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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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| − | &= {-2 a m \over \hbar}{\partial \over \partial x}(x \Psi) \\
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| − | &= {-2 a m \over \hbar}\left({\partial \over \partial x}(x)\Psi + x{\partial \over \partial x}\Psi\right) \\
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| − | &= {-2 a m \over \hbar}\left(\Psi + x {-2 a m x \over \hbar} \Psi \right) \\
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| − | &= {-2 a m \over \hbar}\left(1 + x {-2 a m x \over \hbar} \right) \Psi \\
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| − | &= {-2 a m \over \hbar}\left(1 - {2 a m x^2 \over \hbar} \right) \Psi \\
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| − | &= {2 a m \over \hbar}\left({2 a m x^2 \over \hbar} - 1 \right) \Psi \\
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| − | &= {2 a m \over \hbar}\left({2 a m x^2 \over \hbar} - {\hbar \over \hbar} \right) \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} &= - {\hbar^2 \over 2m} {\partial^2 \Psi \over \partial x^2} + V \Psi \\
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| − | i\hbar (-ai \Psi) &= - {\hbar^2 \over 2m} {2am \over \hbar^2} (2amx^2 - \hbar) \Psi + V \Psi \
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| − | a \hbar \Psi &= - a (2amx^2 - \hbar) \Psi + V \Psi \\
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| − | a \hbar &= - a (2amx^2 - \hbar) + V \\
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| − | V &= a \hbar + a (2amx^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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| − | I caught the mistake. I had lost a negative sign in the x derivative term. [[User:Jgardner|Jgardner]] 01:30, 4 May 2012 (MDT)
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| − | | |
| − | == 1.14.c ==
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| − | | |
| − | <math>
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| − | \begin{align}
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| − | \langle x \rangle
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| − | &= \int_{-\infty}^{+\infty} \Psi^* x \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} x \Psi^* \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} x A^2 e^{-2amx^2/\hbar}\ dx & & \text{Note that }x e^{-bx^2}\text{ is odd.} \\
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| − | &= 0 & & \text{Centered at the origin.} \\
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| − | | |
| − | \langle x^2 \rangle
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| − | &= \int_{-\infty}^{+\infty} \Psi^* x^2 \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} x^2 A^2 e^{-2amx^2/\hbar}\ dx \\
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| − | &= {A^2 \sqrt{\pi} \over 2 \left(2am / \hbar\right)^{3/2}} \\
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| − | &= {\sqrt{2am \pi} \hbar^{3/2} \over 2 \sqrt{\pi \hbar} (2am)^{3/2}} \\
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| − | &= {\hbar \over 4am} & & \text{Looks like it is spread out a bit.}\\
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| − | | |
| − | \langle p \rangle
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| − | &= \int_{-\infty}^{+\infty} \Psi^* (-i \hbar) {\partial \over \partial x} \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} \Psi^* (-i \hbar) {-2 a mx \over \hbar} \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} \Psi^* 2 a m x i \Psi\ dx \\
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| − | &= 2 a m i \int_{-\infty}^{+\infty} x \Psi^* \Psi\ dx & & \text{Odd function. (See above)} \\
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| − | &= 0 & & \text{It looks like it has an equal chance of going left or right.} \\
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| − | | |
| − | \langle p^2 \rangle
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| − | &= \int_{-\infty}^{+\infty} \Psi^* \left((-i \hbar) {\partial \over \partial x}\right)^2 \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} \Psi^* (-\hbar^2) {\partial^2 \over \partial x^2} \Psi\ dx \\
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| − | &= \int_{-\infty}^{+\infty} \Psi^* (-\hbar^2) {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi\ dx \\
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| − | &= -2am \int_{-\infty}^{+\infty} \Psi^* (2 a m x^2 - \hbar) \Psi\ dx \\
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| − | &= -2am \left(
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| − | \int_{-\infty}^{+\infty} \Psi^* 2amx^2 \Psi\ dx -
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| − | \int_{-\infty}^{+\infty} \Psi^* \hbar \Psi\ dx
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| − | \right) \\
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| − | &= -2am \left(
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| − | 2am \int_{-\infty}^{+\infty} \Psi^* x^2 \Psi\ dx -
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| − | \hbar \int_{-\infty}^{+\infty} \Psi^* \Psi\ dx
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| − | \right) & & \text{We've already done these integrals, remember?} \\
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| − | &= -2am ( 2am \langle x^2 \rangle - \hbar) \\
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| − | &= -2am ( 2am {\hbar \over 4am} - \hbar) \\
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| − | &= -2am ( {\hbar \over 2} - \hbar) \\
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| − | &= -2am ( - {\hbar \over 2}) \\
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| − | &= am\hbar & & \text{It seems to be either moving back or forth.} \\
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| − | \end{align}
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| − | </math>
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| − | I have to admit, the result for <math>\langle p^2 \rangle</math> surprises me a great deal. I'd expect it to be non-zero, so that <math>\sigma_p</math> would be non-zero. I obviously made a mistake. [[User:Jgardner|Jgardner]] 01:01, 4 May 2012 (MDT)
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| − | It turns out that I had miscalculated <math>\langle x^2 \rangle</math>, forgetting a factor of 1/2.
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| − | | |
| − | == 1.14.d ==
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| − | | |
| − | <math>
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| − | \begin{align}
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| − | \sigma_x^2 &= \langle x^2 \rangle - \langle x \rangle ^2 \\
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| − | &= {\hbar \over 4am} \\
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| − | \sigma_p^2 &= \langle p^2 \rangle - \langle p \rangle ^2 \\
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| − | &= am\hbar
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| − | \sigma_x \sigma_p &= \sqrt{{\hbar \over 4am}am\hbar} \\
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| − | &= \sqrt{{\hbar^2 \over 4}}
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| − | &= {\hbar \over 2} \ge {\hbar \over 2} \\
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| − | \end{align}
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| − | </math>
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| − | This appears to be the best combination of momentum and position we can hope for given the uncertainty principle. [[User:Jgardner|Jgardner]] 02:02, 4 May 2012 (MDT)
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