_y^2+\rho_z^2) \\ &+ u_{\theta \theta} (\theta_x^2+\theta_y^2+\theta_z^2) \\ &+ u_{\phi \phi} (\phi_x^2+ \phi_y^2+\phi_z^2) \\ &+ u_{\rho} (\rho_{xx}+\rho_{yy}+\rho_{zz}) \\ &+ u_{\theta} (\theta_{xx} + \theta_{yy} + \theta_{zz}) \\ &+ u_{\phi} (\phi_{xx} + \phi_{yy} + \phi_{zz}) \\ &+ 2 u_{\rho \theta} (\rho_x \theta_x + \rho_y \theta_y + \rho_z \theta_z) \\ &+ 2 u_{\rho \phi} (\rho_x \phi_x + \rho_y \phi_y + \rho_z \phi_z) \\ &+ 2 u_{\theta \phi} (\theta_x \phi_x + \theta_y \phi_y + \theta_z \phi_z) \\ \end{aligned}\]

We have already evaluated all the first derivatives in this expression we just have to evaluate the second derivatives and then plug everything in and we will get our result. It turns out all the cross terms vanish.

Restating the results from earlier;

\[\frac{\partial \rho}{\partial x} = \frac{x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho}\]

\[\frac{\partial \rho}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}=\frac{y}{\rho}\]

\[\frac{\partial \rho}{\partial z}=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\rho}\]

\[\frac{\partial \theta}{\partial x} =-\frac{y}{x^2+y^2}=-\frac{\sin(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial z}=0\]

\[\frac{\partial \phi}{\partial x}=\frac{zx}{\rho^2 \sqrt{x^2+y^2}}=\frac{\cos(\phi)\cos(\theta)}{\rho }\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}} =\frac{\cos(\phi)\sin(\theta)}{\rho}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= - \frac{\sin(\phi)}{\rho}\]

\[\begin{aligned} \rho_{xx} &= \frac{\partial}{\partial x} \rho_x \\ &= \frac{\partial}{\partial x} [ x (x^2+y^2+z^2)^{-1/2}] \\ &= (x^2+y^2+z^2)^{-1/2}-\frac{1}{2} x (x^2+y^2+z^2)^{-3/2} (2x)\\ &= \frac{1}{\sqrt{x^2+y^2+z^2}} - \frac{x^2}{(x^2+y^2+z^2)^{3/2}} \\ &= \frac{x^2+y^2+z^2 - x^2}{(x^2+y^2+z^2)^{3/2}} \\ &= \frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}} \\ \end{aligned}\]

By symmetry;

\[\rho_{yy}=\frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}\]

\[\rho_{zz}=\frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}\]

So;

\[\rho_{xx}+\rho_{yy}+\rho_{zz}= \frac{(y^2+z^2)+(x^2+z^2)+(x^2+y^2)}{(x^2+y^2+z^2)^{3/2}}=\frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{3/2}}=\frac{2}{\sqrt{x^2+y^2+z^2}}=\frac{2}{\rho}\]

We calculated earlier;

\[\frac{\partial \theta}{\partial x} =-\frac{y}{x^2+y^2}=-\frac{\sin(\theta)}{\rho \sin(\phi)}\]

So;

\[\begin{aligned} \theta_{xx} &= \frac{\partial}{\partial x} \left [-\frac{y}{x^2+y^2} \right ] \\ &= \frac{\partial}{\partial x} \left [-y(x^2+y^2)^{-1} \right ] \\ &= -y(-1)(x^2+y^2)^{-2} (2x)\\ &= \frac{2xy}{(x^2+y^2)^2} \\ \end{aligned}\]

Similarly;

\[\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{\rho \sin(\phi)}\]

\[\begin{aligned} \theta_{yy} &= \frac{\partial}{\partial y} \left [\frac{x}{x^2+y^2} \right ] \\ &= \frac{\partial}{\partial y} \left [x(x^2+y^2)^{-1} \right ] \\ &= x(-1)(x^2+y^2)^{-2} (2y)\\ &= -\frac{2xy}{(x^2+y^2)^2} \\ \end{aligned}\]

\[\theta_{xx}+\theta_{yy}+\theta_{zz}=0\]

And for the \(\phi\) variables we calculated;

\[\frac{\partial \phi}{\partial x}=\frac{zx}{\rho^2 \sqrt{x^2+y^2}}=\frac{\cos(\phi)\cos(\theta)}{\rho }\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}} =\frac{\cos(\phi)\sin(\theta)}{\rho}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= - \frac{\sin(\phi)}{\rho}\]

Since \(\phi_x\) and \(\phi_y\) have the same form \(\phi_{xx}\) and \(\phi_{yy}\) will have the same form.

\[\begin{aligned} \phi_{xx} &= \frac{\partial}{\partial x} \left [ \frac{zx}{\rho^2 \sqrt{x^2+y^2}} \right ] \\ &= \frac{\partial}{\partial x} \left [ \frac{zx}{(x^2+y^2+z^2) \sqrt{x^2+y^2}} \right ] \\ &= \frac{\partial}{\partial x} \left [ zx (x^2+y^2+z^2)^{-1} (x^2+y^2)^{-1/2} \right ]\\ &= \frac{z}{\rho^2 \sqrt{x^2+y^2}} + zx(-1)(x^2+y^2+z^2)^{-2} (2x)(x^2+y^2)^{-1/2} \\ &+ zx(x^2+y^2+z^2)^{-1} (-1/2)(x^2+y^2)^{-3/2}(2x) \\ &= \frac{z}{\rho^2 \sqrt{x^2+y^2}} - \frac{2zx^2}{\rho^4 \sqrt{x^2+y^2}} -\frac{zx^2}{\rho^2(x^2+y^2)^{3/2}} \\ \end{aligned}\]

By symmetry;

\[\phi_{yy}=\frac{z}{\rho^2 \sqrt{x^2+y^2}} - \frac{2zy^2}{\rho^4 \sqrt{x^2+y^2}} -\frac{zy^2}{\rho^2(x^2+y^2)^{3/2}}\]

\[\begin{aligned} \phi_{xx}+\phi_{yy} &= \frac{2z}{\rho^2 \sqrt{x^2+y^2}} - \frac{2z(x^2+y^2)}{\rho^4 \sqrt{x^2+y^2}} -\frac{z(x^2+y^2)}{\rho^2(x^2+y^2)^{3/2}} \\ &= \frac{2z}{\rho^2 \sqrt{x^2+y^2}} - \frac{2z\sqrt{x^2+y^2}}{\rho^4 } -\frac{z}{\rho^2 \sqrt{x^2+y^2}} \\ &= \frac{z}{\rho^2 \sqrt{x^2+y^2}} - \frac{2z\sqrt{x^2+y^2}}{\rho^4 } \\ \end{aligned}\]

