Chapter 7, Problem 7.5IA

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta x={\left(\langle x^2\rangle -{\langle x\rangle }^2\right)}^{1/2}$ and $\Delta p_x={\left(\langle p_x{}^2\rangle -{\langle p_x\rangle }^2\right)}^{1/2}$ for the ground state of a particle in a box of length $L$ has to be evaluated.  The quantities are to be discussed with reference to the principle of uncertainty.

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 is known as the average value of the given operator.

Answer to Problem 7.5IA

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

The value of $\Delta p_x={\left(\langle p_x{}^2\rangle -{\langle p_x\rangle }^2\right)}^{1/2}$ for the ground state of a particle in a box of length $L$ is $\frac{\hslash \pi }{L}$.

The value of $\Delta x\times \Delta p_x$ is greater than $\frac{\hslash }{2}$, this proves the principle of uncertainty.

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 ground state wavefunction of a particle in a box of length $L$ is shown below.

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

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).

    $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)x^2{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\\ ={\displaystyle \underset{0}{\overset{L}{\int }}\left(2/L\right)x^2{\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(\pi x/L\right)dx}\end{array}$                                                (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 ${\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\sin }^2\left(\pi x/L\right)dx}$ the value of $a$ is $L$ and the value of $k$ is $\left(\frac{\pi }{L}\right)$.  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.

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

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

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

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

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).

    $\begin{array}{c}{\langle }^x={\left({\displaystyle \underset{0}{\overset{L}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)x{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\right)}^2\\ ={\left({\displaystyle \underset{0}{\overset{L}{\int }}\left(2/L\right)x{\sin }^2\left(\pi x/L\right)dx}\right)}^2\\ ={\left(\left(2/L\right){\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(\pi x/L\right)dx}\right)}^2\end{array}$                                             (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 ${\displaystyle \underset{0}{\overset{L}{\int }}x{\sin }^2\left(\pi x/L\right)dx}$ the value of $a$ is $L$ and the value of $k$ is $\left(\frac{\pi }{L}\right)$.  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.

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

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

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

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).

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

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

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.

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

On further solving the above integration,

    $\begin{array}{c}\langle p_x{}^2\rangle =-\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\pi x/L\right)\left(\pi /L\right)\left(-\sin \left(\pi x/L\right)\right)\frac{\text{d}}{\text{d}x}\left(\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\pi x/L\right)\left(\pi /L\right)\sin \left(\pi x/L\right)\frac{\text{d}}{\text{d}x}\left(\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\pi x/L\right){\left(\pi /L\right)}^2\sin \left(\pi x/L\right)dx}\\ =\frac{2{\hslash }^2}{L}{\left(\pi /L\right)}^2{\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(\pi x/L\right)dx}\end{array}$                                 (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 ${\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(\pi x/L\right)dx}$ with ${\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^{2}kxdx}$ the value of $k$ is $\left(\frac{\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}{\sin }^2\left(\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

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

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

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

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

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

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

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 ${\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\pi x/L\right)\cos \left(\pi x/L\right)dx}$ 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 $\left(\frac{\pi }{L}\right)$.  Therefore, the value of ${\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\pi x/L\right)\cos \left(\pi x/L\right)dx}$, using the above integral identity is calculated as shown below.

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

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

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

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

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

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

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

The proof of uncertainty principle is shown below.

    $\Delta x\times \Delta p_x={\left(\frac{L^2}{12}-\frac{L^2}{2{\pi }^2}\right)}^{1/2}\times \frac{\hslash \pi }{L}$                                                                                 (8)

The principle of uncertainty states that the product of $\Delta x\times \Delta p_x$ is greater than or equal to $\frac{\hslash }{2}$.  From equation (8) it is clear that the value of $\Delta x\times \Delta p_x$ is greater than $\frac{\hslash }{2}$ because $L$ will be positive.  This proves the principle of uncertainty.

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta x={\left(\langle x^2\rangle -{\langle x\rangle }^2\right)}^{1/2}$ and $\Delta p_x={\left(\langle p_x{}^2\rangle -{\langle p_x\rangle }^2\right)}^{1/2}$ for the ground state of a harmonic oscillator has to be evaluated.  The quantities are to be discussed with reference to the principle of uncertainty.

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 7.5IA

The value of $\Delta x={\left(\langle x^2\rangle -{\langle x\rangle }^2\right)}^{1/2}$ for the harmonic oscillator is ${\left(\frac{\sqrt{\alpha }}{2}\right)}^{1/2}$.

The value of $\Delta p_x$ for simple harmonic oscillator is ${\left({\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)\right)}^{1/2}$.

The value of $\Delta x\times \Delta p_x$ is a positive quantity this proves the principle of uncertainty.

Explanation of Solution

The normalized ground state wavefunction of a harmonic oscillator is shown below.

