Schr

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 0ltyltL and 0ltzltL gt k1n1?/L
• similarly we need k2n2 ?/L and k3n3 ?/L

• E(h2/2m)(k12k22k32) (h2 ? 2/2mL2)(n12n22n32
  ) E1 (n12n22n32)
• where E1 is ground state energy of 1-d well

2
Particle in a 3-d cube LxLxL

• E (h2 ? 2/2mL2)(n12n22n32) E1 (n12n22n32)
• note n1, n2 and n3 cannot be zero gt ?(x,y,z)0
• lowest energy in the 3-d box is E1,1,13E1
• first excited state has any two ns equal to 1
  and the other equal to 2
• E1,1,2E1,2,1E2,1,16E1
• state is degenerate

3
Particle in a 3-d box

• If the box is such that L1ltL2ltL3 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 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