For \(\phi_z\) we had;

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= - \frac{\sin(\phi)}{\rho}\]

\[\begin{aligned} \phi_{zz} &= \frac{\partial}{\partial z} \left [ - \frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2} \right ] \\ &= \frac{\partial}{\partial z} \left [ - \sqrt{x^2+y^2} (x^2+y^2+z^2)^{-1} \right ] \\ &= \sqrt{x^2+y^2}(x^2+y^2+z^2)^{-2} (2z) \\ &= \frac{2z \sqrt{x^2+y^2}}{\rho^4} \\ \end{aligned}\]

\[\phi_{xx}+\phi_{yy}+\phi_{zz}= \frac{z}{\rho^2 \sqrt{x^2+y^2}}\]

We have;

\[x=\rho \sin(\phi) \cos(\theta)\]

\[y=\rho \sin(\phi) \sin(\theta)\]

\[z=\rho \cos(\phi)\]

\[x^2+y^2 =\rho^2 \sin^2(\phi)\]

\[\sqrt{x^2+y^2} =\rho \sin(\phi)\]

\[\phi_{xx}+\phi_{yy}+\phi_{zz}= \frac{z}{\rho^2 \sqrt{x^2+y^2}}= \frac{\rho \cos(\phi)} {\rho^2 \times \rho \sin(\phi)}=\frac{\cot(\phi)}{\rho^2}\]

So;

\[\phi_{xx}+\phi_{yy}+\phi_{zz} = \frac{\cot(\phi)}{\rho^2}\]

and earlier we calculated;

\[\rho_{xx}+\rho_{yy}+\rho_{zz}= \frac{2}{\rho}\]

\[\theta_{xx}+\theta_{yy}+\theta_{zz}=0\]

and for the first derivatives we calculated earlier;

\[\frac{\partial \rho}{\partial x} = \frac{x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho}\]

\[\frac{\partial \rho}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}=\frac{y}{\rho}\]

\[\frac{\partial \rho}{\partial z}=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\rho}\]

\[\frac{\partial \theta}{\partial x} =-\frac{y}{x^2+y^2}=-\frac{\sin(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{\rho \sin(\phi)}\]

\[\frac{\partial \theta}{\partial z}=0\]

\[\frac{\partial \phi}{\partial x}=\frac{zx}{\rho^2 \sqrt{x^2+y^2}}=\frac{\cos(\phi)\cos(\theta)}{\rho }\]

\[\frac{\partial \phi}{\partial y} = \frac{zy}{\rho^2 \sqrt{x^2+y^2}} =\frac{\cos(\phi)\sin(\theta)}{\rho}\]

\[\frac{\partial \phi}{\partial z} = - \frac{\sqrt{x^2+y^2}}{\rho^2}= - \frac{\sin(\phi)}{\rho}\]

and for the Laplacian we had the expression:

We can calculate now;

\[\rho_x^2+\rho_y^2+\rho_z^2=\frac{x^2+y^2+z^2}{\rho^2}=1\]

\[\theta_x^2+\theta_y^2+\theta_z^2=\frac{x^2+y^2}{(x^2+y^2)^2}=\frac{1}{x^2+y^2}=\frac{1}{\rho^2 \sin^2(\phi)}\]

\[\phi_x^2 + \phi_y^2 + \phi_z^2 = \frac{z^2 x^2}{\rho^4 (x^2+y^2)} + \frac{z^2 y^2}{\rho^4 (x^2+y^2)} + \frac{x^2+y^2}{\rho^4}=\frac{z^2(x^2+y^2)}{\rho^4 (x^2+y^2)}+\frac{x^2+y^2}{\rho^4}=\frac{z^2+x^2+y^2}{\rho^4}=\frac{\rho^2}{\rho^4}=\frac{1}{\rho^2}\]

Now we need to compute the cross terms.

\[\begin{aligned} \rho_x \theta_x + \rho_y \theta_y + \rho_z \theta_z &= \rho_x \theta_x + \rho_y \theta_y \\ &= \frac{x}{\rho} \frac{(-y)}{(x^2+y^2)} + \frac{y}{\rho}\frac{(x)}{(x^2+y^2)} \\ &= 0 \\ \end{aligned}\]

\[\begin{aligned} \rho_x \phi_x + \rho_y \phi_y + \rho_z \phi_z &= \frac{x}{\rho}\frac{zx}{\rho^2 \sqrt{x^2+y^2}} + \frac{y}{\rho} \frac{zy}{\rho^2 \sqrt{x^2+y^2}} + \frac{z}{\rho} \frac{(-\sqrt{x^2+y^2})}{\rho^2} \\ &= \frac{zx^2 + zy^2 }{\rho^3 \sqrt{x^2+y^2}} - \frac{z(x^2+y^2)}{\rho^3 \sqrt{x^2+y^2}} \\ &= 0 \\ \end{aligned}\]

\[\begin{aligned} \theta_x \phi_x + \theta_y \phi_y + \theta_z \phi_z &= \theta_x \phi_x + \theta_y \phi_y \\ &= \frac{-y}{x^2+y^2} \frac{zx}{\rho^2 \sqrt{x^2+y^2}} + \frac{x}{x^2+y^2} \frac{zy}{\rho^2 \sqrt{x^2+y^2}} \\ &= 0 \\ \end{aligned}\]

So the cross terms are all zero and;

We calculated:

\[\rho_x^2+\rho_y^2+\rho_z^2=1\]

\[\theta_x^2+\theta_y^2 + \theta_z^2 = \frac{1}{\rho^2 \sin^2 (\phi)}\]

\[\phi_x^2+\phi_y^2+\phi_z^2 = \frac{1}{\rho^2}\]

\[\rho_{xx}+\rho_{yy}+\rho_{zz}=\frac{2}{\rho}\]

\[\theta_{xx}+\theta_{yy}+\theta_{zz}=0\]

\[\phi_{xx}+\phi_{yy}+\phi_{zz}=\frac{\cot(\phi)}{\rho^2}\]

So we have for the Laplacian;

\[\begin{aligned} \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} &= u_{\rho \rho} + \frac{1}{\rho^2 \sin^2 (\phi)} u_{\theta \theta} + \frac{1}{\rho^2} u_{\phi \phi} + \frac{2}{\rho} u_{\rho} + \frac{\cot(\phi)}{\rho^2} u_{\phi} \\ \end{aligned}\]

Now I want to get this into the form;

\[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho} (\rho^2 u_{\rho}) + \frac{1}{\rho^2 \sin(\phi)} \frac{\partial}{\partial \phi} \left [ \sin(\phi) u_{\phi} \right ] + \frac{1}{\rho^2 \sin^2(\phi)} u_{\theta \theta}\]

