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

I started to add to Devlin's, but I noticed a few places where my notes were a little different, so I decided to post what I have separately and people can compare.

Three points today. Analytical solutions, and a touch on Rodriguez's formula & <math>a_{\pm}</math>, and a touch on the Wag the Dog method.

We want to use the dimensionless form of Schroedinger,
<math> \frac{\partial^2 \psi}{\partial \xi^2} = (\xi^2 - K)\psi(\xi) \ \ \ \ \ \ \ \ (*)</math>
where <math>\xi</math> goes with <math>x</math> and <math>K</math> goes with <math>E</math>. (The book defines them as <math>\xi = \sqrt{\frac{m \omega}{\hbar}} x</math> and <math>K = \frac{2E}{\hbar \omega}</math>.)

We want to find a solution in the form <math>\psi=h(\xi) e^{-\xi^2 / 2}</math> (note that the <math>e^{\xi^2 / 2}</math> term is not normalizable). After differentiating, substituting this form of <math>\psi</math> into (*) gives us
<math>\frac{\partial^2 h(\xi)}{\partial \xi^2} - 2 \xi \frac{\partial h}{\partial \xi} + (K - 1)h = 0 \ \ \ \ \ \ \ \ (* \ *)</math>.
(By the way, this equation was incorrect in my notes, so it may have been incorrect on the board. Also, I had <math>e^{-\xi / 2}</math>, with no square, in my notes, too, for some reason. Maybe an error on my part, but you might want to check your notes.)

To solve, assume h can be represented with a power series:
<math>h(\xi) = \sum_{j=0}^\infty a_j \xi^j</math>.
If we differentiate the series and substitute into (* *), we get
<math>( \ \ )\xi^0 + ( \ \ )\xi + ( \ \ )\xi^2 + \cdots = 0</math>
where the first term is in terms of <math>j_0</math> and <math>j_2</math>, the second in terms of <math>j_1</math> and <math>j_3</math>, etc. In order to be identically zero (true for any <math>\xi</math>), each coefficient must equal zero. In the book it shows the coeffiecients to have the form
<math>(j + 1)(j + 2) a_{j + 2} - 2ja_j + (K - 1)a_j = 0</math>.
This leads to the recurrence relation
<math>a_{j + 2} = \frac{2j + 1 - K}{(j + 1)(j + 2)} a_j</math>.
Once we know <math>a_0</math> and <math>a_1</math>, we know <math>h(\xi)</math>. K is still unknown for now. This is similar to solving a DE any other way, where at least conceptually we are integrating twice and getting an integration constant both times. We will use <math>\psi(\pm \infty) = 0</math> as our boundary values.

We have a problem, however. Recall that <math>e^x = \sum \frac{x^n}{n!}</math>. This means that for large j, our recurrence starts to have terms that resemble terms in the sum for <math>e^{\xi^2}</math>. Which blows up and is not normalizable. The only way to resolve this is if the power series terminates at some point. In the recurrence above, the numerator <math>2j + 1 - K</math> must equal zero at some point. In other words,
<math> K = 2j + 1 \ \ \ \ \ \ \ \ j = 0, 2, \ldots</math>

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