Test the energy-time uncertainty principle for the wave function in Problem$\mathbf{2}\mathbf{.}\mathbf{5}$and the observable x, by calculating${\mathit{\sigma }}_{\mathbf{H}}{\mathit{\sigma }}_{\mathbf{X}}$and$\mathit{d}<x>\mathbf{/}\mathit{d}\mathit{t}$exactly.

The wavefunction is given by$\mathit{\psi }\left(x,0\right)\mathbf{=}\mathit{A}\left[{\psi }_1\left(x\right)+{\psi }_2\left(x\right)\right]$.

First, find the normalization constant as:

$$\begin{gathered} 1={\int }_{-\infty }^{\infty }{\left|\psi \left(x,0\right)\right|}^{2}dx \\ ={\int }_{-\infty }^{\infty }{\psi }^{*}\psi dx={\left|A\right|}^2{\int }_{-\infty }^{\infty }{\left[{\psi }_1\left(x\right)+{\psi }_2\left(x\right)\right]}^{*}\left[{\psi }_1\left(x\right)+{\psi }_2\left(x\right)\right]dx \\ ={\left|A\right|}^2\left[{\int }_{-\infty }^{\infty }{\psi }_1^{*}{\psi }_{1}dx+{\int }_{-\infty }^{\infty }{\psi }_1^{*}{\psi }_{2}dx+{\int }_{-\infty }^{\infty }{\psi }_2^{*}{\psi }_{1}dx+{\int }_{-\infty }^{\infty }{\psi }_2^{*}{\psi }_{2}dx+\right] \end{gathered}$$

Use the orthogonality condition, eliminate the second and the third integrations, whereas the first and the fourth integrations each one of them is equal to one; therefore,

$$\begin{gathered} 1={\left|A\right|}^2\left[1+0+0+1\right] \\ =2{\left|A\right|}^2 \\ A=\sqrt{\frac{1}{2}} \end{gathered}$$

Assume $\omega ={\pi }^2ħ/2m{a}^2$this will lead equation ( ) to become$E_{\mathrm{n}}=n^2\omega ħ$, and therefore:

$$\begin{gathered} \psi \left(x,t\right)=A\left[{\psi }_1\left(x\right)e^{-i{E}_{1}t/ħ}+{\psi }_2\left(x\right)E^{-i{E}_{2}t/ħ}+\right] \\ =\sqrt{\frac{1}{2}}\left[\sqrt{\frac{2}{a}}\sin \left(\frac{\pi }{a}x\right)e^{-i\omega t}+\sqrt{\frac{2}{a}}s\mathrm{in}\left(\frac{2\pi }{a}x\right)e^{-4i\omega t}\right] \\ =\sqrt{\frac{1}{a}}\left[\sin \left(\frac{\pi }{a}x\right)e^{-i\omega t}+s\mathrm{in}\left(\frac{2\pi }{a}x\right)e^{-4i\omega t}\right] \end{gathered}$$

The modulus squared of the wavefunction is:

Now find the expectation value of the position as follow:

The first integral could be solve use integration by parts and assume $u=x$, then $du=dx$, and$dv={\sin }^2\left(\frac{2\pi }{a}x\right)$ and by the use of integration table $v=\frac{x}{2}-\frac{a}{4\pi }\sin \left(\frac{2\pi }{a}x\right)$, therefore:

localid="1656316627360" $$\begin{gathered} ={\int }_0^{a}x{\sin }^2\left(\frac{\pi }{a}x\right)dx=\frac{x^2}{2}-\frac{ax}{4\pi }\sin \left(\frac{2\pi }{a}x\right)\overline{){}_0^a-\left[{\int }_0^a\frac{x}{2}-\frac{a}{4\pi }\sin \left(\frac{2\pi }{a}x\right)dx\right]} \\ ={\left[\frac{x^2}{4}-\frac{ax}{4\pi }\sin \left(\frac{2\pi }{a}x\right)-\frac{a^2}{8{\pi }^2}\cos \left(\frac{2\pi }{a}x\right)\right]}_0^a \\ \frac{a^2}{4} \end{gathered}$$