First;

\[\begin{aligned} \frac{1}{\rho^2} \frac{\partial}{\partial \rho} (\rho^2 u_{\rho}) &= \frac{1}{\rho^2} \left [ \rho^2 u_{\rho \rho} + 2\rho u_{\rho} \right ] \\ &= u_{\rho \rho} + \frac{2}{\rho} u_{\rho} \\ \end{aligned}\]

\[\begin{aligned} \frac{1}{\rho^2 \sin(\phi)} \frac{\partial}{\partial \phi} \left [ \sin(\phi) u_{\phi} \right ] &= \frac{1}{\rho^2 \sin(\phi)} \left [ \sin(\phi) u_{\phi \phi} + \cos(\phi) u_{\phi} \right ] \\ &= \frac{1}{\rho^2} u_{\phi \phi} + \frac{\cot(\phi)}{\rho^2} u_{\phi} \\ \end{aligned}\]

Therefore;

and

\[\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho} \left (\rho^2 \frac{\partial}{\partial \rho} \right ) + \frac{1}{\rho^2 \sin(\phi)} \frac{\partial}{\partial \phi} \left ( \sin(\phi) \frac{\partial}{\partial \phi} \right ) + \frac{1}{\rho^2 \sin^2(\phi)} \frac{\partial^2}{\partial \theta^2}\]

This is equation 4.13 in Intro Quantum Mechanics by Griffiths - the Laplacian in spherical coordinates.

Legendre Polynomials

The Legendre differential equation is;

\[(1-x^2)y''-2xy'+l(l+1)y=0\]

We assume a series solution for \(y\)

\[y=\sum_{n=0}^{\infty} a_n x^n\]

\[y'=\sum_{n=1}^{\infty} n a_n x^{n-1}\]

\[y''=\sum_{n=2}^{\infty}(n-1)na_n x^{n-2}=\sum_{n=0}^{\infty} (n+1)(n+2) a_{n+2} x^n\]

\[x^2 y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^n\]

\[(1-x^2)y''= \sum_{n=0}^{\infty} (n+1)(n+2) a_{n+2} x^n - \sum_{n=2}^{\infty} n(n-1)a_n x^n\]

\[-2xy'=-\sum_{n=1}^{\infty} 2n a_n x^{n}\]

\[l(l+1)y=\sum_{n=0}^{\infty} l(l+1) a_n x^n\]

substituting these in to the Legendre differential equation;

\[\sum_{n=0}^{\infty} (n+1)(n+2) a_{n+2} x^n - \sum_{n=2}^{\infty} n(n-1)a_n x^n -\sum_{n=1}^{\infty} 2n a_n x^{n} + \sum_{n=0}^{\infty} l(l+1) a_n x^n=0\]

Setting the \(nth\) coefficient to zero gives;

\[(n+2)(n+1) a_{n+2} - n (n-1) a_n -2n a_n + l(l+1)a_n = 0\]

\[(n+2)(n+1) a_{n+2} + [l^2+l - n (n-1) -2n ]a_n = 0\]

and

\[(n+2)(n+1) a_{n+2} + [l^2+l - n^2 +n -2n ]a_n = 0\]

therefore

\[(n+2)(n+1) a_{n+2} + [l^2+l - n^2 -n ]a_n = 0\]

Looking at the \(a_n\) coefficient we have;

\[(l-n)(l+n+1)=(l-n)(l+n)+(l-n)=l^2-n^2+(l-n)\]

\[(n+2)(n+1) a_{n+2} + (l-n)(l+n+1)a_n = 0\]

\[a_{n+2}=-\frac{(l-n)(l+n+1)}{(n+2)(n+1)}a_n\]

The general solution of the Legendre differential equation is then a sum of two series containing (as the solution of a second order differential equation should) two constants \(a_0\) and \(a_1\) to be determined by the given initial conditions. Given \(a_0\) we can determine \(a_2, a_4, a_6,...\). Given \(a_1\) we can determine \(a_3, a_5, a_7,...\).

For any integer \(l\) one series terminates giving a polynomial solution; the other series diverges at \(x^2=1\). Negative integers \(l\) give the solutions already obtained with positive integers \(l\). For any integer \(l\) we can set the coefficient (\(a_0\) or \(a_1\)) of the divergent series to zero and get a polynomial solution. If the value of \(a_0\) or \(a_1\) in each polynomial is selected so that \(y=1\) when \(x=1\) the resulting polynomials are called Legendre polynomials written \(P_l(x)\).

Associated Legendre Functions

The goal here is to prove that the solution of the differential equation;

\[(1-x^2)y''-2xy'+\left [l(l+1) - \frac{m^2}{1-x^2} \right ]y=0\]

is the associated Legendre function;

\[y(x)=P_l^m(x)=(1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x)\]

where \(P_l(x)\) is the \(lth\) Legendre polynomial.

First substitute;

\[y=(1-x^2)^{m/2} u\]

Let \(\alpha=m/2\)

\[y(x)=(1-x^2)^{\alpha} u(x)\]

\[y'=\alpha(1-x^2)^{\alpha-1}(-2x) u +(1-x^2)^{\alpha} u'=-2\alpha x (1-x^2)^{\alpha-1}u+(1-x^2)^{\alpha}u'\]

\[y'' = \frac{d}{dx} \left [ -2\alpha x (1-x^2)^{\alpha-1}u \right ] + \frac{d}{dx} (1-x^2)^{\alpha}u'\]

Focusing on the first term in in \(y''\);

\[\begin{aligned} \frac{d}{dx} \left [ -2\alpha x (1-x^2)^{\alpha-1}u \right ] &= -2\alpha (1-x^2)^{\alpha-1} u-2\alpha x (\alpha-1) (1-x^2)^{\alpha-2} (-2x) u - 2\alpha x (1-x^2)^{\alpha-1} u'\\ &= -2\alpha (1-x^2)^{\alpha-1} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-2} u - 2\alpha x (1-x^2)^{\alpha-1} u' \\ \end{aligned}\]

Now looking at the second term in \(y''\);

\[\begin{aligned} \frac{d}{dx} (1-x^2)^{\alpha} u' &= \alpha(1-x^2)^{\alpha-1} (-2x) u' + (1-x^2)^{\alpha} u'' \\ &= -2\alpha x (1-x^2)^{\alpha-1} u' + (1-x^2)^{\alpha}u'' \\ \end{aligned}\]

