Question

Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of ${\mathit{\psi }}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{\psi }}_{\mathbf{2}}$in problem 2.5:$\mathit{\psi }\left(x,0\right)\mathbf{=}\mathit{A}\left[{\psi }_1\left(x\right)+e^{i\varphi }{\psi }_2\left(x\right)\right]$Where $\mathit{\varphi }$is some constant. Find $\mathit{\psi }\left(x,t\right)\mathbf{,}\mathbf{}{\left|\psi \left(x,t\right)\right|}^{\mathbf{2}}$, and $\left(x\right)$, and compare your results with what you got before. Study the special cases $\mathit{\varphi }\mathbf{=}\raisebox{1ex}{$\mathbf{\pi }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{}\mathbf{and}\mathbf{}\mathit{\varphi }\mathbf{=}\mathbf{\pi }$.

Short answer

The value of $\psi \left(x,t\right),{\left|\psi \left(x,t\right)\right|}^2$and $\left(x\right)$are:

$$\begin{gathered} \psi \left(x,t\right)=\frac{1}{\sqrt{a}}{e}^{-i\omega t}\left[\sin \left(\mathrm{\pi x}/\mathrm{a}\right)+\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)e^{-3i\omega t}{e}^{i\varphi }\right] \\ {\left|\psi \left(x,t\right)\right|}^2=\frac{1}{a}\left[{\sin }^2\left(\mathrm{\pi x}/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{\pi x}/\mathrm{a}\right)+2\sin \left(\mathrm{\pi x}/\mathrm{a}\right)\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)\cos \left(3\omega t-\varphi \right)\right] \\ \left(x\right)=\frac{a}{2}\left[1-\frac{32}{9{\mathrm{\pi }}^2}\cos \left(3\omega t-\varphi \right)\right] \end{gathered}$$

Step-by-step solution

The wave function for any subsequent time t

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

$\mathit{\psi }\left(x,t\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{c}}_{\mathbf{n}}{\mathit{\psi }}_{\mathbf{n}}\left(x\right){\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{E}}_{\mathbf{n}}\mathbf{t}\mathbf{/}\mathbf{h}}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{c}}_{\mathbf{n}}{\mathit{\psi }}_{\mathbf{n}}\left(x,t\right)$

Normalize the value of A.

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

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

was found to be

$\psi \left(x,t\right)=\sqrt{\frac{2}{a}}\sum _{n=1}^{\infty }{B}_{n}exp\left(-i\frac{ħ{\mathrm{\pi }}^2{\mathrm{n}}^2}{2m{a}^2}t\right)\sin \frac{n\mathrm{\pi x}}{a},0\le x\le a$

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

$$\begin{gathered} \psi \left(x,0\right)=A\left[{\psi }_1\left(x\right)+e^{i\varphi }{\psi }_2\left(x\right)\right] \\ =A\left(\sqrt{\frac{2}{a}}\sin \frac{\mathrm{\pi x}}{a}+\sqrt{\frac{2}{a}{e}^{i\varphi }}\sin \frac{2\mathrm{\pi x}}{a}\right) \\ =A\sqrt{\frac{2}{a}}\left(\sin \frac{\mathrm{\pi x}}{a}+e^{i\varphi }\sin \frac{2\mathrm{\pi x}}{a}\right) \end{gathered}$$

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

$1={\int }_{-\infty }^{\infty }{\left|\psi \left(x,0\right)\right|}^{2}dx$

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:

$$\begin{gathered} \psi \left(x,0\right)=\sqrt{\frac{2}{a}}\sum _{n=1}^{\infty }{B}_n\sin \frac{n\mathrm{\pi x}}{a} \\ =\sqrt{\frac{2}{a}}{B}_1\sin \frac{\mathrm{\pi x}}{a}+\sqrt{\frac{2}{a}}{B}_2\sin \frac{2\mathrm{\pi x}}{a}+\sqrt{\frac{2}{a}}{B}_3\sin \frac{3\mathrm{\pi x}}{a}+... \end{gathered}$$

Compare the coefficients,

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

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

Use $\omega ={\mathrm{\pi }}^2\mathrm{ħ}/2{\mathrm{ma}}^2$to simplify the result,

$\psi \left(x,t\right)=\frac{1}{\sqrt{a}}{e}^{-i\omega t}\sin \left(\frac{\mathrm{\pi x}}{a}\right)+\frac{1}{\sqrt{a}}{e}^{i\varphi }{e}^{-4i\omega t}\sin \left(\frac{2\mathrm{\pi x}}{a}\right),0\le x\le a$

