91影视

Schrodinger equation is given by,

$i\sout{h}\frac{鈭�蠄}{鈭�t}=\frac{{\sout{h}}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}+V(x,t)蠄(x,t)$

For an infinite square well,

$V(x,t)=V\left(x\right)=\left\{\begin{array}{l}0,0<x<a\\ 鈭�,otherwise\end{array}\right.$

Reducing the partial differential equation to two ordinary differential equations in x and t:

$$\begin{gathered} i\sout{h}\frac{{蠒}^{\text{'}}\left(t\right)}{蠒\left(t\right)}=E \\ -\frac{{\sout{h}}^2}{2m}\frac{{蠄}^{\text{'}\text{'}}\left(x\right)}{蠄\left(x\right)}=E \end{gathered}$$

In the boundary conditions$蠄\left(0\right)=0$and $蠄\left(a\right)=0$, the time-independent Schr枚dinger equation gives normalized solutions of the form,

$$\begin{gathered} {蠄}_n\left(x\right)=\sqrt{\frac{2}{a}}\sin \frac{n\mathrm{蟺虫}}{a} \\ {E}_n=\frac{n^2{\mathrm{蟺}}^2{\sout{\mathrm{h}}}^2}{2m{a}^2} \end{gathered}$$

Using this formula, the solution to the ordinary differential equation in t is ${蠒}_n\left(t\right)=e^{-i{E}_{n}t/h}$. The general solution for $蠄(x,t)$is a linear combination of the product solutions ${蠒}_n\left(t\right){蠄}_n\left(x\right)$for all n.

By using the general solution

$$\begin{gathered} 蠄(x,t)=\underset{n-1}{\overset{鈭�}{鈭�}}{c}_n{蠒}_n\left(t\right){蠄}_n\left(x\right) \\ 蠄(x,t)=\underset{n-1}{\overset{鈭�}{鈭�}}{c}_n\sqrt{\frac{2}{a}}exp\left(-i\frac{n^2{\mathrm{蟺}}^2\sout{\mathrm{h}}}{2m{a}^2}\right)\sin \frac{n\mathrm{蟺虫}}{a} \end{gathered}$$

at t = 0

$$\begin{gathered} 蠄(x,0)=\underset{n-1}{\overset{鈭�}{鈭�}}{c}_n\sqrt{\frac{2}{a}}\sin \frac{n\mathrm{蟺虫}}{a} \\ 蠄(x,0)=A{\sin }^3\frac{\mathrm{蟺虫}}{a} \\ 蠄(x,0)=A{\left(\frac{e^{inx/a}-e^{-inx/a}}{2i}\right)}^3 \\ 蠄(x,0)=A\left(\frac{e^{3inx/a}-3e^{-inx/a}+3e^{-inx/a}-e^{-3inx/a}}{8i^3}\right) \\ 蠄(x,0)=A\left[\frac{3}{4}\left(\frac{e^{inx/a}-e^{-inx/a}}{2i}\right)-\frac{1}{4}\left(\frac{e^{3inx/a}-e^{-3inx/a}}{2i}\right)\right] \\ 蠄(x,0)=\frac{3A}{4}\sin \frac{\mathrm{蟺虫}}{a}-\frac{A}{4}\sin \frac{3\mathrm{蟺虫}}{a} \\ 蠄(x,0)=\frac{3A}{4}\sqrt{\frac{a}{2}}{蠄}_1\left(x\right)-\frac{A}{4}\sqrt{\frac{a}{2}}{蠄}_3\left(x\right) \end{gathered}$$

Comparing the coefficients,

$$\begin{gathered} c_1\sqrt{\frac{2}{a}}=\frac{3A}{4},n=1 \\ {c}_1\sqrt{\frac{2}{a}}=\frac{A}{4},n=3 \\ {c}_1\sqrt{\frac{2}{a}}=0,n鈮�1\&n鈮�3 \end{gathered}$$

