===== Nov 02 (Mon) Main Topics in Chap 4, Separation variables for Spherical coordinate =====
** Responsible party: Captain America, David Hilbert's hat   ** 

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Please try to include the following

  * main points understood, and expand them - what is your understanding of what the points were.
    * expand these points by including many of the details the class discussed.
  * main points which are not clear.  - describe what you have understood and what the remain questions surrounding the point(s).
    * Other classmates can step in and clarify the points, and expand them. 
  * How the main points fit with the big picture of QM.  Or what is not clear about how today's points fit in in a big picture.
  * wonderful tricks which were used in the lecture.\\
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**Class input on main points of the beginning of Chapter 4:**
  * <math>\nabla^2</math> spherical coordinates
  * <math>L^2, L_z</math>, and why they have discrete values
  * The Hydrogen atom model
  * How spin affects this
  * Degenerate energies and states
  * Quantum numbers
    * Where are they from?
    * What do they do?

**Using 3-D Coordinates:**

From the one-dimensional Schrodinger Equation:
<math> [-\frac{\hbar^2}{2m}\frac{ \partial^2}{ \partial x^2} + V(x)]\psi=E \psi</math>

The kinetic energy term, <math>-\frac{\hbar^2}{2m}\frac{\partial^2}{ \partial x^2}</math>,  must model the 3-Dimensional kinetic energy of the system, and therefore turns into:

<math> [-\frac{\hbar^2}{2m}\nabla^2+ V(x,y,z)]\psi=E \psi</math>

Where <math>\nabla^2</math> is equal to <math> \frac{ \partial^2}{ \partial x^2} + \frac{  \partial^2}{ \partial y^2} + \frac{ \partial^2}{ \partial z^2}</math>. You can see that if you restrict this 3 dimensional case to one dimension, our original one dimensional Schrodinger equation comes out. 


**Separation of Variables**

Using spherical coordinates will be useful for future problems that we will be solving, so it is necessary to transform the Schrodinger equation into spherical coordinates. Using the relations <math> x = r sin \theta cos \phi </math>, <math> y = r sin\theta sin\phi</math>, and <math>z = r cos\theta </math>, you can derive the Laplacian in spherical coordinates. 

The Laplacian will take the form of:
<math>\nabla^2=\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2}{\partial \phi^2})</math>

Plugging this into the Schrodinger equation we get:

<math>-\frac{\hbar^2}{2 m} [\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial \psi}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial \psi}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2 \psi}{\partial \phi^2})] + V \psi = E \psi</math>






Suppose that the wavefunction is a separable solution. It has the form:

<math>\psi(r,\theta,\phi) = R(r)Y(\theta,\phi). </math>

What we want to do is plug this form into the Schrodinger equation, and use the fact that 

<math> \frac{\partial \psi}{\partial r} =  \frac{\partial (RY)}{\partial r} = Y \frac{\partial R}{\partial r} </math>, <math> \frac{\partial \psi}{\partial \theta} =  \frac{\partial (RY)}{\partial \theta} = R \frac{\partial Y}{\partial \theta} </math>, <math> \frac{\partial \psi}{\partial \phi} =  \frac{\partial (RY)}{\partial \phi} = R \frac{\partial Y}{\partial \phi} </math>

To reduce the Schrodinger equation into one side dependent on r,<math> \frac{\partial R}{\partial r} </math> and the other side dependent on <math> \phi </math>, <math> \theta </math>, <math> \frac{\partial Y}{\partial \theta} </math>, and <math> \frac{\partial Y}{\partial \phi} </math>. Because each side of the equation is dependent on R or Y alone, you can set both sides equal to a constant and turn the Schrodinger equation into a set of solvable differential equations. 

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