2.6.3.1 Solution

Question: Verify that the sets of quantum numbers shown in the spectrum figure 2.15 do indeed produce the indicated energy levels.

Answer: The generic expression for the energy is

$$E_{n_xn_yn_z} = \frac{2n_x+2n_y+2n_z+3}2\; \hbar\omega$$

or defining $N=n_x+n_y+n_z$,

$$E_{n_xn_yn_z} = \frac{2N+3}2\; \hbar\omega$$

Now for the bottom level, $n_x=n_y=n_z=0$, so $N=n_x+n_y+n_z=0$, this state has energy $\frac 32\hbar\omega$.

Similarly, in each of the three sets of the second energy level in figure 2.15, the three quantum numbers $n_x$, $n_y$, and $n_z$ add up to $N=1$, giving this state energy $\frac 52\hbar\omega$.

For the third energy level, the three quantum numbers of each set add up to $N=2$, giving energy $\frac 72\hbar\omega$, and for the fourth set, the quantum numbers in each of the ten sets add up to $N=3$ for an energy $\frac 92\hbar\omega$.