Chapter 28, Problem 57P

(a)

To determine

The expectation value of $x$.

Answer to Problem 57P

The expectation value of $x$ is $\frac{L}{2}$.

Explanation of Solution

Write the equation for the expectation value of $x$.

  $\langle x\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}{\Psi }_2^{\ast }}x{\Psi }_{2}dx$        (I)

Here, $\langle x\rangle$ is the expectation value of $x$ , ${\Psi }_2$ is the given wave function, ${\Psi }_2^{\ast }$ is the complex conjugate of the given wave function and $L$ is the length of the square well.

Write the expression of the given wave function.

  ${\Psi }_2\left(x\right)=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)$        (II)

Write the expression for the complex conjugate of the given wave function.

  ${\Psi }_2^{\ast }\left(x\right)=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)$        (III)

Put equations (II) and (III) in equation (I) and rearrange it.

  $\begin{array}{c}\langle x\rangle ={\displaystyle \underset{0}{\overset{L}{\int }}\left(\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)\right)x\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)dx}\\ =\frac{2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\left({\sin }^2\left(\frac{2\pi x}{L}\right)\right)xdx}\\ =\frac{2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\left(\frac{1-\cos 2\left(\frac{2\pi x}{L}\right)}{2}\right)xdx}\\ =\frac{1}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\left(1-\cos \frac{4\pi x}{L}\right)xdx}\end{array}$

Integrate the above equation.

  $\begin{array}{c}\langle x\rangle =\frac{1}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\left(x-x\cos \frac{4\pi x}{L}\right)dx}\\ =\frac{1}{L}{\left[\frac{x^2}{2}\right]}_0^L-\frac{1}{L}\frac{L^2}{16{\pi }^2}{\left[\frac{4\pi x}{L}\sin \frac{4\pi x}{L}+\cos \frac{4\pi x}{L}\right]}_0^L\\ =\frac{1}{L}\frac{L^2}{2}-0-\frac{1}{L}\frac{L^2}{16{\pi }^2}\frac{4\pi L}{L}\sin \frac{4\pi L}{L}+\cos \frac{4\pi L}{L}-0-\cos 0\\ =\frac{L}{2}\end{array}$

Conclusion:

Therefore, the expectation value of $x$ is $\frac{L}{2}$.

(b)

To determine

The probability of finding the particle near $\frac{L}{2}$ by calculating the probability that the particle lies in the range $0.490L\le x\le 0.510L$.

Answer to Problem 57P

The probability of finding the particle near $\frac{L}{2}$ by calculating the probability that the particle lies in the range $0.490L\le x\le 0.510L$ is $5.26\times {10}^{-5}$.

Explanation of Solution

Write the equation for the probability that the particle lies in the range $0.490L\le x\le 0.510L$.

  $P={\displaystyle \underset{0.490L}{\overset{0.510L}{\int }}{\Psi }_2^{\ast }}{\Psi }_{2}dx$        (IV)

Here, $P$ is the probability.

Put equations (II) and (III) in equation (IV) and rearrange it.

  $\begin{array}{c}P={\displaystyle \underset{0.490L}{\overset{0.510L}{\int }}\left(\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)\right)\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)}dx\\ =\frac{2}{L}{\displaystyle \underset{0.490L}{\overset{0.510L}{\int }}{\sin }^2\left(\frac{2\pi x}{L}\right)dx}\\ =\frac{2}{L}{\displaystyle \underset{0.490L}{\overset{0.510L}{\int }}\left(\frac{1-\cos 2\left(\frac{2\pi x}{L}\right)}{2}\right)dx}\\ =\frac{1}{L}{\displaystyle \underset{0.490L}{\overset{0.510L}{\int }}\left(1-\cos \frac{4\pi x}{L}\right)dx}\end{array}$

Integrate the above equation.

