91影视

A wave function is a mathematical function that connects the location of an electron in space (defined by x, y, and z coordinates) to the amplitude of its wave, which corresponds to the energy of the electron.

(a)

Consider that

$\begin{array}{c}\mathrm{蠄}(\mathrm{x},0)=\mathrm{A}\left[{\mathrm{蠄}}_1\left(\mathrm{x}\right)+{\mathrm{蠄}}_2\left(\mathrm{x}\right)\right]|\mathrm{蠄}{|}^2\\ =|\mathrm{A}{|}^2\left[({{\mathrm{蠄}}_1}^{*}+{\mathrm{蠄}}_2*)({\mathrm{蠄}}_1+{\mathrm{蠄}}_2)\right]\\ =|\mathrm{A}{|}^2[{\left|{\mathrm{蠄}}_1\right|}^2+{\left|{\mathrm{蠄}}_2\right|}^2+{{\mathrm{蠄}}_1}^{*}{\mathrm{蠄}}_2+{\mathrm{蠄}}_2{*}_1]\end{array}$

Normalization condition is,

$\begin{array}{c}鈭�|\mathrm{蠄}{|}^2\mathrm{dx}=1\\ |\mathrm{A}{|}^2鈭�({\left|{\mathrm{蠄}}_1\right|}^2+{\left|{\mathrm{蠄}}_2\right|}^2+{\mathrm{蠄}}_1^{*}{\mathrm{蠄}}_2+{\mathrm{蠄}}_2^{*}{\mathrm{蠄}}_1)\mathrm{dx}=1\mathrm{n}\end{array}$

The ${\mathrm{蠄}}_1$and ${\mathrm{蠄}}_2$ are orthonormal states,

$鈭�{\left|{\mathrm{蠄}}_1\right|}^2\mathrm{dx}=1$and $鈭�{\left|{\mathrm{蠄}}_2\right|}^2\mathrm{dx}=1$

And

$鈭�{\mathrm{蠄}}_1^{*}{\mathrm{蠄}}_2\mathrm{dx}=0$and $鈭�{\mathrm{蠄}}_2^{*}{\mathrm{蠄}}_1\mathrm{dx}=0$

$\begin{array}{c}\left|\mathrm{A}{|}^2\right(2)=1\\ \mathrm{A}=\frac{1}{\sqrt{2}}\end{array}$

Therefore, the normalization of $\mathrm{蠄}(\mathrm{x},0)$is$\mathrm{A}=\frac{1}{\sqrt{2}}$

(b)

The expression for

$\mathrm{蠄}(\mathrm{x},\mathrm{t})=\frac{1}{\sqrt{2}}\left[{\mathrm{蠄}}_1{\mathrm{e}}^{鈭�\left(\frac{{\displaystyle {\mathrm{E}}_{\mathrm{t}}\mathrm{t}}}{{\displaystyle \mathrm{n}}}\right)}+{\mathrm{蠄}}_2{\mathrm{e}}^{鈭�\left(\frac{{\displaystyle {\mathrm{E}}_{,}\mathrm{t}}}{{\displaystyle \mathrm{n}}}\right)}\right]$

Let ${\mathrm{E}}_{\mathrm{n}}={\mathrm{n}}^2\mathrm{鈩�}\mathrm{蝇}$

${\mathrm{E}}_1=\mathrm{鈩�}\mathrm{蝇}$

And,

${\mathrm{E}}_2=4\mathrm{鈩�}\mathrm{蝇}$

Where, $\mathrm{蝇}=\frac{{\mathrm{蟺}}^2\mathrm{鈩�}}{2{\mathrm{ma}}^2}$

So now have an endless square well utilising the wave function.

$\begin{array}{c}{\mathrm{蠄}}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{a}}}\sin \frac{\mathrm{n蟺虫}}{\mathrm{a}}\\ \mathrm{蠄}(\mathrm{x},\mathrm{t})=\frac{1}{\sqrt{2}}\left[\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�\mathrm{颈胃迟}}+\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�\mathrm{i}\text{ifex}}\right]\\ =\frac{1}{\sqrt{2}}\left[\sqrt{\frac{2}{\mathrm{a}}}\right]\left[\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�\mathrm{颈胃迟}}+\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�\mathrm{itat}}\right]\\ =\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{鈭�\mathrm{iex}}\left[\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�3\mathrm{iex}}\right]\end{array}$

Further solving above equation,

$\begin{array}{c}\left|\mathrm{蠄}\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+\sin \frac{2\mathrm{蟺虫}}{\mathrm{a}}{\mathrm{e}}^{鈭�3\mathrm{sit}}\right]\left[\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+\sin \frac{2\mathrm{蟺虫}}{\mathrm{a}}{\mathrm{e}}^{3\mathrm{ses}}\right]\\ \left|\mathrm{蠄}\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)+\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)[{\mathrm{e}}^{鈭�3\sin }+{\mathrm{e}}^{3\sin }]\right]\\ \left|\mathrm{蠄}\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)+2\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\left(\cos 3\mathrm{蝇t}\right)\right]\end{array}$

