22.214.171.124 Solution harmc-b
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.
The generic expression for the energy is
or defining ,
Now for the bottom level, 0, and since the three quantum numbers , , and cannot be negative, the only way that can be zero is if all three numbers are zero. If any one of , , or would be positive, then so would be . So there is only one state 0.
For the second energy level, 1. To get a nonzero sum , you must have one of , , and to be nonzero, but not more than 1, or would be more than 1 too. Also, if one of , , and is 1, then the other two must be 0 or their sum would still be greater than 1. That means that precisely one of , , and must be 1 and the other two 0. There are three possibilities for the one that is 1, , , or , resulting in the three different sets of quantum numbers shown in the spectrum figure 4.1.
For the third energy level, 2, the maximum value value any one of , , and 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 , , and 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, 3, the maximum value value that any one of , , and 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 3. That gives the tenth set.
A less intuitive, but more general, procedure is to simply derive the number of different eigenstates for a given mathematically and show that it agrees with the figure: The possible values that the quantum number can have in order for not to exceed are in the range from 0 to , and for each such value of , must be in the range from 0 to for no to exceed . For each such acceptable pair of values and , there is exactly one allowed value . So there is exactly one state for each acceptable pair of values and . Which means that the number of states is, using summation symbols,
The sum within the parentheses is 1, and then the remaining sum is an arithmetic series [1, 21.1], producing
Substituting in 0, you get 1, a single state, for 1, 3, three states, for 2 six states and for 3 ten states. So the spectrum 4.1 shows all states.