  $\begin{array}{c}P=\frac{1}{L}{\left[x\right]}_{0.490L}^{0.510L}-\frac{1}{L}\frac{L}{4\pi }{\left[\sin \frac{4\pi x}{L}\right]}_{0.490L}^{0.510L}\\ =\frac{1}{L}\left(0.510L-0.490L\right)-\frac{1}{4\pi }\left[\sin \left(\frac{4\pi \times 0.510L}{L}\right)-\sin \left(\frac{4\pi \times 0.490L}{L}\right)\right]\\ =0.020-\frac{1}{4\pi }\left(\sin 2.04\pi -\sin 1.96\pi \right)\\ =5.26\times {10}^{-5}\end{array}$

Conclusion:

Therefore, the probability of finding the particle near $\frac{L}{2}$ by calculating the probability that the particle lies in the range $0.490L\le x\le 0.510L$ is $5.26\times {10}^{-5}$ .

(c)

To determine

The probability of finding the particle near $\frac{L}{4}$ by calculating the probability that the particle lies in the range $0.240L\le x\le 0.260L$ .

Answer to Problem 57P

The probability of finding the particle near $\frac{L}{4}$ by calculating the probability that the particle lies in the range $0.240L\le x\le 0.260L$ is $3.99\times {10}^{-2}$.

Explanation of Solution

Write the equation for the probability that the particle lies in the range $0.240L\le x\le 0.260L$ .

  $P={\displaystyle \underset{0.240L}{\overset{0.260L}{\int }}{\Psi }_2^{\ast }}{\Psi }_{2}dx$        (V)

Put equations (II) and (III) in equation (IV) and rearrange it.

  $\begin{array}{c}P={\displaystyle \underset{0.240L}{\overset{0.260L}{\int }}\left(\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)\right)\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)}dx\\ =\frac{2}{L}{\displaystyle \underset{0.240L}{\overset{0.260L}{\int }}{\sin }^2\left(\frac{2\pi x}{L}\right)dx}\\ =\frac{2}{L}{\displaystyle \underset{0.240L}{\overset{0.260L}{\int }}\left(\frac{1-\cos 2\left(\frac{2\pi x}{L}\right)}{2}\right)dx}\\ =\frac{1}{L}{\displaystyle \underset{0.240L}{\overset{0.260L}{\int }}\left(1-\cos \frac{4\pi x}{L}\right)dx}\end{array}$

Integrate the above equation.

  $\begin{array}{c}P=\frac{1}{L}{\left[x\right]}_{0.240L}^{0.260L}-\frac{1}{L}\frac{L}{4\pi }{\left[\sin \frac{4\pi x}{L}\right]}_{0.240L}^{0.260L}\\ =\frac{1}{L}\left(0.260L-0.240L\right)-\frac{1}{4\pi }\left[\sin \left(\frac{4\pi \times 0.260L}{L}\right)-\sin \left(\frac{4\pi \times 0.240L}{L}\right)\right]\\ =3.99\times {10}^{-2}\end{array}$

Conclusion:

Therefore, the probability of finding the particle near $\frac{L}{4}$ by calculating the probability that the particle lies in the range $0.240L\le x\le 0.260L$ is $3.99\times {10}^{-2}$ .

(d)

To determine

The argument for the statement that the result of part (a) does not contradict the result of part (b) and part (c).

Answer to Problem 57P

It is more probable to find the particle either near $x=L/4$ or near $x=3L/4$ than at the center where the probability is zero and the symmetry of the wave function ensures that the average position is $x=L/2$ .

Explanation of Solution

Probability density is the relative probability per unit volume that the particle will be found at any given point in the volume. The probability density for the given function with $n=2$ is shown in figure 28.22b. From the figure it is clear that the it is more probable to find the particle either near $x=L/4$ or near $x=3L/4$ than at the center where the probability is zero.

The expectation value of $x$ is the average position at which a particle will be found after many measurements. The given wave function is a function of sine which is symmetric. The symmetry of the wave function ensures that the average position will be at $x=L/2$ .

Conclusion:

Thus, it is more probable to find the particle either near $x=L/4$ or near $x=3L/4$ than at the center where the probability is zero and the symmetry of the wave function ensures that the average position is $x=L/2$.