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>