91影视

Initial wave function is given as:

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

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

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

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

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

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

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

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

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

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

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