91Ó°ÊÓ

To frame $\mathrm{ψ}\left(x,t\right)$, tack onto each term its characteristic time dependence $exp\left(-i{E}_{n}t/h\right)$. Equation 2.17 is,

$\mathit{ψ}\left(x,t\right)\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{c}}_{\mathbf{n}}{\mathit{ψ}}_{\mathbf{n}}\left(x\right){\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{E}}_{\mathbf{n}}\mathbf{t}\mathbf{/}\mathbf{h}}\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{c}}_{\mathbf{n}}{\mathit{ψ}}_{\mathbf{n}}\left(x,t\right)$

Problem 2.3 gives the general solution to the Schrödinger equation for the infinite square well potential

$V\left(x\right)=\left\{\begin{array}{ll}0& \mathrm{if}0≤\mathrm{x}≤\mathrm{a}\\ ∞& \mathrm{otherwise}\end{array}\right.$

was found to be

$\mathrm{ψ}\left(x,t\right)=\sqrt{\frac{2}{a}}\underset{n=1}{\overset{∞}{∑}}{B}_{n}exp\left(-i\frac{\mathrm{Ä\S }{\mathrm{Ï€}}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{n\mathrm{Ï€³\ae }}{a},0≤x≤a$

The coefficients $B_n$are determined by using the provided initial condition,

$$\begin{gathered} \mathrm{ψ}\left(x,0\right)=A\left[{\mathrm{ψ}}_1\left(x\right)+e^{i\mathrm{Ï•}}{\mathrm{ψ}}_2\left(x\right)\right] \\ =A\left(\sqrt{\frac{2}{a}}\sin \frac{\mathrm{Ï€³\ae }}{a}+\sqrt{\frac{2}{a}{e}^{i\mathrm{Ï•}}}\sin \frac{2\mathrm{Ï€³\ae }}{a}\right) \\ =A\sqrt{\frac{2}{a}}\left(\sin \frac{\mathrm{Ï€³\ae }}{a}+e^{i\mathrm{Ï•}}\sin \frac{2\mathrm{Ï€³\ae }}{a}\right) \end{gathered}$$

First, normalize the initial wave function to find A by using the condition:

$1={∫}_{-∞}^{∞}{\left|\mathrm{ψ}\left(x,0\right)\right|}^{2}dx$

Next, solve for A to get the value as $\frac{1}{\sqrt{2}}$.

For t=0 in the general solution can be written as follows:

$$\begin{gathered} \mathrm{ψ}\left(x,0\right)=\sqrt{\frac{2}{a}}\underset{n=1}{\overset{∞}{∑}}{B}_n\sin \frac{n\mathrm{Ï€³\ae }}{a} \\ =\sqrt{\frac{2}{a}}{B}_1\sin \frac{\mathrm{Ï€³\ae }}{a}+\sqrt{\frac{2}{a}}{B}_2\sin \frac{2\mathrm{Ï€³\ae }}{a}+\sqrt{\frac{2}{a}}{B}_3\sin \frac{3\mathrm{Ï€³\ae }}{a}+... \end{gathered}$$

Compare the coefficients,

role="math" localid="1658128694684" $$\begin{gathered} \sqrt{\frac{2}{a}}{B}_1=\frac{1}{\sqrt{a}} \\ {B}_1=\frac{1}{\sqrt{2}} \\ \sqrt{\frac{2}{a}}{B}_2=\frac{1}{\sqrt{a}}{e}^{i\mathrm{Ï•}} \\ {B}_2=\frac{1}{\sqrt{2}}{e}^{i\mathrm{Ï•}} \\ \sqrt{\frac{2}{a}}{B}_n=0 \\ \mathrm{for}\mathrm{n}â‰\yen 3\mathrm{then}{\mathrm{B}}_{\mathrm{n}}=0 \end{gathered}$$