Do the same thing to the third integration and get the same result again. For the second integration use identity $\sin \left(\alpha \right)\sin \left(\beta \right)=1/2\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha -\beta \right)\right]$then perform an integration by parts and obtain the result as:

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

Therefore, the expectation value of the position will be:

The expectation value of the momentum could be found by the use of the expectation value of the position, as:

Given below is two possible energies’ $E_1$and $E_2$, where,

$E_1=\frac{{\pi }^2{ħ}^2}{2m{a}^2}{E}_2=\frac{{\left(2\right)}^2{\pi }^2{ħ}^2}{2m{a}^2}$and each one has a probability of one half (i.e., $p_1=p_2=1/2$), therefore, the average energy (i.e., the expectation value of the energy) is:

The expression for${\sigma }_H^2$is given by:

The expression foris given by:

$H^2\psi =\frac{1}{\sqrt{2}}\left[\left(H^2{\psi }_1\right)e^{-E_{1}t/ħ}+\left(H^2{\psi }_2\right)e^{-E_{n}t/ħ}\right]$

Where,

$$\begin{gathered} H{\psi }_1=E_1{\psi }_1 \\ {H}^2{\psi }_1=E_{1}H{\psi }_1 \\ =E_1^2{\psi }_1 \end{gathered}$$

And $H^2{\psi }_2=E_2^2{\psi }_2$.

The expectation value ofwill be:

localid="1656347269478"

Substitute all the known values in the expression of ${\sigma }_H^2$.

$$\begin{gathered} {\sigma }_H^2=\frac{1}{2}\left(E_1^2+E_2^2\right)-\frac{1}{4}{\left(E_1+E_2\right)}^2 \\ =\frac{1}{4}\left(2E_1^2+2E_2^2-E_2^2-E_1^2-2E_1{E}_2-E_2^2\right) \\ =\frac{1}{4}\left(E_1^2-2E_1{E}_2+E_2^2\right) \\ =\frac{1}{4}{\left(E_2-E_1\right)}^2 \\ Thus:{\sigma }_H=\frac{1}{2}\left(E_2-E_1\right) \end{gathered}$$

Find the expectation value of $x^2$, which is given by:

Solve the steps as given below,

And from problem 2.4 obtain the result as :

From the calculated values the expectation value of $x^2$will be,

Calculate ${\sigma }_x^2$as:

It is given that:

$$\begin{gathered} \mathrm{The}\mathrm{uncertainty}\mathrm{principle}\mathrm{holds}\mathrm{if}: \\ {\left(\frac{3}{4}\right)}^2\left[\frac{1}{3}-\frac{5}{4{\mathrm{\pi }}^2}-{\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2{\cos }^2\left(3\omega t\right)\right]\ge {\left(\frac{8}{3{\mathrm{\pi }}^2}\right)}^2{\sin }^2\left(3\omega t\right) \\ \mathrm{Further}\mathrm{solving}\mathrm{above}\mathrm{equation}\mathrm{as}, \\ \frac{1}{3}-\frac{5}{4{\mathrm{\pi }}^2}\ge {\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2{\cos }^2\left(3\mathrm{\omega t}\right)+{\left(\frac{4}{3}\frac{8}{3{\mathrm{\pi }}^2}\right)}^2{\sin }^2\left(3\mathrm{\omega t}\right)={\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2 \\ \frac{1}{3}-\frac{5}{4{\mathrm{\pi }}^2}\ge {\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2\left({\cos }^2\left(3\mathrm{\omega t}\right)+{\sin }^2\left(3\mathrm{\omega t}\right)\right) \\ \frac{1}{3}-\frac{5}{4{\mathrm{\pi }}^2}\ge {\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2 \\ \mathrm{Solving}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{values}: \\ \frac{1}{3}-\frac{5}{4{\mathrm{\pi }}^2}=0.20668{\left(\frac{32}{9{\mathrm{\pi }}^2}\right)}^2=0.12978 \\ \mathrm{Thus},\mathrm{the}\mathrm{uncertainty}\mathrm{principle}\mathrm{holds}\mathrm{the}\mathrm{condition}. \end{gathered}$$