Chapter 41, Problem 63CP

(a)

To determine

The energy of the state.

Answer to Problem 63CP

The energy of the state will be $\frac{3\hslash \omega }{2}$.

Explanation of Solution

Write the Schrodinger equation for a simple harmonic oscillator.

  $\frac{-{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+\frac{1}{2}m{\omega }^2{x}^2\psi =E\psi$

Here, $\hslash$ is the reduced Plank’s constant, $m$ is the mass of article, $\omega$ is the angular frequency of oscillator, and $E$ is the energy of oscillator.

Write the given wave function.

  $\psi =Bx{e}^{-\left(m\omega /2\hslash \right)x^2}$

Differentiate the above equation with respect to $x$.

  $\begin{array}{c}\frac{d\psi }{dx}=\frac{d\left(Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right)}{dx}\\ =B{e}^{-\left(m\omega /2\hslash \right)x^2}+Bx\left(-\frac{m\omega }{2\hslash }\right)2x{e}^{-\left(m\omega /2\hslash \right)x^2}\\ =B{e}^{-\left(m\omega /2\hslash \right)x^2}-B\left(\frac{m\omega }{\hslash }\right)x^2{e}^{-\left(m\omega /2\hslash \right)x^2}\end{array}$

Differentiate again the above equation.

  $\begin{array}{c}\frac{d^2\psi }{d{x}^2}=\frac{d^2\left(B{e}^{-\left(m\omega /2\hslash \right)x^2}-B\left(\frac{m\omega }{\hslash }\right)x^2{e}^{-\left(m\omega /2\hslash \right)x^2}\right)}{d{x}^2}\\ =Bx\left(-\frac{m\omega }{\hslash }\right)x{e}^{-\left(m\omega /2\hslash \right)x^2}-B\left(\frac{m\omega }{\hslash }\right)2x{e}^{-\left(m\omega /2\hslash \right)x^2}-B\left(\frac{m\omega }{\hslash }\right)x^2\left(-\frac{m\omega }{\hslash }\right)x{e}^{-\left(m\omega /2\hslash \right)x^2}\\ =-3B\left(\frac{m\omega }{\hslash }\right)x{e}^{-\left(m\omega /2\hslash \right)x^2}+B{\left(\frac{m\omega }{\hslash }\right)}^2{x}^3{e}^{-\left(m\omega /2\hslash \right)x^2}\end{array}$

Conclusion:

Rewrite the Schrodinger equation by substituting the equations for $\frac{d\psi }{dx}$ and $\frac{d^2\psi }{d{x}^2}$.

  $\begin{array}{c}\frac{-{\hslash }^2}{2m}\left[-3B\left(\frac{m\omega }{\hslash }\right)x{e}^{-\left(m\omega /2\hslash \right)x^2}+B{\left(\frac{m\omega }{\hslash }\right)}^2{x}^3{e}^{-\left(m\omega /2\hslash \right)x^2}\right]+\frac{1}{2}m{\omega }^2{x}^2\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]=E\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]\\ \left(\frac{3\hslash \omega }{2}\right)\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]+\bcancel{\left(-\frac{1}{2}m{\omega }^2{x}^2\right)\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]}+\bcancel{\left(\frac{1}{2}m{\omega }^2{x}^2\right)\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]}=E\left[Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right]\\ \left(\frac{3\hslash \omega }{2}\right)\left(Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right)=E\left(Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right)\end{array}$

Compare the coefficients of $Bx{e}^{-\left(m\omega /2\hslash \right)x^2}$.

    $E=\frac{3\hslash \omega }{2}$

Therefore, the energy of the state will be $\frac{3\hslash \omega }{2}$.

(b)

To determine

The position at which there is no any possibility to find a particle.

Answer to Problem 63CP

Particle cannot be found at $x=0$.

Explanation of Solution

The possibility of finding a particle depends on the value of wave function. A particle cannot be found at a position where wave function vanishes. The wave function vanishes at $x=0$. That means, there will be no any probability foe finding a particle at $x=0$.

Therefore, the particle cannot be found at $x=0$.

(c)

To determine

The most probable position for finding the particle.

Answer to Problem 63CP

The maximum probability for particles are at $x=\pm \sqrt{\frac{\hslash }{m\omega }}$.

Explanation of Solution

Point of maximum probability is a point of maxima.

Write the condition to be satisfied at the point of maximum probability of finding the particle.

  $\frac{d\psi }{dx}=0$

Conclusion:

Substitute $Bx{e}^{-\left(m\omega /2\hslash \right)x^2}$ for $\psi$ in the above equation.

