Responsible party: East End, Devlin

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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.

Devlin:

• Use dimensionless form of DE

<math>let \xi = \sqr(\frac{m\omega}{\hbar}) x </math> and <math> K=\frac{2E}{\hbar\omega}. (C) </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^(\frac{-1}{2} \xi^2) </math>.

• Behavior at large

<math> \psi\approx h(\xi)e^(\frac{-1}{2} \xi^2) </math>

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

• 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 = \frac{2j+1-K)a_j}{(j+1)(j+2)} </math>

Now all we need to know is <math> a_0 </math> and <math> a_1 </math> 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 \frac{2a_j}{j}. </math>

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

(WHere C is an arbitrary constant) The solution clearly goes asymptotic.

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>

Going back to equation (C), we now have

<math> E_n=(n+\frac{1}{2})\hbar\omega. </math>

This shows the quantization of energy. In general, <math> h_n(\xi) </math> will be a polynomial of degree <math> n </math>. It will involve only even powers if n is an even interger and vice versa.

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