Chapter 11, Problem 11.19E

Interpretation Introduction

Interpretation:

The value of $\langle p_x\rangle$ for ${\Psi }_0$ and ${\Psi }_1$ of a harmonic oscillator is to be calculated. Whether the calculated values make sense or not is to be stated.

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 11.19E

The value of $\langle p_x\rangle$ for ${\Psi }_0$ and ${\Psi }_1$ of a harmonic oscillator is $0$. The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity. Therefore, the calculated value makes sense.

Explanation of Solution

The general wavefunction of harmonic oscillator is expressed as,

$\Psi ={\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2^{n}n!}\right)}^{\frac{1}{2}}\cdot H_n\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}$ …(1)

Substitute the value of $n=0$ in equation (1).

$\begin{array}{rll}{\Psi }_0& =& {\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2^0\times 0!}\right)}^{\frac{1}{2}}\cdot H_0\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\\ & =& {\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot H_0\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\end{array}$

Substitute the value of $n=1$ in equation (1).

$\begin{array}{rll}{\Psi }_1& =& {\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot {\left(\frac{1}{2^{1}1!}\right)}^{\frac{1}{2}}\cdot H_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\\ & =& {\left(\frac{\alpha }{4\pi }\right)}^{\frac{1}{4}}\cdot H_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\end{array}$

The value of $\langle p_x\rangle$ is calculated by the formula.

$\langle p_x\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\Psi }^{\ast }\cdot -i\hslash \frac{\partial }{\partial x}\Psi dx}$ …(2)

Substitute the value of ${\Psi }_0$ in equation (2).

$\langle p_x\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot H_0\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{\ast }\cdot -i\hslash \frac{\partial }{\partial x}\left({\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{4}}\cdot H_0\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}$

From Table 11.1, substitute the value of $H_0\left({\alpha }^{1/2}x\right)=1$ in the above equation.

$\begin{array}{rll}\langle p_x\rangle & =& {\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(e^{-\alpha x^2/2}\right)}^{\ast }\left(-i\hslash \frac{\partial }{\partial x}\left(e^{-\alpha x^2/2}\right)\right)dx}\\ & =& -i\hslash {\left(\frac{\alpha }{\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(e^{-\alpha x^2/2}\right)}^{\ast }\left(-\frac{2\alpha x}{2}{e}^{-\alpha x^2/2}\right)dx}\\ & =& \frac{i\hslash a^{\frac{3}{2}}}{{\pi }^{\frac{1}{2}}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x{e}^{-\alpha x^2}dx}\end{array}$…(3)

The above equation show that $e^{-\alpha x^2}$ is an even function whereas $x$ is an odd function. Therefore, the overall function is odd.

The integration of odd function going from $-\infty \text{to}+\infty$ and centered at $0$ is expressed as,

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

Therefore, equation (3) becomes,

$\langle p_x\rangle =0$

Substitute the value of ${\Psi }_1$ in equation (2).

$\langle p_x\rangle ={\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left({\left(\frac{\alpha }{4\pi }\right)}^{\frac{1}{4}}\cdot H_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)}^{\ast }\cdot -i\hslash \frac{\partial }{\partial x}\left({\left(\frac{\alpha }{4\pi }\right)}^{\frac{1}{4}}\cdot H_1\left({\alpha }^{1/2}x\right)\cdot e^{-\alpha x^2/2}\right)dx}$

From Table 11.1, substitute the value of $H_1\left({\alpha }^{1/2}x\right)=2{\alpha }^{1/2}x$ in the above equation.

$\begin{array}{rll}\langle p_x\rangle & =& {\left(\frac{\alpha }{4\pi }\right)}^{\frac{1}{2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(2{\alpha }^{1/2}x\cdot e^{-\alpha x^2/2}\right)}^{\ast }\cdot -i\hslash \frac{\partial }{\partial x}\left(2{\alpha }^{1/2}x\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& -i\hslash \frac{4{\alpha }^{3/2}}{2{\pi }^{1/2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{\left(x\cdot e^{-\alpha x^2/2}\right)}^{\ast }\cdot \frac{\partial }{\partial x}\left(x\cdot e^{-\alpha x^2/2}\right)dx}\\ & =& -i\hslash \frac{4{\alpha }^{3/2}}{2{\pi }^{1/2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}x\cdot e^{-\alpha x^2/2}\left[e^{-\alpha x^2/2}-\alpha x^2{e}^{-\alpha x^2/2}\right]dx}\\ & =& i\hslash \frac{4{\alpha }^{5/2}}{2{\pi }^{1/2}}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}\left(x\cdot e^{-\alpha x^2}-\alpha x^3\cdot e^{-\alpha x^2}\right)dx}\end{array}$

The above equation shows that $e^{-\alpha x^2}$ is an even function whereas $x^3$ and $x$ are odd functions. Therefore, the overall function is odd.

The integration of odd function going from $-\infty \text{to}+\infty$ and centered at $0$ is expressed as,

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

Therefore, equation (3) becomes,

$\langle p_x\rangle =0$

Hence, the value of $\langle p_x\rangle$ for ${\Psi }_0$ and ${\Psi }_1$ of a harmonic oscillator is $0$.

The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity.

Conclusion

The value of $\langle p_x\rangle$ for ${\Psi }_0$ and ${\Psi }_1$ of a harmonic oscillator is $0$. The average momentum of harmonic oscillator is zero because the movement of mass takes place back and forth in both directions and momentum is a vector quantity. Therefore, the calculated value makes sense.