A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states:

$\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{A}\mathbf{\left[}{\mathbf{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathbf{\psi }}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}$

You can look up the series

$\frac{\mathbf{1}}{{\mathbf{1}}^{\mathbf{6}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{3}}^{\mathbf{6}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{5}}^{\mathbf{6}}}\mathbf{+}\mathbf{\cdots }\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{6}}}{\mathbf{960}}$

and

$\frac{\mathbf{1}}{{\mathbf{1}}^{\mathbf{4}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{3}}^{\mathbf{4}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{5}}^{\mathbf{4}}}\mathbf{+}\mathbf{\cdots }\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{4}}}{\mathbf{96}}$

in math tables. under "Sums of Reciprocal Powers" or "Riemann Zeta Function."

(a) Normalize $\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}$ . (That is, find A. This is very easy, if you exploit the orthonormality of ${\mathbf{\psi }}_{\mathbf{1}}$and ${\mathbf{\psi }}_2$ Recall that, having ${\mathbf{\psi }}_{}$normalized at , $\mathbf{t}\mathbf{=}\mathbf{0}$ , you can rest assured that is stays normalized—if you doubt this, check it explicitly after doing part(b).

(b) Find $\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$and ${\mathbf{\left|}\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{\right|}}^{\mathbf{2}}$Express the latter as a sinusoidal function of time. To simplify the result, let $\mathbf{\omega }\mathbf{\equiv }\raisebox{1ex}{${\mathbf{\pi }}^{\mathbf{2}}\mathbf{\hslash }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}{\mathbf{ma}}^{\mathbf{2}}$}\right.$

c)Compute $\mathbf{⟨}\mathbf{x}\mathbf{⟩}$ . Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation?(If your amplitude is greater than $\raisebox{1ex}{$\mathbf{a}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$ , go directly to jail.

(d) Compute $\mathbf{⟨}\mathbf{p}\mathbf{⟩}$

(e) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of$\mathbf{H}$.How does it compare with E₁ and E₂

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

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

Normalization condition is,

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

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

$\int {\left|{\mathrm{\psi }}_1\right|}^2\mathrm{dx}=1$and $\int {\left|{\mathrm{\psi }}_2\right|}^2\mathrm{dx}=1$

And

$\int {\mathrm{\psi }}_1^{*}{\mathrm{\psi }}_2\mathrm{dx}=0$and $\int {\mathrm{\psi }}_2^{*}{\mathrm{\psi }}_1\mathrm{dx}=0$

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

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

(b)

The expression for

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

Let ${\mathrm{E}}_{\mathrm{n}}={\mathrm{n}}^2\mathrm{\hslash }\mathrm{\omega }$

${\mathrm{E}}_1=\mathrm{\hslash }\mathrm{\omega }$

And,

${\mathrm{E}}_2=4\mathrm{\hslash }\mathrm{\omega }$

Where, $\mathrm{\omega }=\frac{{\mathrm{\pi }}^2\mathrm{\hslash }}{2{\mathrm{ma}}^2}$

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

$\begin{array}{c}{\mathrm{\psi }}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{a}}}\sin \frac{\mathrm{n\pi x}}{\mathrm{a}}\\ \mathrm{\psi }(\mathrm{x},\mathrm{t})=\frac{1}{\sqrt{2}}\left[\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{i\theta t}}+\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{i}\text{ifex}}\right]\\ =\frac{1}{\sqrt{2}}\left[\sqrt{\frac{2}{\mathrm{a}}}\right]\left[\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{i\theta t}}+\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{itat}}\right]\\ =\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\mathrm{iex}}\left[\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-3\mathrm{iex}}\right]\end{array}$

Further solving above equation,

$\begin{array}{c}\left|\mathrm{\psi }\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+\sin \frac{2\mathrm{\pi x}}{\mathrm{a}}{\mathrm{e}}^{-3\mathrm{sit}}\right]\left[\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+\sin \frac{2\mathrm{\pi x}}{\mathrm{a}}{\mathrm{e}}^{3\mathrm{ses}}\right]\\ \left|\mathrm{\psi }\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)+\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)[{\mathrm{e}}^{-3\sin }+{\mathrm{e}}^{3\sin }]\right]\\ \left|\mathrm{\psi }\right(\mathrm{x},\mathrm{t}){|}^2=\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)+2\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)\left(\cos 3\mathrm{\omega t}\right)\right]\end{array}$

