Consider a particle in a rigid box of length L. For each of the states $\mathrm{n}=1,\mathrm{n}=2,$and $\mathrm{n}=3$:

a. Sketch graphs of ${\left|\mathrm{\psi }\left(\mathrm{x}\right)\right|}^2$. Label the points $\mathrm{x}=0$and $\mathrm{x}=\mathrm{L}$.

b. Where, in terms of L, are the positions at which the particle is most likely to be found?

c. Where, in terms of L, are the positions at which the particle is least likely to be found?

d. Determine, by examining your ${\left|\mathrm{\psi }\left(\mathrm{x}\right)\right|}^2$graphs, if the probability of finding the particle in the left one-third of the box is less than, equal to, or greater than $\frac{1}{3}$. Explain your reasoning.

e. Calculate the probability that the particle will be found in the left one-third of the box

We have given,

One dimensional box of length L.

We have to sketch the graph of${\left|\mathrm{\psi }\left(\mathrm{x}\right)\right|}^2$for different value of n.

We have to find the positon where the particle can mostly found.

From the graph we can see that particle will be most likely be found in between$\frac{\mathrm{L}}{2}$, for $\mathrm{n}=1$. Since there the probability density is at the peak. Similarly for$\mathrm{n}=2$, the particle will mostly found at$\frac{\mathrm{L}}{4},\frac{3L}{4}$and for the$\mathrm{n}=3$, the particle most likely be found at$\frac{\mathrm{L}}{6,}\frac{L}{2},\frac{5L}{6}$.

We have to find the least found distances in box for the particles.

From the probability density graph we can say that where the curve will have most less height where the particle will least found.

so, for $\mathrm{n}=1$,particle will least found at the edges of the wall.

For $\mathrm{n}=2$, the particle will least found at the edges and at the center of the box.

For $\mathrm{n}=3$, the particle will least found at $\frac{\mathrm{L}}{3},\frac{2L}{3}$and at the edges.

We have to find the probability for the left side of the box for one third part in each cases.

For $\mathrm{n}=1$,the probability will be,

$$\begin{gathered} {\mathrm{P}}_1={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_1={\int }_0^{\frac{\mathrm{L}}{3}}\left|\frac{2}{\mathrm{L}}{\sin }^2\frac{\mathrm{\pi x}}{\mathrm{L}}\right|\mathrm{dx} \\ {\mathrm{P}}_1=\frac{1}{3} \end{gathered}$$

then it is equal to one third.

For $\mathrm{n}=2$, the probability will be,

$$\begin{gathered} {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}\frac{2}{\mathrm{L}}{\sin }^2\frac{2\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx} \\ {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}\frac{1}{\mathrm{L}}(1-\cos \frac{4\mathrm{\pi x}}{\mathrm{L}})\mathrm{dx} \\ {\mathrm{P}}_2=\frac{1}{\mathrm{L}}{\left(\mathrm{x}-\sin \frac{4\mathrm{\pi x}}{\mathrm{L}}\times \frac{\mathrm{L}}{4\mathrm{\pi }}\right)}_0^{\frac{\mathrm{L}}{3}} \\ {\mathrm{P}}_2=\frac{1}{3}+\frac{\sqrt{3}}{8\mathrm{\pi }} \end{gathered}$$

it is greater than one third.

For $\mathrm{n}=3$, the probability will be,

$$\begin{gathered} {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}\frac{2}{\mathrm{L}}{\sin }^2\frac{3\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx} \\ {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}\frac{1}{\mathrm{L}}(1-\cos \frac{6\mathrm{\pi x}}{\mathrm{L}})\mathrm{dx} \\ {\mathrm{P}}_3=\frac{1}{\mathrm{L}}{\left(\mathrm{x}-\sin \frac{6\mathrm{\pi x}}{\mathrm{L}}\times \frac{\mathrm{L}}{6\mathrm{\pi }}\right)}_0^{\frac{\mathrm{L}}{3}} \\ {\mathrm{P}}_3=\frac{1}{3} \end{gathered}$$

it is equal to one third.

We have to find probability of the particle in one third part.

$$\begin{gathered} {\mathrm{P}}_1={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_1={\int }_0^{\frac{\mathrm{L}}{3}}\left|\frac{2}{\mathrm{L}}{\sin }^2\frac{\mathrm{\pi x}}{\mathrm{L}}\right|\mathrm{dx} \\ {\mathrm{P}}_1=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}\frac{2}{\mathrm{L}}{\sin }^2\frac{2\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx} \\ {\mathrm{P}}_2={\int }_0^{\frac{\mathrm{L}}{3}}\frac{1}{\mathrm{L}}(1-\cos \frac{4\mathrm{\pi x}}{\mathrm{L}})\mathrm{dx} \\ {\mathrm{P}}_2=\frac{1}{\mathrm{L}}{\left(\mathrm{x}-\sin \frac{4\mathrm{\pi x}}{\mathrm{L}}\times \frac{\mathrm{L}}{4\mathrm{\pi }}\right)}_0^{\frac{\mathrm{L}}{3}} \\ {\mathrm{P}}_2=\frac{1}{3}+\frac{\sqrt{3}}{8\mathrm{\pi }} \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}{\left|\mathrm{\psi }\left(\mathrm{X}\right)\right|}^2\mathrm{dx} \\ {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}\frac{2}{\mathrm{L}}{\sin }^2\frac{3\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx} \\ {\mathrm{P}}_3={\int }_0^{\frac{\mathrm{L}}{3}}\frac{1}{\mathrm{L}}(1-\cos \frac{6\mathrm{\pi x}}{\mathrm{L}})\mathrm{dx} \\ {\mathrm{P}}_3=\frac{1}{\mathrm{L}}{\left(\mathrm{x}-\sin \frac{6\mathrm{\pi x}}{\mathrm{L}}\times \frac{\mathrm{L}}{6\mathrm{\pi }}\right)}_0^{\frac{\mathrm{L}}{3}} \\ {\mathrm{P}}_3=\frac{1}{3} \end{gathered}$$