[DutchGirl13]

I am asked to find the degeneracies of the first four energy levels for a particle in a 3D box with dimensions a=b=1.5c.
For energy, I have the expression E_nxnynz=(h^2/8m)*(n_x²/a² +n_y²/b²+ n_z²/c²), which simplifies into E_nxnynz=(h^2/8ma²)*(n_x²+n_y²+ 2.25 n_z²). The first energy level, E₁₁₁ is 4.25M, where M= (h^2/8ma²). The next level, variants of 211 have two arrangements that result in 7.25. The 221 variants have two arrangements that result in 14M. The 311 variants have two arrangements that result in 12.25M. Therefore the first level has onefold degeneracy, while the other three have twofold. However, the back of the book gives the answer that only 211 is doubly degenerate, with the rest being singly degenerate. Could someone lead me through the right process? This is 3-32 from MacQuarrie second edition.

[juanrga]

I am asked to find the degeneracies of the first four energy levels for a particle in a 3D box with dimensions a=b=1.5c.
For energy, I have the expression E_nxnynz=(h^2/8m)*(n_x²/a² +n_y²/b²+ n_z²/c²), which simplifies into E_nxnynz=(h^2/8ma²)*(n_x²+n_y²+ 2.25 n_z²). The first energy level, E₁₁₁ is 4.25M, where M= (h^2/8ma²). The next level, variants of 211 have two arrangements that result in 7.25. The 221 variants have two arrangements that result in 14M. The 311 variants have two arrangements that result in 12.25M. Therefore the first level has onefold degeneracy, while the other three have twofold. However, the back of the book gives the answer that only 211 is doubly degenerate, with the rest being singly degenerate. Could someone lead me through the right process? This is 3-32 from MacQuarrie second edition.

I do not know why you think that 221 has double degeneracy. Neither I know from where you get 14M.

Filling the values of n_x, n_y, and n_z in your above formula, the first seven levels are

111:      M (1 + 1 + 2.25)

211:      M (4 + 1 + 2.25)
121:      M (1 + 4 + 2.25)

221:      M (4 + 4 + 2.25)

112:      M (1 + 1 + 9)

131:      M (1 + 9 + 2.25)
311:      M (9 + 1 + 2.25)

122:      M (1 + 4 + 9)
212:      M (4 + 1 + 9)

113:      M (1 + 1 + 20.25)

The book is correct: for the first four, 111 and 221 are not degenerate and 211 has a double degeneracy.