Hence, the value as a sinusoidal function of $\mathrm{蠄}(\mathrm{x},\mathrm{t})$is

$\frac{{\mathrm{e}}^{鈭�\mathrm{sant}}}{\sqrt{\mathrm{a}}}\left[\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right){\mathrm{e}}^{鈭�3\mathrm{ett}}\right]$and of$\left|\mathrm{蠄}\right(\mathrm{x},\mathrm{t}){|}^2$is$\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)+2\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\left(\cos 3\mathrm{蝇t}\right)\right]$

(c)

The expression for

Now,

$\begin{array}{l}<\mathrm{x}>=鈭�\mathrm{x}\left|\mathrm{蠄}\right(\mathrm{x},\mathrm{t}){|}^2\mathrm{dx}\\ =\frac{1}{\mathrm{a}}{鈭�}_0^{\mathrm{a}}\mathrm{x}\left[{\sin }^2\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)+2\cos \left(3\mathrm{蝇t}\right)\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\right]\mathrm{dx}\end{array}$

Solving individual term as,

$\begin{array}{c}{鈭�}_0^{\mathrm{a}}{\mathrm{xsin}}^2\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)={鈭�}_0^{\mathrm{a}}\mathrm{x}\frac{\left[1鈭�\cos \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\right]}{2}\\ =\frac{1}{2}\left[{鈭�}_0^{\mathrm{a}}\left(\mathrm{x}鈭�\mathrm{xcos}\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\right)\mathrm{dx}\right]\\ ={\frac{1}{2}\left\{\frac{{\mathrm{x}}^2}{2}鈭�\mathrm{x}\frac{\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)}{\left(\frac{2\mathrm{蟺}}{\mathrm{a}}\right)}鈭�\frac{\cos \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)}{\frac{4{\mathrm{蟺}}^2}{{\mathrm{a}}^2}}\right\}|}_0^{\mathrm{a}}\\ ={\left[\frac{{\mathrm{x}}^2}{4}鈭�\frac{\mathrm{xsin}\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)}{\left(\frac{4\mathrm{蟺}}{\mathrm{a}}\right)}鈭�\frac{\cos \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)}{\left(8\frac{{\mathrm{蟺}}^2}{{\mathrm{a}}^2}\right)}\right]}_0^{\mathrm{a}}\\ =\left[\frac{{\mathrm{a}}^2}{4}鈭�0鈭�0\right]\\ =\frac{{\mathrm{a}}^2}{4}\end{array}$

$鈭�{\mathrm{xsin}}^2\left(\frac{\mathrm{n蟺虫}}{\mathrm{a}}\right)=\frac{{\mathrm{a}}^2}{4}$,This is independent of $\mathrm{n蝇t}$

So, there you have,

$\begin{array}{c}{鈭�}_0^{\mathrm{a}}{\mathrm{xsin}}^2\left(\frac{\mathrm{n蟺虫}}{\mathrm{a}}\right)={鈭�}_0^{\mathrm{a}}{\mathrm{xsin}}^2\left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\\ =\frac{{\mathrm{a}}^2}{4}\end{array}$

Now,

${鈭�}_0^{\mathrm{a}}\mathrm{x}\left(\sin \frac{\mathrm{蟺虫}}{\mathrm{a}}\sin \frac{2\mathrm{蟺虫}}{\mathrm{a}}\mathrm{dx}\right)2\cos \left(3\mathrm{蝇t}\right)\mathrm{dx}=2\cos \left(3\mathrm{蝇t}\right){鈭�}_0^{\mathrm{a}}\mathrm{xsin}\left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{蟺虫}}{\mathrm{a}}\right)\mathrm{dx}$

Substitute $2\mathrm{sinAsinB}=\cos (\mathrm{A}鈭�\mathrm{B})鈭�\cos (\mathrm{A}+\mathrm{B})$, in the above integral

