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The particle-wave equation in an infinity square well is given as:\

$\mathit{蠄}\left(x,t\right)\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}{\mathit{c}}_{\mathbf{n}}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}\mathbf{e}}\left(-i\frac{n^2{蟺}^2\sout{h^2}}{2m{a}^2}T\right)\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{n}\mathbf{蟺}\mathbf{x}}{\mathbf{a}}$

The wave function鈥檚 original form is$\mathrm{唯}\left(\mathrm{x},0\right)$, The quantum revival time is defined such that,

localid="1658230974607" $唯\mathit{(}\mathit{x}\mathit{,}\mathit{0}\mathit{)}\mathit{=}唯\mathit{(}\mathit{x}\mathit{,}\mathit{T}\mathit{)}$

$$\begin{gathered} \underset{\mathrm{n}=1}{\overset{鈭�}{鈭�}}{c}_{\mathrm{n}}\sqrt{\frac{2}{a}}\sin \frac{n蟺x}{a}=\underset{\mathrm{n}=1}{\overset{鈭�}{鈭�}}{c}_n\sqrt{\frac{2}{a}}{e}^{\left(-\mathrm{i}\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2{\mathrm{ma}}^2}\right)}\sin \frac{\mathrm{苍蟺虫}}{\mathrm{a}} \\ {e}^{\left(-\mathrm{i}\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2{\mathrm{ma}}^2}\right)}=1 \\ \cos \left(-\mathrm{i}\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2{\mathrm{ma}}^2}\right)-i\sin \left(-\mathrm{i}\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2{\mathrm{ma}}^2}\right)=1 \end{gathered}$$

Therefore,

$\cos \left(-\mathrm{i}\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2{\mathrm{ma}}^2}\mathrm{T}\right)=1$

And,

$-i\sin \left(\frac{{\mathit{n}}^{\mathit{2}}{\mathit{蟺}}^{\mathit{2}}\sout{{\mathit{h}}^{\mathit{2}}}}{\mathit{2}\mathit{m}{\mathit{a}}^{\mathit{2}}}T\right)=0$

Both the equations are satisfied if,

$\frac{{蟺}^2{\sout{h}}^2}{2m{a}^2}T=2q蟺,$

Since $n^2$is an integer.

Because we want the shortest revival time, q = 1.

$$\begin{gathered} \frac{{蟺}^2{\sout{h}}^2}{2m{a}^2}T=2蟺 \\ T=\frac{4m{a}^2}{蟺\sout{h}} \end{gathered}$$

Thus, the wave function of a particle in the infinite square well returns to its original form after a quantum revival time.

b)

Classically, a particle bouncing back and forth between the walls of an infinite square well goes a distance 2a before it reaches its initial state again.

So,

$$\begin{gathered} 2a=vT \\ T=\frac{2a}{v} \end{gathered}$$

Since,

$$\begin{gathered} E=PE+KE \\ E=0+\frac{1}{2}m{v}^2 \\ E=\frac{1}{2}m{v}^2 \\ v=+\sqrt{\frac{2E}{m}} \end{gathered}$$

Substituting this into the formula for T,

$$\begin{gathered} T=\frac{2a}{\sqrt{{\displaystyle \frac{2E}{m}}}} \\ T=a\sqrt{\frac{2m}{E}} \end{gathered}$$

Thus, classical revival time is calculated as: $T=a\sqrt{\frac{2m}{E}}$.

For the quantum and classical revival times to be equal, the energy would be

$$\begin{gathered} \frac{4m{a}^2}{蟺\sout{h}}=a\sqrt{\frac{2m}{E}} \\ \sqrt{E}=a\sqrt{2m}\frac{蟺\sout{h}}{4m{a}^2} \\ E=a^2\left(2m\right)\frac{{蟺}^2{\sout{h}}^2}{16m^2{a}^4} \\ E=\frac{{蟺}^2{\sout{h}}^2}{8m{a}^2} \end{gathered}$$

$\frac{{蟺}^2{\sout{h}}^2}{8m{a}^2}$is the value of energy for the quantum and classical revival times will be equal.