$$\begin{gathered} \mathrm{ψ}\left(x,t\right)=\sqrt{\frac{2}{a}}\underset{n=1}{\overset{∞}{∑}}{B}_{n}exp\left(-i\frac{\mathrm{Ä\S }{\mathrm{Ï€}}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{n\mathrm{Ï€³\ae }}{a} \\ =\sqrt{\frac{2}{a}}{B}_{1}exp\left(-i\frac{\mathrm{Ä\S }{\mathrm{Ï€}}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{\mathrm{Ï€³\ae }}{a}+\sqrt{\frac{2}{a}}{B}_{2}exp\left(-i\frac{\mathrm{Ä\S }{\mathrm{Ï€}}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{2\mathrm{Ï€³\ae }}{a} \\ =\frac{1}{\sqrt{a}}exp\left(-i\frac{\mathrm{Ä\S }{\mathrm{Ï€}}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{\mathrm{Ï€³\ae }}{a}+\frac{1}{\sqrt{a}}{e}^{i\mathrm{Ï•}}exp\left(-i\frac{2\mathrm{Ä\S }{\mathrm{Ï€}}^2}{m{a}^2}t\right)\sin \frac{2\mathrm{Ï€³\ae }}{a} \end{gathered}$$

Use $\mathrm{Ó\neg }={\mathrm{Ï€}}^2\mathrm{Ä\S }/2{\mathrm{ma}}^2$to simplify the result,

$\mathrm{ψ}\left(x,t\right)=\frac{1}{\sqrt{a}}{e}^{-i\mathrm{Ó\neg }t}\sin \left(\frac{\mathrm{Ï€³\ae }}{a}\right)+\frac{1}{\sqrt{a}}{e}^{i\mathrm{Ï•}}{e}^{-4i\mathrm{Ó\neg }t}\sin \left(\frac{2\mathrm{Ï€³\ae }}{a}\right),0≤x≤a$

Writing the solution in terms of the eigenstates,

$$\begin{gathered} \mathrm{Ó\neg }\left(x,t\right)=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{2}{a}}\sin \frac{\mathrm{Ï€³\ae }}{a}\right)e^{-i\mathrm{Ó\neg }t}+\frac{e^{i\mathrm{Ï•}}}{\sqrt{2}}\left(\sqrt{\frac{2}{a}}\sin \frac{2\mathrm{Ï€³\ae }}{a}\right)e^{-4i\mathrm{Ó\neg }t} \\ =\frac{1}{\sqrt{2}}{\mathrm{Ó\neg }}_1\left(x\right)e^{-i\mathrm{Ó\neg }t}+\frac{e^{i\mathrm{Ï•}}}{\sqrt{2}}{\mathrm{ψ}}_2\left(x\right)e^{-4i\mathrm{Ó\neg }t} \end{gathered}$$

Therefore the value is $\mathrm{ψ}\left(x,t\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{a}$}\right.e^{-i\mathrm{Ó\neg }t}\left[\sin \left(\frac{\mathrm{Ï€}}{a}x\right)+\sin \left(\frac{2\mathrm{Ï€}}{a}x\right)e^{-i3\mathrm{Ó\neg }t}{e}^{i\mathrm{Ï•}}\right]$ .

Substitute in the given function values:

$$\begin{gathered} {\mathrm{ψ}}_1\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{\mathrm{π}}{a}x\right); \\ {\mathrm{ψ}}_2\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{2\mathrm{π}}{a}x\right) \end{gathered}$$

$$\begin{gathered} \mathrm{ψ}\left(x,t\right)=\frac{1}{\sqrt{2}}\sqrt{\frac{2}{a}}\left[\sin \left(\mathrm{Ï€³\ae }/\mathrm{a}\right)e^{-i\mathrm{Ó\neg }t}+e^{i\mathrm{Ï•}}\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)e^{-4i\mathrm{Ó\neg }t}\right] \\ =\frac{1}{a}\left[\begin{array}{c}{\sin }^2\left(\mathrm{Ï€³\ae }/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)+\sin \left(\mathrm{Ï€³\ae }/\mathrm{a}\right).\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)e^{3i\mathrm{Ó\neg }t}{e}^{-i\mathrm{Ï•}}\\ +\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)e^{3i\mathrm{Ó\neg }t}{e}^{-i\mathrm{Ï•}}.\sin \left(\mathrm{Ï€³\ae }/\mathrm{a}\right)\end{array}\right] \\ =\frac{1}{a}\left[\begin{array}{c}{\sin }^2\left(\mathrm{Ï€³\ae }/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)\\ +\sin \left(\mathrm{Ï€³\ae }/\mathrm{a}\right)\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)\left(e^{3i\mathrm{Ó\neg }t}{e}^{-i\mathrm{Ï•}}+e^{-3i\mathrm{Ó\neg }t}{e}^{-i\mathrm{Ï•}}\right)\end{array}\right] \end{gathered}$$

