Chapter 10, Problem 10.31E

Interpretation Introduction

(a)

Interpretation:

The given wavefunction is to be normalized over the indicated range.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.31E

The normalized wavefunction is $\Psi =\sqrt{5}{x}^2$.

Explanation of Solution

The given wavefunction is $\Psi =x^2$.

Assume the normalization constant of the wavefunction as $N$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{0}{\overset{1}{\int }}{\left(N\Psi \right)}^{*}N\Psi }dx=1$

Where,

• $N$ is the normalization constant.

• $\Psi$ is the wavefunction.

Substitute the value of $\Psi$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(N{x}^2\right)}^{*}N{x}^2}dx& =& 1\\ N^2{\displaystyle \underset{0}{\overset{1}{\int }}{x}^4}dx& =& 1\\ N^2{\left[\frac{x^5}{5}\right]}_0^1& =& 1\\ N^2& =& 5\end{array}$

Solve the above equation.

$N=\pm \sqrt{5}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\Psi =\sqrt{5}{x}^2$.

Conclusion

The normalized wavefunction is $\Psi =\sqrt{5}{x}^2$.

Interpretation Introduction

(b)

Interpretation:

The given wavefunction is to be normalized over the indicated range.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.31E

The normalized wavefunction is $\Psi =\sqrt{30}\frac{1}{x}$.

Explanation of Solution

The given wavefunction is $\Psi =1/x$.

Assume the normalization constant of the wavefunction is $N$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{5}{\overset{6}{\int }}{\left(N\Psi \right)}^{*}N\Psi }dx=1$

Where,

• $N$ is the normalization constant.

• $\Psi$ is the wavefunction.

Substitute the value of $\Psi$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{5}{\overset{6}{\int }}{\left(N\frac{1}{x}\right)}^{*}N\frac{1}{x}}dx& =& 1\\ N^2{\displaystyle \underset{5}{\overset{6}{\int }}\frac{1}{x^2}}dx& =& 1\\ N^2{\displaystyle \underset{5}{\overset{6}{\int }}\frac{x^{-2+1}}{-2+1}}& =& 1\\ -N^2{\left[\frac{1}{x}\right]}_5^6& =& 1\end{array}$

Solve the above equation.

$\begin{array}{rll}-N^2\left[\frac{1}{6}-\frac{1}{5}\right]& =& 1\\ N^2\left[\frac{1}{30}\right]& =& 1\\ N^2& =& 30\\ N& =& \pm \sqrt{30}\end{array}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\Psi =\sqrt{30}\frac{1}{x}$.

Conclusion

The normalized wavefunction is $\Psi =\sqrt{30}\frac{1}{x}$.

Interpretation Introduction

(c)

Interpretation:

The given wavefunction is to be normalized over the indicated range.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.31E

The normalized wavefunction is $\Psi =\sqrt{\frac{2}{\pi }}\cos x$.

Explanation of Solution

The given wavefunction is $\Psi =\cos x$.

Assume the normalization constant of the wavefunction is $N$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\left(N\Psi \right)}^{*}N\Psi }dx=1$

Where,

• $N$ is the normalization constant.

• $\Psi$ is the wavefunction.

Substitute the value of $\Psi$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\left(N\cos x\right)}^{*}N\cos x}dx& =& 1\\ N^2{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}x}dx& =& 1\end{array}$

Function ${\cos }^{2}x$ is even, therefore the range will be $0$ to $\frac{\pi }{2}$.

$2N^2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}x}dx=1$

From Appendix 1,

${\displaystyle \int {\cos }^{2}bxdx}=\frac{x}{2}+\frac{1}{4b}\sin \left(2bx\right)$

From this relation, the above equation becomes,

$\begin{array}{rll}2N^2{\left[\frac{x}{2}+\frac{1}{4}\left(\sin 2x\right)\right]}_0^{\frac{\pi }{2}}& =& 1\\ 2N^2\left[\frac{\pi }{4}+\frac{1}{4}\left(\sin 2\left(\frac{\pi }{2}\right)\right)-\left(\frac{0}{4}+\frac{1}{4}\left(\sin 2\left(0\right)\right)\right)\right]& =& 1\\ 2N^2\left[\frac{\pi }{4}+\frac{1}{4}\left(\sin \pi \right)\right]& =& 1\end{array}$

Since $\sin n\pi =0$, the above equation becomes,

$\begin{array}{rll}2N^2\left[\frac{\pi }{4}\right]& =& 1\\ N^2\left(\frac{\pi }{2}\right)& =& 1\\ N& =& \pm \sqrt{\frac{2}{\pi }}\end{array}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\Psi =\sqrt{\frac{2}{\pi }}\cos x$.

Conclusion

The normalized wavefunction is $\Psi =\pm \sqrt{\frac{2}{\pi }}\cos x$.

Interpretation Introduction

(d)

Interpretation:

The given wavefunction is to be normalized over the indicated range.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.31E

The normalized wavefunction is $\Psi =a^{-\frac{3}{2}}{e}^{-\frac{r}{a}}$.

