A particle is in the ground level of a box that extends from x= 0 to x= L. (a) What is the probability of finding the particle in the region between 0 and L/4? Calculate this by integrating ${\left|\mathrm{\psi }\left(\mathrm{x}\right)\right|}^2\mathrm{dx}$ , where c is normalized, from x= 0 to x= L/4. (b) What is the probability of finding the particle in the region x= L/4 to x= L/2 ? (c) How do the results of parts (a) and (b) compare? Explain. (d) Add the probabilities calculated in parts (a) and (b). (e) Are your results in parts (a), (b), and (d) consistent with Fig. 40.12b? Explain.

The probability distribution between two points is given as,

$\underset{\mathrm{x}1}{\overset{\mathrm{x}2}{\int }}{\left|\mathrm{\psi }\right|}^2\mathrm{dx}$

Also, the ground state wave function for particle in a box is given as,

${\mathrm{\psi }}_1=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{\mathrm{TTX}}{\mathrm{L}}$

Now, use standard relations as,

$$\begin{gathered} \left(1\right){\sin }^2\mathrm{\theta }=\frac{1}{2}\left(1-\cos 2\mathrm{\theta }\right) \\ \left(2\right)\int \cos \mathrm{axdx}=\frac{1}{\mathrm{\alpha }}\sin \mathrm{ax} \end{gathered}$$

Thus, the probability distribution is,

$$\begin{gathered} \underset{\mathrm{x}1}{\overset{\mathrm{x}2}{\int }}{\left|\mathrm{\psi }\right|}^2\mathrm{dx}=\frac{2}{\mathrm{L}}\underset{0}{\overset{\mathrm{L}/4}{\int }}{\sin }^2\frac{\mathrm{TTX}}{\mathrm{L}}\mathrm{dx} \\ =\frac{2}{\mathrm{L}}\underset{0}{\overset{\mathrm{L}/4}{\int }}\frac{1}{2}\left(1-\cos \frac{2\mathrm{TTX}}{\mathrm{L}}\right)\mathrm{dx} \\ =\frac{1}{\mathrm{L}}{\overline{)\left(\mathrm{x}-\frac{\mathrm{L}}{2\mathrm{TT}}\sin \frac{2\mathrm{TTX}}{\mathrm{L}}\right)}}_0^{\mathrm{L}/4} \\ =\frac{1}{4}-\frac{1}{2\mathrm{TT}} \\ =0.0908 \end{gathered}$$

(b) Repeat the same process just changing the limits of the integration,

$$\begin{gathered} \underset{\mathrm{x}1}{\overset{\mathrm{x}2}{\int }}{\left|\mathrm{\psi }\right|}^2\mathrm{dx}=\frac{2}{\mathrm{L}}\underset{L/4}{\overset{\mathrm{L}/2}{\int }}{\sin }^2\frac{\mathrm{TTX}}{\mathrm{L}}\mathrm{dx} \\ =\frac{2}{\mathrm{L}}\underset{L/4}{\overset{\mathrm{L}/2}{\int }}\frac{1}{2}\left(1-\cos \frac{2\mathrm{TTX}}{\mathrm{L}}\right)\mathrm{dx} \\ =\frac{1}{\mathrm{L}}{\overline{)\left(\mathrm{x}-\frac{\mathrm{L}}{2\mathrm{TT}}\sin \frac{2\mathrm{TTX}}{\mathrm{L}}\right)}}_{L/4}^{\mathrm{L}/2} \\ =\frac{1}{4}-\frac{1}{2\mathrm{TT}} \\ =0.409 \end{gathered}$$

(c) There is greater probability of the particle being at the middle point or near the middle point of the box than at the edge.

The probabilities add up to,

$$\begin{gathered} 0.409+0.0908 \\ =0.5 \\ =\frac{1}{2} \end{gathered}$$

Since, the particle cannot escape the box, the probability of finding it anywhere inside the box is unity. Therefore, it is equally likely to find the particle between x = 0 to x = L/2 and x = L/2 and x = L with a probability each of $\frac{1}{2}$ .

Refer to the Fig. 40.12b from the text book.

It is concluded that the results are consistent with it as the probability is greater at the centre of the box and also the probability distribution is symmetric about the centre too.