Chapter 7, Problem 7.6IA

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta x$ and $\Delta p_x$ for the $n\text{th}$ state of a particle in a box of length $L$ have to be evaluated.

Concept Introduction:

The wave function represents the exact position of the electron in an atom.  The wave function is represented by $\psi$ in the quantum mechanics.  Expectation value of an operator is known as the average value of the given operator.

Answer to Problem 7.6IA

The value of $\Delta x$ for the $n\text{th}$ state of a particle in a box of length $L$ is ${\left(\frac{L^2}{12}-\frac{L^2}{2n^2{\pi }^2}\right)}^{1/2}$.

The value of $\Delta p_x$ for the $n\text{th}$ state of a particle in a box of length $L$ is $\frac{\hslash n\pi }{L}$.

Explanation of Solution

The value of $\Delta x$ is calculated by the following formula.

    $\Delta x={\left(\langle x^2\rangle -{\langle x\rangle }^2\right)}^{1/2}$        (1)

The normalized $n\text{th}$ state wavefunction of a particle in a box of length $L$ is shown below.

    $\psi ={\left(\frac{2}{L}\right)}^{1/2}\sin \left(\frac{n\pi x}{L}\right)$

The expectation value of $\langle x^2\rangle$ of a particle in a box of length $x=0$ to $x=L$ is calculated by the following formula.

    $\langle x^2\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}{\psi }^{\ast }{x}^2\psi dx}$        (2)

Where,

$\psi$ is the normalized wave function.

${\psi }^{\ast }$ is the conjugated complex of the normalized wave function.

The wave function is real.  So, ${\psi }^{\ast }$ is equal to $\psi$.

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (1).

    $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)x^2{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)dx}\\ ={\displaystyle \underset{0}{\overset{L}{\int }}\left(2/L\right)x^2{\sin }^2\left(n\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(n\pi x/L\right)dx}\end{array}$        (3)

The value of ${\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\sin }^2\left(kx\right)dx}$ is $\frac{a^3}{6}-\left(\frac{a^2}{4k}-\frac{1}{8k^3}\right)\sin \left(2ka\right)-\frac{a}{4k^2}\cos 2ka$.

Compare the integral ${\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\sin }^2\left(kx\right)dx}$ with ${\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(n\pi x/L\right)dx}$ the value of $a$ is $L$ and the value of $k$ is $\left(\frac{n\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(n\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(n\pi x/L\right)dx}=\left(\begin{array}{c}\frac{L^3}{6}-\left(\frac{L^2}{4\left(n\pi /L\right)}-\frac{1}{8{\left(n\pi /L\right)}^3}\right)\sin \left(2\left(n\pi /L\right)L\right)\\ -\frac{L}{4{\left(n\pi /L\right)}^2}\cos 2\left(n\pi /L\right)L\end{array}\right)\\ =\left(\begin{array}{c}\frac{L^3}{6}-\left(\frac{L^2}{4\left(n\pi /L\right)}-\frac{1}{8{\left(n\pi /L\right)}^3}\right)\sin \left(2n\pi \right)\\ -\frac{L^3}{4n^2{\pi }^2}\cos 2n\pi \end{array}\right)\\ =\left(\begin{array}{c}\frac{L^3}{6}-\left(\frac{L^2}{4\left(n\pi /L\right)}-\frac{1}{8{\left(n\pi /L\right)}^3}\right)\left(0\right)-\\ \frac{L^3}{4n^2{\pi }^2}\left(1\right)\end{array}\right)\left(\begin{array}{l}\because \sin \left(2n\pi \right)=0\text{and}\\ \text{cos}\left(2n\pi \right)=1\end{array}\right)\\ =\frac{L^3}{6}-\frac{L^3}{4n^2{\pi }^2}\end{array}$

Substitute the value of ${\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(n\pi x/L\right)dx}$ in equation (3).

    $\begin{array}{c}\langle x^2\rangle =\left(2/L\right)\left(\frac{L^3}{6}-\frac{L^3}{4n^2{\pi }^2}\right)\\ =\left(2/L\right)\frac{L}{2}\left(\frac{L^2}{3}-\frac{L^2}{2n^2{\pi }^2}\right)\\ =\left(\frac{L^2}{3}-\frac{L^2}{2n^2{\pi }^2}\right)\end{array}$

Therefore, the value of $\langle x^2\rangle$ is $\left(\frac{L^2}{3}-\frac{L^2}{2n^2{\pi }^2}\right)$.

