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) \\ &= {-2 a m \over \hbar}{\partial \over \partial x}(x \Psi) \\ &= {-2 a m \over \hbar}\left({\partial \over \partial x}(x)\Psi + x{\partial \over \partial x}\Psi\right) \\ &= {-2 a m \over \hbar}\left(\Psi + x {-2 a m x \over \hbar} \Psi \right) \\ &= {-2 a m \over \hbar}\left(1 + x {-2 a m x \over \hbar} \right) \Psi \\ &= {-2 a m \over \hbar}\left(1 - {2 a m x^2 \over \hbar} \right) \Psi \\ &= {2 a m \over \hbar}\left({2 a m x^2 \over \hbar} - 1 \right) \Psi \\ &= {2 a m \over \hbar}\left({2 a m x^2 \over \hbar} - {\hbar \over \hbar} \right) \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} {2am \over \hbar^2} (2amx^2 - \hbar) \Psi + V \Psi \ 
                          a \hbar \Psi &= - a (2amx^2 - \hbar) \Psi + V \Psi \\
                               a \hbar &= - a (2amx^2 - \hbar) + V \\
                                     V &= a \hbar + a (2amx^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)

I caught the mistake. I had lost a negative sign in the x derivative term. Jgardner 01:30, 4 May 2012 (MDT)

1.14.c

<math> \begin{align} \langle x \rangle

&= \int_{-\infty}^{+\infty} \Psi^* x \Psi\ dx \\
&= \int_{-\infty}^{+\infty} x \Psi^* \Psi\ dx \\
&= \int_{-\infty}^{+\infty} x A^2 e^{-2amx^2/\hbar}\ dx & & \text{Note that }x e^{-bx^2}\text{ is odd.} \\
&= 0 & & \text{Centered at the origin.} \\

\langle x^2 \rangle

&= \int_{-\infty}^{+\infty} \Psi^* x^2 \Psi\ dx \\
&= \int_{-\infty}^{+\infty} x^2 A^2 e^{-2amx^2/\hbar}\ dx \\
&= {A^2 \sqrt{\pi} \over \left(2am / \hbar\right)^{3/2}} \\
&= {\sqrt{2am \pi} \hbar^{3/2} \over \sqrt{\pi \hbar} (2am)^{3/2}} \\
&= {\hbar \over 2am} & & \text{Looks like it is spread out a bit.}\\

\langle p \rangle

&= \int_{-\infty}^{+\infty} \Psi^* (-i \hbar) {\partial \over \partial x} \Psi\ dx \\
&= \int_{-\infty}^{+\infty} \Psi^* (-i \hbar) {-2 a mx \over \hbar} \Psi\ dx \\
&= \int_{-\infty}^{+\infty} \Psi^* 2 a m x i \Psi\ dx \\
&= 2 a m i \int_{-\infty}^{+\infty} x \Psi^* \Psi\ dx & & \text{Odd function. (See above)} \\
&= 0 & & \text{It looks like it has an equal chance of going left or right.} \\

\langle p^2 \rangle

&= \int_{-\infty}^{+\infty} \Psi^* \left((-i \hbar) {\partial \over \partial x}\right)^2 \Psi\ dx \\
&= \int_{-\infty}^{+\infty} \Psi^* (-\hbar^2) {\partial^2 \over \partial x^2} \Psi\ dx \\
&= \int_{-\infty}^{+\infty} \Psi^* (-\hbar^2) {2 a m \over \hbar^2} (2 a m x^2 - \hbar) \Psi\ dx \\
&= -2am \int_{-\infty}^{+\infty} \Psi^* (2 a m x^2 - \hbar) \Psi\ dx \\
&= -2am \left(
  \int_{-\infty}^{+\infty} \Psi^* 2amx^2 \Psi\ dx - 
  \int_{-\infty}^{+\infty} \Psi^* \hbar \Psi\ dx
\right) \\
&= -2am \left(
  2am \int_{-\infty}^{+\infty} \Psi^* x^2 \Psi\ dx - 
  \hbar \int_{-\infty}^{+\infty} \Psi^* \Psi\ dx
\right) & & \text{We've already done these integrals, remember?} \\
&= -2am ( 2am \langle x^2 \rangle - \hbar) \\
&= -2am ( 2am {\hbar \over 2am} - \hbar) \\
&= -2am ( \hbar - \hbar) \\
&= 0 & & \text{That particle is really staying put!}\\

\end{align} </math>

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. Jgardner 01:01, 4 May 2012 (MDT)