Q14. Particle in a cubic box: energy levels and degeneracy

(i) Energy eigenvalues (8 marks)

1. V=0 inside (0<x,y,z<a), V=∞ outside. -(ħ²)/(2m)∇²ψ=Eψ.
2. Separate: ψ=X(x)Y(y)Z(z), E=Eₓ+E_y+E_z; each factor obeys the one-dimensional equation -(ħ²)/(2m)(d²X)/(dx²)=EₓX with X(0)=X(a)=0: X=√(2a)sin(nₓπ x)/(a), Eₓ=(nₓ²h²)/(8ma²), nₓ=1,2,3,…
3. Similarly for y and z (same length a). Therefore ψ_(nₓn_yn_z)=(2a)^(3/2)sin(nₓπ x)/(a)sin(n_yπ y)/(a)sin(n_zπ z)/(a),=(h²)/(8ma²)(nₓ²+n_y²+n_z²)
4. Lowest level (1,1,1): E=(3h²)/(8ma²) (non-degenerate, zero-point energy).

(ii) Degeneracy (8 marks)

Definition: the degeneracy g of an energy level is the number of linearly independent wavefunctions (quantum states) that have that energy.

1. E=(3h²)/(4ma²)=(6h²)/(8ma²): nₓ²+n_y²+n_z²=6: (2,1,1),(1,2,1),(1,1,2). No other combination: g=3.
2. E=(7h²)/(4ma²)=(14h²)/(8ma²): n²=14: (3,2,1) and its 3!=6 permutations (all different): g=6.

(iii) Degeneracy and symmetry (4 marks)

• The cube has the symmetry of Oₕ: the three axes are equivalent, so the Hamiltonian is unchanged when x,y,z are interchanged. Wavefunctions that differ only by this permutation, e.g. (2,1,1), (1,2,1), (1,1,2), must have the same energy: this is a symmetry (essential) degeneracy; the degeneracy equals the dimension of an irreducible representation of Oₕ.
• If the box is distorted (a≠ b≠ c) the symmetry falls and the degeneracy is lifted: E=(h²)/(8m)((nₓ²)/(a²)+(n_y²)/(b²)+(n_z²)/(c²)).
• Some degeneracies are accidental: e.g. (5,1,1) and (3,3,3) with n²=27 have the same energy without a symmetry link.