$\begin{array}{c}{鈭�}_0^{\mathrm{a}}\mathrm{x}\left(\sin \frac{\mathrm{蟺虫}}{\mathrm{a}}\sin \frac{2\mathrm{蟺虫}}{\mathrm{a}}\mathrm{dx}\right)2\cos \left(3\mathrm{蝇t}\right)\mathrm{dx}\\ =\cos \left(3\mathrm{蝇t}\right){鈭�}_0^{\mathrm{a}}\mathrm{x}\left[\cos \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)鈭�\cos \left(\frac{3\mathrm{蟺虫}}{\mathrm{a}}\right)\right]\mathrm{dx}\\ =\cos \left(3\mathrm{蝇t}\right)\left[\mathrm{x}\frac{\sin \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)}{\left(\frac{\mathrm{蟺}}{\mathrm{a}}\right)}+\frac{{\mathrm{a}}^2}{{\mathrm{蟺}}^2}\cos \left(\frac{\mathrm{蟺虫}}{\mathrm{a}}\right)鈭�\mathrm{x}\frac{\sin \left(\frac{3\mathrm{蟺虫}}{\mathrm{a}}\right)}{\left(\frac{3\mathrm{蟺}}{\mathrm{a}}\right)}鈭�\frac{{\mathrm{a}}^2}{9{\mathrm{蟺}}^2}\cos \left(\frac{3\mathrm{蟺虫}}{\mathrm{a}}\right)\right]\\ =\cos \left(3\mathrm{蝇t}\right)\left[\frac{鈭�2{\mathrm{a}}^2}{{\mathrm{蟺}}^2}+\frac{2{\mathrm{a}}^2}{9{\mathrm{蟺}}^2}\right]\\ =\cos \left(3\mathrm{蝇t}\right)\left(\frac{鈭�2{\mathrm{a}}^2}{{\mathrm{蟺}}^2}\right)\left[1鈭�\frac{1}{9}\right]\\ =鈭�\cos \left(3\mathrm{蝇t}\right)\frac{16{\mathrm{a}}^2}{9{\mathrm{蟺}}^2}\end{array}$

Replace these values in the integral above.

$\begin{array}{c}<\mathrm{x}>=\frac{1}{\mathrm{a}}\left[\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{a}}^2}{4}鈭�\frac{16{\mathrm{a}}^2}{9{\mathrm{蟺}}^2}\cos 3\mathrm{蝇t}\right]\\ <\mathrm{x}>=\frac{\mathrm{a}}{2}\left[1鈭�\frac{32}{9{\mathrm{蟺}}^2}\cos \left(3\mathrm{蝇t}\right)\right]\end{array}$

This is a function that oscillates.

Amplitude $=\frac{32}{9{\mathrm{蟺}}^2}\left(\frac{\mathrm{a}}{2}\right)$

Angular frequency $=3\mathrm{蝇}$

$=\frac{3{\mathrm{蟺}}^2\mathrm{鈩�}}{2{\mathrm{ma}}^2}$

Therefore, the value ofis $\frac{\mathrm{a}}{2}\left[1鈭�\frac{32}{9{\mathrm{蟺}}^2}\cos \left(3\mathrm{蝇t}\right)\right]$, amplitude is$\frac{32}{9{\mathrm{蟺}}^2}\left(\frac{\mathrm{a}}{2}\right)$and angular frequency is$\frac{3{\mathrm{蟺}}^2\mathrm{h}}{2{\mathrm{ma}}^2}$.

(d)

The expression for

Insert $\frac{\mathrm{a}}{2}\left[1鈭�\frac{32}{9{\mathrm{蟺}}^2}\cos \left(3\mathrm{蝇t}\right)\right]$for in respective equation

Substitute $\frac{{\mathrm{蟺}}^2\mathrm{h}}{2{\mathrm{ma}}^2}$ for in respective equation

Substitute $\frac{{\mathrm{蟺}}^2\mathrm{h}}{2{\mathrm{ma}}^2}$for $\mathrm{蝇}$in equation (7).

Thus, the value of is$\frac{8\mathrm{h}}{3\mathrm{a}}\sin \left(3\mathrm{蝇t}\right)$

(e)

Consider that ${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2}$

$鈭�{\mathrm{E}}_1=\frac{{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2},\text{鈥夆赌夆赌�}{\mathrm{E}}_2=\frac{2{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{{\mathrm{ma}}^2}$

Since $\mathrm{蠄}(\mathrm{x},0)=\mathrm{A}\left[{\mathrm{蠄}}_1\left(\mathrm{x}\right)+{\mathrm{蠄}}_2\left(\mathrm{x}\right)\right]$

The probability of ${\mathrm{E}}_1$ and ${\mathrm{E}}_2$ is equal, and the total probability is 1, so

${\mathrm{P}}_1={\mathrm{P}}_2=\frac{1}{2}$

$\begin{array}{l}<\mathrm{H}>=\frac{1}{2}({\mathrm{E}}_1+{\mathrm{E}}_2)\\ =\frac{1}{2}\left[\frac{{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2}+\frac{4{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2}\right]\\ =\frac{1}{2}\left[\frac{5{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2}\right]\\ =\frac{5{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{4{\mathrm{ma}}^2}\end{array}$

Therefore,

${\mathrm{E}}_1<\mathrm{H}<{\mathrm{E}}_2$

Thus, the value of energies are$\frac{{\mathrm{蟺}}^2{\mathrm{h}}^2}{2{\mathrm{ma}}^2}$and$\frac{2{\mathrm{蟺}}^2{\mathrm{h}}^2}{{\mathrm{ma}}^2}$,their probability of occurrence is$\frac{1}{2}$for both, the value of is$\frac{5{\mathrm{蟺}}^2{\mathrm{h}}^2}{4{\mathrm{ma}}^2}$and its comparison with ${\mathrm{E}}_1$and${\mathrm{E}}_2$ is${\mathrm{E}}_1<\mathrm{H}<{\mathrm{E}}_2$.