4.1.3.1 So­lu­tion harmc-a

Ques­tion:

Ver­ify that the sets of quan­tum num­bers shown in the spec­trum fig­ure 4.1 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 $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_x+n_y+n_z$,

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

Now for the bot­tom level, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, so $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_x+n_y+n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, this state has en­ergy $\frac 32\hbar\omega$.

Sim­i­larly, in each of the three sets of the sec­ond en­ergy level in fig­ure 4.1, the three quan­tum num­bers $n_x$, $n_y$, and $n_z$ add up to $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, giv­ing this state en­ergy $\frac 52\hbar\omega$.

For the third en­ergy level, the three quan­tum num­bers of each set add up to $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, giv­ing en­ergy $\frac 72\hbar\omega$, and for the fourth set, the quan­tum num­bers in each of the ten sets add up to $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 for an en­ergy $\frac 92\hbar\omega$.