Degeneracy and Symmetry

Let's expand our model of a particle in one-dimensional potential well into three-dimensional potential well which has the following size 0<x<a, 0<y<b, 0<z<c and therefore our particle will have the following energy states inside of this well:

E = h²/8m(n1²/a² + n2²/b² + n3²/c²)

where n1, n2 and n3 are quantum states along x-, y- and z-axis correspondingly (ni = 1,2,3,...).
The wavefunction y(x,y,z) = yx(x) . yy(y) ·yz(z) can be expanded as follows:

y(x,y,z) = (8/abc)½ .sin(n1πx/a) ·sin(n2πy/b) ·sin(n3πz/c)

If all well dimensions are the same, i.e. a = b = c then we will have particular case here since we can obtain the same energy values for different quantum numbers:

E = h²/8ma²(n12 + n22 + n32) = h²/8ma²· K2 = E1 · K2

Since we obtain the same energy values for different quantum numbers it's time to talk about state symmetry and degeneracy:

All combinations of n1, n2 and n3 that finally give the same K value have the same energy, correspondingly. Obviously corresponding wavefunctions are different for each combination. So we are talking about degenerated energy level in this case. The degeneracy degree that is assigned as g corresponds to the number of independent wavefunctions which give the same energy. Eventually we obtain, for instance, for the first six energy levels (where E1 = h²/8ma):

Energy | Combinations (n1, n2, n3) | Degeneracy g
3 E1 | (1, 1, 1) | 1
6 E1 | (2, 1, 1), (1, 2, 1) (1, 1, 2) | 3
9 E1 | (2, 2, 1) (2, 1, 2) (1, 2, 2) | 3
11 E1 | (3, 1, 1) (1, 3, 1) (1, 1, 3) | 3
12 E1 | (2, 2, 2) | 1
14 E1 | (1, 2, 3) (1, 3, 2) ( 2, 3, 1) (2, 1, 3) ( 3, 1, 2) (3, 2, 1) | 6

The level degeneracy can always be explained by symmetry considerations. One can see the wavefunctions for a particle in two-dimensional potential well on the next illustration (see fig. 1).

yn1n2 = (4/a b)½ sin(πx/a n1) · sin(πy/a n2)
E = h²/8ma² (n12 + n22)

For (n1 = 2; n2 = 1 and n1 =; n2 = 2) the energy levels are twofold degenerated. The wavefunctions y21 and y12 are asymmetric (i.e. they change its sign upon reflection relatively x-axis (y21) and y-axis (y12)). The rotation on 90° transfers one wavefunction into another one.

Fig.1: Wavefunctions of a particle in two-dimensional potential well and its contour diagrams.

If there is a symmetry point then the stationary levels are described by wavefunctions that have either positive (y(A) = +y(A')) or negative (y(A) = -y(A')) parity, i.e. the wavefunction sign either stays the same (positive parity) or changes (if coordinates reflect upon symmetry plane (for instance, x ®-x) when x = 0 is symmetry axis).

However: The energy of corresponding level is higher the more frequent wavefunction changes its sign.

Fig.2: The degeneracy density g(E) as a function of energy E.

The particle which is squeezed in the the well can take only definite discrete energy values. It's quite easy to calculate the state numbers d N(E) in the range [E, E + dE] and then find the energy state density g(E) = dN/dE, i.e. the amount of states per range of energy unit:

g(E) = 2πV(2m³)½/h³. E½

And then one can obtain the state sum Q from which it's also possible to derive all other thermodynamical quantities:

Q = V/h³ (2πmkT)3/2

Moreover we've received very important quantity: number of state transitions.