Chapter 43, Problem 37P

To determine

To show that the energy of the electron in a three dimensional box of edge length $L$ and volume $L^3$ having wave function of $\psi =A\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\sin \left(k_{z}z\right)$ is $E=\frac{{\hslash }^2{\pi }^2}{2m_e{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$.

Answer to Problem 37P

It is showed that the energy of the electron in a three dimensional box of edge length $L$ and volume $L^3$ having wave function of $\psi =A\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\sin \left(k_{z}z\right)$ is $E=\frac{{\hslash }^2{\pi }^2}{2m_e{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$.

Explanation of Solution

Write the wave function of the particle given in question.

  $\psi =A\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\sin \left(k_{z}z\right)$                                                                              (I)

Here, $\psi$ is the total wave function of the particle, $k_x$ is the wave vector in $x$ direction, $k_y$ is the wave vector in $y$ direction and $k_z$ is the wave vector in $z$ direction.

The electron moves in a cub of length $L$.

Substitute $L$ for $x,y\text{and}\text{z}$ in above equation to get wave function at boundary.

  $\psi =A\sin \left(k_{x}L\right)\sin \left(k_{y}L\right)\sin \left(k_{z}L\right)$                                                                            (II)

At boundary the wave function vanishes. Therefore, at $x=L$, $\psi =0$.

Substitute $0$ for $\psi$ in equation (II) to get $k_x,k_y$ and $k_z$ values.

  $0=A\sin \left(k_{x}L\right)\sin \left(k_{y}L\right)\sin \left(k_{z}L\right)$                                                                           (III)

  $\begin{array}{l}\sin \left(k_{x}L\right)=0\text{or}\sin \left(k_{y}L\right)=0\text{or}\sin \left(k_{z}L\right)=0\\ k_{x}L=n_x\pi k_{y}L=n_y\pi k_{z}L=n_z\pi \\ k_x=\frac{n_x\pi }{L}{k}_y=\frac{n_y\pi }{L}{k}_y=\frac{n_y\pi }{L}\end{array}$

Solve above equation for $k_x$.

  $\begin{array}{c}\sin \left(k_{x}L\right)=0\\ k_{x}L=n_x\pi \text{where}{n}_x=1,2,\mathrm{3...}\\ k_x=\frac{n_x\pi }{L}\end{array}$

Solve above equation for $k_y$.

  $\begin{array}{c}\sin \left(k_{y}L\right)=0\\ k_{y}L=n_y\pi \text{where}{n}_x=1,2,\mathrm{3...}\\ k_y=\frac{n_y\pi }{L}\end{array}$

Solve above equation for $k_z$.

  $\begin{array}{c}\sin \left(k_{z}L\right)=0\\ k_{z}L=n_z\pi \text{where}{n}_x=1,2,\mathrm{3...}\\ k_z=\frac{n_z\pi }{L}\end{array}$

Substitute $\frac{n_x\pi }{L}$ for $k_x$, $\frac{n_y\pi }{L}$ for $k_y$ and $\frac{n_z\pi }{L}$ for $k_z$ in equation (I) to get $\psi$.

  $\psi =A\sin \left(\frac{n_x\pi }{L}x\right)\sin \left(\frac{n_y\pi }{L}y\right)\sin \left(\frac{n_z\pi }{L}z\right)$                                                         (IV)

Write the three dimensional Schrodinger equation for the electron moving in a cubical box of potential $U$.

  $\frac{{\hslash }^2}{2m_e}\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}\right)=\left(U-E\right)\psi$                                                                 (V)

Here, $m_e$ is the mass of the electron, $U$ is the potential energy, $E$ is the energy of the electron.

In the problem, the electron is moving in zero potential.

Substitute $0$ for $U$ in equation (V).

  $\begin{array}{l}\frac{{\hslash }^2}{2m_e}\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}\right)=\left(0-E\right)\psi \\ \frac{{\hslash }^2}{2m_e}\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}\right)=-E\psi \end{array}$                                                                    (V)

Substitute $\sin \left(\frac{n_x\pi }{L}L\right)\sin \left(\frac{n_y\pi }{L}L\right)\sin \left(\frac{n_z\pi }{L}L\right)$ for $\psi$ in equation (V) to get $E$.

  $\frac{{\hslash }^2}{2m_e}\left(\begin{array}{l}\frac{{\partial }^2\sin \left(\frac{n_x\pi }{L}L\right)\sin \left(\frac{n_y\pi }{L}L\right)\sin \left(\frac{n_z\pi }{L}L\right)}{\partial x^2}\\ +\frac{{\partial }^2\sin \left(\frac{n_x\pi }{L}L\right)\sin \left(\frac{n_y\pi }{L}L\right)\sin \left(\frac{n_z\pi }{L}L\right)}{\partial y^2}\\ +\frac{{\partial }^2\sin \left(\frac{n_x\pi }{L}L\right)\sin \left(\frac{n_y\pi }{L}L\right)\sin \left(\frac{n_z\pi }{L}L\right)}{\partial z^2}\end{array}\right)=-E\sin \left(\frac{n_x\pi }{L}L\right)\sin \left(\frac{n_y\pi }{L}L\right)\sin \left(\frac{n_z\pi }{L}L\right)$

Simplify above equation.

  $\frac{{\hslash }^2}{2m_e}\left(-{\left(\frac{n_x\pi }{L}\right)}^2-{\left(\frac{n_y\pi }{L}\right)}^2-{\left(\frac{n_z\pi }{L}\right)}^2\right)\sin \left(\frac{n_x\pi }{L}x\right)\sin \left(\frac{n_y\pi }{L}y\right)\sin \left(\frac{n_z\pi }{L}z\right)=-E\sin \left(\frac{n_x\pi }{L}x\right)\sin \left(\frac{n_y\pi }{L}y\right)\sin \left(\frac{n_z\pi }{L}z\right)$

Conclusion:

Substitute $\psi$ for $\sin \left(\frac{n_x\pi }{L}x\right)\sin \left(\frac{n_y\pi }{L}y\right)\sin \left(\frac{n_z\pi }{L}z\right)$ in above equation and rearrange to get $E$.

  $\frac{{\hslash }^2}{2m_e}\left(-{\left(\frac{n_x\pi }{L}\right)}^2-{\left(\frac{n_y\pi }{L}\right)}^2-{\left(\frac{n_z\pi }{L}\right)}^2\right)\psi =-E\psi$

Equate the coefficient of above equation to get $E$.

  $\begin{array}{c}-E=\frac{{\hslash }^2}{2m_e}\left(-{\left(\frac{n_x\pi }{L}\right)}^2-{\left(\frac{n_y\pi }{L}\right)}^2-{\left(\frac{n_z\pi }{L}\right)}^2\right)\\ E=\frac{{\hslash }^2{\pi }^2}{2m_e{L}^2}\left(n_x^2+n_y^2+n_y^2\right)\text{where }{n}_x,n_y,n_z=1,2,3,\mathrm{...}\end{array}$

Therefore, it is showed that the energy of the electron in a three dimensional box of edge length $L$ and volume $L^3$ having wave function of $\psi =A\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\sin \left(k_{z}z\right)$ is $E=\frac{{\hslash }^2{\pi }^2}{2m_e{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$.