Therefore combining these terms we get the following expression for \(y''\);

\[\begin{aligned} y'' &= -2\alpha (1-x^2)^{\alpha-1} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-2} u - 2\alpha x (1-x^2)^{\alpha-1} u'-2\alpha x (1-x^2)^{\alpha-1} u' + (1-x^2)^{\alpha}u'' \\ &= -2\alpha (1-x^2)^{\alpha-1} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-2} u - 4\alpha x (1-x^2)^{\alpha-1} u' + (1-x^2)^{\alpha}u'' \\ \end{aligned}\]

Restating the results we have derived thus far;

\[y=(1-x^2)^{\alpha} u\]

\[y'=-2\alpha x (1-x^2)^{\alpha-1}u+(1-x^2)^{\alpha}u'\]

\[y''=-2\alpha (1-x^2)^{\alpha-1} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-2} u - 4\alpha x (1-x^2)^{\alpha-1} u' + (1-x^2)^{\alpha}u''\]

We plug these in to the differential equation;

\[(1-x^2)y''-2xy'+\left [l(l+1) - \frac{m^2}{1-x^2} \right ]y=0\]

\[\begin{aligned} (1-x^2)y'' &= -2\alpha (1-x^2)^{\alpha} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-1} u - 4\alpha x (1-x^2)^{\alpha} u' + (1-x^2)^{\alpha+1}u'' \\ \end{aligned}\]

\[\begin{aligned} -2xy' &= 4\alpha x^2 (1-x^2)^{\alpha-1}u-2x(1-x^2)^{\alpha}u' \\ \end{aligned}\]

\[\begin{aligned} l(l+1) y &= l(l+1)(1-x^2)^{\alpha} u \\ \end{aligned}\]

\[\begin{aligned} -\frac{m^2}{1-x^2} y &= -m^2(1-x^2)^{\alpha-1} u \\ \end{aligned}\]

Combining these together we can rewrite the differential equation as;

\[\begin{aligned} -2\alpha (1-x^2)^{\alpha} u + 4\alpha (\alpha-1)x^2 (1-x^2)^{\alpha-1} u - 4\alpha x (1-x^2)^{\alpha} u' + (1-x^2)^{\alpha+1}u'' \\ +4\alpha x^2 (1-x^2)^{\alpha-1}u-2x(1-x^2)^{\alpha}u' + l(l+1)(1-x^2)^{\alpha} u -m^2(1-x^2)^{\alpha-1} u \\ =0 \end{aligned}\]

We can factor out \((1-x^2)^{\alpha-1}\);

\[(1-x^2)^{\alpha-1} \Phi=0\]

where

\[\begin{aligned} \Phi = \\ -2\alpha (1-x^2) u + 4\alpha (\alpha-1)x^2 u - 4\alpha x (1-x^2) u' + (1-x^2)^{2}u'' \\ +4\alpha x^2 u-2x(1-x^2)u' + l(l+1)(1-x^2) u -m^2 u \end{aligned}\]

\[\begin{aligned} \Phi = \\ [-2\alpha (1-x^2) + 4\alpha (\alpha-1)x^2+4\alpha x^2+ l(l+1)(1-x^2)-m^2 ] u \\ +[- 4\alpha x (1-x^2) -2x(1-x^2)]u' \\ + (1-x^2)^{2}u'' \\ \end{aligned}\]

Looking at the coefficient of \(u'\) for the moment;

\[\begin{aligned} - 4\alpha x (1-x^2) -2x(1-x^2) &= -2x[2\alpha(1-x^2) + (1-x^2)] \\ &= -2x (2\alpha + 1)(1-x^2) \end{aligned}\]

Now looking at the coefficient of \(u\) we have;

\[\begin{aligned} &-2\alpha (1-x^2) + 4\alpha (\alpha-1)x^2+4\alpha x^2+ l(l+1)(1-x^2)-m^2\\&= -2\alpha (1-x^2) + (4\alpha^2-4\alpha)x^2+4\alpha x^2+ l(l+1)(1-x^2)-m^2 \\&=-2\alpha (1-x^2) + 4\alpha^2 x^2-4\alpha x^2+4\alpha x^2+ l(l+1)(1-x^2)-m^2 \\&= -2\alpha (1-x^2) + 4\alpha^2 x^2+ l(l+1)(1-x^2)-m^2 \\ \end{aligned}\]

Recall \(\alpha=m/2\) so \(2\alpha=m\) and \(4\alpha^2=m^2\) therefore

therefore the coefficient of \(u\) becomes;

\[\begin{aligned} &-2\alpha (1-x^2) + 4\alpha^2 x^2+ l(l+1)(1-x^2)-m^2\\&= -m(1-x^2)+m^2 x^2 + l(l+1)(1-x^2)-m^2 \\&= -m+mx^2+m^2 x^2 + l(l+1)(1-x^2)-m^2 \\&= -m(m+1)+mx^2+m^2 x^2 + l(l+1)(1-x^2) \\ &= -m(m+1)+m(m+1)x^2 + l(l+1)(1-x^2) \\ &= [l(l+1) -m(m+1)](1-x^2) \end{aligned}\]

Plugging these expressions for the coefficient of \(u\) and \(u'\) into the previous expression for \(\Phi\) we have;

\[\begin{aligned} &\Phi \\&= [-2\alpha (1-x^2) + 4\alpha (\alpha-1)x^2+4\alpha x^2+ l(l+1)(1-x^2)-m^2 ] u \\ & +[- 4\alpha x (1-x^2) -2x(1-x^2)]u' \\& + (1-x^2)^{2}u'' \\ &= [l(l+1) -m(m+1)](1-x^2)u-2x (2\alpha + 1)(1-x^2)u'+(1-x^2)^{2}u'' \end{aligned}\]

Recall the differential equation written in terms of \(u\) took the form;

\[(1-x^2)^{\alpha-1}\Phi=0\]

dividing by \((1-x^2)^{\alpha-1}\) we get \(\Phi=0\) and then we can again divide by \((1-x^2)\) to get;

\[[l(l+1) -m(m+1)]u-2x (2\alpha + 1)u'+(1-x^2)u''=0\]

but recall \(2\alpha=m\) so

\[[l(l+1) -m(m+1)]u-2x (m + 1)u'+(1-x^2)u''=0\]

or rearranging;

\[(1-x^2)u''-2(m+1)xu'+[l(l+1) -m(m+1)]u=0\]

which is Equation 10.3 in Boas Ch 12.10

For \(m=0\) this is Legendre’s equation with solutions \(P_l(x)\).

