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| classes:2009:fall:phys4101.001:lec_notes_1021 [2009/10/23 22:57] – yk | classes:2009:fall:phys4101.001:lec_notes_1021 [2009/10/23 23:09] (current) – yk |
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| Based on these self consistency checks which work out, we keep(better word?) this generalized probability interpretation of <math>|C_n|^2</math>. | Based on these self consistency checks which work out, we keep(better word?) this generalized probability interpretation of <math>|C_n|^2</math>. |
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| **Proof of orthonormality of infinite square well and simple harmonic oscillator(this part was not mentioned in class)** | **Proof of orthonormality of infinite square well and simple harmonic oscillator(this part was not mentioned in class)** These are very complicated, so it's amazing that the proof for general case is much easier than these individual cases. |
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| * infinite square well | * infinite square well |
| The wave function is <math>\psi_n(x)=\sqrt{\frac{2}{L}} \sin\frac{n \pi x}{L} </math> | The wave function is <math>\psi_n(x)=\sqrt{\frac{2}{L}} \sin\frac{n \pi x}{L} </math> |
| The left hand side <math>e^{2ts} \int_{-\infty}^{+\infty} e^{-(t+s-\xi)^2} d\xi =e^{2ts} \int_{-\infty}^{+\infty} e^{-z^2} dz =e^{2ts}sqrt{\pi}=sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!} </math> | The left hand side <math>e^{2ts} \int_{-\infty}^{+\infty} e^{-(t+s-\xi)^2} d\xi =e^{2ts} \int_{-\infty}^{+\infty} e^{-z^2} dz =e^{2ts}sqrt{\pi}=sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!} </math> |
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| So we get <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{t^m s^n}{m!n!} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi</math> | So we get <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{t^m s^n}{m!n!} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi</math>. |
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| If <math>n \neq m </math>, the left hand side is independent of m, while the right hand side is dependent of m. So in order to satisfy the equation the integral must be zero. <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi =0</math> | This equality should hold for any values of //t// and //s//. Since both sides of the equation is polynomials of //t// and //s//, the coefficients for <math>t^m s^n</math> must equal to each other. |
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| If n=m then <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}= \sum_{n=0}^{\infty} \frac{(ts)^n}{2(n!)} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi </math>, so <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! </math> | If <math>n \neq m </math>, the left hand side has no corresponding term (coefficient = zero), while the right hand side has those terms. So in order to satisfy the equation the integral must be zero. <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi =0</math> |
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| | If n=m then <math>sqrt{\pi} \frac{(2ts)^n}{n!}= \frac{(ts)^n}{(n!)^2} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi </math>, so <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! </math> |
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| So <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! \delta_m_n </math> | So <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! \delta_m_n </math> |