Question 2.6: A free particle, which is initially localized in the range ,...

First we need to normalize \Psi (x,0) :

1=\int_{-\infty }^{\infty }\left|\Psi (x,0)\right| ^{2}dx=\left|A\right| ^{2}\int _{-a}^{a}dx=2a\left|A\right| ^{2} \Rightarrow A=\frac{1}{\sqrt{2a} }

Next we calculate\phi(k) , using Equation 2.104 [\phi (k)=\frac{1}{\sqrt{2\pi} } \int_{-\infty}^{\infty}{\psi (x,0)e^{-ikx}} dx]:

\phi (k)=\frac{1}{\sqrt{2\pi} }\frac{1}{\sqrt{2a} } \int_{-a}^{a}{e^{-ikx}} dx=\frac{1}{2\sqrt{\pi a} } \frac{e^{-ikx}}{-ik}\mid ^{a}_{-a}

=\frac{1}{k\sqrt{\pi a} }\left(\frac {e ^{ika}-e^{-ika}}{2i}\right)=\frac{1}{\sqrt {\pi a} }\frac{\sin(ka) }{k}

Finally, we plug this back into Equation 2.101 [\Psi (x,t)=\frac{1}{ \sqrt{2\pi } }\int_{-\infty }^{\infty }{\phi (k) }e^{i(kx-\frac{\hbar k^{2}}{2m} t)}dk]:

\Psi(x,t)=\frac{1}{\pi \sqrt{2a} }\int_{-\infty }^{\infty }{\frac{\sin(ka)}{k} }e^{i(kx-\frac{\hbar k^{2}}{2m} t)}dk           (2.105)

Unfortunately, this integral cannot be solved in terms of elementary functions, though it can of course be evaluated numerically (Figure 2.8). (There are, in fact, precious few cases in which the integral for \Psi (x,t)(Equation 2.101)  can be carried out explicitly; see Problem 2.21 for a particularly beautiful example.)

In Figure 2.9 I have plotted \Psi (x,0) and  \phi (k) . Note that for small a, \Psi (x,0) is narrow (in , while is broad (in , and vice versa for large a. But k is related to momentum, by Equation 2.97 [P=\hslash k], so this is a manifestation of the uncertainty principle: the position can be well defined (small a) , or the momentum (large a) , but not both.