Chapter 11, Problem 11F.7P

Interpretation Introduction

Interpretation:

The transition probability for the transition $n=1\to n=2$ is non-zero while the transition probability for the transition $n=1\to n=3$ is zero has to be shown.

Concept introduction:

An electronic state of energy has its own vibrational states.  The energy between the electronic states is large, followed by vibrational states and then rotational states.  During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Answer to Problem 11F.7P

The transition probability for the transition $n=1\to n=2$ is non-zero while the transition probability for the transition $n=1\to n=3$ is zero has been rightfully shown.

Explanation of Solution

The normalized wavefunction for a one-dimensional box is given below.

    ${\psi }_n=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$

Where,

• $n$ is the quantum number of the energy state.
• $L$ is the length of the box.

The wavefunction for a one-dimensional box with $n=1$ is given below.

    ${\psi }_1=\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)$

The wavefunction for a one-dimensional box with $n=2$ is given below.

    ${\psi }_2=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)$

The expression for transition probability ($\mu$) is given below.

    ${\mu }_q={\displaystyle \int {\psi }_{\text{1}}^{*}\mu {\psi }_2\text{d}\tau }$        (1)

Where

• $\mu$ is the magnitude of the dipole moment.
• ${\psi }_{\text{1}}^{*}$ is the conjugate of ${\psi }_1$.

The wavefunction, ${\psi }_1$ is real. Therefore, the value of ${\psi }_{\text{1}}^{*}$ is equal to ${\psi }_1$.

The value of $\mu$ is equal to $-ex$.

Substitute ${\psi }_2$, $\mu$ and ${\psi }_{\text{1}}^{*}$ in equation (1).

    $\begin{array}{c}{\mu }_q={\displaystyle \underset{0}{\overset{L}{\int }}\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\left(-ex\right)\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\text{d}x}\\ =-\frac{2e}{L}{\displaystyle \underset{0}{\overset{L}{\int }}x\sin \left(\frac{\pi x}{L}\right)\sin \left(\frac{\pi x}{L}\right)\text{d}x}\\ =-\frac{2e}{L}\times \frac{1}{2}{\displaystyle \underset{0}{\overset{L}{\int }}x\left(\cos \left(\frac{\pi x}{L}\right)-\cos \left(\frac{3\pi x}{L}\right)\right)\text{d}x}\text{as}\text{sinAsinB=}\frac{1}{2}\left[\cos \left(\text{A}-\text{B}\right)-\cos \left(\text{A+B}\right)\right]\\ =-\frac{2e}{L}\times \frac{1}{2}\left({\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{\pi x}{L}\right)\text{d}x-{\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{3\pi x}{L}\right)\text{d}x}}\right)\end{array}$

The equation given above is solved further as shown below.

    $\begin{array}{c}{\mu }_q=-\frac{2e}{L}\times \frac{1}{2}\left({\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{\pi x}{L}\right)\text{d}x-{\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{3\pi x}{L}\right)\text{d}x}}\right)\\ =-\frac{e}{L}\left[\begin{array}{l}\left(\frac{L}{\pi }{\left[x\sin \left(\frac{\pi x}{L}\right)\right]}_0^L-\frac{L}{\pi }{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\frac{\pi x}{L}\right)\text{d}x}\right)\\ -\left(\frac{L}{3\pi }{\left[x\sin \left(\frac{3\pi x}{L}\right)\right]}_0^L-\frac{L}{3\pi }{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\frac{3\pi x}{L}\right)\text{d}x}\right)\end{array}\right]\\ =-\frac{e}{L}\left[\left(0+\frac{L^2}{{\pi }^2}{\left[\cos \left(\frac{\pi x}{L}\right)\right]}_0^L\right)-\left(0+\frac{L^2}{9{\pi }^2}{\left[\cos \left(\frac{3\pi x}{L}\right)\right]}_0^L\right)\right]\\ =-\frac{e}{L}\left[\left(\frac{L^2}{{\pi }^2}\left(\cos \pi -\cos 0\right)\right)-\frac{L^2}{9{\pi }^2}\left(\cos 3\pi -\cos 0\right)\right]\end{array}$        (2)

The equation (2) was solved further as shown below.

    $\begin{array}{c}{\mu }_q=-\frac{e}{L}\left[\left(\frac{L^2}{{\pi }^2}\left(\cos \pi -\cos 0\right)\right)-\frac{L^2}{9{\pi }^2}\left(\cos 3\pi -\cos 0\right)\right]\\ =-\frac{e}{L}\times \frac{L^2}{{\pi }^2}\left[\left(\cos \pi -\cos 0\right)-\frac{1}{9}\left(\cos 3\pi -\cos 0\right)\right]\\ =-\frac{eL}{{\pi }^2}\left[\left(-1-1\right)-\frac{1}{9}\left(-1-1\right)\right]\\ =\frac{16eL}{{\pi }^2}\end{array}$

Therefore, the transition probability for the transition $n=1\to n=2$ is non-zero.

