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

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

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

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 n3 E 1 1 1 49 1 1 2 61 1 2 1 76 1 1 3 81 1 2 2 88 1 2 3 108 1 1 4 109 1 3 1 121 1 3 2 133 1 2 4 136