    $\psi ={\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}$

The expectation value of $\langle x^2\rangle$ of a harmonic oscillator is calculated by the following formula.

    $\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\psi }^{\ast }{x}^2\psi dx}$                                                                                                     (9)

Where,

• $\psi$ is the normalized ground state wave function for harmonic oscillator.
• ${\psi }^{\ast }$ 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 (9).

    $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}{x}^2{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}dx}\\ ={\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}{x}^2{\text{e}}^{-x^2/2}dx}\\ ={\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}\end{array}$                                                                    (10)

The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ is $\frac{\overline{)\frac{n+1}{2}}}{a^{\left(\frac{n+1}{2}\right)}}$ if $n$ is even.

Compare the integral ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ with ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ the value of $a$ is $1$ and the value of $n$ is $2$.  Therefore, the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$, using the above integral identity is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}=\frac{\overline{)\frac{2+1}{2}}}{1^{\left(\frac{2+1}{2}\right)}}\\ =\overline{)\frac{3}{2}}\\ =\frac{\sqrt{\pi }}{2}\end{array}$

Substitute the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ in equation (10).

    $\begin{array}{c}\langle x^2\rangle ={\left(\frac{\alpha }{\pi }\right)}^{1/2}\frac{\sqrt{\pi }}{2}\\ =\frac{\sqrt{\alpha }}{2}\end{array}$

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

The expectation value of ${\langle x\rangle }^2$ of a harmonic oscillator is calculated by the following formula.

    ${\langle x\rangle }^2={\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\psi }^{\ast }{x}^2\psi dx}\right)}^2$                                                                                              (11)

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

    $\begin{array}{c}{\langle }^x={\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}x{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}dx}\right)}^2\\ ={\left({\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}x{\text{e}}^{-x^2/2}dx}\right)}^2\\ ={\left({\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{\text{e}}^{-x^2}dx}\right)}^2\end{array}$                                                                 (12)

The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ is $0$ if $n$ is odd.

Compare the integral ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ with ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ the value of $a$ is $1$ and the value of $n$ is $1$. 

Substitute the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ in equation (12).

    $\begin{array}{c}{\langle }^x={\left({\left(\frac{\alpha }{\pi }\right)}^{1/2}\times 0\right)}^2\\ =0\end{array}$

Therefore, the value of ${\langle x\rangle }^2$ is $0$.

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

    $\begin{array}{c}\Delta x={\left(\frac{\sqrt{\alpha }}{2}-0\right)}^{1/2}\\ ={\left(\frac{\sqrt{\alpha }}{2}\right)}^{1/2}\end{array}$

Hence, the value of $\Delta x={\left(\langle x^2\rangle -{\langle x\rangle }^2\right)}^{1/2}$ for the harmonic oscillator is ${\left(\frac{\sqrt{\alpha }}{2}\right)}^{1/2}$.

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

    $\begin{array}{c}\langle p_x{}^2\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}{p}_x{}^2{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}dx}\\ ={\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}{\left(\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)}^2{\text{e}}^{-x^2/2}dx}\left(\because p_x=\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)\\ ={\left(\frac{\hslash }{i}\right)}^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}}{\text{d}x}{\text{e}}^{-x^2/2}\right)dx}\left(\because \frac{\text{d}}{\text{d}x}\sin x=\cos x\right)\\ =-{\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left({\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left(-x^2/2\right)\right)dx}\left(\because {\text{i}}^2=-1\right)\end{array}$

On further solving the above integration,

    $\begin{array}{c}\langle p_x{}^2\rangle =-{\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left({\text{e}}^{-x^2/2}\left(-x\right)\right)dx}\\ ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left({\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left(x\right)+x\frac{\text{d}}{\text{d}x}\left({\text{e}}^{-x^2/2}\right)\right)dx}\\ ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left({\text{e}}^{-x^2/2}+x{\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left(-x^2/2\right)\right)dx}\\ ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left({\text{e}}^{-x^2/2}+x{\text{e}}^{-x^2/2}\left(-x\right)\right)dx}\end{array}$                                      (13)

Separate the integrals of equation (13) as shown below.

    $\langle p_x{}^2\rangle ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2}dx}-{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}\right)$                                                               (14)

The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ is $\frac{\overline{)\frac{n+1}{2}}}{a^{\left(\frac{n+1}{2}\right)}}$ if $n$ is even.

Compare the integral ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ with ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ the value of $a$ is $1$ and the value of $n$ is $2$.  The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}=\frac{\overline{)\frac{2+1}{2}}}{1^{\left(\frac{2+1}{2}\right)}}\\ =\overline{)\frac{3}{2}}\\ =\frac{\sqrt{\pi }}{2}\end{array}$

Therefore, the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ is $\frac{\sqrt{\pi }}{2}$.