The wavefunction for a one-dimensional box with $n=3$ is given below.

    ${\psi }_2=\sqrt{\frac{2}{L}}\sin \left(\frac{3\pi x}{L}\right)$

The expression for transition probability ($\mu$) is given below.

    ${\mu }_q={\displaystyle \int {\psi }_{\text{1}}^{*}\mu {\psi }_3\text{d}\tau }$        (1)

Where

• $\mu$ is the magnitude of the dipole moment.
• ${\psi }_{\text{1}}^{*}$ is the conjugate of ${\psi }_1$.

The wavefunction, ${\psi }_1$ is real. Therefore, the value of ${\psi }_1$ is equal to ${\psi }_{\text{1}}^{*}$.

The value of $\mu$ is equal to $-ex$.

Substitute ${\psi }_3$, $\mu$ and ${\psi }_{\text{1}}^{*}$ in equation (1).

    $\begin{array}{c}{\mu }_q={\displaystyle \underset{0}{\overset{L}{\int }}\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right)\left(-ex\right)\sqrt{\frac{2}{L}}\sin \left(\frac{3\pi x}{L}\right)\text{d}x}\\ =-\frac{2e}{L}{\displaystyle \underset{0}{\overset{L}{\int }}x\sin \left(\frac{\pi x}{L}\right)\sin \left(\frac{3\pi x}{L}\right)\text{d}x}\\ =-\frac{2e}{L}\times \frac{1}{2}{\displaystyle \underset{0}{\overset{L}{\int }}x\left(\cos \left(\frac{2\pi x}{L}\right)-\cos \left(\frac{4\pi x}{L}\right)\right)\text{d}x}\text{as}\text{sinAsinB=}\frac{1}{2}\left[\cos \left(\text{A}-\text{B}\right)-\cos \left(\text{A+B}\right)\right]\\ =-\frac{2e}{L}\times \frac{1}{2}\left({\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{2\pi x}{L}\right)\text{d}x-{\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{4\pi x}{L}\right)\text{d}x}}\right)\end{array}$

The equation given above is solved further as shown below.

    $\begin{array}{c}{\mu }_q=-\frac{2e}{L}\times \frac{1}{2}\left({\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{2\pi x}{L}\right)\text{d}x-{\displaystyle \underset{0}{\overset{L}{\int }}x\cos \left(\frac{4\pi x}{L}\right)\text{d}x}}\right)\\ =-\frac{e}{L}\left[\begin{array}{l}\left(\frac{L}{2\pi }{\left[x\sin \left(\frac{2\pi x}{L}\right)\right]}_0^L-\frac{L}{2\pi }{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\frac{2\pi x}{L}\right)\text{d}x}\right)\\ -\left(\frac{L}{4\pi }{\left[x\sin \left(\frac{4\pi x}{L}\right)\right]}_0^L-\frac{L}{4\pi }{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\frac{4\pi x}{L}\right)\text{d}x}\right)\end{array}\right]\\ =-\frac{e}{L}\left[\left(0+\frac{L^2}{4{\pi }^2}{\left[\cos \left(\frac{2\pi x}{L}\right)\right]}_0^L\right)-\left(0+\frac{L^2}{16{\pi }^2}{\left[\cos \left(\frac{4\pi x}{L}\right)\right]}_0^L\right)\right]\\ =-\frac{e}{L}\left[\left(\frac{L^2}{4{\pi }^2}\left(\cos 2\pi -\cos 0\right)\right)-\frac{L^2}{16{\pi }^2}\left(\cos 4\pi -\cos 0\right)\right]\end{array}$        (3)

The equation (3) was solved further as shown below.

    $\begin{array}{c}{\mu }_q=-\frac{e}{L}\left[\left(\frac{L^2}{4{\pi }^2}\left(\cos 2\pi -\cos 0\right)\right)-\frac{L^2}{16{\pi }^2}\left(\cos 4\pi -\cos 0\right)\right]\\ =-\frac{e}{L}\times \frac{L^2}{4{\pi }^2}\left[\left(\cos 2\pi -\cos 0\right)-\frac{1}{4}\left(\cos 4\pi -\cos 0\right)\right]\\ =-\frac{eL}{{\pi }^2}\left[\left(0-0\right)-\frac{1}{4}\left(0-0\right)\right]\\ =0\end{array}$

Therefore, the transition probability for the transition $n=1\to n=3$ is zero.