Differentiate 10.3 obtaining;

\[\frac{d}{dx} \left [ (1-x^2)u''\right ] = -2x u'' + (1-x^2)u'''\]

\[\begin{aligned} \frac{d}{dx} \left [ -2(m+1) xu' \right ] &= -2(m+1)u' -2(m+1)x u'' \\ &= -2(m+1)[u'+xu''] \\ \end{aligned}\]

\[\begin{aligned} \frac{d}{dx} \left [ l(l+1)-m(m+1) \right ] u &= \left [ l(l+1)-m(m+1) \right ] u'\\ \end{aligned}\]

so differentiating;

\[(1-x^2)u''-2(m+1)xu'+[l(l+1) -m(m+1)]u=0\]

we obtain

\[-2x u'' + (1-x^2)u'''-2(m+1)u'-2(m+1)xu''+[ l(l+1)-m(m+1) ] u'=0\]

combining \(u''\) terms;

\[\begin{aligned} & (1-x^2)u'''+[-2x -2(m+1)x]u'' -2(m+1)u'+[ l(l+1)-m(m+1) ] u' = 0 \\ & \implies (1-x^2)u'''+-2x[1 +(m+1)]u'' + [ -2(m+1)+l(l+1)-m(m+1)]u' = 0\\ & \implies (1-x^2)u'''+-2x(m+2)u'' + [l(l+1)+(-m-2)(m+1)]u' = 0 \\ & \implies (1-x^2)u'''+-2(m+2)xu'' + [l(l+1)-(m+1)(m+2)]u' = 0 \\ & \implies (1-x^2)(u')''+-2[(m+1)+1]x(u')' + [l(l+1)-(m+1)(m+2)]u' = 0 \end{aligned}\]

But this is just 10.3 with \(u'\) in place of \(u\) and \((m+1)\) in place of \(m\).

In other words if \(P_l(x)\) is a solution of (10.3) with \(m=0\), then \(P_l'(x)\) is a solution of (10.3) with \(m=1\), \(P_l''(x)\) is a solution with \(m=2\) and in general for integral \(m\), \(0\leq m \leq l\)

\[\frac{d^m}{dx^m} P_l(x)\]

is a solution of 10.3.

Then

\[u(x)=\frac{d^m}{dx^m} P_l(x)\]

is a solution of 10.3 so;

\[y=(1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x)\]

is a solution of 10.1.

The functions

\[y=(1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x)\]

are called associated Legendre functions and are denoted by;

\[P_l^m(x)=(1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x)\]

Angular Equation

\[\sin(\theta) \frac{d}{d\theta} \left ( \sin(\theta) \frac{d\Theta}{d\theta} \right ) + [l(l+1)\sin^2(\theta)-m^2]\Theta = 0\]

Let \(x=\cos(\theta)\) then

\[\frac{dx}{d\theta}=-\sin(\theta)\]

Let \(\Theta(\theta)=y(x)\) then

\[\frac{d\Theta}{d\theta}=\frac{dy}{dx}\frac{dx}{d\theta}=-\sin(\theta)\frac{dy}{dx}\]

and

\[\sin(\theta)\frac{d\Theta}{d\theta}=-\sin^2(\theta)\frac{dy}{dx}\]

Since \(x=\cos(\theta)\) we have

\[\sin^2(\theta)=1-\cos^2(\theta)=1-x^2\]

and

\[\sin(\theta)\frac{d\Theta}{d\theta}=-\sin^2(\theta)\frac{dy}{dx}=-(1-x^2)\frac{dy}{dx}\]

taking the derivative wrt \(\theta\) we have

\[\begin{aligned} \frac{d}{d\theta} \left ( \sin(\theta)\frac{d\Theta}{d\theta} \right ) &= \frac{d}{d\theta} \left [ -(1-x^2)\frac{dy}{dx} \right ] \\ &= -\left [ \frac{d}{d\theta} (1-x^2) \right ] \frac{dy}{dx} -(1-x^2) \frac{d}{d\theta} \frac{dy}{dx} \\ &= -\left [ \frac{d}{dx} (1-x^2) \frac{dx}{d\theta}\right ] \frac{dy}{dx} -(1-x^2) \frac{d}{dx} \frac{dy}{dx} \frac{dx}{d\theta} \\ &= -\left [ (-2x) (-\sin(\theta))\right ] \frac{dy}{dx} -(1-x^2) \frac{d^2 y}{dx^2} (-\sin(\theta)) \\ &= -2x\sin(\theta) y'(x) +(1-x^2)\sin(\theta) y''(x) \end{aligned}\]

\[\begin{aligned} \sin(\theta)\frac{d}{d\theta} \left ( \sin(\theta)\frac{d\Theta}{d\theta} \right ) &= -2x\sin^2(\theta) y'(x) +(1-x^2)\sin^2(\theta) y''(x) \\ &= -2x (1-x^2) y'(x) +(1-x^2)^2 y''(x) \end{aligned}\]

Therefore the differential equation becomes;

\[\begin{aligned} \sin(\theta)\frac{d}{d\theta} \left ( \sin(\theta)\frac{d\Theta}{d\theta} \right ) + [l(l+1)\sin^2(\theta)-m^2]\Theta = 0 \\ \implies -2x (1-x^2) y'(x) +(1-x^2)^2 y''(x) + [l(l+1)(1-x^2) - m^2]y(x) = 0 \\ \implies -2x y'(x) +(1-x^2) y''(x) + \left [l(l+1) - \frac{m^2}{1-x^2} \right ]y(x) = 0 \\ \implies (1-x^2) y''(x) -2x y'(x) + \left [l(l+1) - \frac{m^2}{1-x^2} \right ]y(x) = 0 \end{aligned}\]

This is the associated Legendre differential equation.

The solution is the associated Legendre function

\[y=P_l^m(x)=(1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x)\]

Recall \(x=\cos(\theta)\)

Clearly we can multiply the solution by any constant and we will still have a solution.

So \(\Theta(\theta)=AP_l^m(\cos(\theta))\) (recall \(\Theta(\theta)=y(x)\))

Laguerre Polynomials

The goal in this section is to show that the \(nth\) Laguerre polynomial defined by;

\[y=L_n(x)=\sum_{m=0}^n (-1)^m {n \choose m} \frac{x^m}{m!}\]

is the solution of the differential equation;

\[xy''+(1-x)y'+ny=0\]

Taking derivatives of \(L_n(x)\) we have;

\[\begin{aligned} y' &= \sum_{m=1}^n (-1)^m {n \choose m } \frac{mx^{m-1}}{m!} \\ &= \sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!} \\ \end{aligned}\]

and the second derivative of \(L_n(x)\) is;

\[\begin{aligned} y'' &= \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{m x^{m-1}}{m!} \\ &= \sum_{m=0}^{n-2} (-1)^{m+2} {n \choose m+2} \frac{x^m}{m!} \\ &= \sum_{m=0}^{n-2} (-1)^m {n \choose m+2} \frac{x^m}{m!} \end{aligned}\]

Now I compute each term in the differential equation;

\[xy''+(1-x)y'+ny=0\]

