Chapter 41, Problem 57AP

(a)

To determine

The expectation value ${\langle x\rangle }_0$.

Answer to Problem 57AP

The expectation value ${\langle x\rangle }_0$ is $0$.

Explanation of Solution

Write the expectation relation.

  ${\langle x\rangle }_0={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }_0{\left(x\right)}^{*}x{\psi }_0\left(x\right)dx}$                                                                                 (I)

Here, ${\langle x\rangle }_0$ is the expectation value and $\psi \left(x\right)$ is the wave function

Conclusion:

Substitute ${\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}$ for ${\psi }_0\left(x\right)$ and ${\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}$ for ${\psi }_0{\left(x\right)}^{*}$ and simplify.

  $\begin{array}{c}{\langle x\rangle }_0={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left({\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}\right)x\left({\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}\right)dx}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}dx}\end{array}$

It is an odd function. The integration of odd function goes to zero.

    $\begin{array}{c}{\langle x\rangle }_0={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2}dx}\\ =0\end{array}$

Therefore, the expectation value ${\langle x\rangle }_0$ is $0$.

(b)

To determine

The expectation value ${\langle x\rangle }_1$.

Answer to Problem 57AP

The expectation value ${\langle x\rangle }_1$ is $0$.

Explanation of Solution

Rewrite the expectation relation.

  ${\langle x\rangle }_1={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }_0{\left(x\right)}^{*}x{\psi }_0\left(x\right)dx}$

Here, ${\langle x\rangle }_1$ is the expectation value.

Conclusion:

Substitute ${\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}$ for ${\psi }_0\left(x\right)$ and ${\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}$ for ${\psi }_0{\left(x\right)}^{*}$ and simplify.

  $\begin{array}{c}{\langle x\rangle }_1={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left({\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}\right)x\left({\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}\right)dx}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}dx}\end{array}$

It is an odd function. The integration of odd function goes to zero.

    ${\langle x\rangle }_1=0$

Therefore, the expectation value ${\langle x\rangle }_1$ is $0$.

(c)

To determine

The expectation value ${\langle x\rangle }_{01}$.

Answer to Problem 57AP

The expectation value ${\langle x\rangle }_{01}$ is $\frac{1}{\sqrt{2a}}$.

Explanation of Solution

Rewrite the expectation relation.

  ${\langle x\rangle }_{01}={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\psi }_0{\left(x\right)}^{*}x{\psi }_0\left(x\right)dx}$

Here, ${\langle x\rangle }_{01}$ is the expectation value.

Write the given wave function.

  ${\psi }_{01}\left(x\right)=\frac{1}{\sqrt{2}}\left[{\psi }_0\left(x\right){\psi }_1\left(x\right)\right]$

Conclusion:

Substitute $\frac{1}{\sqrt{2}}\left[{\psi }_0\left(x\right)+{\psi }_1\left(x\right)\right]$ for ${\psi }_0\left(x\right)$ and $\frac{1}{\sqrt{2}}\left[{\psi }_0\left(x\right)+{\psi }_1\left(x\right)\right]$ for ${\psi }_0{\left(x\right)}^{*}$ and simplify.

  $\begin{array}{c}{\langle x\rangle }_{01}={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2}}\left[{\psi }_0\left(x\right)+{\psi }_1\left(x\right)\right]x\frac{1}{\sqrt{2}}\left[{\psi }_0\left(x\right)+{\psi }_1\left(x\right)\right]dx}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\frac{1}{2}{\left({\psi }_0+{\psi }_1\right)}^{2}dx}\\ =\frac{1}{2}{\langle x\rangle }_0+\frac{1}{2}{\langle x\rangle }_1+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }_0\left(x\right){\psi }_1\left(x\right)dx}\end{array}$

First two term is an odd function. The integration of odd function goes to zero.

    $\begin{array}{c}{\langle x\rangle }_{01}=0+0+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }_0\left(x\right){\psi }_1\left(x\right)dx}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }_0\left(x\right){\psi }_1\left(x\right)dx}\end{array}$

Substitute ${\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2/2}$ for ${\psi }_0\left(x\right)$ and ${\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}$ for ${\psi }_1\left(x\right)$ in the above expression and integrate.

  $\begin{array}{c}{\langle x\rangle }_{01}={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left({\left(\frac{a}{\pi }\right)}^{1/2}{e}^{-a{x}^2/2}\right)}\left({\left(\frac{4a^3}{\pi }\right)}^{1/2}{x}^2{e}^{-a{x}^2/2}\right)dx\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left(\frac{a}{\pi }\right)}^{1/4}{e}^{-a{x}^2/2}{\left(\frac{4a^3}{\pi }\right)}^{1/4}x{e}^{-a{x}^2/2}dx}\\ =2{\left(\frac{2a^2}{\pi }\right)}^{1/2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-a{x}^2}dx}\\ =2{\left(\frac{2a^2}{\pi }\right)}^{1/2}\frac{1}{4}{\left(\frac{\pi }{a^3}\right)}^{1/2}\\ =\frac{1}{\sqrt{2a}}\end{array}$

Therefore, the expectation value ${\langle x\rangle }_{01}$ is $\frac{1}{\sqrt{2a}}$.