The expectation value of ${\langle x\rangle }^2$ of a particle in a box of length in range of $x=0$ to $x=L$ is calculated by the following formula.

    ${\langle x\rangle }^2={\left({\displaystyle \underset{0}{\overset{L}{\int }}{\psi }^{\ast }x\psi dx}\right)}^2$        (4)

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (1).

    $\begin{array}{c}{\langle x\rangle }^2={\left({\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)x{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)dx}\right)}^2\\ ={\left({\displaystyle \underset{0}{\overset{L}{\int }}\left(2/L\right)x{\sin }^2\left(n\pi x/L\right)dx}\right)}^2\\ ={\left(\left(2/L\right){\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(n\pi x/L\right)dx}\right)}^2\end{array}$        (5)

The value of ${\displaystyle \underset{0}{\overset{a}{\int }}x{\sin }^2\left(kx\right)dx}$ is $\frac{a^2}{4}-\frac{1}{4k}\left(a\sin \left(2ka\right)\right)-\frac{1}{8k^2}\left(\cos 2ka-1\right)$.

Compare the integral ${\displaystyle \underset{0}{\overset{a}{\int }}x{\sin }^2\left(kx\right)dx}$ with ${\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(n\pi x/L\right)dx}$ the value of $a$ is $L$ and the value of $k$ is $\left(\frac{n\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(n\pi x/L\right)dx}=\left(\begin{array}{c}\frac{L^2}{4}-\frac{1}{4\times \frac{n\pi }{L}}\left(L\sin \left(2\times \frac{n\pi }{L}\times L\right)\right)-\\ \frac{1}{8\times {\left(\frac{n\pi }{L}\right)}^2}\left(\cos \left(2\times \frac{n\pi }{L}\times L\right)-1\right)\end{array}\right)\\ =\frac{L^2}{4}-\frac{L}{4n\pi }\left(L\sin \left(2n\pi \right)\right)-\frac{L^2}{8n^2{\pi }^2}\left(\cos \left(2n\pi \right)-1\right)\\ =\frac{L^2}{4}-\frac{L}{4n\pi }\left(0\right)-\frac{L^2}{32{\pi }^2}\left(1-1\right)\left(\because \sin \left(2n\pi \right)=0\text{and}\text{cos}\left(2n\pi \right)=1\right)\\ =\frac{L^2}{4}\end{array}$

Substitute the value of ${\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(n\pi x/L\right)dx}$ in the expectation value of $x$ in equation (5).

    $\begin{array}{c}{\langle x\rangle }^2={\left(\left(2/L\right)\left(\frac{L^2}{4}\right)\right)}^2\\ ={\left(\frac{L}{2}\right)}^2\\ =\frac{L^2}{4}\end{array}$

Therefore, the value of ${\langle x\rangle }^2$ is $\frac{L^2}{4}$.

Substitute the value of $\langle x^2\rangle$ and ${\langle x\rangle }^2$ in equation (1).

    $\begin{array}{c}\Delta x={\left(\left(\frac{L^2}{3}-\frac{L^2}{2n^2{\pi }^2}\right)-\frac{L^2}{4}\right)}^{1/2}\\ ={\left(\frac{L^2}{3}-\frac{L^2}{2n^2{\pi }^2}-\frac{L^2}{4}\right)}^{1/2}\\ ={\left(\frac{L^2}{12}-\frac{L^2}{2n^2{\pi }^2}\right)}^{1/2}\end{array}$

Hence, the value of $\Delta x$ for the ground state of a particle in a box of length $L$ is ${\left(\frac{L^2}{12}-\frac{L^2}{2n^2{\pi }^2}\right)}^{1/2}$.

The value of $\Delta p_x$ is calculated by the following formula.

    $\Delta p_x={\left(\langle p_x{}^2\rangle -{\langle p_x\rangle }^2\right)}^{1/2}$        (6)

The expectation value of the momentum of the electron is calculated by the following expression.