Compare the integral ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^0{\text{e}}^{-x^2}dx}$ with ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ the value of $a$ is $1$ and the value of $n$ is $0$.  The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^0{\text{e}}^{-x^2}dx}$ is calculated as shown below.

    $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^0{\text{e}}^{-x^2}dx}=\frac{\overline{)\frac{0+1}{2}}}{1^{\left(\frac{0+1}{2}\right)}}\\ =\overline{)\frac{1}{2}}\\ =\sqrt{\pi }\end{array}$

Therefore, the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^0{\text{e}}^{-x^2}dx}$ is $\sqrt{\pi }$.

Substitute the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^0{\text{e}}^{-x^2}dx}$ and ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{\text{e}}^{-x^2}dx}$ in equation (14).

    $\begin{array}{c}\langle p_x{}^2\rangle ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\sqrt{\pi }-\frac{\sqrt{\pi }}{2}\right)\\ ={\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)\end{array}$

Therefore, the value of $\langle p_x{}^2\rangle$ for the ground state harmonic oscillator is ${\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)$.

The value of ${\langle p_x\rangle }^2$ is calculated as shown below.

    $\begin{array}{c}{\langle }^{p_x}={\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}{p}_x{\left(\frac{\alpha }{\pi }\right)}^{1/4}{\text{e}}^{-x^2/2}dx}\right)}^2\\ ={\left({\left(\frac{\alpha }{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left(\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right){\text{e}}^{-x^2/2}dx}\right)}^2\left(\because p_x=\frac{\hslash }{i}\frac{\text{d}}{\text{d}x}\right)\\ ={\left(\frac{\hslash }{i}\right)}^2\left(\frac{\alpha }{\pi }\right){\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left(\frac{\text{d}}{\text{d}x}{\text{e}}^{-x^2/2}\right)dx}\right)}^2\left(\because \frac{\text{d}}{\text{d}x}\sin x=\cos x\right)\\ =-{\hslash }^2\left(\frac{\alpha }{\pi }\right){\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2/2}\left({\text{e}}^{-x^2/2}\frac{\text{d}}{\text{d}x}\left(-x^2/2\right)\right)dx}\right)}^2\left(\because {\text{i}}^2=-1\right)\end{array}$

On further solving the above integration as shown below,

    $\begin{array}{c}{\langle }^{p_x}=-{\hslash }^2\left(\frac{\alpha }{\pi }\right){\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2}\left(-x\right)dx}\right)}^2\\ =-{\hslash }^2\left(\frac{\alpha }{\pi }\right){\left({\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\text{e}}^{-x^2}xdx}\right)}^2\end{array}$                                                                     (15)

The value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ is $0$ if $n$ is odd.

Compare the integral ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{\text{e}}^{-x^2}dx}$ with ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^n{\text{e}}^{-a{x}^2}dx}$ the value of $a$ is $1$ and the value of $n$ is $1$.  Therefore, the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{\text{e}}^{-x^2}dx}$ is $0$.

Substitute the value of ${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{\text{e}}^{-x^2}dx}$ in equation (15).

    $\begin{array}{c}{\langle }^{p_x}=-{\hslash }^2\left(\frac{\alpha }{\pi }\right){\left(0\right)}^2\\ =0\end{array}$

Therefore, the value of ${\langle p_x\rangle }^2$ for the harmonic oscillator is $0$.

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

    $\begin{array}{c}\Delta p_x={\left({\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)-0\right)}^{1/2}\\ ={\left({\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)\right)}^{1/2}\\ =\hslash {\left(\frac{\alpha }{\pi }\right)}^{1/4}{\left(\frac{\sqrt{\pi }}{2}\right)}^{1/2}\end{array}$

Hence, the value of $\Delta p_x$ for simple harmonic oscillator is ${\left({\hslash }^2{\left(\frac{\alpha }{\pi }\right)}^{1/2}\left(\frac{\sqrt{\pi }}{2}\right)\right)}^{1/2}$.

The proof of uncertainty principle is shown below.

    $\begin{array}{c}\Delta x\times \Delta p_x={\left(\frac{\sqrt{\alpha }}{2}\right)}^{1/2}\times \hslash {\left(\frac{\alpha }{\pi }\right)}^{1/4}{\left(\frac{\sqrt{\pi }}{2}\right)}^{1/2}\\ =\frac{\hslash \sqrt{\alpha }}{{\left(2\right)}^{1/4}}\end{array}$                                                             (16)

The principle of uncertainty states that the product of $\Delta x\times \Delta p_x$ is greater than or equal to $\frac{\hslash }{2}$.  From equation (16) it is clear that the value of $\Delta x\times \Delta p_x$ is a positive quantity.  This means that there is some uncertainty in the measurement of change in position and change in momentum simultaneously.  This proves the principle of uncertainty.