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

$\frac{{\mathrm{e}}^{-\mathrm{sant}}}{\sqrt{\mathrm{a}}}\left[\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right){\mathrm{e}}^{-3\mathrm{ett}}\right]$and of$\left|\mathrm{\psi }\right(\mathrm{x},\mathrm{t}){|}^2$is$\frac{1}{\mathrm{a}}\left[{\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)+2\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)\left(\cos 3\mathrm{\omega t}\right)\right]$

(c)

The expression for

Now,

$\begin{array}{l}<\mathrm{x}>=\int \mathrm{x}\left|\mathrm{\psi }\right(\mathrm{x},\mathrm{t}){|}^2\mathrm{dx}\\ =\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{a}}\mathrm{x}\left[{\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)+{\sin }^2\left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)+2\cos \left(3\mathrm{\omega t}\right)\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)\right]\mathrm{dx}\end{array}$

Solving individual term as,

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

$\int {\mathrm{xsin}}^2\left(\frac{\mathrm{n\pi x}}{\mathrm{a}}\right)=\frac{{\mathrm{a}}^2}{4}$,This is independent of $\mathrm{n\omega t}$

So, there you have,

$\begin{array}{c}{\int }_0^{\mathrm{a}}{\mathrm{xsin}}^2\left(\frac{\mathrm{n\pi x}}{\mathrm{a}}\right)={\int }_0^{\mathrm{a}}{\mathrm{xsin}}^2\left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)\\ =\frac{{\mathrm{a}}^2}{4}\end{array}$

Now,

${\int }_0^{\mathrm{a}}\mathrm{x}\left(\sin \frac{\mathrm{\pi x}}{\mathrm{a}}\sin \frac{2\mathrm{\pi x}}{\mathrm{a}}\mathrm{dx}\right)2\cos \left(3\mathrm{\omega t}\right)\mathrm{dx}=2\cos \left(3\mathrm{\omega t}\right){\int }_0^{\mathrm{a}}\mathrm{xsin}\left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{a}}\right)\mathrm{dx}$

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

$\begin{array}{c}{\int }_0^{\mathrm{a}}\mathrm{x}\left(\sin \frac{\mathrm{\pi x}}{\mathrm{a}}\sin \frac{2\mathrm{\pi x}}{\mathrm{a}}\mathrm{dx}\right)2\cos \left(3\mathrm{\omega t}\right)\mathrm{dx}\\ =\cos \left(3\mathrm{\omega t}\right){\int }_0^{\mathrm{a}}\mathrm{x}\left[\cos \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)-\cos \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)\right]\mathrm{dx}\\ =\cos \left(3\mathrm{\omega t}\right)\left[\mathrm{x}\frac{\sin \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)}{\left(\frac{\mathrm{\pi }}{\mathrm{a}}\right)}+\frac{{\mathrm{a}}^2}{{\mathrm{\pi }}^2}\cos \left(\frac{\mathrm{\pi x}}{\mathrm{a}}\right)-\mathrm{x}\frac{\sin \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)}{\left(\frac{3\mathrm{\pi }}{\mathrm{a}}\right)}-\frac{{\mathrm{a}}^2}{9{\mathrm{\pi }}^2}\cos \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)\right]\\ =\cos \left(3\mathrm{\omega t}\right)\left[\frac{-2{\mathrm{a}}^2}{{\mathrm{\pi }}^2}+\frac{2{\mathrm{a}}^2}{9{\mathrm{\pi }}^2}\right]\\ =\cos \left(3\mathrm{\omega t}\right)\left(\frac{-2{\mathrm{a}}^2}{{\mathrm{\pi }}^2}\right)\left[1-\frac{1}{9}\right]\\ =-\cos \left(3\mathrm{\omega t}\right)\frac{16{\mathrm{a}}^2}{9{\mathrm{\pi }}^2}\end{array}$

Replace these values in the integral above.

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

This is a function that oscillates.

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

Angular frequency $=3\mathrm{\omega }$

$=\frac{3{\mathrm{\pi }}^2\mathrm{\hslash }}{2{\mathrm{ma}}^2}$

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

(d)

The expression for

Insert $\frac{\mathrm{a}}{2}\left[1-\frac{32}{9{\mathrm{\pi }}^2}\cos \left(3\mathrm{\omega t}\right)\right]$for in respective equation

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

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

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

(e)

Consider that ${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{ma}}^2}$

$\therefore {\mathrm{E}}_1=\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{ma}}^2},{\mathrm{E}}_2=\frac{2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{ma}}^2}$

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

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

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

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

Therefore,

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

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