Chapter 11, Problem 11.22E

Interpretation Introduction

(a)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is identically zero.

Explanation of Solution

The wavefunction ${\Psi }_1$ of a harmonic oscillator is expressed as,

${\Psi }_1={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

The wavefunction ${\Psi }_2$ of a harmonic oscillator is expressed as,

${\Psi }_2={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{8}\right)}^{\frac{1}{2}}{H}_2\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_1$ and ${\Psi }_2$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_1{}^{*}{\Psi }_{2}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{8}\right)}^{\frac{1}{2}}{H}_2\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& \frac{1}{4}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_1\left({\alpha }^{1/2}x\right)\cdot H_2\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2}dx}\end{array}$

The integration of harmonic oscillator is solved using Hermite polynomials. The integral of Hermite polynomials is solved by the formula given in Table 11.2.

$\begin{array}{rllll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_a{\left(\xi \right)}^{*}{H}_b\left(\xi \right)e^{-{\xi }^2}d\xi }& =& 0(\text{When}a\ne b,\xi & =& {\alpha }^{1/2}x)\\ & =& 2^{a}a!{\pi }^{1/2}(\text{When}a& =& b,\xi & =& {\alpha }^{1/2}x)\end{array}$

In the given integral, the wavefunctions are ${\Psi }_1$ and ${\Psi }_3$. Their Hermite polynomials are $H_1$ and $H_2$. The value of $a$ and $b$ is not same. Hence, the given integral is identically zero.

Conclusion

The given integral is identically zero.

Interpretation Introduction

(b)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is identically zero.

Explanation of Solution

The wavefunction ${\Psi }_1$ of a harmonic oscillator is expressed as,

${\Psi }_1={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_1$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_1{}^{*}\widehat{x}{\Psi }_{1}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}\widehat{x}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& \frac{1}{2}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{H}_1{\left({\alpha }^{1/2}x\right)}^2\cdot e^{-\alpha x^2}dx}\end{array}$

In the above equation, $H_1{\left({\alpha }^{1/2}x\right)}^2$ and $e^{-\alpha x^2}$ are even functions whereas $x$ is an odd function. The multiplication of odd function with even function gives an odd function. The integration of odd function from $-\infty$ to $+\infty$ centered at zero is $0$. Therefore, the given integral is identically zero.

Conclusion

The given integral is identically zero.

Interpretation Introduction

(c)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is not identically zero.

Explanation of Solution

The wavefunction ${\Psi }_1$ of a harmonic oscillator is expressed as,

${\Psi }_1={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_1$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_1{}^{*}{\widehat{x}}^2{\Psi }_{1}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}{\widehat{x}}^2\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& \frac{1}{2}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{x}^2{H}_1{\left({\alpha }^{1/2}x\right)}^2\cdot e^{-\alpha x^2}dx}\end{array}$

In the above equation, $x^2$, $H_1{\left({\alpha }^{1/2}x\right)}^2$ and $e^{-\alpha x^2}$ are even functions. The multiplication of an even function with another even function gives an even function. The integration of even function from $-\infty$ to $+\infty$ centered at zero is not equal to zero. Therefore, the given integral is not identically zero.

Conclusion

The given integral is not identically zero.

Interpretation Introduction

(d)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is identically zero.

Explanation of Solution

The wavefunction ${\Psi }_1$ of a harmonic oscillator is expressed as,

${\Psi }_1={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

The wavefunction ${\Psi }_3$ of a harmonic oscillator is expressed as,

${\Psi }_3={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{48}\right)}^{\frac{1}{2}}{H}_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_1$ and ${\Psi }_3$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_1{}^{*}{\Psi }_{3}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{48}\right)}^{\frac{1}{2}}{H}_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& {\left(\frac{1}{96}\right)}^{\frac{1}{2}}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_1\left({\alpha }^{1/2}x\right)\cdot H_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2}dx}\end{array}$

The integration of harmonic oscillator is solved using Hermite polynomials. The integral of Hermite polynomials is solved by the formula given in Table 11.2.

$\begin{array}{rllll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_a{\left(\xi \right)}^{*}{H}_b\left(\xi \right)e^{-{\xi }^2}d\xi }& =& 0(\text{When}a\ne b,\xi & =& {\alpha }^{1/2}x)\\ & =& 2^{a}a!{\pi }^{1/2}(\text{When}a& =& b,\xi & =& {\alpha }^{1/2}x)\end{array}$

In the given integral, the wavefunctions are ${\Psi }_1$ and ${\Psi }_3$. Their Hermite polynomials are $H_1$ and $H_2$. The value of $a$ and $b$ is not same. Hence, the given integral is identically zero.

Conclusion

The given integral is identically zero.

Interpretation Introduction

(e)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is not identically zero.

Explanation of Solution

The wavefunction ${\Psi }_3$ of a harmonic oscillator is expressed as,

${\Psi }_3={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{48}\right)}^{\frac{1}{2}}{H}_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_3$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_3{}^{*}{\Psi }_{3}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{48}\right)}^{\frac{1}{2}}{H}_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{48}\right)}^{\frac{1}{2}}{H}_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& \frac{1}{48}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_3\left({\alpha }^{1/2}x\right)\cdot H_3\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2}dx}\end{array}$

The integration of harmonic oscillator is solved using Hermite polynomials. The integral of Hermite polynomials is solved by the formula given in Table 11.2.

$\begin{array}{rllll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{H}_a{\left(\xi \right)}^{*}{H}_b\left(\xi \right)e^{-{\xi }^2}d\xi }& =& 0(\text{When}a\ne b,\xi & =& {\alpha }^{1/2}x)\\ & =& 2^{a}a!{\pi }^{1/2}(\text{When}a& =& b,\xi & =& {\alpha }^{1/2}x)\end{array}$

In the given integral, the wavefunctions are ${\Psi }_1$ and ${\Psi }_3$. Their Hermite polynomials are $H_1$ and $H_2$. The value of $a$ and $b$ is same. Hence, the given integral is not identically zero.

Conclusion

The given integral is not identically zero.

Interpretation Introduction

(f)

Interpretation:

Whether the given integrals involving harmonic oscillator wavefunctions is identically zero, not identically zero or indeterminate is to be determined.

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 11.22E

The given integral is indeterminate.

Explanation of Solution

The wavefunction ${\Psi }_1$ of a harmonic oscillator is expressed as,

${\Psi }_1={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$

Substitute the value of ${\Psi }_1$ in the given integral.

$\begin{array}{rll}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }_1{}^{*}\widehat{V}{\Psi }_{1}dx}& =& {\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{*}\widehat{V}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}{H}_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& \frac{1}{2}{\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}V\cdot H_1\left({\alpha }^{1/2}x\right)\cdot H_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2}dx}\end{array}$

In the above equation, $H_1{\left({\alpha }^{1/2}x\right)}^2$ and $e^{-\alpha x^2}$ are even functions. However, whether $V$ is odd or even function or not is not known. Hence, the given integral is indeterminate unless the value of $V$ is given.

Conclusion

The given integral is indeterminate.