:-h_bar/2m d2ψ/dx2 + V(x)ψ = Eψ (1) |
:- |
where h_bar is Plancks constant divided by 2π, m is the mass of the particle, ψ is the wavefunction that we want to find, V(x) is a function describing the potential at each point x and E is the energy. For the case of the particle in a 1-dimensional box of length L, the potential is zero inside the box, but rises abruptly to infinity at x = 0 and x = L. Thus for the region inside the box V(x) = 0 and Equation 1 reduces to: |
where |
:-h_bar/2m d2ψ/dx2 = Eψ (2) |
:- |
:E = k2 h_bar2 / 2m (3) |
:E = k2 |
:En = n2h_bar2π2/2mL2 =n2h2/8mL2 n = 1,2,... (9) |
:En = n2 |
:-h_bar/2m (d2ψ/dx2 + d2ψ/dy2) = Eψ (11) |
:- |
:-h_bar/2m (Yd2X/dx2 + Xd2Y/dy2) = EXY (13) |
:- |
:-h_bar/2m (X"/X + Y"/Y) = E (14) |
:- |
:-h_bar/2m X"/X = Ex and -h_bar/2m Y"/Y = Ey (15) |
:- |
:-h_bar/2m d2X/dx2 = ExX (16) |
:- |
:-h_bar/2m d2Y/dy2 = EyY (17) |
:- |
:Enx,ny = {(nx/Lx)2 + (ny/Ly)2 + (nz/Lz)2} h2/8m (23) |
:Enx,ny,nz = {(nx/Lx)2 + (ny/Ly)2 + (nz/Lz)2} h2/8m (23) |