Introduction to Quantum Mechanics/Chapter 1/Problems

1.14

1.14.a

To find A, all we have to do is normalize the wave function.

<math> \begin{align} 1 &= \int_{-\infty}^{+\infty} \Psi^*\Psi\ dx \\

 &= \int_{-\infty}^{+\infty} A e^{-a[(mx^2/\hbar)-it]} A e^{-a[(mx^2/\hbar)+it]}\ dx \\
 &= \int_{-\infty}^{+\infty} A^2 e^{-2a(mx^2/\hbar)}\ dx \\
 &= A^2 \int_{-\infty}^{+\infty} e^{-2a(mx^2/\hbar)}\ dx & & \text{I had to look up how to solve this integral.}\\

\end{align} </math>

How do you solve the integral above?

First, we substitute <math>b = {2am \over \hbar}</math> for simplicity. (These are all constants.)

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>.

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>

<math>

\begin{align} \int_{-\infty}^{+\infty} e^{-bx^2}\ dx

 &= \int_{-\infty}^{+\infty} {1 \over \sqrt{b}} e^{-y^2}\ dx \\
 &= \sqrtTemplate:\pi \over b  \\
 &= \sqrtTemplate:\pi \hbar \over 2am \\

1 &= A^2 \sqrtTemplate:\pi \hbar \over 2am \\ A^2 &= \sqrtTemplate:2am \over \pi \hbar \\ A &= \sqrt[4]Template:2am \over \pi \hbar \\ \end{align} </math>

1.14.b

The derivatives are not that hard. <math> \begin{align} {\partial \over \partial t} \Psi &= {\partial \over \partial t}\left( A e^{-a[(mx^2/\hbar)+it]} \right ) \\ &= {\partial \over \partial t}\left( A e^{-a(mx^2/\hbar)}e^{-ait} \right ) \\ &= A e^{-a(mx^2/\hbar)}{\partial \over \partial t} e^{-ait} \\ &= A e^{-a(mx^2/\hbar)}(-ai) e^{-ait} \\ &= -ai A e^{-a[(mx^2/\hbar)+it]} \\ &= -ai \Psi \\

{\partial \over \partial x} \Psi &= {\partial \over \partial x}\left( A e^{-a[(mx^2/\hbar)+it]} \right ) \\ &= A e^{-ait} {\partial \over \partial x} e^{-a(mx^2/\hbar)} \\ &= 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) \\ &= A e^{-ait} e^{-a(mx^2/\hbar)} [-a(2mx/\hbar)] \\ &= {-2 a mx \over \hbar} A e^{-a[(mx^2/\hbar)+it]} \\ &= {-2 a mx \over \hbar} \Psi \\

{\partial^2 \over \partial x^2} \Psi &= {\partial \over \partial x}{\partial \over \partial x} \Psi \\ &={\partial \over \partial x}\left( {-2 a m x \over \hbar} \Psi \right) \\ &={\partial \over \partial x}\left( {-2 a m x \over \hbar}\right) \Psi + {-2 a m x \over \hbar}{\partial \over \partial x}\Psi \\ &= {-2 a m \over \hbar} \Psi + \left({-2 a m x \over \hbar}\right)^2 \Psi \\ &= {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi \\ \end{align} </math>

Let's plug & chug.

<math> \begin{align} i\hbar {\partial \Psi \over \partial t}

 &= - {\hbar^2 \over 2m} {\partial^2 \Psi \over \partial x^2} + V \Psi \\

i\hbar (-ai \Psi)

 &= {\hbar^2 \over 2m}   {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi + V \Psi \\ 

a \hbar \Psi

 &= a (2 a m x^2 - \hbar) \Psi + V \Psi \\

a \hbar

 &= a (2 a m x^2 - \hbar) + V \\

V &= a \hbar - a (2 a m x^2 - \hbar) \\

 &= -2 a^2 m x^2

\end{align} </math>

I'm pretty sure I've made several mistakes. I have to go over this carefully to see where. Jgardner 22:18, 3 May 2012 (MDT)