Responsible party: Captain America, David Hilbert's hat

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Main class wiki page: home

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.
  

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®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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