Responsible party: East End, Devlin

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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.

Devlin:

• Use dimensionless form of DE

<math>let \xi = \sqr [(mx\omega}/\hbar)] </math> and <math> K=(2E)/(\hbar\omega). </math>

Then we can use the dimensionless form of the Schrodinger <math> \frac{\partial^2}{\partial x^2}=(\xi^2-K)\psi(x) </math>

We can think of <math> \xi </math> as approximately <math> x </math> and also <math> \psi </math> as approximately <math> e^-(\xi)^2/2) </math>.

• Behavior at large

<math> \psi\approx h(\xi)e^(-\xi^2/2) </math>

We use this and hope that <math> h(\xi) </math> is simpler than <math> \psi(\xi) </math>

1. Substitute <math> \psi </math> into

<math> \frac{\partial^2}{\partial x^2}=(\xi^2-K)\psi(x). </math>

Differentiate and then Schrodinger's equation becomes

<math> \frac{\partial^2h(\xi)}{\partial \xi^2}=-2h\xi(\xi)+(K-1)(h(\xi)=0 </math> (A)

• Use power series to find a solution.

Assume the solution can be found in the form <math> h(\xi)=\sum a_j \xi^j. </math>.

Differentiate each term: <math> \frac{\partial h}{partial\xi}=\sum a_j\xi^j </math>

Differentiate once more:

<math> \frac{\partial^2h}{partial\xi^2}=\sum (j+1)(j+2)a_j+2\xi^j </math>

• Recursive Equation

Plug equation into (A) and we get a recursive equation that can be illustrated like this:

<math> (blah)(\xi)^0+ (blahblah)(\xi)^1 +(moreblah)(\xi)^2+…=0. </math>

Since the equation needs to hold true for all <math> \xi </math>, the 'blahs' must equal zero. We now have this equation

<math> a_j+2 = (2j+1-K)a_j/[(j+1)(j+2)] </math>

Now all we need to know is a_0 and a_1 and we can find all a.

• Normalize

This is good, but not all the solutions that are found are normalizable. For example, at very large j, the formula is <math> a_j+2 \approx 2a_j/j.

Then the solution is <math> a_j\approx \frac{C}{(j/2)!}. </math> Which makes <math> h(\xi)\approx Ce^\xi^2. </math>

The solution blows up. To get normalizable solutions, we need the power series to terminate. To do this, we need the numerator to go to zero. That is <math> 2j+1-K=0, </math> so <math>K=2j+1.</math>

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