Therefore the value of ${\left|\mathrm{ψ}\left(x,t\right)\right|}^2$is:

$\frac{1}{a}\left[{\sin }^2\left(\frac{\mathrm{Ï€}}{a}x\right)+{\sin }^2\left(\frac{2\mathrm{Ï€}}{a}x\right)+2\sin \left(\frac{\mathrm{Ï€}}{a}x\right)\sin \left(\frac{2\mathrm{Ï€}}{a}x\right)\cos \left(3\mathrm{Ó\neg }t-\mathrm{Ï•}\right)\right]$

Also, the value of$\left(x\right)=\left[1-\frac{32}{9{\mathrm{Ï€}}^2}\cos \left(3\mathrm{Ó\neg }t-\mathrm{Ï•}\right)\right]$ .

This amounts physically to starting the clock at a different time (i.e., shifting the t=0 point).

The energy levels are given by

$E_n=\left(\mathrm{Ä\S }\mathrm{Ó\neg }\right)n^2$

So that

$\mathrm{ψ}\left(x,t\right)=A\left[{\mathrm{ψ}}_1\left(x\right)e^{-i\mathrm{Ó\neg }t}+e^{i\mathrm{Ï•}}{\mathrm{ψ}}_2\left(x\right)e^{-4i\mathrm{Ó\neg }t}\right]$

Now, If $\mathrm{Ï•}=\raisebox{1ex}{$\mathrm{Ï€}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Write,

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

Then $\cos \left(3\mathrm{Ó\neg }t-\mathrm{Ï•}\right)=\sin \left(3\mathrm{Ó\neg }t\right)$

$\left(x\right)$starts at$\frac{a}{2}$

If $\mathrm{Ï•}=\mathrm{Ï€}$, then

$\mathrm{ψ}\left(x,0\right)=A\left[{\mathrm{ψ}}_1\left(x\right)-{\mathrm{ψ}}_2\left(x\right)\right]$

then $\cos \left(3\mathrm{Ó\neg }t-\mathrm{Ï•}\right)=-\cos \left(3\mathrm{Ó\neg }t\right)$;

$\left(x\right)$starts at $\frac{a}{2}\left(1+\frac{32}{9{\mathrm{Ï€°ù}}^2}\right)$

Thus, the value of $\mathrm{ψ}\left(x,t\right),{\left|\mathrm{ψ}\left(x,t\right)\right|}^2$, and $\left(x\right)$are as follows:

$\mathrm{ψ}\left(x,t\right)=\frac{1}{\sqrt{a}}{e}^{-i\mathrm{Ó\neg }t}\left[\sin \left(\mathrm{Ï€³\ae }/\mathrm{a}\right)+\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)e^{-3i\mathrm{Ó\neg }t}{e}^{i\mathrm{Ï•}}\right]$

${\left|\mathrm{ψ}\left(x,t\right)\right|}^2=\frac{1}{a}{\sin }^2\left(\mathrm{Ï€³\ae }/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)+2\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)\sin \left(2\mathrm{Ï€³\ae }/\mathrm{a}\right)\cos \left(3\mathrm{Ó\neg }t/\mathrm{Ï•}\right)$

And

$\left(x\right)=\frac{a}{2}\left[1-\frac{32}{9{\mathrm{Ï€}}^2}\cos \left(3\mathrm{Ó\neg }t-\mathrm{Ï•}\right)\right]$