Schrödinger Wave Function Of A Particle In A Box

Assume a box contains a particle and that the interior corresponds to:

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

for some L_x, L_y and L_z. Assume also that the potential energy is zero for the interior and unbounded for the exterior. The Hamiltonian eigenfunctions are:

[A_xsin(k_xx) + B_xcos(k_xx)][A_ysin(k_yy) + B_ycos(k_yy)][A_zsin(k_zz) + B_zcos(k_zz)]

for some A_x, B_x, A_y, B_y, A_z, B_z, k_x, k_y and k_z.  The corresponding eigenvalues are:

[ħ² / (2m)] (k_x² + k_y² + k_z²)

where m is the mass of the particle.  Because the Hamiltonian eigenfunctions are zero at the edges:

B_x = B_y = B_z = 0, k_xL_x = n_xπ, k_yL_y = n_yπ and k_zL_z = n_zπ

for some integers n_x, n_y and n_z.  Because the volume integrals of the squares over all space are one:

A_x = √ 2 / L_x, A_y = √ 2 / L_y and A_z = √ 2 / L_z.

The Hamiltonian eigenfunctions become solutions to the Schrödinger equation when the corresponding exponential time functions are added. Therefore, solutions to the Schrödinger equation consist of combinations of functions of the following form:

sin(jπx / L_x) sin(kπy / L_y) sin(lπz / L_z) e^{-iE_jklt / ħ}

where:

E_jkl = (π² ħ²)(j² / L_x² + k² / L_y² + l² / L_z²) / (2m)

for integers j, k and l. The coefficients must be chosen to give the correct initial Schrödinger wave function.