First we have for \(xy''\);

\[\begin{aligned} xy'' &= \sum_{m=0}^{n-2} (-1)^m {n \choose m+2} \frac{x^{m+1}}{m!}\\ &= \sum_{m=1}^{n-1} (-1)^{m-1} {n \choose m+1} \frac{x^m}{(m-1)!} \\ &= \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{(m-1)!} \end{aligned}\]

next \((1-x)y'\)

\[\begin{aligned} (1-x)y' &= \sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!} -\sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^{m+1}}{m!} \\ &= \sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!} -\sum_{m=1}^{n} (-1)^{m} {n \choose m} \frac{x^{m}}{(m-1)!} \end{aligned}\]

Finally for \(ny\) we have;

\[\begin{aligned} ny &= n\sum_{m=0}^n (-1)^m {n \choose m}\frac{x^m}{m!} \\ \end{aligned}\]

Combining these terms we get;

\[\begin{aligned} & xy''+(1-x)y'+ny \\ &= \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{(m-1)!} + \sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!} -\sum_{m=1}^{n} (-1)^{m} {n \choose m} \frac{x^{m}}{(m-1)!} \\ & + n\sum_{m=0}^n (-1)^m {n \choose m}\frac{x^m}{m!} \\ \end{aligned}\]

We want to show this expression is equal to zero so \(L_n(x)\) is a solution of the differential equation. To do this I will simplify it into a sum from \(m=1\) to \(m=n-1\) with remainder terms, show that the remainder terms equal zero and show that the sum is equal to zero. To figure out the remainder terms we have to evaluate the terms of some of the sums for \(m=0\) and \(m=n\).

For the sum;

\[\sum_{m=0}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!}\]

when \(m=0\) since \({n \choose 1}=n\) we get;

\[(-1)^{0+1} {n \choose 0+1} \frac{x^0}{0!} = -n\]

For the sum;

\[-\sum_{m=1}^{n} (-1)^{m} {n \choose m} \frac{x^{m}}{(m-1)!}\]

when \(m=n\) since \({n \choose n}=1\) we get;

\[-(-1)^n {n \choose n} \frac{x^n}{(n-1)!}=(-1)^{n+1} \frac{x^n}{(n-1)!}\]

For the sum;

\[n\sum_{m=0}^n (-1)^m {n \choose m}\frac{x^m}{m!}\]

when \(m=0\) since \({n \choose 0}=1\) we get;

\[n(-1)^0 {n \choose 0} \frac{x^0}{0!}=n\]

and for this same sum;

\[n\sum_{m=0}^n (-1)^m {n \choose m}\frac{x^m}{m!}\]

when \(m=n\) we get

\[n(-1)^n {n \choose n} \frac{x^n}{n!}=n (-1)^n \frac{x^n}{n!}=(-1)^n\frac{x^n}{(n-1)!}\]

Combining we have four remainder terms;

\[-n +(-1)^{n+1}\frac{x^n}{(n-1)!} +n + (-1)^n \frac{x^n}{(n-1)!}=0\]

so the remainder terms are zero and we are just left with the sums from \(m=1\) to \(m=n-1\)

\[\begin{aligned} & xy''+(1-x)y'+ny \\ &= \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{(m-1)!} + \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \frac{x^m}{m!} -\sum_{m=1}^{n-1} (-1)^{m} {n \choose m} \frac{x^{m}}{(m-1)!} \\ & + n\sum_{m=1}^{n-1} (-1)^m {n \choose m}\frac{x^m}{m!} \\ &= \sum_{m=1}^{n-1} (-1)^{m+1} {n \choose m+1} \left [ \frac{1}{(m-1)!}+\frac{1}{m!} \right ] x^m + \sum_{m=1}^{n-1} (-1)^m {n \choose m} \left [ \frac{n}{m!} - \frac{1}{(m-1)!} \right ] x^m \end{aligned}\]

The coefficient of \(x^m\) in this sum is;

\[\begin{aligned} & C_m \\ &= (-1)^{m+1} {n \choose m+1} \left [ \frac{1}{(m-1)!}+\frac{1}{m!} \right ] + (-1)^m {n \choose m} \left [ \frac{n}{m!} - \frac{1}{(m-1)!} \right ] \\ &= (-1)^{m+1} {n \choose m+1} \frac{1}{m!} (1+m) + (-1)^m {n \choose m} \frac{1}{m!}(n-m)\\ &= (-1)^m {n \choose m} \frac{1}{m!}(n-m) - (-1)^{m} {n \choose m+1} \frac{1}{m!} (1+m) \\ &= (-1)^m \frac{1}{m!} \left [ {n \choose m} (n-m) - {n \choose m+1} (1+m) \right ] \end{aligned}\]

We use a standard binomial coefficient identity - the forward form of the Pascal recursion;

\[{n \choose m+1} = \frac{(n-m)}{m+1} {n \choose m}\]

therefore;

\[\begin{aligned} & C_m \\ &= (-1)^m \frac{1}{m!} \left [ {n \choose m} (n-m) - {n \choose m+1} (1+m) \right ] \\ &= (-1)^m \frac{1}{m!} \left [ {n \choose m} (n-m) - \frac{(n-m)}{m+1} {n \choose m} (1+m) \right ]\\ &= (-1)^m \frac{1}{m!} \left [ {n \choose m} (n-m) - (n-m) {n \choose m} \right ] \\ &= 0 \end{aligned}\]

Therefore \(C_m=0\).

Therefore the entire sum and so we have;

\[xy''+(1-x)y'+ny=0\]

where

\[y=L_n(x)=\sum_{m=0}^n (-1)^m {n \choose m} \frac{x^m}{m!}\]

so a solution of this differential equation is indeed the Laguerre polynomial.

Associated Laguerre Polynomials

The goal in this section is to show that a solution of the differential equation;

\[xy'' + (k+1-x)y' + ny=0\]

is given by the associated Laguerre polynomial;

\[y(x)=L_n^k(x) = (-1)^k \frac{d^k}{dx^k} L_{n+k}(x)\]

where \(L_{n+k}(x)\) is the \((n+k)th\) Laguerre polynomial.

The Laguerre polynomial \(L_n(x)\) satisfies;

\[xy''+(1-x)y'+ny=0\]

Replace \(n\) by \(n+k\) to get the differential equation satisfied by the Laguerre polynomial \(L_{n+k}\);

\[xL_{n+k}''+(1-x)L_{n+k}'+(n+k)L_{n+k}=0\]

Let \(f(x)=L_{n+k}(x)\) then;

\[xf''+(1-x)f'+(n+k)f=0\]

We will differentiate this equation \(k\) times using the Leibniz rule for differentiating products to prove our original goal.

