Particle in 3D Cubic Rigid Box

6.3 Particle in 3D Cubic Rigid Box

6.3.1 Schrödinger Equation

Consider a particle of mass $m\,$ and energy $E\,$ constrained to move inside a three dimensional cubic box of sides of length equal to a parallel to the $x,\, y,\, z$ -axes respectively. Suppose there is no force acting on the particle in the box. For its position $(x,\, y,\, z)$ given by

    $0< \,x \,< \,a$.

    $0< \,y \,< \,a$.

    $0< \,z \,< \,a$.


The appropriate potential is $V(x,\,y,\,z) = 0$ inside and $V(x,\,y,\,z) = \infty$ outside the box

Particle wave function $\psi(x,\,y,\,z,\,t)$ the wave vector $k\,$, and the angular frequency, $\omega\,$, are related to the momentum, $P\,$, and energy, $E\,$ of the particle as follows

    $P=\hbar k$, and $E=\hbar \,\omega$,


For the motion of the particle inside the box the Schrödinger's time independent equation

    $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\psi=-\frac{2m}{\hbar^2}E \psi$,---------------(1)


It is a partial differential equation and can be solved by the technique of separation of variables. It is assumed that the function $\psi\,$ can be written as a product of three functions, $X,\, Y$ and $Z\,$ each of which depends on only one of the coordinates. The wave-function satisfies the boundary condition that it must be zero at the edges of the box.

    $\psi(x,\,y,\,z)=X(x)Y(y)Z(z)$.---------------(2)


The factors of the wave-function satisfy the boundary conditions

    $X(0) \,= \,X \,(a) \,= \,0$ ---------------(3a)


    $Y(0) \,= \,Y \,(a) \,= \,0$ ---------------(3b)


    $Z(0) \,= \,Z \,(a) \,= \,0$ ---------------(3c)


Substituting () into Eq. (), and rearranging,

Since the Hamiltonian is the sum of three terms with totally separate variables,

Substitute this into the Schrodinger equation to give

    $\frac{X^{''}}{X}+\frac{Y^{''}}{Y}+\frac{Z^{''}}{Z}=-\frac{2m}{\hbar^2}E$,---------------(4)


where ' denotes a derivative with respect to argument. It is evident that the only way in which the above equation can be satisfied at all points within the box is if

    $\frac{X^{''}}{X} \,= \,-k_x^2$, ---------------(5a)


    $\frac{Y^{''}}{Y} \,= \,-k_y^2$, ---------------(5b)


    $\frac{Z^{''}}{Z} \,= \,-k_z^2$, ---------------(5c)


where $k_x^2\,$, $k_y^2\,$, and $k_z^2\,$ are spatial constants. The right-hand sides of the above equations must contain negative, rather than positive, spatial constants, because it would not otherwise be possible to satisfy the boundary conditions.

6.3.2 General Formulae for Energy Eigen Values and Eigen functions

The solutions to the above equations which are properly normalized, and satisfy the boundary conditions, are

    $X(x) \,= \,\sqrt{\frac{2}{a}}\,sin\,(k_x \,x)$, ---------------(6a)


    $Y(y) \,= \,\sqrt{\frac{2}{a}}\,sin\,(k_y \,y)$, ---------------(6b)


    $Z(z) \,= \,\sqrt{\frac{2}{a}}\,sin\,(k_z \,z)$, ---------------(6c)


    $k_x \,= \,\frac{l_x\pi}{a}$, ---------------(7a)


    $k_y \,= \,\frac{l_y\pi}{a}$, ---------------(7b)


    $k_z \,= \,\frac{l_z\pi}{a}$. ---------------(7c)


$l_x\,$, $l_y\,$ and $l_z\,$ are positive integers

    $E=\frac{l^2 \pi^2 \hbar^2}{2 m a^2}$. ---------------(8)


where

    $l^2=l_x^2+l_y^2+l_z^2$. ---------------(9)


6.3.3 Degenerate & Nondegenerate Energy States

$\ast$ The energy of the particle in the ground state is given $E_{111}\,$ by

    $E_{111}=3\pi^2\hbar^2/2 m L^2$


Ground state has only one wave function. No other state has this amount of energy. The Ground state and the energy level are said to be non-degenerate.

$\ast$ In the 3-D cubical box potential the energy of a state depends upon the sum of the squares of the quantum numbers. The particle having a particular value of energy in the excited state has several different stationary states or wave functions. These states and energy eigenvalues are said to be degenerate.

$\ast$ For the first excited state, three combinations of the quantum numbers $(n_x,\, n_y, \, n_z )$ are $(2,\,1,\,1),\, (1,2,1),\, (1,1,2)$. The sum of squares of the quantum numbers in each combination is same (equal to 6). Each state has same energy $= 6 \pi^2\hbar^ 2/ 2 m L^2$. Corresponding to these combinations three different eigen functions and three different states are possible. The first excited state is said to be three-fold or triply degenerate,

Degree of Degeneracy- The number of independent wave functions for the stationary states of an energy level is called as the degree of degeneracy of the energy level. The value of energy levels with the corresponding combinations and sum of squares of the quantum numbers as well as the degree of degeneracy are depicted in the table

$n^2 \,= \, n_x^2+n_y^2+n_z^2$