Chapter 35, Problem 37P

(a)

To determine

The total classical energy.

Answer to Problem 37P

The total classical energy is $\langle E\rangle =\langle \frac{1}{2}m\left({\omega }_0^2\langle x^2\rangle +\left(P_x^2\right)\right)\rangle$ .

Explanation of Solution

Formula used:

The expression for $U\langle x\rangle$ total classical energy given by,

  $U\langle x\rangle =\frac{1}{2}m{\omega }_0{x}^2$

The expression for $K$ is given by,

  $K=\frac{1}{2}{p}_x^2/m$

The expression for total classical energy is given as,

  $E=U\langle x\rangle +K$

Calculation:

The expression for total classical energy is calculated as,

  $\begin{array}{c}E=U\langle x\rangle +K\\ =\frac{1}{2}m{\omega }_0^2\langle x^2\rangle +\frac{\langle P_x^2\rangle }{2m}\\ =\frac{1}{2}m\left({\omega }_0^2\langle x^2\rangle +\langle P_x^2\rangle \right)\end{array}$

Conclusion:

Therefore, the total classical energy is $E=\langle \frac{1}{2}m\left({\omega }_0^2\langle x^2\rangle +\left(P_x^2\right)\right)\rangle$ .

(b)

To determine

The value of $\langle \Delta x\rangle$ and ${\langle \Delta p_x\rangle }^2$ .

Answer to Problem 37P

The value of ${\langle \Delta x\rangle }^2=\langle x^2\rangle -{\langle x\rangle }^2$ and ${\langle \Delta p_x\rangle }^2=\langle p^2\rangle -{\langle p\rangle }^2$.

Explanation of Solution

Formula used:

The expression for $\langle \Delta x^2\rangle$ is written as,

  $\langle \Delta x^2\rangle =\left[{\left(x-\langle x\rangle \right)}^2\right]$

The expression for $\langle \Delta p^2\rangle$ is written as,

  $\langle \Delta p^2\rangle =\left[{\left(p-\langle p\rangle \right)}^2\right]$

Calculation:

The value of $\langle \Delta x^2\rangle$ is calculated as,

  $\begin{array}{c}\langle \Delta x^2\rangle =\left[{\left(x-\langle x\rangle \right)}^2\right]\\ =\left[x^2-2x\langle x\rangle -{\langle x\rangle }^2\right]\left(\therefore \langle x\rangle =0\right)\\ =\langle x^2\rangle \\ =\frac{\hslash }{2m{\omega }_0}\end{array}$

The value of $\langle \Delta p^2\rangle$ is calculated as,

  $\begin{array}{c}\langle \Delta p^2\rangle =\left[{\left(p-\langle p\rangle \right)}^2\right]\\ =\left[p^2-2p\langle p\rangle -{\langle p\rangle }^2\right]\left(\therefore \langle p\rangle =0\right)\\ =\langle p^2\rangle \\ =\frac{1}{2}\hslash m{\omega }_0\end{array}$

Conclusion:

Therefore, ${\langle \Delta x\rangle }^2=\langle x^2\rangle -{\langle x\rangle }^2$ and ${\langle \Delta p_x\rangle }^2=\langle p^2\rangle -{\langle p\rangle }^2$ .

(c)

To determine

The value of $\langle x\rangle$ , $\langle p_x\rangle$ must be equals to 0.

Answer to Problem 37P

The value of $\langle x\rangle$ , $\langle p_x\rangle$ will be 0.

Explanation of Solution

Formula used:

The expression for $\langle \Delta x^2\rangle$ is written as,

  $\langle \Delta x^2\rangle =\left[{\left(x-\langle x\rangle \right)}^2\right]$

The expression for $\langle \Delta p^2\rangle$ is written as,

  $\langle \Delta p^2\rangle =\left[{\left(p-\langle p\rangle \right)}^2\right]$

The value of $\langle x\rangle$ is defined as

  $\langle x\rangle ={{\displaystyle \int x\left|\psi \right|}}^{2}dx$

It gives the value of $\langle x\rangle$ is equal to 0 because of the integral of the odd function for the ground state for the excited state of the harmonic oscillator.

Hence, $\left(\langle x\rangle =0\right)$ and $\left(\langle p\rangle =0\right)$ .

Conclusion:

Therefore, the value of $\langle x\rangle$ , $\langle p_x\rangle$ will be 0.

