Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 30 : Quantum Mechanical Concepts
with $ k_x^2+k_y^2+k_z^2=0$.
  The solutions of the above with boundary condition (i.e. wanishing of wavefunction at the walls) gives
 
$\displaystyle \psi_x(x) = A\sin(k_xx)$
  where $ k_x = n_x\pi/L$, $ n_x$ being any non-zero positive integer. Thus the complete solution (with normalization constant) is
 
$\displaystyle \psi(x,y,z) = \frac{2^{3/2}}{\sqrt{V}}\sin k_xx\sin k_yy\sin k_zz$
  and the energy
 
$\displaystyle E = \frac{\hbar^2\pi^2}{2m}\left(\frac{n_x^2}{L^2}+\frac{n_y^2}{L^2}+ \frac{n_z^2}{L^2}\right)\eqno (B)$
  where $ V= L^3$ is the volume of the crystal.
 
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