Question

Some degeneracies are easy to understand on the basis of symmetry in the physical situation. Others are surprising, or “accidental”. In the states given in Table 7.1, which degeneracies, if any, would you call accidental and why?

Short answer

The way that the initial three have similar energy is grasped by the cubic balance. Nonetheless, the way that (3,3,3) has a similar energy to the initial three is not perceived as evenness, so it is viewed as an "accidental" degeneracy.

Step-by-step solution

Step 1:Define degeneracy

"Degeneracy" signifies more than one quantum state with the equivalent strongly characterized energy.

The permitted energies for a molecule in the cubic boundless potential are.

${\mathrm{E}}_{\mathrm{nx},\mathrm{ny},\mathrm{nz}}=({\mathrm{n}}_{\mathrm{x}}^2+{\mathrm{n}}_{\mathrm{y}}^2+{\mathrm{n}}_{\mathrm{z}}^2)\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

Where n_x,n_y,n_z are integers, m is the mass of the particle, and L is the lengthin which we are studying the behavior of the particle.

The accidental degeneracy in Table 7.1.

Because of cubic evenness, all potential blends of any three numbers $\left(n_x,n_y,n_z\right)$have similar energy.

For instance (2,1,1) , (1,2,1) and (1,1,2) structure a three-crease degenerate state.

When we take a gander at Table 7.1 cautiously, see that (5,1,1) , (1,5,1) , (1,1,5) and (3,3,3) structure a four-overlap degenerate state.

The way that the initial three have similar energy is grasped by the cubic balance. Nonetheless, the way that (3,3,3) has a similar energy to the initial three is not perceived as evenness, so it is viewed as an "accidental" degeneracy.