The Leibniz rule for differentiating products states;

\[\frac{d^k}{dx^k} (f g) = \sum_{m=0}^k {k \choose m} \left [ \frac{d^m}{dx^m} f \right ] \left [ \frac{d^{k-m}}{dx^{k-m}} g\right ]\]

Therefore;

\[\frac{d^k}{dx^k} (xf'') = \sum_{m=0}^k {k \choose m} \left [ \frac{d^m}{dx^m} x \right ] \left [ \frac{d^{k-m}}{dx^{k-m}} f'' \right ]\]

Since for \(m \geq 2\) we have;

\[\frac{d^m}{dx^m} x=0\]

\[\frac{d^k}{dx^k} (xf'')= {k \choose 0} \left [ \frac{d^0}{dx^0} x \right ] \left [ \frac{d^{k-0}}{dx^{k-0}} f''\right ]+ {k \choose 1} \left [ \frac{d}{dx} x \right ] \left [ \frac{d^{k-1}}{dx^{k-1}} f''\right ]=x f^{(k+2)}+kf^{(k+1)}\]

Similarly;

\[\frac{d^k}{dx^k} (1-x) f' = \sum_{m=0}^k {k \choose m} \left [ \frac{d^m}{dx^m} (1-x) \right ] \left [ \frac{d^{k-m}}{dx^{k-m}} f' \right ]\]

but for \(m\geq 2\)

\[\frac{d^m}{dx^m} (1-x)=0\]

therefore;

\[\frac{d^k}{dx^k} (1-x) f' = {k \choose 0} \left [ \frac{d^0}{dx^0} (1-x) \right ] \left [ \frac{d^{k-0}}{dx^{k-0}} f' \right ]+ {k \choose 1} \left [ \frac{d}{dx} (1-x) \right ] \left [ \frac{d^{k-1}}{dx^{k-1}} f' \right ] = (1-x) f^{(k+1)} - k f^{(k)}\]

Summarising we have;

\[\frac{d^k}{dx^k} (xf'')= x f^{(k+2)}+kf^{(k+1)}\]

\[\frac{d^k}{dx^k} (1-x) f'=(1-x) f^{(k+1)} - k f^{(k)}\]

Therefore when we differentiate the equation;

\[xf''+(1-x)f'+(n+k)f=0\]

satisfied by \(f=L_{n+k}\), \(k\) times we get;

\[\frac{d^k}{dx^k} (xf'') + \frac{d^k}{dx^k} (1-x) f' + (n+k) f^{(k)}=0\]

or plugging in;

\[x f^{(k+2)}+kf^{(k+1)}+(1-x) f^{(k+1)} - k f^{(k)} + (n+k) f^{(k)}=0\]

simplifying;

\[x (f^{(k)})''+(k+1-x) (f^{(k)})' +n f^{(k)}=0\]

Recall \(f=L_{n+k}\) so

\[f^{(k)} (x)= \frac{d^k}{dx^k} L_{n+k}(x)\]

If we multiply by \((-1)^k\) which is constant wrt \(x\) and let

\[y=(-1)^k f^{(k)}=(-1)^k \frac{d^k}{dx^k} L_{n+k}\]

then we get;

\[xy''+(k+1-x)y'+ny=0\]

where

\[y=(-1)^k \frac{d^k}{dx^k} L_{n+k}=L_n^k(x)\]

Therefore a solution of the differential equation;

\[xy''+(k+1-x)y'+ny=0\]

is the associated Laguerre polynomial;

\[y=(-1)^k \frac{d^k}{dx^k} L_{n+k}=L_n^k(x)\]

Radial Equation for Hydrogen Atom

\[\rho \frac{d^2 v}{d\rho^2} + 2(l+1-\rho) \frac{dv}{d\rho} + [\rho_0- 2(l+1)]v=0\]

Let \(x=2\rho\) so \(\rho=x/2\)

\[\frac{dv}{d\rho}=\frac{dv}{dx}\frac{dx}{d\rho}=2\frac{dv}{dx}\]

\[\frac{d}{d\rho} = \frac{dx}{d\rho} \frac{d}{dx} = 2 \frac{d}{dx}\]

\[\frac{d^2v}{d\rho^2} = \frac{d}{d\rho} \left ( 2\frac{dv}{dx}\right ) = 2 \frac{d}{dx} \left ( 2 \frac{dv}{dx} \right ) = 4 \frac{d^2v}{dx^2}\]

Substitute these into the following equation;

\[\begin{aligned} \rho \frac{d^2 v}{d\rho^2} + 2(l+1-\rho) \frac{dv}{d\rho} + [\rho_0- 2(l+1)]v=0 \\ \implies \left (\frac{x}{2} \right ) \left ( 4 \frac{d^2v}{dx^2} \right ) + 2(l+1-x/2) \left [ 2 \frac{dv}{dx} \right ]+[\rho_0 - 2(l+1)] v=0\\ \implies 2x \frac{d^2v}{dx^2} + (4(l+1)-2x) \frac{dv}{dx} + [\rho_0- 2(l+1)]v=0 \\ \implies x \frac{d^2v}{dx^2} + (2(l+1)-x) \frac{dv}{dx} + \frac{1}{2}[\rho_0- 2(l+1)]v=0 \\ \end{aligned}\]

So the equation;

\[\rho \frac{d^2 v}{d\rho^2} + 2(l+1-\rho) \frac{dv}{d\rho} + [\rho_0- 2(l+1)]v=0\]

can be written in terms of \(x=2\rho\) as;

\[x \frac{d^2v}{dx^2} + (2(l+1)-x) \frac{dv}{dx} + \frac{1}{2}[\rho_0- 2(l+1)]v=0\]

Let \(k=2l+1\) then \(k+1 = 2(l+1)\)

Let \(n'=\frac{\rho_0-2(l+1)}{2}\)

then we can write the differential equation as;

\[x \frac{d^2v}{dx^2} + (k+1-x) \frac{dv}{dx} + n'v=0\]

This is the associated Laguerre differential equation.

