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| − | = 1.14 =
| + | #REDIRECT [[Introduction to Quantum Mechanics/Chapter 1]] |
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| − | == 1.14.a ==
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| − | To find A, all we have to do is normalize the wave function.
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| − | <math>
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| − | \begin{align}
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| − | 1 &= \int_{-\infty}^{+\infty} \Psi^*\Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} A e^{-a[(mx^2/\hbar)-it]} A e^{-a[(mx^2/\hbar)+it]}\ dx \\
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| − | &= \int_{-\infty}^{+\infty} A^2 e^{-2a(mx^2/\hbar)}\ dx \\
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| − | &= A^2 \int_{-\infty}^{+\infty} e^{-2a(mx^2/\hbar)}\ dx & & \text{I had to look up how to solve this integral.}\\
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| − | \end{align}
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| − | </math>
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| − | How do you solve the integral above?
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| − | First, we substitute <math>b = {2am \over \hbar}</math> for simplicity. (These are all constants.)
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| − | Next, we note that this looks like the Gaussian integral. Although no definite integral exists, <math>\int_{-\infty}^{+\infty} e^-{x^2}\ dx = \sqrt{\pi}</math>.
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| − | Now all we have to do is eliminate the coefficient of x in the exponent. We can do this with integration by substitution, using <math>y^2 = b x^2,\ x = {y \over \sqrt{b}},\ dx = {1 \over \sqrt{b}}dy</math>
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| − | :<math>
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| − | \begin{align}
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| − | \int_{-\infty}^{+\infty} e^{-bx^2}\ dx
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| − | &= \int_{-\infty}^{+\infty} {1 \over \sqrt{b}} e^{-y^2}\ dx \\
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| − | &= \sqrt{{\pi \over b}} \\
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| − | &= \sqrt{{\pi \hbar \over 2am}} \\
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| − | 1 &= A^2 \sqrt{{\pi \hbar \over 2am}} \\
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| − | A^2 &= \sqrt{{2am \over \pi \hbar}} \\
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| − | A &= \sqrt[4]{{2am \over \pi \hbar}} \\
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| − | \end{align}
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| − | </math>
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| − | == 1.14.b ==
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| − | The derivatives are not that hard.
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| − | <math>
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| − | \begin{align}
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| − | {\partial \over \partial t} \Psi
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| − | &= {\partial \over \partial t}\left( A e^{-a[(mx^2/\hbar)+it]} \right ) \\
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| − | &= {\partial \over \partial t}\left( A e^{-a(mx^2/\hbar)}e^{-ait} \right ) \\
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| − | &= A e^{-a(mx^2/\hbar)}{\partial \over \partial t} e^{-ait} \\
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| − | &= A e^{-a(mx^2/\hbar)}(-ai) e^{-ait} \\
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| − | &= -ai A e^{-a[(mx^2/\hbar)+it]} \\
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| − | &= -ai \Psi \\
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| − | {\partial \over \partial x} \Psi
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| − | &= {\partial \over \partial x}\left( A e^{-a[(mx^2/\hbar)+it]} \right ) \\
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| − | &= A e^{-ait} {\partial \over \partial x} e^{-a(mx^2/\hbar)} \\
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| − | &= A e^{-ait} e^{-a(mx^2/\hbar)} {\partial \over \partial x} [-a(mx^2/\hbar)] & & \text{Chain rule.} (f(g))' = f'(g)g';\ f = e^x;\ g = -a(mx^2/\hbar) \\
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| − | &= A e^{-ait} e^{-a(mx^2/\hbar)} [-a(2mx/\hbar)] \\
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| − | &= {-2 a mx \over \hbar} A e^{-a[(mx^2/\hbar)+it]} \\
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| − | &= {-2 a mx \over \hbar} \Psi \\
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| − | {\partial^2 \over \partial x^2} \Psi &= {\partial \over \partial x}{\partial \over \partial x} \Psi \\
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| − | &={\partial \over \partial x}\left( {-2 a m x \over \hbar} \Psi \right) \\
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| − | &={\partial \over \partial x}\left( {-2 a m x \over \hbar}\right) \Psi
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| − | + {-2 a m x \over \hbar}{\partial \over \partial x}\Psi \\
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| − | &= {-2 a m \over \hbar} \Psi + \left({-2 a m x \over \hbar}\right)^2 \Psi \\
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| − | &= {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi \\
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| − | \end{align}
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| − | </math>
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| − | Let's plug & chug.
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| − | <math>
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| − | \begin{align}
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| − | i\hbar {\partial \Psi \over \partial t}
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| − | &= - {\hbar^2 \over 2m} {\partial^2 \Psi \over \partial x^2} + V \Psi \\
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| − | i\hbar (-ai \Psi)
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| − | &= {\hbar^2 \over 2m} {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi + V \Psi \\
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| − | a \hbar \Psi
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| − | &= a (2 a m x^2 - \hbar) \Psi + V \Psi \\
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| − | a \hbar
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| − | &= a (2 a m x^2 - \hbar) + V \\
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| − | V &= a \hbar - a (2 a m x^2 - \hbar) \\
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| − | &= -2 a^2 m x^2
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| − | \end{align}
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| − | </math>
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| − | I'm pretty sure I've made several mistakes. I have to go over this carefully to see where. [[User:Jgardner|Jgardner]] 22:18, 3 May 2012 (MDT)
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