A particle in the infinite square well (Equation 2.22) has the initial wave function Ψ (x, 0) = A sin3(πx/a) (0 ≤ x ≤ a). Determine A, find Ψ(x, t), and calculate 〈x〉as a function of time. What is the expectation value of the energy? Hint: sinnθ and cosnθ can be reduced, by repeated application of the
trigonometric sum formulas, to linear combinations of sin(mθ) and cos(mθ), with m = 0, 1, 2, . . ., n.

Schrodinger equation is given by,

$i\sout{h}\frac{\partial \psi }{\partial t}=\frac{{\sout{h}}^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}+V(x,t)\psi (x,t)$

For an infinite square well,

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

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

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

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

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

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

By using the general solution

$$\begin{gathered} \psi (x,t)=\sum _{n-1}^{\infty }{c}_n{\varphi }_n\left(t\right){\psi }_n\left(x\right) \\ \psi (x,t)=\sum _{n-1}^{\infty }{c}_n\sqrt{\frac{2}{a}}exp\left(-i\frac{n^2{\mathrm{\pi }}^2\sout{\mathrm{h}}}{2m{a}^2}\right)\sin \frac{n\mathrm{\pi x}}{a} \end{gathered}$$

at t = 0

$$\begin{gathered} \psi (x,0)=\sum _{n-1}^{\infty }{c}_n\sqrt{\frac{2}{a}}\sin \frac{n\mathrm{\pi x}}{a} \\ \psi (x,0)=A{\sin }^3\frac{\mathrm{\pi x}}{a} \\ \psi (x,0)=A{\left(\frac{e^{inx/a}-e^{-inx/a}}{2i}\right)}^3 \\ \psi (x,0)=A\left(\frac{e^{3inx/a}-3e^{-inx/a}+3e^{-inx/a}-e^{-3inx/a}}{8i^3}\right) \\ \psi (x,0)=A\left[\frac{3}{4}\left(\frac{e^{inx/a}-e^{-inx/a}}{2i}\right)-\frac{1}{4}\left(\frac{e^{3inx/a}-e^{-3inx/a}}{2i}\right)\right] \\ \psi (x,0)=\frac{3A}{4}\sin \frac{\mathrm{\pi x}}{a}-\frac{A}{4}\sin \frac{3\mathrm{\pi x}}{a} \\ \psi (x,0)=\frac{3A}{4}\sqrt{\frac{a}{2}}{\psi }_1\left(x\right)-\frac{A}{4}\sqrt{\frac{a}{2}}{\psi }_3\left(x\right) \end{gathered}$$

Comparing the coefficients,

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

Hence,

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

Here the wave function is:

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

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,

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

In terms of eigenstates,

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

The expectation value of energy can be calculated as:

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

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, $E=\frac{9{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{10m{a}^2}$