4.1.3.2 So­lu­tion harmc-b

Ques­tion:

Ver­ify that there are no sets of quan­tum num­bers miss­ing in the spec­trum fig­ure 4.1; the listed ones are the only ones that pro­duce those 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, $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and since the three quan­tum num­bers $n_x$, $n_y$, and $n_z$ can­not be neg­a­tive, the only way that $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_x+n_y+n_z$ can be zero is if all three num­bers are zero. If any one of $n_x$, $n_y$, or $n_z$ would be pos­i­tive, then so would be $N$. So there is only one state $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

For the sec­ond en­ergy level, $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. To get a nonzero sum $N$, you must have one of $n_x$, $n_y$, and $n_z$ to be nonzero, but not more than 1, or $N$ would be more than 1 too. Also, if one of $n_x$, $n_y$, and $n_z$ is 1, then the other two must be 0 or their sum $N$ would still be greater than 1. That means that pre­cisely one of $n_x$, $n_y$, and $n_z$ must be 1 and the other two 0. There are three pos­si­bil­i­ties for the one that is 1, $n_x$, $n_y$, or $n_z$, re­sult­ing in the three dif­fer­ent sets of quan­tum num­bers shown in the spec­trum fig­ure 4.1.

For the third en­ergy level, $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, the max­i­mum value value any one of $n_x$, $n_y$, and $n_z$ could pos­si­bly have is 2, but then the other two must be zero. That leads to the first three sets of quan­tum num­bers shown in the spec­trum. If the max­i­mum value among $n_x$, $n_y$, and $n_z$ is not 2 but 1, then a sec­ond one must also be 1, or they would not add up to 2. So in this case you have two of them 1 and the third 0. There are three pos­si­bil­i­ties for the one that is 0, pro­duc­ing the last three sets of quan­tum num­bers shown in the spec­trum 4.1 at this en­ergy level.

For the fourth en­ergy level, $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, the max­i­mum value value that any one of $n_x$, $n_y$, and $n_z$ could have is 3, with the other two 0, pro­duc­ing the first three sets of quan­tum num­bers. If the max­i­mum value is 2, then the other two quan­tum num­bers must add up to 1, which means one of them is 1 and the other 0. There are three pos­si­bil­i­ties for which quan­tum num­ber is 2, and for each of these there are two pos­si­bil­i­ties for which of the other two is 1. That pro­duces the next six sets of quan­tum num­bers. Fi­nally, if the max­i­mum value is 1, then both other num­bers will have to be 1 too to add up to $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. That gives the tenth set.

A less in­tu­itive, but more gen­eral, pro­ce­dure is to sim­ply de­rive the num­ber of dif­fer­ent eigen­states $S$ for a given $N$ math­e­mat­i­cally and show that it agrees with the fig­ure: The pos­si­ble val­ues that the quan­tum num­ber $n_x$ can have in or­der for $n_x+n_y+n_z$ not to ex­ceed $N$ are in the range from 0 to $N$, and for each such value of $n_x$, $n_y$ must be in the range from 0 to $N-n_x$ for $n_x+n_y+n_z$ no to ex­ceed $N$. For each such ac­cept­able pair of val­ues $n_x$ and $n_y$, there is ex­actly one al­lowed value $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $N-n_x-n_y$. So there is ex­actly one state for each ac­cept­able pair of val­ues $n_x$ and $n_y$. Which means that the num­ber of states is, us­ing sum­ma­tion sym­bols,

$$S = \sum_{n_x=0}^N\left(\sum_{n_y=0}^{N-n_x} 1 \right)$$

The sum within the paren­the­ses is $(N-n_x+1)$ $\times$ 1, and then the re­main­ing sum is an arith­metic se­ries [1, 21.1], pro­duc­ing

$$S=(N+1)(N+2)/2.$$

Sub­sti­tut­ing in $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, you get $S$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, a sin­gle state, for $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $S$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, three states, for $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 six states and for $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 ten states. So the spec­trum 4.1 shows all states.