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| classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 11:32] – x500_moore616 | classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 23:56] (current) – x500_moore616 |
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| *Use dimensionless form of DE | *Use dimensionless form of DE |
| <math>let \xi = \sqrt{\frac{m\omega}{\hbar}} x </math> | let <math> \xi = \sqrt{\frac{m\omega}{\hbar}} x </math> |
| and <math> K=\frac{2E}{\hbar\omega}. (C) </math> | and <math> K=\frac{2E}{\hbar\omega}. \ \ \ \ \ \ \ \ (C) </math> |
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| Then we can use the dimensionless form of the Schrodinger | Then we can use the dimensionless form of the Schrodinger |
| <math> \frac{\partial^2 \psi}{\partial x^2}=(\xi^2-K)\psi </math> | <math> \frac{\partial^2 \psi}{\partial x^2}=(\xi^2-K)\psi </math> |
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| 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>. | 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>. |
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| Differentiate and then Schrodinger's equation becomes | Differentiate and then Schrodinger's equation becomes |
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| <math> \frac{\partial^2h(\xi)}{\partial \xi^2}=-2h\xi(\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> |
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| Differentiate once more: | Differentiate once more: |
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| <math> \frac{\partial^2h}{\partial\xi^2}=\sum (j+1)(j+2)a_j+2\xi^j </math> | <math> \frac{\partial^2h}{\partial\xi^2}=\sum (j+1)(j+2)a_{j+2}\xi^j </math> |
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| <math> | <math> |
| a_j+2 = \frac{2j+1-K)a_j}{(j+1)(j+2)} </math> | a_{j+2} = \frac{2j+1-K)a_j}{(j+1)(j+2)} </math> |
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| Now all we need to know is <math> a_0 </math> and <math> a_1 </math> and we can find all a. | Now all we need to know is <math> a_0 </math> and <math> a_1 </math> and we can find all a. |
| This is good, but not all the solutions that are found are normalizable. For example, at very large j, the formula is | This is good, but not all the solutions that are found are normalizable. For example, at very large j, the formula is |
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| <math> a_(j+2) \approx \frac{2a_j}{j}. </math> | <math> a_{j+2} \approx \frac{2a_j}{j}. </math> |
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| Then the solution is | Then the solution is |
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| 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, \\ | 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> | <math> K = 2j + 1 \ \ \ \ \ \ \ \ j = 0, 2, \ldots</math>. |
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| | 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>. |
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| | 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. |
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| | 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. |
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| | == 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. |
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| | == 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>. |
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| | == Conclusion == |
| | Okay, hopefully between our two sets of notes you will be okay. Sorry I took so long to make my contribution. |
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| | 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). |
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| | Peace. EE. |
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