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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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| − | == Comments ==
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| − | This is actually a really good example that you can think hard about to really understand what is going on in the wave function.
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| − | Since the wave function is complex, you can think of it as <math>\Psi = \Psi_{real} + \Psi_{imag} i</math>. You can see how <math>r^2 = \Psi_{real}^2 + \Psi_{imag}^2</math> is constant over time for any given point. The time derivative has the wave rotating with a period of <math>2 \pi / a</math> clockwise. The time derivative for the complex conjugate is moving at the same speed backwards.
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| − | The overall shape of the wave over x is simply the Gaussian. It smooths out to 0 in either direction, with a nice rounded bump in the middle. If it weren't for the potential pushing it towards the origin, it would spread out over time.
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| − | Speaking of the potential, it is zero at the origin, and squeezes with parabolic energy as you move out. This is the force not of gravity near the surface of the earth (which doesn't vary based on how far from the surface you move), but a spring. The farther away you go, the harder it pushes you back to the middle.
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| − | If you plug in the time and space derivatives into the schrodinger equation, you get something that looks like this:
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| − | <math>
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| − | \begin{align}
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| − | i \hbar {\partial \over \partial t} \Psi &= -{\hbar \over 2m} {\partial^2 \over \partial x^2} \Psi + V \Psi \\
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| − | a \hbar \Psi &= a\hbar(1 - 2max^2)\Psi + a\hbar2max^2 \Psi \\
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| − | \Psi &= (1 - 2max^2)\Psi + 2max^2 \Psi \\
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| − | \end{align}
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| − | </math>
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| − | The left side is how the wave function changes over time. There is a natural turning that forces the wave function to rotate through imaginary space -- the i factor.
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| − | The first term on the right is how the wave function wants to spread out over space. The wave function has a natural repulsiveness to itself. (Or is this the momentum term---if it is moving, it wants to keep on moving?)
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| − | The last term on the right is how the potential, the external force, is forcing the wave function to stick together around the origin.
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| − | The wave function described is a sort of steady state. It has reached a point where it no longer changes over time (except for a constant spinning through the imaginary space.)
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| − | You can imagine taking a single point in time. You can see how every point along the x axis has a force <math>F = - {\partial V \over \partial x}</math>.
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