Chapter 7, Problem 7B.8P

(a)

Interpretation Introduction

Interpretation:

The probability of the finding of the particle between $x=0$ and $x=L/2$, $y=0$ and $y=L/2$ has to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.8P

The probability of finding the particle between $x=0$ and $x=L/2$, $y=0$ and $y=L/2$ is $\frac{L^3}{4}$.

Explanation of Solution

The probability of finding the particle between $x=0$ and $x=L/2$, $y=0$ and $y=L/2$ is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{y=0}{\overset{y=L/2}{\int }}{\displaystyle \underset{x=0}{\overset{x=L/2}{\int }}{\psi }^2}}\text{d}x\text{d}y$                                                                      (1)

The given wave function is shown below.

    $\psi =\left(2/L\right)\sin \left(\pi x/L\right)\sin \left(\pi y/L\right)$

Substitute the value of wavefunction in equation (1) and separate the integrals as shown below.

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{y=0}{\overset{y=L/2}{\int }}{\displaystyle \underset{x=0}{\overset{x=L/2}{\int }}{\left(\left(2/L\right)\sin \left(\pi x/L\right)\sin \left(\pi y/L\right)\right)}^2}}\text{d}x\text{d}y\\ ={\left(2/L\right)}^2{\displaystyle \underset{y=0}{\overset{L/2}{\int }}{\sin }^2\left(\pi x/L\right)\text{d}y{\displaystyle \underset{x=0}{\overset{L/2}{\int }}{\sin }^2\left(\pi y/L\right)}}\text{d}x\end{array}$

Use the identity shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration,

    $\begin{array}{c}\text{Probability}={\left(2/L\right)}^2{\displaystyle \underset{y=0}{\overset{L/2}{\int }}\left(\frac{1-\cos 2\left(\pi y/L\right)}{2}\right)\text{d}y{\displaystyle \underset{x=0}{\overset{L/2}{\int }}\left(\frac{1-\cos 2\left(\pi x/L\right)}{2}\right)}}\text{d}x\\ =\frac{1}{4}\left(4/L^2\right){\displaystyle \underset{y=0}{\overset{L/2}{\int }}\left(1-\cos 2\left(\pi y/L\right)\right)\text{d}y{\displaystyle \underset{x=0}{\overset{L/2}{\int }}1-\cos 2\left(\pi x/L\right)}}\text{d}x\\ =L^2{\left[y-\frac{\sin 2\left(\pi y/L\right)}{2\left(\pi /L\right)}\right]}_{y=0}^{L/2}{\left[x-\frac{\sin 2\left(\pi x/L\right)}{2\left(\pi /L\right)}\right]}_{x=0}^{L/2}\\ =L^2\left(\begin{array}{c}\left[\left(\frac{L}{2}-\frac{\sin 2\left(\pi L/2L\right)}{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 2\left(\pi \times 0/L\right)}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{L}{2}-\frac{\sin 2\left(\pi \times L/2L\right)}{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 2\left(\pi \times 0/L\right)}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\end{array}$

On further solving the above equation,

    $\begin{array}{c}\text{Probability}=L^2\left(\begin{array}{c}\left[\left(\frac{L}{2}-\frac{\sin 2\left(\pi L/2L\right)}{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 2\left(\pi \times 0/L\right)}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{L}{2}-\frac{\sin 2\left(\pi \times L/2L\right)}{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 2\left(\pi \times 0/L\right)}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\\ =L^2\left(\begin{array}{c}\left[\left(\frac{L}{2}-\frac{\sin \pi }{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 0}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{L}{2}-\frac{\sin \pi }{2\left(\pi /L\right)}\right)-\left(0-\frac{\sin 0}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\\ =L^2\left(\begin{array}{c}\left[\left(\frac{L}{2}-0\right)-\left(0-0\right)\right]\\ \left[\left(\frac{L}{2}-0\right)-\left(0-0\right)\right]\end{array}\right)\left(\because \sin n\pi =0;n=0,1,2,\mathrm{3....}\right)\\ =L^2\left(\frac{L}{2}\times \frac{L}{2}\right)\end{array}$

On further solving the above integration,

    $\text{Probability}=\frac{L^3}{4}$

Hence, the probability of finding the particle between $x=0$ and $x=L/2$, $y=0$ and $y=L/2$ is $\frac{L^3}{4}$.

(b)

Interpretation Introduction

Interpretation:

The probability of finding the particle between $x=L/4$ and $x=3L/4$, $y=L/4$ and $y=3L/4$ is to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.8P

The probability of finding the particle between $x=L/4$ and $x=3L/4$, $y=L/4$ and $y=3L/4$ is $L^2{\left(\frac{L}{2}+\frac{L}{\pi }\right)}^2$.

Explanation of Solution

The probability of finding the particle between $x=L/4$ and $x=3L/4$, $y=L/4$ and $y=3L/4$ is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{y=L/4}{\overset{y=3L/4}{\int }}{\displaystyle \underset{x=L/4}{\overset{x=3L/4}{\int }}{\psi }^2}}\text{d}x\text{d}y$                                                                    (2)

The given wave function is shown below.

    $\psi =\left(2/L\right)\sin \left(\pi x/L\right)\sin \left(\pi y/L\right)$

Substitute the value of wavefunction in equation (2) and separate the integrals as shown below.

