A free particle has the initial wave function
$\mathbf{\psi }\left(x,0\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{a}\left|x\right|}\mathbf{,}$

where A and a are positive real constants.

(a)Normalize$\mathbf{\psi }\left(x,0\right)\mathbf{.}$

(b) Find$\mathbf{\varphi }\left(k\right)$.

(c) Construct $\mathbf{\psi }\left(x,t\right)$,in the form of an integral.

(d) Discuss the limiting cases very large, and a very small.

Initial wave function is given as:

$\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{a}\mathbf{\left|}\mathbf{x}\mathbf{\right|}}\mathbf{,}$

Where A and a are positive real constants.

In quantum physics, a wave function is a variable quantity that mathematically explains a particle's wave properties. The magnitude of a particle's wave function at a certain point in space and time depends on how likely it is that the particle was there at the time.

To find the constant A, use the normalizing condition, where

$$\begin{gathered} 1={\int }_{-\infty }^{\infty }\overline{)\mathrm{\psi }}{\overline{)\left(\mathrm{x},0\right)}}^2\mathrm{dx} \\ =2{\left|\mathrm{A}\right|}^2{\int }_0^{\infty }{\mathrm{e}}^{-2\mathrm{ax}\mathrm{dx}} \end{gathered}$$

But the integrand is an even function over a symmetric region, so it is written as,

$$\begin{gathered} 1=2{\left|A\right|}^2{\overline{)\frac{e^{-2ax}}{-2a}}}_0^{\infty } \\ =\frac{{\left|A\right|}^2}{a} \\ A=\sqrt{a} \end{gathered}$$

Hence the value of A is,$\sqrt{a}$.

$$\begin{gathered} \mathrm{To}\mathrm{find},\mathrm{\varphi }\left(\mathrm{k}\right)\mathrm{it}\mathrm{is}\mathrm{expressed}\mathrm{as}, \\ \mathrm{\varphi }\left(\mathrm{k}\right)=\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-\mathrm{a}\left|\mathrm{x}\right|}{\mathrm{e}}^{-\mathrm{ikx}}\mathrm{dx} \\ =\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-\mathrm{a}\left|\mathrm{x}\right|}\left(\cos \mathrm{kx}-\mathrm{i}\sin \mathrm{kx}\right)\mathrm{dx} \end{gathered}$$

The cosine integrand is even, and the sine is odd, so the latter vanishes and

$$\begin{gathered} \mathrm{\varphi }\left(\mathrm{k}\right)=2\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{ax}}\cos \mathrm{kxdx} \\ =\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{ax}}\left({\mathrm{e}}^{\mathrm{ikx}}+{\mathrm{e}}^{-\mathrm{ikx}}\right)\mathrm{dx} \\ =\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\int }_0^{\infty }\left({\mathrm{e}}^{\left(\mathrm{ik}-\mathrm{a}\right)\mathrm{x}}+{\mathrm{e}}^{-\left(\mathrm{ik}+\mathrm{a}\right)\mathrm{x}}\right)\mathrm{dx} \\ =\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}{\overline{)\left[\frac{{\mathrm{e}}^{\left(\mathrm{ik}-\mathrm{a}\right)\mathrm{x}}}{\mathrm{ik}-\mathrm{a}}+\frac{{\mathrm{e}}^{-\left(\mathrm{ik}+\mathrm{a}\right)\mathrm{x}}}{-\left(\mathrm{ik}+\mathrm{a}\right)}\right]}}_0^{\infty } \\ \mathrm{\varphi }\left(\mathrm{k}\right)=\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}\left(\frac{-1}{\mathrm{ik}-\mathrm{a}}+\frac{1}{\mathrm{ik}+\mathrm{a}}\right) \\ =\frac{\mathrm{A}}{\sqrt{2\mathrm{\pi }}}\frac{-\mathrm{ik}-\mathrm{a}+\mathrm{ik}-\mathrm{a}}{-{\mathrm{k}}^2-{\mathrm{a}}^2} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{A}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ \mathrm{\varphi }\left(\mathrm{k}\right)=\sqrt{\frac{\mathrm{a}}{2\mathrm{\pi }}}\frac{2\mathrm{a}}{{\mathrm{k}}^2+{\mathrm{a}}^2} \\ \mathrm{Hence}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\varphi }\left(\mathrm{k}\right)\mathrm{is},\sqrt{\frac{\mathrm{a}}{2\mathrm{\pi }}}\frac{2\mathrm{a}}{{\mathrm{k}}^2+{\mathrm{a}}^2}. \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{function}\mathrm{of}\mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)\mathrm{is}\mathrm{expressed}\mathrm{as}, \\ \mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }\mathrm{\varphi }\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{i}\left(\mathrm{kx}-\frac{\mathrm{hk}}{2\mathrm{m}}\mathrm{t}\right)}\mathrm{dk} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\varphi }\left(\mathrm{k}\right)\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ \mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }\left(\frac{\sqrt{\mathrm{a}}}{2\mathrm{\pi }}\frac{2\mathrm{a}}{{\mathrm{k}}^2+{\mathrm{a}}^2}\right){\mathrm{e}}^{\mathrm{i}\left(\mathrm{kx}-\frac{{\mathrm{hk}}^2}{2\mathrm{m}}\mathrm{t}\right)}\mathrm{dk} \\ \mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}2\sqrt{\frac{{\mathrm{a}}^3}{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }\frac{1}{{\mathrm{k}}^2+{\mathrm{a}}^2}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{kx}-\frac{{\mathrm{hk}}^2}{2\mathrm{m}}\mathrm{t}\right)}\mathrm{dk} \\ =\frac{{\mathrm{a}}^{3/2}}{\mathrm{\pi }}{\int }_{-\infty }^{\infty }\frac{1}{{\mathrm{k}}^2+{\mathrm{a}}^2}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{kx}-\frac{{\mathrm{hk}}^2}{2\mathrm{m}}\mathrm{t}\right)}\mathrm{dk} \\ \mathrm{Hence}\mathrm{the}\mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)\mathrm{is}\mathrm{expressed}\mathrm{as},\frac{{\mathrm{a}}^{3/2}}{\mathrm{\pi }}{\int }_{-\infty }^{\infty }\frac{1}{{\mathrm{k}}^2+{\mathrm{a}}^2}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{kx}-\frac{{\mathrm{hk}}^2}{2\mathrm{m}}\mathrm{t}\right)}\mathrm{dk} \end{gathered}$$

While $\varphi \left(k\right)\cong \sqrt{2/\pi a}$is broad and at for big a,$\psi \left(x,0\right)$ is a sharp, narrow spike; position is well-defined, but momentum is poorly defined. $\mathrm{\varphi }\left(\mathrm{k}\right)\cong \left(\sqrt{2{\mathrm{a}}^3/\mathrm{\pi }}\right)/{\mathrm{k}}^2$is a sharp, narrow spike for small a, while $\psi \left(x,0\right)$is a broad and at; position is poorly defined but momentum is clearly defined