    $\begin{array}{c}\langle p_x{}^2\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)p_x{}^2{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right){\left(\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)}^2\sin \left(n\pi x/L\right)dx}\left(\because p_x=\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)\\ =\frac{2{\hslash }^2}{{\text{i}}^{2}L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}}{\text{d}x}\sin \left(n\pi x/L\right)\right)dx}\left(\because \frac{\text{d}}{\text{d}x}\sin x=\cos x\right)\\ =-\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\left(n\pi /L\right)\frac{\text{d}}{\text{d}x}\cos \left(n\pi x/L\right)dx}\left(\because {\text{i}}^2=-1\right)\end{array}$

On further solving the above integration,

    $\begin{array}{c}\langle p_x{}^2\rangle =-\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\left(n\pi /L\right)\left(-\sin \left(n\pi x/L\right)\right)\frac{\text{d}}{\text{d}x}\left(n\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\left(n\pi /L\right)\sin \left(n\pi x/L\right)\frac{\text{d}}{\text{d}x}\left(n\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right){\left(n\pi /L\right)}^2\sin \left(n\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\left(n\pi /L\right)}^2{\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(n\pi x/L\right)dx}\end{array}$        (7)

The value of ${\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^{2}kxdx}$ is $\frac{1}{2}a-\frac{1}{4k}\sin 2ka$.

Compare the integral ${\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(n\pi x/L\right)dx}$ with ${\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^{2}kxdx}$ the value of $k$ is $\left(\frac{n\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(n\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(n\pi x/L\right)dx}=\frac{1}{2}L-\frac{1}{4\left(\frac{n\pi }{L}\right)}\sin 2\left(\frac{n\pi }{L}\right)L\\ =\frac{L}{2}-\frac{L}{4n\pi }\sin 2n\pi \left(\because \sin 2n\pi =0\right)\\ =\frac{L}{2}\end{array}$

Substitute the value of ${\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(n\pi x/L\right)dx}$ in equation (7).

    $\begin{array}{c}\langle p_x{}^2\rangle =\frac{2{\hslash }^2}{L}{\left(n\pi /L\right)}^2\frac{L}{2}\\ =\frac{{\hslash }^2{n}^2{\pi }^2}{L^2}\end{array}$

Therefore, the value of $\langle p_x{}^2\rangle$ for the normalized wavefunction ${\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)$ is $\frac{{\hslash }^2{n}^2{\pi }^2}{L^2}$.

The expectation value of the square of the momentum of the particle is calculated by the following expression.

    $\begin{array}{c}{\langle p_x\rangle }^2={\left({\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)p_x{\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)dx}\right)}^2\\ ={\left(\frac{2\hslash }{\text{i}L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\frac{\text{d}}{\text{d}x}\sin \left(n\pi x/L\right)dx}\right)}^2\left(\because {\widehat{p}}_x=\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)\\ ={\left(\frac{2\hslash }{\text{i}L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)\times \frac{\text{d}}{\text{d}x}\left(n\pi x/L\right)dx}\right)}^2\left(\because \frac{\text{d}}{\text{d}x}\sin x=\cos x\right)\\ ={\left(\frac{2\hslash }{\text{i}L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)\times \left(n\pi /L\right)dx}\right)}^2\end{array}$

The value of ${\displaystyle \underset{0}{\overset{a}{\int }}\sin \text{A}x\cos \text{B}xdx}$ is $\frac{\cos \left(\text{B}-\text{A}\right)a-1}{2\left(\text{B}-\text{A}\right)}-\frac{\cos \left(\text{A}+\text{B}\right)a-1}{2\left(\text{A}+\text{B}\right)}$.

Compare the integral ${\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)dx}$ with ${\displaystyle \underset{0}{\overset{a}{\int }}\sin \text{A}x\cos \text{B}xdx}$ the value of $a$ is $L$ and the values of $\text{A}$ and $\text{B}$ are $\left(\frac{n\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

$\begin{array}{c}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)dx}=\frac{\cos \left(\left(\left(n\pi /L\right)-\left(n\pi /L\right)\right)L\right)-1}{2\left(\left(n\pi /L\right)-\left(n\pi /L\right)\right)}-\frac{\cos \left(\left(\left(n\pi /L\right)+\left(n\pi /L\right)\right)L\right)-1}{2\left(\left(n\pi /L\right)+\left(n\pi /L\right)\right)}\\ =\frac{\cos \left(0\right)-1}{2\left(\left(n\pi /L\right)-\left(n\pi /L\right)\right)}-\frac{\cos \left(2n\pi \right)-1}{2\left(\left(n\pi /L\right)+\left(n\pi /L\right)\right)}\\ =\frac{1-1}{2\left(\left(n\pi /L\right)-\left(n\pi /L\right)\right)}-\frac{1-1}{2\left(2n\pi /L\right)}\left(\because \cos 0=1\text{and}\cos 2n\pi =1\right)\\ =0\end{array}$

Substitute the value of ${\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(n\pi x/L\right)\cos \left(n\pi x/L\right)dx}$ in equation (2).