(d)

To determine

The value of $\langle E\rangle =\frac{1}{2}m{\omega }^2{Z}^2+\raisebox{1ex}{${\hslash }^2$}\!\left/ \!\raisebox{-1ex}{$8mZ$}\right.$ .

Answer to Problem 37P

The value of $\langle E\rangle =\frac{1}{2}m{\omega }^2{Z}^2+\raisebox{1ex}{${\hslash }^2$}\!\left/ \!\raisebox{-1ex}{$8mZ$}\right.$ .

Explanation of Solution

Given:

The value of $\Delta p_x\Delta x=\frac{{\hslash }^2}{2}$ .

Formula used:

The expression for energy can be as,

  $\langle E\rangle =\frac{1}{2}m{\omega }^2\langle x^2\rangle +\frac{\langle \Delta P_x^2\rangle }{2m}$

Calculation:

The expression for energy can be calculate as,

  $\begin{array}{c}\langle E\rangle =\frac{1}{2}m{\omega }^2\langle x^2\rangle +\frac{\langle \Delta P_x^2\rangle }{2m}\\ =\frac{1}{2}m{\omega }^2\langle x^2\rangle +\frac{1}{2m}\left[\frac{{\hslash }^2}{4\langle x^2\rangle }\right]\\ =\frac{1}{2}m{\omega }^2\langle x^2\rangle +\frac{1}{8m}\left[\frac{{\hslash }^2}{\langle x^2\rangle }\right]\\ =\frac{1}{2}m{\omega }^2{Z}^2+\frac{{\hslash }^2}{8mZ}\end{array}$

Conclusion:

Therefore, the value of energy will be $\langle E\rangle =\frac{1}{2}m{\omega }^2{Z}^2+\raisebox{1ex}{${\hslash }^2$}\!\left/ \!\raisebox{-1ex}{$8mZ$}\right.$ .

(e)

To determine

The value of. $Z$

Answer to Problem 37P

The value of the value of $Z=\frac{\hslash }{\sqrt{2}m\omega }$ .

Explanation of Solution

Formula used:

The expression for $\langle E\rangle$ is given by,

  $\langle E\rangle =\frac{1}{2}m{\omega }^2{Z}^2+\raisebox{1ex}{${\hslash }^2$}\!\left/ \!\raisebox{-1ex}{$8mZ$}\right.$

Calculation:

The value of $Z$ can be calculated as,

  $\begin{array}{c}\frac{d\langle E\rangle }{dZ}=\frac{d}{dZ}\left(\frac{1}{2}m{\omega }^{2}Z+\frac{1}{8m}\left[\frac{{\hslash }^2}{Z}\right]\right)\\ 0=\frac{1}{2}m{\omega }^2-\frac{{\hslash }^2}{4m{Z}^2}\\ \frac{1}{2}m{\omega }^2=\frac{{\hslash }^2}{4m{Z}^2}\\ Z=\frac{\hslash }{\sqrt{2}m\omega }\end{array}$

Conclusion:

Therefore, the value of $Z=\frac{\hslash }{\sqrt{2}m\omega }$ .

(f)

To determine

The minimum energy will be $\langle E_{\min .}\rangle =\frac{1}{2}\hslash {\omega }_0$ .

Answer to Problem 37P

The minimum energy is equal to $\langle E_{\min .}\rangle =\frac{1}{2}\hslash {\omega }_0$ .

Explanation of Solution

Formula used:

The expression for energy can be written as,

  $\langle E\rangle =\frac{1}{2}m{\omega }_0^{2}Z+\frac{1}{8m}\left[\frac{{\hslash }^2}{Z}\right]$

Calculation:

The minimum energy at minimum point $Z=\frac{\hslash }{2m{\omega }_0}$ can be calculate as,

  $\begin{array}{c}\langle E_{\min .}\rangle =\frac{1}{2}m{\omega }_0^2\left(\frac{\hslash }{2m{\omega }_0}\right)+\frac{{\hslash }^2}{8m}\left(\frac{2m{\omega }_0}{\hslash }\right)\\ =\frac{\hslash {\omega }_0}{4}+\frac{\hslash {\omega }_0}{4}\\ =\frac{1}{2}\hslash {\omega }_0\end{array}$

Conclusion:

Therefore, the minimum energy is equal to $\langle E_{\min .}\rangle =\frac{1}{2}\hslash {\omega }_0$ .