Particle in 3D Rectangular Rigid Box

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6.2 Particle in 3D Rectangular Rigid Box

6.2.1 Schrödinger Equation in Cartesian Co-ordinates

Consider a particle of mass m\, and energy E\, constrained to move in a three-dimensional rectangular potential well having sides of length equal to L_x,\, L_y,\, L_z parallel to the x,\, y,\, z -axes respectively. Suppose there is no force acting on the particle in the box. The appropriate potential is V(x,\,y,\,z) = 0 for its position at a point (x,\, y,\, z) given by

    0< \, x \, < \, L_x
    0< \, y \, < \, L_y
    0< \, z \, < \, L_z

and V(x,\,y,\,z) = \infty outside the box

Fig. Particle in 3-D rectangular box

The total kinetic energy T\, is given by


    T=-\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)=-\frac{\hbar^2}{2m}\triangledown^2\,........... ---------------(1)

\triangledown^2\, is called as the Laplacian operator

    -\frac{\hbar^2}{2m}\triangledown^2 \psi=E \psi

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

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

It is a partial differential equation. 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 x,\, y,\, z respectively. The equation can be solved by the technique of separation of variables.

    \psi=X(x).\,Y(y).\,Z(z) ---------------(3)

The Hamiltonian is the sum of three independent one dimensional Hamiltonians in totally separate variables

    H= H_x+H_y+H_z\,.---------------(4)

    [H_x+H_y+H_z]\psi(x,\,y,\,z)=E \psi(x,\,y,\,z) ---------------(5)

Substituting for \psi\, the Schrödinger's equation in the separated variables is

    X^{''}YZ+XY^{''}Z+XY^{''}Z+\frac{2mE}{\hbar^2}XYZ=0 ---------------(6)

The terms X^{''},\, Y^{''},\, Z^{''} are ordinary derivatives instead of the partial derivatives as each of the functions X,\, Y,\, Z, is a function of one variable only. Dividing the equation by XYZ


Each term of this equation depends on a different variable x,\, y, or z\, and the three variables x,\, y, and z\, are independent. The term k^2=\frac{2mE}{\hbar^2} is a constant for a particular value of T\, kinetic energy. Velocity of the particle is a vector quantity. Velocity and corresponding kinetic energy can be resolved in three components along three co-ordinate axes x,\, y,\, z.

     E \,= E_x+E_y+E_z ---------------(8)

The only way for the equation to remain valid for all of x,\, y, and z\, in the interval is for each term to be constant. Therefore with separation of variables, the Schrödinger's equation separates into three independent equations in x,\, y,\, z respectively as follows

    \frac{-\hbar^2}{2m}\left(\frac{\partial^2 X(x)}{\partial x^2}\right)  \,=E_x X \,(x)
    \frac{-\hbar^2}{2m}\left(\frac{\partial^2 Y(y)}{\partial y^2}\right)  \,=E_y Y \,(y)
    \frac{-\hbar^2}{2m}\left(\frac{\partial^2 Z(z)}{\partial z^2}\right)  \,=E_z Z \,(z) ---------------(9)

These differential equations have the solution:

    X(x)  \,=  A_x \,sin \,(k_x x)+B_x \,cos\, (k_x x)        \,\,\,\,E_x =\frac{\hbar^2 k_x^2}{2m} ---------------(10a)
    Y(y)  \,=  A_y \,sin \,(k_y y)+B_y \,cos\, (k_y y)        \,\,\,\,E_y =\frac{\hbar^2 k_y^2}{2m} ---------------(10b)
    Z(z) \,= A_z \,sin \,(k_z z) +B_z \,cos\, (k_z z)          \,\,\,\,E_z =\frac{\hbar^2 k_z^2}{2m} ---------------(10c)

Applying the boundary conditions

\psi(x=0)=0\rArr\, B_x=0     \,\,\,\,\,\,\,\,\psi(x=L_x)=0\rArr \, k_x L_x=n_x \pi ---------------(10a)

\psi(y=0)=0\rArr \,B_y=0     \,\,\,\,\,\,\,\,\psi(y=L_y)=0\rArr \, k_y L_y=n_y \pi ---------------(10b)

\psi(z=0)=0\rArr \,B_z=0     \,\,\,\,\,\,\,\,\psi(z=L_z)=0\rArr \, k_z L_z=n_z \pi ---------------(11c)

The resulting x-components of eigenfunctions are of the form

    \Psi_n=\sqrt{\frac{2}{L}}\,sin\,\left(\frac{n\pi x}{L}\right) \,\,\,\,\mathbf{n}=1,\,2,\,3,\,4....

The solution for y,\, z components have the same form. The total normalized eigenfunctions for the motion of the particle in the box are given by

    \psi(x,\,y,\,z)=\sqrt{\frac{8}{L_x L_y L_z}}sin(k_x x)\,sin\, (k_y y)\,sin(k_z z)

    k_x=\frac{\pi n_x}{L_x} ,\, k_y=\frac{\pi n_y}{L_y} ,\, k_z=\frac{\pi n_z}{L_z} ---------------(12) 
    n_x= 1,\, 2,\, 3,\, 4\, ....
    n_y= 1,\, 2,\, 3,\, 4 \,....
    n_z= 1,\, 2,\, 3,\, 4\, .... ---------------(13)

The eigenfunctions are zero outside the box. The eigenvalues of energy E\, are given by

    E=\frac{\hbar^2(k_x^2+k_y^2+k_z^2)}{2m}=\frac{\hbar^2\pi^2}{2m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\right) ---------------(14)

These eigenvalues of the energy E\, are called as the energy levels of the particle. They form quantized or discrete energy spectrum.

\ast The integers n_x,\, n_y,\,  n_z are called as the quantum numbers. They are required to specify each stationary state. If the sign of the quantum numbers is changed, then there is no change in the energy and in the wave function except for the minus sign. Hence all the stationary states are given by the positive integral values of n_x,\, n_y,\,  n_z.

\ast The value of any of the quantum numbers n_x,\, n_y,\,  n_z cannot be zero. If any one is taken as zero then \psi (x,\,y,\,z) = 0. It implies that the particle does not exist in the box

\ast The lowest possible energy is obtained for n_{x,} =1,\, n_{y,}=1,\,   n_z =1. It is called as the ground state energy and it depends on the values of L_x,\, L_y,\, L_z


The wave function corresponding to ground state energy is called as ground state wave function and denoted by \Psi_{111}\,

\ast The excited state energy levels E_{nx,\,ny,\,nz} are obtained by substituting different positive integer values for n_x,\,n_y,\,n_z and the corresponding eigen functions \Psi_{nx,\,ny,\,nz} are labeled by these quantum numbers

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