(a) Find the Fourier transform for \phi (k)= \begin{cases} A(a-|k|) & |k|\leq a, \\ 0, & |k|\gt a. \end{cases}
where a is a positive parameter and A is a normalization factor to be found.
(b) Calculate the uncertainties Δx and Δp and check whether they satisfy the uncertainty principle.
(a) The normalization factor A can be found at once:
1=\int_{-\infty }^{+\infty }{|\phi (k)|^2dk=|A|^2\int_{-a}^{0}{(a+k)^2dk+|A|^2} }\int_{0}^{a}{(a-k)^2dk}=2|A|^2\int_{0}^{a}{(a-k)^2dk}=2|A|^2\int_{0}^{a}{(a^2-2ak+k^2)}dk
=\frac{2a^3}{3}|A|^2, (1.197)
which yields A=\sqrt{3/(2a^3)} . The shape of \phi (k)=\sqrt{3/(2a^3)}(a-|k| ) is displayed in Figure 1.18.
Now, the Fourier transform of \phi (k) is
\psi _0(x)=\frac{1}{\sqrt{2\pi } }\int_{-\infty }^{+\infty }{\phi (k)}e^{ikx}dk=\frac{1}{\sqrt{2\pi } }\sqrt{\frac{3}{2a^3} }\left[\int_{-a }^{0 }(a+k)e^{ikx}dk+\int_{0 }^{a }(a-k)e^{ikx}dk\right]
=\frac{1}{\sqrt{2\pi } }\sqrt{\frac{3}{2a^3} }\left[\int_{-a }^{0 }ke^{ikx}dk-\int_{0 }^{a }ke^{ikx}dk+a\int_{-a }^{a }e^{ikx}dk\right] . (1.198)
Using the integrations
\int_{-a }^{0 }ke^{ikx}dk=\frac{a}{ix}e^{-iax} +\frac{1}{x^2}\left(1-e^{-iax}\right) , (1.199)
\int_{0 }^{a }ke^{ikx}dk=\frac{a}{ix}e^{iax} +\frac{1}{x^2}\left(e^{iax}-1\right) , (1.200)
\int_{-a }^{a }e^{ikx}dk=\frac{1}{ix}\left(e^{iax}-e^{-iax}\right)=\frac{2\sin (ax)}{x} , (1.201)
and after some straightforward calculations, we end up with
\psi _0(x)=\frac{4}{x^2}\sin ^2 \left(\frac{ax}{2} \right). (1.202)
As shown in Figure 1.18, this wave packet is localized: it peaks at x = 0 and decreases gradually as x increases. We can verify that the maximum of \psi _0(x) occurs at x = 0; writing \psi _0(x) as a^2(ax/2)^{-2}\sin ^2(ax/2) and since \lim _{x\rightarrow 0}\sin (bx)/(bx)\rightarrow 1, we obtain \psi _0(0)=a^2.
(b) Figure 1.18a is quite suggestive in defining the half-width of \phi (k):\Delta k=a (hence the momentum uncertainty is \Delta p=\hbar a). By defining the width as \Delta k=a, we know with full certainty that the particle is located between -a\leq k\leq a; according to Figure 1.18a, the probability of finding the particle outside this interval is zero, for \phi (k) vanishes when |k|\gt a.
Now, let us find the width \Delta x of \psi _0(x). since \sin (a\pi /2a)=1,\psi _0(\pi /a)=4a^2/\pi ^2, and that \psi _0(0)=a^2, we can obtain from (1.202) that \psi _0(\pi /a)=4a^2/\pi ^2=4/\pi ^2\psi _0(0), or
\frac{\psi _0(\pi /a)}{\psi _0(0)}=\frac{4}{\pi ^2}. (1.203)
This suggests that \Delta x=\pi /a: when x=\pm \Delta x=\pm \pi /a the wave packet \psi _0(x) drops to 4/\pi ^2 from its maximum value \psi _0(0)=a^2. In sum, we have \Delta x=\pi /a and \Delta k =a; hence
\Delta x \Delta k= \pi (1.204)
or
\Delta x \Delta p= \pi \hbar , (1.205)
since \Delta k= \Delta p / \hbar . In addition to satisfying Heisenberg’s uncertainty principle (1.57) \Delta x\Delta p_x\geq \frac{\hbar }{2}, \Delta y\Delta p_y\geq \frac{\hbar }{2} ,\Delta z\Delta p_z\geq \frac{\hbar }{2}. , this relation shows that the product \Delta x \Delta p is higher than \hbar/2:\Delta x\Delta p\gt \hbar/2. The wave packet (1.202) therefore offers a clear illustration of the general statement outlined above; namely, only Gaussian wave packets yield the lowest limit to Heisenberg’s uncertainty principle \Delta x\Delta p= \hbar/2 (see (1.114)) \Delta x\Delta p= \frac{\hbar}{2}. . All other wave packets, such as (1.202), yield higher values for the product \Delta x \Delta p.