Explanation of Solution

The given wavefunction is $\Psi =e^{-\frac{r}{a}}$.

Assume the normalization constant of the wavefunction is $N$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N\Psi \right)}^{*}N\Psi }d\tau =1$

Where,

• $N$ is the normalization constant.

• $\Psi$ is the wavefunction.

Substitute the value of $\Psi$ and $d\tau$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{0}{\overset{\infty }{\int }}N{e}^{-\frac{r}{a}}N{e}^{-\frac{r}{a}}4r^{2}dr}& =& 1\\ N^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\frac{2r}{a}}4r^{2}dr}& =& 1\end{array}$

Assume,

$\begin{array}{rll}\frac{2r}{a}& =& u\\ dr& =& \frac{a}{2}du\end{array}$

Substitute the value of $\frac{2r}{a}$ in the above equation.

$\begin{array}{rll}{N}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-u}4{\left(\frac{au}{2}\right)}^2\frac{a}{2}du}& =& 1\\ N^2\left(\frac{a^3}{2}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{u}^2{e}^{-u}du}& =& 1\end{array}$

From Appendix 1,

${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-bx}dx}=\frac{n!}{b^{n+1}}$

From this relation, the above equation becomes,

$\begin{array}{rll}{N}^2\left(\frac{a^3}{2}\right)2!& =& 1\\ N^2{a}^3& =& 1\\ N& =& \frac{1}{\sqrt{a^3}}\\ & =& \pm a^{-\frac{3}{2}}\end{array}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\Psi =a^{-\frac{3}{2}}{e}^{-\frac{r}{a}}$.

Conclusion

The normalized wavefunction is $\Psi =a^{-\frac{3}{2}}{e}^{-\frac{r}{a}}$.

Interpretation Introduction

(e)

Interpretation:

The given wavefunction is to be normalized over the indicated range.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.31E

The normalized wavefunction is $\Psi ={\left(2\right)}^{\frac{1}{4}}{a}^{-\frac{3}{4}}{e}^{-\frac{r^2}{a}}$.

Explanation of Solution

The given wavefunction is $\Psi =e^{-\frac{r^2}{a}}$.

Assume the normalization constant of the wavefunction is $N$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(N\Psi \right)}^{*}N\Psi }d\tau =1$

Where,

• $N$ is the normalization constant.

• $\Psi$ is the wavefunction.

Substitute the value of $\Psi$ and $d\tau$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}N{e}^{-\frac{r^2}{a}}N{e}^{-\frac{r^2}{a}}4r^{2}dr}& =& 1\\ N^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{2r^2}{a}}4r^{2}dr}& =& 1\end{array}$ … (1)

Assume,

$\begin{array}{rll}\frac{2r^2}{a}& =& u\\ r& =& \sqrt{\frac{au}{2}}\end{array}$

Differentiate equation $\frac{2r^2}{a}=u$.

$\begin{array}{rll}\frac{2\cdot 2rdr}{a}& =& du\\ dr& =& \frac{a}{4r}du\end{array}$

Substitute value of $r$ in the above equation.

$\begin{array}{rll}dr& =& \frac{a}{4\sqrt{\frac{au}{2}}}du\\ & =& \frac{\sqrt{2}{a}^{\frac{1}{2}}{u}^{-\frac{1}{2}}}{4}du\end{array}$

Substitute the value of $r$, $dr$ and $u$ in equation (1).

$\begin{array}{rll}{N}^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-u}4{\left(\sqrt{\frac{au}{2}}\right)}^2\frac{\sqrt{2}{a}^{\frac{1}{2}}{u}^{-\frac{1}{2}}}{4}du}& =& 1\\ N^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-u}\left(\frac{au}{2}\right)\sqrt{2a}{u}^{-\frac{1}{2}}du}& =& 1\end{array}$

It is known that,

${\displaystyle \underset{-a}{\overset{a}{\int }}f\left(x\right)dx}=2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx}$

Using this relation, the above equation becomes,

$2\frac{N^2\sqrt{2}{a}^{\frac{3}{2}}}{2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{u}^{\frac{1}{2}}{e}^{-u}du}=1$

From Appendix 1,

${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-bx}dx}=\frac{n!}{b^{n+1}}$

Using this relation, the above equation becomes,

$\begin{array}{rll}{N}^2\sqrt{2}\times \frac{1}{2}{a}^{\frac{3}{2}}& =& 1\\ N^2\frac{1}{\sqrt{2}}{a}^{\frac{3}{2}}& =& 1\\ N^2& =& \sqrt{2}{a}^{-\frac{3}{2}}\\ N& =& {\left(2\right)}^{\frac{1}{4}}{a}^{-\frac{3}{4}}\end{array}$

Therefore, the normalized wavefunction is $\Psi ={\left(2\right)}^{\frac{1}{4}}{a}^{-\frac{3}{4}}{e}^{-\frac{r^2}{a}}$.

Conclusion

The normalized wavefunction is $\Psi ={\left(2\right)}^{\frac{1}{4}}{a}^{-\frac{3}{4}}{e}^{-\frac{r^2}{a}}$.