The solution is the associated Laguerre function \(v(x)=L_{n'}^k(x)\)

Since \(\rho_0 =2n\)

\[n' = \frac{2n-2(l+1)}{2}=n-l-1\]

and \(k=2l+1\)

so the solution of the original differential equation is

\[v(\rho)=L_{n-l-1}^{2l+1}(2\rho)\]

Radial Equation

\[-\frac{\hbar^2}{2m}\frac{d^2 u}{dr^2}+ \left [-\frac{e^2}{4\pi \epsilon_0} \frac{1}{r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right ]u=Eu\]

\[\kappa =\frac{\sqrt{-2mE}}{\hbar}\]

\[\kappa^2 =-\frac{2mE}{\hbar^2}\]

\[\frac{\kappa^2}{E}=-\frac{2m}{\hbar^2}\]

\[-\frac{\hbar^2}{2m}=\frac{E}{\kappa^2}\]

\[E=-\frac{\hbar^2 \kappa^2}{2m}\]

\[\frac{1}{E}=-\frac{2m}{\hbar^2 \kappa^2}\]

\[-\frac{\hbar^2}{2m}\frac{d^2 u}{dr^2}+ \left [-\frac{e^2}{4\pi \epsilon_0} \frac{1}{r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right ]u=Eu\]

\[\frac{E}{\kappa^2}\frac{d^2 u}{dr^2}+ \left [-\frac{e^2}{4\pi \epsilon_0} \frac{1}{r} -\frac{E}{\kappa^2} \frac{l(l+1)}{r^2} \right ]u=Eu\]

Multiply by;

\[\frac{1}{E}=-\frac{2m}{\hbar^2 \kappa^2}\]

\[\frac{1}{\kappa^2}\frac{d^2 u}{dr^2}+ \left [\left ( -\frac{2m}{\hbar^2 \kappa^2} \right ) \left ( -\frac{e^2}{4\pi \epsilon_0} \frac{1}{r} \right ) -\frac{1}{\kappa^2} \frac{l(l+1)}{r^2} \right ]u=u\]

\[\frac{1}{\kappa^2}\frac{d^2 u}{dr^2}= \left [1 - \left ( \frac{m e^2}{2 \pi \epsilon_0 \hbar^2 \kappa } \frac{1}{\kappa r} \right ) + \frac{l(l+1)}{(\kappa r)^2} \right ]u\]

Let

\[\rho = \kappa r\]

\[\rho_0 =\frac{me^2}{2\pi \epsilon_0 \hbar^2 \kappa}\]

then

\[\frac{du}{dr}=\frac{du}{d\rho}\frac{d\rho}{dr}=\kappa \frac{du}{d\rho}\]

\[\frac{d^2u}{dr^2} = \frac{d}{dr} \left [ \kappa \frac{du}{d\rho} \right ]= \kappa \left [ \frac{d^2 u}{d\rho^2} \frac{d\rho}{dr} \right ] = \kappa^2 \frac{d^2 u}{d\rho^2}\]

\[\frac{d^2 u}{d\rho^2} = \frac{1}{\kappa^2} \frac{d^2u} {dr^2}\]

\[\frac{d^2 u}{d\rho^2} = \left [ 1 - \frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2} \right ] u\]

\[u(\rho)=\rho^{l+1}e^{-\rho}v(\rho)\]

\[\begin{aligned} \frac{du}{d\rho} \\&= (l+1)\rho^l e^{-\rho} v+ \rho^{l+1}(-e^{-\rho})v + \rho^{l+1}e^{-\rho} \frac{dv}{d\rho} \\ &= \rho^l e^{-\rho} \left [ (l+1-\rho) v + \rho \frac{dv}{d\rho} \right ] \end{aligned}\]

\[\begin{aligned} \frac{d^2u}{d\rho^2} \\&= l\rho^{l-1}\rho^{-l} \frac{du}{d\rho}-\frac{du}{d\rho} + \rho^l e^{-\rho} \frac{d}{d\rho} \left [ (l+1-\rho) v + \rho \frac{dv}{d\rho} \right ] \\ &= l\rho^{l-1}\rho^{-l} \frac{du}{d\rho}-\frac{du}{d\rho} + \rho^l e^{-\rho} \left [ (l+1) \frac{dv}{d\rho}-v-\rho\frac{dv}{d\rho} + \frac{dv}{d\rho} + \rho \frac{d^2 v }{d\rho^2} \right ]\\ &= l\rho^{-1} \frac{du}{d\rho}-\frac{du}{d\rho} + \rho^l e^{-\rho} \left [ (l+2-\rho) \frac{dv}{d\rho}-v + \rho \frac{d^2 v }{d\rho^2} \right ] \end{aligned}\]

\[\frac{du}{d\rho}=\rho^l e^{-\rho} \left [ (l+1-\rho) v + \rho \frac{dv}{d\rho} \right ]\]

\[\begin{aligned} \frac{d^2u}{d\rho^2} \\ &= \rho^l e^{-\rho} \left [ \frac{l(l+1)}{\rho}v-lv+l\frac{dv}{d\rho}+ (l+2-\rho) \frac{dv}{d\rho}-v + \rho \frac{d^2 v }{d\rho^2} \right ] \\& -\rho^l e^{-\rho} \left [ (l+1-\rho) v + \rho \frac{dv}{d\rho} \right ] \\ &= \rho^l e^{-\rho} \left [ \frac{l(l+1)}{\rho} v -(l+1)v + (2l +2 - \rho) \frac{dv}{d\rho} + \rho \frac{d^2 v}{d\rho^2} \right ] \\& -\rho^l e^{-\rho} \left [ (l+1-\rho) v + \rho \frac{dv}{d\rho} \right ] \\&= \rho^l e^{-\rho} \left [ \frac{l(l+1)}{\rho} v -(2l + 2 -\rho) v + (2l+2 - 2 \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2}\right ] \\&= \rho^l e^{-\rho} \left \{ \left [ \frac{l(l+1)}{\rho} -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2}\right \} \end{aligned}\]

Use;

\[u(\rho) = \rho^{l+1}e^{-\rho} v\]

\[\frac{d^2 u}{d\rho^2} = \left [ 1 - \frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2} \right ] u\]

\[\frac{d^2 u}{d\rho^2} =\rho^l e^{-\rho} \left \{ \left [ \frac{l(l+1)}{\rho} -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2}\right \}\]

\[\begin{aligned} & \rho^l e^{-\rho} \left \{ \left [ \frac{l(l+1)}{\rho} -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2}\right \} = \left [ 1 - \frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2} \right ] \rho^{l+1} e^{-\rho} v\\& \implies \left [ \frac{l(l+1)}{\rho} -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2} = \left [ 1 - \frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2} \right ] \rho v \\&\implies \left [ \frac{l(l+1)}{\rho} -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2} = \left [ \rho - \rho_0 + \frac{l(l+1)}{\rho} \right ] v \\&\implies \left [ -2l - 2 +\rho) \right ]v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2} = \left [ \rho - \rho_0 + \right ] v \\&\implies ( -2l - 2 +\rho_0) v + 2(l+1 - \rho) \frac{dv}{d\rho} +\rho \frac{d^2 v}{d\rho^2} = 0 \\&\implies \rho \frac{d^2 v}{d\rho^2}+ 2(l+1 - \rho) \frac{dv}{d\rho}+ [\rho_0 -2(l+1)] v = 0 \end{aligned}\]