Chapter 7, Problem 7C.11P

(a)

Interpretation Introduction

Interpretation:

The expectation values $\langle x\rangle$, $\langle x^2\rangle$, $\langle p_x\rangle$, and $\langle p_x^2\rangle$ have to be determined.

Concept introduction:

The expectation value is the average value of a large number of measurements of an observable.  The expectation value of a measurable parameter is calculated by the formula shown below.

  $\langle \Omega \rangle ={\displaystyle \int {\psi }^{*}\widehat{\left(\Omega \right)}\psi \text{d}\tau }$

Where,

• $\psi$ is the wave function.
• $\Omega$ is an observable.

Answer to Problem 7C.11P

The value of $\langle x\rangle$ is $\underset{\_}{0}$.  The value of $\langle x^2\rangle$ is $\frac{1}{4a}$.  The value of $\langle p_x\rangle$ is $\underset{\_}{0}$.  The value of $\langle p_x^2\rangle$ is $a{\hslash }^2$.

Explanation of Solution

The value of $\psi$ is given below.

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

Where,

• $a$ is a constant.

The expectation value of $x$ for a particle is given by the expression.

    $\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }^{*}\widehat{\left(x\right)}\psi \text{d}x}$        (2)

Substitute the value of $\psi$ in equation (2).

    $\begin{array}{c}\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)}^{*}\widehat{\left(x\right)}\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(e^{-2a{x}^2}\right)x\text{d}x}\end{array}$        (3)

As the integral in equation (3) is odd, therefore, the value of the integrand will be zero.

    $\begin{array}{c}\langle x\rangle ={\left(\frac{2a}{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(e^{-2a{x}^2}\right)x\text{d}x}\\ =\underset{\_}{0}\end{array}$

Therefore, the expectation value of $\langle x\rangle$ is $\underset{\_}{0}$.

The expectation value of $x^2$ for a particle is given by the expression.

    $\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }^{*}\widehat{\left(x^2\right)}\psi \text{d}x}$        (4)

Substitute the value of $\psi$ in equation (4).

    $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)}^{*}\widehat{\left(x^2\right)}\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(e^{-2a{x}^2}\right)x^2\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \frac{1}{2}{\left(\frac{\pi }{{\left(2a\right)}^3}\right)}^{1/2}\left(\because {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{e}^{-k{x}^2}\text{d}x=\frac{1}{2}}{\left(\frac{\pi }{k^3}\right)}^{1/2}\right)\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \frac{1}{2a}\times {\left(\frac{2\pi }{4a}\right)}^{1/2}\end{array}$        (5)

Equation (5) can be solved further to determine the value of $\langle x^2\rangle$ as shown below.

    $\begin{array}{c}\langle x^2\rangle ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \frac{1}{2a}\times {\left(\frac{2\pi }{4a}\right)}^{1/2}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \frac{1}{8a}\times {\left(\frac{2\pi }{a}\right)}^{1/2}\\ =\frac{1}{4a}\end{array}$

Therefore, the value of $\langle x^2\rangle$ is $\frac{1}{4a}$.

The momentum operator, $\widehat{p_x}$ is given by the expression.

    $\widehat{p_x}=\frac{\hslash }{\text{i}}\frac{\text{d}}{\text{d}x}$        (6)

Where,

• $\hslash$ is a constant.

The expectation value of $\widehat{p_x}$ for a particle is given by the expression.

    $\langle p_x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }^{*}\left(\widehat{p_x}\right)\psi \text{d}x}$        (7)

Substitute the values of $\psi$ and equation (6) in equation (7).

    $\begin{array}{c}\langle p_x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)}^{*}\left(\frac{\hslash }{\text{i}}\frac{\text{d}}{\text{d}x}\right)\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(\frac{h}{\text{i}}\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(e^{-a{x}^2}\right)\left(\frac{\text{d}}{\text{d}x}\right)\left(e^{-a{x}^2}\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(\frac{h}{\text{i}}\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(e^{-a{x}^2}\right)\left(-2ax\right)\left(e^{-a{x}^2}\right)\text{d}x}\\ =\left(-2a\right)\times {\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(\frac{h}{\text{i}}\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(e^{-2a{x}^2}\right)\text{d}x}\end{array}$        (8)

As the integral in equation (8) is odd, therefore, the value of the integrand will be zero.

    $\begin{array}{c}\langle p_x\rangle =\left(-2a\right)\times {\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(\frac{h}{\text{i}}\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(e^{-2a{x}^2}\right)\text{d}x}\\ =\underset{\_}{0}\end{array}$

Therefore, the expectation value of $\langle p_x\rangle$ is $\underset{\_}{0}$.

The momentum operator, $\widehat{p_x^2}$ is given by the expression.

    $\widehat{p_x^2}=-{\hslash }^2\frac{{\text{d}}^2}{\text{d}{x}^2}$        (9)

The expectation value of $\widehat{p_x^2}$ for a particle is given by the expression.

    $\langle p_x^2\rangle ={\displaystyle \underset{0}{\overset{\text{L}}{\int }}{\psi }^{*}\left(\widehat{p_x^2}\right)\psi \text{d}x}$        (10)

Substitute the value of $\psi$ and equation (9) in equation (10).

