91Ó°ÊÓ

In quantum mechanics, the particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle free to move in a small space surrounded by impenetrable barriers.

When a particle is confined to a box of infinite potential well with

potential$U\left(x\right)$.

The expression for the probability of finding a particle inside the box is as follow:

${∫}_{x_1}^{x_2}p\left(x\right)={∫}_{x_1}^{x_2}{A}^2{\sin }^2\left(\frac{n\mathrm{π}}{L}x\right)dx$

Here, $p\left(x\right)$ is the probability of the practice finding a particle inside the box, A is the normalization constant, L is the length of the box,n is the ground state of the practice.

(a)

Find the probability of the practice inside the box having the limits x=0 and x=0.25L as follow:

${∫}_{x_1}^{x_2}p\left(x\right)={∫}_{x_1}^{x_2}{A}^2{\sin }^2\left(\frac{n\mathrm{π}}{L}x\right)dx$

Substitute $\sqrt{\frac{2}{L}}$for A and 1 for n

$$\begin{gathered} {∫}_{x_1}^{x_2}p\left(x\right)={∫}_0^{\frac{L}{4}}{\left(\sqrt{\frac{2}{L}}\right)}^2{\sin }^2\left(\frac{\mathrm{π}}{L}x\right)dx \\ =\frac{2}{L}{∫}_0^{\frac{L}{4}}\left(\frac{1-{\cos }^2\left({\displaystyle \frac{\mathrm{π}}{L}}x\right)}{2}\right)dx \\ =\frac{2}{L}\left({∫}_0^{\frac{L}{4}}\left(\frac{1}{2}\right)dx-{∫}_0^{\frac{L}{4}}\left(\frac{{\cos }^2\left({\displaystyle \frac{\mathrm{π}}{L}}x\right)}{2}\right)dx\right) \\ =\frac{2}{L}\left(\frac{1}{2}\left(\frac{L}{4}\right)-\left(\frac{1}{2}\right)\frac{\sin 2\left({\displaystyle \frac{\mathrm{π}}{L}}\left({\displaystyle \frac{L}{4}}\right)-0\right)}{\left({\displaystyle \frac{2\mathrm{π}}{L}}\right)}\right) \end{gathered}$$

Further simplification will give,

$$\begin{gathered} {∫}_0^{\frac{L}{4}}P\left(x\right)=\frac{2}{L}\left(\frac{L}{8}-\frac{L}{4\mathrm{π}}\right) \\ =\frac{1}{4}-\frac{1}{2\mathrm{π}} \\ =0.091 \end{gathered}$$

Hence, the value is 0.091.

(b)

Find the probability of the particle inside the box having the limits $x=\frac{3L}{4}$and x=L as follows:

${∫}_{x_1}^{x_2}p\left(x\right)={∫}_{x_1}^{x_2}{A}^2{\sin }^2\left(\frac{n\mathrm{π}}{L}x\right)d$

Substitute $\sqrt{\frac{2}{L}}$for A and 1 forn

$$\begin{gathered} {∫}_{\frac{3L}{4}}^{L}P\left(x\right)={∫}_{\frac{3L}{4}}^L{\left(\sqrt{\frac{2}{L}}\right)}^2{\sin }^2\left(\frac{\mathrm{π}}{L}x\right)dx \\ =\frac{2}{L}{∫}_{\frac{3L}{4}}^L\left(\frac{1-\cos 2\left({\displaystyle \frac{\mathrm{π}}{L}}x\right)}{2}\right)dx \\ =\frac{2}{L}\left({∫}_{\frac{3L}{4}}^L\left(\frac{1}{2}\right)dx-{∫}_{\frac{3L}{4}}^L\left(\frac{\cos 2\left({\displaystyle \frac{\mathrm{π}}{L}}x\right)}{2}\right)dx\right) \\ =\frac{2}{L}\left(\frac{1}{2}\left(L-\frac{3L}{4}\right)-\left(\frac{1}{2}\right)\frac{\sin 2\left({\displaystyle \frac{\mathrm{π}}{L}}\left(L-{\displaystyle \frac{3L}{4}}\right)-0\right)}{\left({\displaystyle \frac{2\mathrm{π}}{L}}\right)}\right) \end{gathered}$$

Further simplification will give,

$$\begin{gathered} {∫}_{\frac{3L}{4}}^{L}P\left(x\right)=\frac{2}{L}\left(\frac{L}{8}-\frac{L}{4\mathrm{π}}\right) \\ =\frac{1}{4}-\frac{1}{2\mathrm{π}} \\ =0.091 \end{gathered}$$

Hence, the value is 0.091.

(c)

Find the probability of the particle inside the box having the limits $x=\frac{L}{4}$and $x=\frac{3L}{4}$as follows:

The total probability to find the particle inside a box is 1.

Therefore, the probability between the limits is,

$$\begin{gathered} p+0.091+0.091=1 \\ p=0.82 \end{gathered}$$

Hence, the value is 0.82.