  $\begin{array}{c}\frac{d\psi }{dx}=\frac{d\left(Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right)}{dx}\\ =B{e}^{-\left(m\omega /2\hslash \right)x^2}-B\left(\frac{m\omega }{\hslash }\right)x^2{e}^{-\left(m\omega /2\hslash \right)x^2}=0\\ 1-\left(\frac{m\omega }{\hslash }\right)x^2=0\end{array}$

Rewrite the above relation in terms of $x$.

    $x=\pm \sqrt{\frac{\hslash }{m\omega }}$

Therefore, the maximum probabilities for particles are at $x=\pm \sqrt{\frac{\hslash }{m\omega }}$.

(d)

To determine

The value of $B$ required for the normalization of wave function.

Answer to Problem 63CP

$B$ is ${\left(\frac{4m^3{\omega }^3}{\pi {\hslash }^3}\right)}^{1/4}$.

Explanation of Solution

Write the normalization condition.

  ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\psi \right|}^{2}dx}=1$

Conclusion:

Substitute $Bx{e}^{-\left(m\omega /2\hslash \right)x^2}$ for $\psi$ in the above equation.

  $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{B}^2{x}^2{e}^{-\left(m\omega /\hslash \right)x^2}dx}=1\\ 2B^2\frac{1}{4}\sqrt{\frac{\pi }{{\left(m\omega /\hslash \right)}^3}}=1\\ B^2\frac{{\pi }^{1/2}}{2}{\left(\frac{\hslash }{m\omega }\right)}^{3/2}=1\end{array}$

Rewrite the above relation in terms of $B$.

  $\begin{array}{c}B=\frac{2^{1/2}}{{\pi }^{1/4}}{\left(\frac{m\omega }{\hslash }\right)}^{3/4}\\ ={\left(\frac{4m^3{\omega }^3}{\pi {\hslash }^3}\right)}^{1/4}\end{array}$

Therefore, $B$ is ${\left(\frac{4m^3{\omega }^3}{\pi {\hslash }^3}\right)}^{1/4}$.

(e)

To determine

The classical probability of finding the particle in the width $\delta$ centred at $x=2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}$.

Answer to Problem 63CP

Classical probability is zero in the width $\delta$.

Explanation of Solution

Write the equation for potential energy.

  $U=\frac{1}{2}m{\omega }^2{x}^2$

Here, $U$ is the potential energy.

Conclusion:

Substitute $2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}$ for $x$ in the above equation.

  $\begin{array}{c}U=\frac{1}{2}m{\omega }^2{\left(2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}\right)}^2\\ =2\hslash \omega \end{array}$

The total energy of oscillator is $\frac{3}{2}\hslash \omega$. The calculated result, $2\hslash \omega$ is greater than $\frac{3}{2}\hslash \omega$. That means the probability of finding the particle in width $\delta$ is zero.

Therefore, the classical probability is zero in the width $\delta$.

(f)

To determine

The real probability of finding the particle in the width $\delta$ centred at $x=2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}$.

Answer to Problem 63CP

The real probability in width $\delta$ is $8\delta {\left(\frac{m\omega }{\hslash \pi }\right)}^{1/2}{e}^{-4}$.

Explanation of Solution

Write the equation to calculate the real probability.

  $P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\psi \right|}^{2}dx}$

Here, $P$ is the probability.

Conclusion:

Substitute $Bx{e}^{-\left(m\omega /2\hslash \right)x^2}$ for ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left|\psi \right|}$ and $\delta$ for $x$ in the above equation to find $P$.

  $\begin{array}{c}P={\left(Bx{e}^{-\left(m\omega /2\hslash \right)x^2}\right)}^2\delta \\ =B^2{x}^2{e}^{-\left(m\omega /2\hslash \right)x^2}\delta \end{array}$

Rewrite the above relation by substituting ${\left(\frac{4m^3{\omega }^3}{\pi {\hslash }^3}\right)}^{1/2}$ for $B$ and $2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}$ for $x$ in the above equation to find $P$.

    $\begin{array}{c}P={\left(\frac{4m^3{\omega }^3}{\pi {\hslash }^3}\right)}^{1/2}{\left(2{\left(\frac{\hslash }{m\omega }\right)}^{\frac{1}{2}}\right)}^2{e}^{-\left(m\omega /2\hslash \right)x^2}\delta \\ =\delta \frac{2}{{\pi }^{1/2}}\left(\frac{m^{3/2}{\omega }^{3/2}}{{\hslash }^{3/2}}\right)\left(\frac{4\hslash }{m\omega }\right)e^{-\left(m\omega /\hslash \right)4\left(\hslash /m\omega \right)}\\ =8\delta {\left(\frac{m\omega }{\hslash \pi }\right)}^{1/2}{e}^{-4}\end{array}$

Therefore, the real probability in width $\delta$ is $8\delta {\left(\frac{m\omega }{\hslash \pi }\right)}^{1/2}{e}^{-4}$.