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{y=L/4}{\overset{y=3L/4}{\int }}{\displaystyle \underset{x=L/4}{\overset{x=3L/4}{\int }}{\left(\left(2/L\right)\sin \left(\pi x/L\right)\sin \left(\pi y/L\right)\right)}^2}}\text{d}x\text{d}y\\ ={\left(2/L\right)}^2{\displaystyle \underset{y=L/4}{\overset{3L/4}{\int }}{\sin }^2\left(\pi x/L\right)\text{d}y{\displaystyle \underset{x=0}{\overset{3L/4}{\int }}{\sin }^2\left(\pi y/L\right)}}\text{d}x\end{array}$

Use the identity shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration,

    $\begin{array}{c}\text{Probability}={\left(2/L\right)}^2{\displaystyle \underset{y=L/4}{\overset{3L/4}{\int }}\left(\frac{1-\cos 2\left(\pi y/L\right)}{2}\right)\text{d}y{\displaystyle \underset{x=L/4}{\overset{3L/4}{\int }}\left(\frac{1-\cos 2\left(\pi x/L\right)}{2}\right)}}\text{d}x\\ =\frac{1}{4}\left(4/L^2\right){\displaystyle \underset{y=L/4}{\overset{3L/4}{\int }}\left(1-\cos 2\left(\pi y/L\right)\right)\text{d}y{\displaystyle \underset{x=L/4}{\overset{3L/4}{\int }}1-\cos 2\left(\pi x/L\right)}}\text{d}x\\ =L^2{\left[y-\frac{\sin 2\left(\pi y/L\right)}{2\left(\pi /L\right)}\right]}_{y=L/4}^{3L/4}{\left[x-\frac{\sin 2\left(\pi x/L\right)}{2\left(\pi /L\right)}\right]}_{x=L/4}^{3L/4}\\ =L^2\left(\begin{array}{c}\left[\left(\frac{3L}{4}-\frac{\sin 2\left(3\pi L/4L\right)}{2\left(\pi /L\right)}\right)-\left(\frac{L}{4}-\frac{\sin 2\left(\pi \times L/4L\right)}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{3L}{4}-\frac{\sin 2\left(3\pi \times L/4L\right)}{2\left(\pi /L\right)}\right)-\left(\frac{L}{4}-\frac{\sin 2\left(\pi \times L/4L\right)}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\end{array}$

On further solving the above equation,

    $\begin{array}{c}\text{Probability}=L^2\left(\begin{array}{c}\left[\left(\frac{3L}{4}-\frac{\sin \left(3\pi /2\right)}{2\left(\pi /L\right)}\right)-\left(\frac{L}{4}-\frac{\sin \left(\pi /2\right)}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{3L}{4}-\frac{\sin \left(3\pi /2\right)}{2\left(\pi /L\right)}\right)-\left(\frac{L}{4}-\frac{\sin \left(\pi /2\right)}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\\ =L^2\left(\begin{array}{c}\left[\left(\frac{3L}{4}-\frac{-1}{2\left(\pi /L\right)}\right)-\left(\frac{L}{4}-\frac{1}{2\left(\pi /L\right)}\right)\right]\\ \left[\left(\frac{L}{2}-\frac{-1}{2\left(\pi /L\right)}\right)-\left(0-\frac{1}{2\left(\pi /L\right)}\right)\right]\end{array}\right)\left(\begin{array}{l}\because \sin \left(3\pi /2\right)=-1\\ \text{and}\sin \left(\pi /2\right)=1\end{array}\right)\\ =L^2\left(\begin{array}{c}\left[\left(\frac{3L}{4}+\frac{L}{2\pi }\right)-\left(\frac{L}{4}-\frac{L}{2\pi }\right)\right]\\ \left[\left(\frac{3L}{4}+\frac{L}{2\pi }\right)-\left(\frac{L}{4}-\frac{L}{2\pi }\right)\right]\end{array}\right)\\ =L^2\left(\left(\frac{3L}{4}+\frac{L}{2\pi }-\frac{L}{4}+\frac{L}{2\pi }\right)\left(\frac{3L}{4}+\frac{L}{2\pi }-\frac{L}{4}+\frac{L}{2\pi }\right)\right)\end{array}$

On further solving the above integration,

    $\begin{array}{c}\text{Probability}=L^2\left(\left(\frac{3L}{4}+\frac{L}{2\pi }-\frac{L}{4}+\frac{L}{2\pi }\right)\left(\frac{3L}{4}+\frac{L}{2\pi }-\frac{L}{4}+\frac{L}{2\pi }\right)\right)\\ =L^2\left(\left(\frac{2L}{4}+\frac{2L}{2\pi }\right)\left(\frac{2L}{4}+\frac{2L}{2\pi }\right)\right)\\ =L^2\left(\left(\frac{L}{2}+\frac{L}{\pi }\right)\left(\frac{L}{2}+\frac{L}{\pi }\right)\right)\\ =L^2{\left(\frac{L}{2}+\frac{L}{\pi }\right)}^2\end{array}$

Hence, the probability of finding the particle between $x=L/4$ and $x=3L/4$, $y=L/4$ and $y=3L/4$ is $L^2{\left(\frac{L}{2}+\frac{L}{\pi }\right)}^2$.