Question: An electron is trapped in a cubic 3D well. In the states $\left(n_x,n_y,n_z\right)$= (a) (2,1,1) (b) (1,2,1)(c) (1,1,2), what is the probability of finding the electron in the region $\left(\mathbf{0}⩽\mathbf{x}⩽\mathbf{L},\mathbf{L}\mathbf{/}\mathbf{3}⩽\mathbf{y}⩽\mathbf{2}\mathbf{L}\mathbf{/}\mathbf{3},\mathbf{0}⩽\mathbf{z}⩽\mathbf{L}\right)$. Discus any difference in these results.

The state of electron in (a) is $\left(n_x,n_y,n_z\right)=\left(2,1,1\right)$

The state of electron in (b) is $\left(n_x,n_y,n_z\right)=\left(1,2,1\right)$

The state of electron in (c) is $\left(n_x,n_y,n_z\right)=\left(1,1,2\right)$

The normalization constant is the constant which is multiplied to non-negative area to become unity.

The probability of finding an electron in the region is given by:

$p\left(n_x,n_y,n_z\right)=\underset{x=0,y=L/3,z=0}{\overset{x=L,y=2L/3,z=L}{\int }}{\left[{\left(\frac{2}{L}\right)}^{3/2}\sin \left(\frac{n_x\pi x}{L}\right)\sin \left(\frac{n_y\pi y}{L}\right)\sin \left(\frac{n_z\pi z}{L}\right)\right]}^{2}dxdydz......\left(1\right)$

Substitute all the values in the above equation (1).

$$\begin{gathered} p\left(2,1,1\right)=\underset{x=0,y=L/3,z=0}{\overset{x=L,y=2L/3,z=L}{\int }}{\left[{\left(\frac{2}{L}\right)}^{3/2}\sin \left(\frac{\left(2\right)\pi x}{L}\right)\sin \left(\frac{\left(1\right)\pi y}{L}\right)\sin \left(\frac{\left(1\right)\pi z}{L}\right)\right]}^{2}dxdydz \\ p\left(2,1,1\right)={\left(\frac{2}{L}\right)}^3\left[\underset{x=0}{\overset{x=L}{\int }}{\sin }^2\left(\frac{2\pi x}{L}\right)dx\right]\left[\underset{y=L/3}{\overset{y=2L/3}{\int }}{\sin }^2\left(\frac{\pi y}{L}\right)dy\right]\left[\underset{z=0}{\overset{z=L}{\int }}{\sin }^2\left(\frac{\pi z}{L}\right)dz\right] \\ p\left(2,1,1\right)={\left(\frac{2}{L}\right)}^3{\left(\frac{L}{2}\right)}^2\left[\left(\frac{\left(2L/3\right)-\left(L/3\right)}{2}\right)-\frac{L}{4\pi }\left[\sin \left(\frac{2\pi \left(2L/3\right)}{L}\right)-\sin \left(\frac{2\pi \left(L/3\right)}{L}\right)\right]\right] \\ p\left(2,1,1\right)=\frac{2}{L}\left(\frac{L}{6}+\frac{L}{4\pi }\sqrt{3}\right) \end{gathered}$$

$$\begin{gathered} p\left(2,1,1\right)=\left(\frac{1}{3}-\frac{1}{4\pi }\sqrt{3}\right) \\ p\left(2,1,1\right)=0.196 \end{gathered}$$

Therefore, the probability of finding trapped electron for states (1,2,1) is 0.196.