1. Schrödinger Equation in 3d
Consider a cubic “box” in which an electron of mass m is confined
outside the cube, we have U(x,y,z)= and inside U(x,y,z)=0
hence electron has (x,y,z)=0 on all faces of the cube
(x,y,z)=Asin(k1x)sin(k2y)sin(k3z)
(L,y,z)=0 for all 0<y<L and 0<z<L => k1=n1/L
similarly we need k2=n2 /L and k3=n3 /L
E=(ħ2/2m)(k12+k22+k32) = (ħ2  2/2mL2)(n12+n22+n32) =E1 (n12+n22+n32)
where E1 is ground state energy of 1-d well

2. Particle in a 3-d cube LxLxL
E= (ħ2  2/2mL2)(n12+n22+n32) =E1 (n12+n22+n32)
note: n1, n2 and n3 cannot be zero => (x,y,z)=0
lowest energy in the 3-d box is E1,1,1=3E1
first excited state has any two n’s equal to 1 and the other equal to 2
E1,1,2=E1,2,1=E2,1,1=6E1
state is degenerate

3. Particle in a 3-d box
If the box is such that L1<L2<L3 then the degeneracy is lifted

4. A particle is confined to a three-dimensional box that has sides L1, L2 = 2L1, and L3 = 3L1. Give the quantum numbers n1, n2, n3 that correspond to the lowest ten quantum states of this box.
E = (h2/8mL12 )(n12 + n22/4 + n32/9)
= (h2/288mL12 )(36n12 + 9n22 + n32 ).
The energies in units of h2/288mL12 are listed in the following table.
n1 n2 n E