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--- classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/27 16:47] – x500_bast0052
+++ classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 23:56] (current) – x500_moore616
@@ Line 18: / Line 18: @@
 === Main Points ===
+Devlin:
-Steps to the Analytic Method:
+==Steps to solving the SHO using the Analytic Method:==
-)  Use dimensionless form of DE
+  *Use dimensionless form of DE
+let <math> \xi = \sqrt{\frac{m\omega}{\hbar}} x </math>
-<math>let \xi = \sqr [(mx\omega}/\hbar)] </math>
+and <math> K=\frac{2E}{\hbar\omega}. \ \ \ \ \ \ \ \  (C) </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>
+<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^<sup>-(\xi)^2/2)</sup> </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>.
-.)  <math> \psi\approx h(\xi)e^(-\xi^2/2) </math>
+  *Behavior at large
-We use this and hope that <math> h(\xi) </math> is simpler than <math> \psi(\xi) </math>
+<math> \psi \approx h(\xi)e^{\frac{-1}{2} \xi^2} </math>
-.)  Substitute <math> \psi </math> into
+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}=-2\xih(\xi)+(K-1)(h(\xi)=0 </math> (A)
+<math> \frac{\partial^2h(\xi)}{\partial \xi^2}=-2h\xi(\xi)+(K-1)(h(\xi)=0 \ \ \ \ \ \ \ \ (A) </math>
-.)  Use power series to find a solution.
+  *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>.
-<math> h(\xi)=\sum a\sub n * \xi^n. </math>.
+Differentiate each term:
+<math> \frac{\partial h}{\partial\xi}=\sum a_j\xi^j </math>
-Differentiate each term twice and then plug that into (A) and we get a recursive equation that can be illustrated like this:
+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
+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.
+=== East End 9/28/09 ===
+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.
+== Analytical solutions ==
+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>.
+The lecture didn't go into as much detail as the book, and it seems to me that the professor was simply trying to help explain what the book was saying.  Anyway, as an example, let //K// = 5.  Then //j// = 2 will give <math>a_{j + 2} = a_4 = 0</math>, even if <math>a_0 \not= 0</math>.  Or, if //K// = 3, then //j// = 1 and <math>a_3 = 0</math> even if <math>a_1 \not= 0</math>.
+One of the even series or the odd series will have all zero terms.  And there is actually only one unknown, since we get <math>a_2</math> from knowing <math>a_0</math>, or we get <math>a_3</math> from knowing <math>a_1</math>.  The one unknown is found by normalizing.  Anyway, to get past the handwaving in class, I will refer to the book and quote the final result.  If you work through the first few j values (or n values really, which are the values of j that terminate the series) you will eventually get \\
+<math>\psi_n(x) = \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n(\xi) e^{- \xi^2 / 2}</math>, \\
+where <math>H_n</math> are the //Hermite polynomials//.  See the table on page 56 of the text for the first few.
+And, going back a half step, substituting the values for //K//, based on the terminating value of //j//, //n//, <math>K = 2n + 1</math> into the definition of //K// involving E, we get \\
+<math>E_n = \left(n + \frac{1}{2} \right) \hbar \omega \ \ \ \ \ \ \ \ n = 0, 1, 2, \ldots</math>. \\
+This is the same quantization of energy we already know.
+== Wag the dog ==
+This simply refers to showing that only the allowed values of //K//, and hence //E//, lead to normalizable solutions.  If you make plots of <math>\psi</math> vs. //x// (right?), and pick various values of //E// (which depends on K), the only ones where the values at the far right and left go to zero are the allowed values.  See page 55 for an example where the author selects values of //E// near 0.5.  The graphs clearly show curves that would not be normalizable.  It turns out that right at an //E// of 0.5 the ends of the graph do approach 0 and lead to a normalizable solution.
+== Rodriguez's formula ==
+I'm not going to include this.  He decided to leave it for the "motivated student."  Basically it relates the Hermite polynomials to the ladder operator <math>a_+</math>.
+== Conclusion ==
+Okay, hopefully between our two sets of notes you will be okay.  Sorry I took so long to make my contribution.
+The difference between the two methods, as far as I can tell, is that one is simple (ladder operators), but are only convenient for the first few //n// values.  The analytical solution involving Hermite polynomials is a little more complicated, especially in it's derivation, but for large n can save a lot of time because you don't have to find all the previous solutions (smaller-n solutions).
+Peace.  EE.

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