    $\begin{array}{c}\langle p_x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)\left(-{\hslash }^2\frac{{\text{d}}^2}{\text{d}{x}^2}\right)\left({\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(-{\hslash }^2\right)\times {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2}\frac{\text{d}}{\text{d}x}\left(e^{-a{x}^2}\left(-2ax\right)\right)\text{d}x}\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times {\hslash }^2\times \left(2a\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2}\left(e^{-a{x}^2}+\left(-2a{x}^2{e}^{-a{x}^2}\right)\right)}\text{d}x\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times {\hslash }^2\times \left(2a\right)\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2}{e}^{-a{x}^2}\text{d}x-{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}2a{x}^2{e}^{-a{x}^2}{e}^{-a{x}^2}\text{d}x}}\right)\end{array}$        (11)

The equation (11) can be solved further as shown below.

  $\begin{array}{c}\langle p_x^2\rangle ={\left(\frac{2a}{\pi }\right)}^{1/2}\times \left(-{\hslash }^2\right)\times \left(-2a\right)\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2}{e}^{-a{x}^2}\text{d}x-2a{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{e}^{-2a{x}^2}\text{d}x}}\right)\\ ={\left(\frac{2a}{\pi }\right)}^{1/2}\times {\hslash }^2\times \left(2a\right)\left({\left(\frac{\pi }{2a}\right)}^{1/2}-2a\left(\frac{1}{2}{\left(\frac{\pi }{{\left(2a\right)}^3}\right)}^{1/2}\right)\right)\left(\because {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{e}^{-k{x}^2}\text{d}x=\frac{1}{2}}{\left(\frac{\pi }{k^3}\right)}^{1/2}\right)\\ =2a{\hslash }^2\left({\left(\frac{2a}{\pi }\right)}^{1/2}\times {\left(\frac{\pi }{2a}\right)}^{1/2}-\left(2a\times {\left(\frac{2a}{\pi }\right)}^{1/2}\times \frac{1}{8a}\times {\left(\frac{2\pi }{a}\right)}^{1/2}\right)\right)\\ =2a{\hslash }^2\left(1-\frac{1}{2}\right)\end{array}$

The above equation can be solved further as shown below.

    $\langle p_x^2\rangle =a{\hslash }^2$

Therefore, the value of $\langle p_x^2\rangle$ is $a{\hslash }^2$.

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta p_x$ and $\Delta x$ have to be determined.

Concept introduction:

According to the Heisenberg uncertainty principle, it is impossible to specify simultaneously, with arbitrary precision, both the linear momentum and the position of a particle.  According to it if the position is known, the momentum cannot be predicted simultaneously. $\Delta x$ represents the uncertainty in position and $\Delta p_x$ represents the uncertainty in momentum.

Answer to Problem 7C.11P

The value of $\Delta x$ is $\frac{1}{2a^{1/2}}$.  The value of $\Delta p_x$ is $a^{1/2}\hslash$.

Explanation of Solution

The expression to determine $\Delta x$ value is given below.

    $\Delta x={\left\{\langle x^2\rangle -{\langle x\rangle }^2\right\}}^{1/2}$        (12)

The value of $\langle x\rangle$ is $0$.

The value of $\langle x^2\rangle$ is $\frac{1}{4a}$.

Substitute the values of $\langle x\rangle$ and $\langle x^2\rangle$ in equation (12).

    $\begin{array}{c}\Delta x={\left\{\frac{1}{4a}-0\right\}}^{1/2}\\ =\frac{1}{2a^{1/2}}\end{array}$

Therefore, the value of $\Delta x$ is $\frac{1}{2a^{1/2}}$.

The expression to determine $\Delta p_x$ value is given below.

    $\Delta p_x={\left\{\langle p_x^2\rangle -\langle p_x^2\rangle \right\}}^{1/2}$        (13)

The value of $\langle p_x\rangle$ is $0$.

The value of $\langle p_x^2\rangle$ is $a{\hslash }^2$.

Substitute the values of $\langle p_x\rangle$ and $\langle p_x^2\rangle$ in equation (13).

    $\begin{array}{c}\Delta p_x={\left\{a{\hslash }^2-0\right\}}^{1/2}\\ =a^{1/2}\hslash \end{array}$

Therefore, the value of $\Delta p_x$ is $a^{1/2}\hslash$.

(c)

Interpretation Introduction

Interpretation:

Whether the value for the product, $\Delta p_x\Delta x$ is consistent with the predictions from the uncertainty principle or not has to be determined.

Concept introduction:

According to the Heisenberg uncertainty principle, it is impossible to specify simultaneously, with arbitrary precision, both the linear momentum and the position of a particle.  According to it if the position is known, the momentum cannot be predicted simultaneously.

Answer to Problem 7C.11P

The value for the product, $\Delta p_x\Delta x$ is consistent with the predictions from the uncertainty principle.

Explanation of Solution

The value of $\Delta x$ is $\frac{1}{2a^{1/2}}$.

The value of $\Delta p_x$ is $a^{1/2}\hslash$.

Therefore, the value of $\Delta p_x\Delta x$ is determined below.

    $\begin{array}{c}\Delta p_x\Delta x=\frac{1}{2a^{1/2}}\times a^{1/2}\hslash \\ =\frac{\hslash }{2}\end{array}$

This is in accordance with the uncertainty principle according to which the expression given below holds true.

    $\Delta p_q\Delta q\ge \frac{\hslash }{2}$