Writing the solution in terms of the eigenstates,

$$\begin{gathered} \omega \left(x,t\right)=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{2}{a}}\sin \frac{\mathrm{\pi x}}{a}\right)e^{-i\omega t}+\frac{e^{i\varphi }}{\sqrt{2}}\left(\sqrt{\frac{2}{a}}\sin \frac{2\mathrm{\pi x}}{a}\right)e^{-4i\omega t} \\ =\frac{1}{\sqrt{2}}{\omega }_1\left(x\right)e^{-i\omega t}+\frac{e^{i\varphi }}{\sqrt{2}}{\psi }_2\left(x\right)e^{-4i\omega t} \end{gathered}$$

Therefore the value is $\psi \left(x,t\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{a}$}\right.e^{-i\omega t}\left[\sin \left(\frac{\mathrm{\pi }}{a}x\right)+\sin \left(\frac{2\mathrm{\pi }}{a}x\right)e^{-i3\omega t}{e}^{i\varphi }\right]$ .

Substitute in the given function values:

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

$$\begin{gathered} \psi \left(x,t\right)=\frac{1}{\sqrt{2}}\sqrt{\frac{2}{a}}\left[\sin \left(\mathrm{\pi x}/\mathrm{a}\right)e^{-i\omega t}+e^{i\varphi }\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)e^{-4i\omega t}\right] \\ =\frac{1}{a}\left[\begin{array}{c}{\sin }^2\left(\mathrm{\pi x}/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{\pi x}/\mathrm{a}\right)+\sin \left(\mathrm{\pi x}/\mathrm{a}\right).\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)e^{3i\omega t}{e}^{-i\varphi }\\ +\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)e^{3i\omega t}{e}^{-i\varphi }.\sin \left(\mathrm{\pi x}/\mathrm{a}\right)\end{array}\right] \\ =\frac{1}{a}\left[\begin{array}{c}{\sin }^2\left(\mathrm{\pi x}/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{\pi x}/\mathrm{a}\right)\\ +\sin \left(\mathrm{\pi x}/\mathrm{a}\right)\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)\left(e^{3i\omega t}{e}^{-i\varphi }+e^{-3i\omega t}{e}^{-i\varphi }\right)\end{array}\right] \end{gathered}$$

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

$\frac{1}{a}\left[{\sin }^2\left(\frac{\mathrm{\pi }}{a}x\right)+{\sin }^2\left(\frac{2\mathrm{\pi }}{a}x\right)+2\sin \left(\frac{\mathrm{\pi }}{a}x\right)\sin \left(\frac{2\mathrm{\pi }}{a}x\right)\cos \left(3\omega t-\varphi \right)\right]$

Also, the value of$\left(x\right)=\left[1-\frac{32}{9{\mathrm{\pi }}^2}\cos \left(3\omega t-\varphi \right)\right]$ .

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

Calculate the values by assigning ϕ values.

The energy levels are given by

$E_n=\left(ħ\omega \right)n^2$

So that

$\psi \left(x,t\right)=A\left[{\psi }_1\left(x\right)e^{-i\omega t}+e^{i\varphi }{\psi }_2\left(x\right)e^{-4i\omega t}\right]$

Now, If $\varphi =\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Write,

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

Then $\cos \left(3\omega t-\varphi \right)=\sin \left(3\omega t\right)$

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

If $\varphi =\mathrm{\pi }$, then

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

then $\cos \left(3\omega t-\varphi \right)=-\cos \left(3\omega t\right)$;

$\left(x\right)$starts at $\frac{a}{2}\left(1+\frac{32}{9{\mathrm{\pi r}}^2}\right)$

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

$\psi \left(x,t\right)=\frac{1}{\sqrt{a}}{e}^{-i\omega t}\left[\sin \left(\mathrm{\pi x}/\mathrm{a}\right)+\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)e^{-3i\omega t}{e}^{i\varphi }\right]$

${\left|\psi \left(x,t\right)\right|}^2=\frac{1}{a}{\sin }^2\left(\mathrm{\pi x}/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{\pi x}/\mathrm{a}\right)+{\sin }^2\left(2\mathrm{\pi x}/\mathrm{a}\right)+2\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)\sin \left(2\mathrm{\pi x}/\mathrm{a}\right)\cos \left(3\omega t/\mathrm{\varphi }\right)$

And

$\left(x\right)=\frac{a}{2}\left[1-\frac{32}{9{\mathrm{\pi }}^2}\cos \left(3\omega t-\varphi \right)\right]$