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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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