    $\begin{array}{c}{\langle p_x\rangle }^2={\left(\frac{2\hslash }{\text{i}L}\times 0\times \left(\pi /L\right)\right)}^2\\ =0\end{array}$

Therefore, the value of ${\langle p_x\rangle }^2$ for the normalized wavefunction ${\left(2/L\right)}^{1/2}\sin \left(n\pi x/L\right)$ is $0$.

Substitute the value of ${\langle p_x\rangle }^2$ and $\langle p_x{}^2\rangle$ in equation (7).

    $\begin{array}{c}\Delta p_x={\left(\frac{{\hslash }^2{n}^2{\pi }^2}{L^2}-0\right)}^{1/2}\\ =\frac{\hslash n\pi }{L}\end{array}$

Hence, the value of $\Delta p_x$ is $\frac{\hslash n\pi }{L}$.

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta x$ and $\Delta p_x$ for the $\nu \text{th}$ state of a harmonic oscillator have to be evaluated.

Concept Introduction:

A system that experiences a restoring force when displaced from its equilibrium position is known as the harmonic oscillator.  The restoring force experienced by the system is directly proportional to the displacement.  Masses connected to the spring and the pendulums are the examples of harmonic oscillator.

Answer to Problem 7.6IA

The value of $\Delta x$ for the $\nu \text{th}$ state of a harmonic oscillator is ${\left(\nu +\frac{1}{2}\right)}^{1/2}{\left(\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}\right)}^{1/2}$.

The value of $\Delta p_x$ for $\nu \text{th}$ state of a simple harmonic oscillator is ${\left(\frac{1}{2}\hslash {\left(k_{\text{f}}m\right)}^{1/2}\right)}^{1/2}$.

Explanation of Solution

The value of $\langle x\rangle$ for the $\nu \text{th}$ state of a harmonic oscillator is zero.  Therefore, the value of ${\langle x\rangle }^2$ for the $\nu \text{th}$ state of a harmonic oscillator is zero.

The value of $\langle x^2\rangle$ for the $\nu \text{th}$ state of a harmonic oscillator is $\left(\nu +\frac{1}{2}\right)\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}$.  The value of $k_{\text{f}}$ represents the force constant and $m$ represents the mass of particle.

Substitute the value of ${\langle x\rangle }^2$ and $\langle x^2\rangle$ in equation (1).

    $\begin{array}{c}\Delta x={\left(\left(\nu +\frac{1}{2}\right)\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}-0\right)}^{1/2}\\ ={\left(\left(\nu +\frac{1}{2}\right)\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}\right)}^{1/2}\\ ={\left(\nu +\frac{1}{2}\right)}^{1/2}{\left(\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}\right)}^{1/2}\end{array}$

Hence, the value of $\Delta x$ for the $\nu \text{th}$ state of a harmonic oscillator is ${\left(\nu +\frac{1}{2}\right)}^{1/2}{\left(\frac{\hslash }{{\left(m{k}_{\text{f}}\right)}^{1/2}}\right)}^{1/2}$.

The value of $\langle p_x\rangle$ for the $\nu \text{th}$ state of a harmonic oscillator is zero.  Therefore, the value of ${\langle p_x\rangle }^2$ for the $\nu \text{th}$ state of a harmonic oscillator is zero.

The value of $\langle p_x^2\rangle$ for the $\nu \text{th}$ state of a harmonic oscillator is $\frac{1}{2}\hslash {\left(k_{\text{f}}m\right)}^{1/2}$.  The value of $k_{\text{f}}$ represents the force constant and $m$ represents the mass of particle.

Substitute the value of ${\langle p_x\rangle }^2$ and $\langle p_x^2\rangle$ in equation (6).

    $\begin{array}{c}\Delta p_x={\left(\frac{1}{2}\hslash {\left(k_{\text{f}}m\right)}^{1/2}-0\right)}^{1/2}\\ ={\left(\frac{1}{2}\hslash {\left(k_{\text{f}}m\right)}^{1/2}\right)}^{1/2}\end{array}$

Hence, the value of $\Delta p_x$ for $\nu \text{th}$ state of a simple harmonic oscillator is ${\left(\frac{1}{2}\hslash {\left(k_{\text{f}}m\right)}^{1/2}\right)}^{1/2}$.