Hence,

localid="1658294538499" $蠄(x,t)=\frac{3A}{4}\exp \left(-i\frac{{\mathrm{蟺}}^2\sout{\mathrm{h}}}{2m{a}^2}t\right)\sin \frac{\mathrm{蟺虫}}{a}-\frac{A}{4}\exp \left(-i\frac{9{\mathrm{蟺}}^2\sout{\mathrm{h}}}{2m{a}^2}t\right)\sin \frac{3\mathrm{蟺虫}}{a}$

Here the wave function is:

$$\begin{gathered} 1={鈭�}_0^a{\left|蠄(x,0)\right|}^{2}dx \\ 1={鈭�}_0^a{\left[\frac{3A}{4}\sqrt{\frac{a}{2}}{蠄}_1\left(x\right)-\frac{A}{4}\sqrt{\frac{a}{2}}{蠄}_3\left(x\right)\right]}^{2}dx \\ 1={鈭�}_0^a\left\{\frac{9A^2}{16}\frac{a}{2}{\left[{蠄}_1\left(x\right)\right]}^2-2\frac{3A}{4}\frac{A}{4}\frac{a}{2}{蠄}_1\left(x\right){蠄}_3\left(x\right)+\frac{A^2}{16}\frac{a}{2}{\left[{蠄}_3\left(x\right)\right]}^2\right\}dx \end{gathered}$$

Using the orthonormality of eigenstates to evaluate this integral,

$$\begin{gathered} 1=\frac{5a{A}^2}{16} \\ A=\frac{4}{\sqrt{5a}} \end{gathered}$$

Hence, the wave function becomes,

$蠄(x,t)=\frac{1}{\sqrt{5a}}\left[3\exp \left(-i\frac{{\mathrm{蟺}}^2\sout{\mathrm{h}}}{2m{a}^2}t\right)\sin \frac{\mathrm{蟺虫}}{a}-\exp \left(-i\frac{9{\mathrm{蟺}}^2\sout{\mathrm{h}}}{2m{a}^2}t\right)\sin \frac{3\mathrm{蟺虫}}{a}\right]$

In terms of eigenstates,

$$\begin{gathered} 蠄(x,t)=\frac{3A}{4}\sqrt{\frac{a}{2}}{蠄}_1\left(x\right)e^{-i{E}_{1}t/\sout{h}}-\frac{A}{4}\sqrt{\frac{a}{2}}{蠄}_3\left(x\right)e^{-i{E}_{3}t/\sout{h}} \\ 蠄(x,t)=\frac{3}{\sqrt{10}}{蠄}_1\left(x\right)e^{-i{E}_{1}t/\sout{h}}-\frac{1}{\sqrt{10}}{蠄}_3\left(x\right)e^{-i{E}_{3}t/\sout{h}} \end{gathered}$$

The expectation value of energy can be calculated as:

$$\begin{gathered} E=\underset{n}{鈭�}{\left|c_n\right|}^2{E}_n \\ E={\left|c_1\right|}^2\left(E_1\right)+{\left|c_3\right|}^2\left(E_3\right) \\ E=E_1{\left|\frac{3}{\sqrt{10}}\right|}^2+E_3{\left|-\frac{1}{\sqrt{10}}\right|}^2 \\ E=\frac{9}{10}{E}_1+\frac{1}{10}{E}_3 \\ E=\frac{9}{10}脳\frac{{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2m{a}^2}+\frac{1}{10}脳\frac{9{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{2m{a}^2} \\ E=\frac{9{\mathrm{蟺}}^2\sout{{\mathrm{h}}^2}}{10m{a}^2} \end{gathered}$$

The expectation value of x can be calculated here as:

Splitting up the integral and evaluating it, we get,

Calculated values are :$A=\frac{4}{\sqrt{5a}}$

role="math" localid="1658296781760"

Expectation value for energy, $E=\frac{9{\mathrm{蟺}}^2{\sout{\mathrm{h}}}^2}{10m{a}^2}$