Must the total orbital angular momentum quantum number $L$ be less than the principal quantum number $n$? If so, why?

I am studying LS coupling and term symbols. In my textbook, there is an exercise:

Why is it impossible for a $$2\ ^{2}\text{D}_{3/2}$$ state to exist?

The answer says, the total orbital angular momentum quantum number must less than the principal quantum number. But in my opinion, considering the electron configuration, $$1s^{2}2s^{2}2p^{2}$$, if the two electrons in $$2p$$, the outer subshell, have quantum numbers $$(1, 1/2)$$ and $$(1, -1/2)$$ respectively which are in the term of $$(m_{l}, m_{s})$$, $$m_{l}$$ is the magnetic quantum number, and $$m_{s}$$ is the spin magnetic quantum number, then the total orbital angular quantum number is $$1+1=2$$ which is equal to its principal quantum number. This example is conflict against the answer.

Which is wrong, my example or the answer in the textbook?

• Not sure if I understand you, but are you summing the states of two different electrons? $l \leq n$ for a single electron. – FGSUZ Jan 16 at 15:57
• If you have an even number of electrons, as you do in carbon, then the total spin (and therefore the total angular momentum) must be integers instead of half-integers, which rules out your example. More generally, your statement "then the total orbital angular quantum number is 1+1=2" isn't particularly accurate: the addition of angular momenta in QM is a complex subject, and 1+1=2 is only one possibility among several; if this is unfamiliar, then you should take a long step back and study that topic from the ground up. – Emilio Pisanty Jan 16 at 17:16
• For a single electron, I know it is always true that $l < n$. I want to confirm that whether the total orbital angular momentum quantum number, usually use $L$ as the symbol, is always less than $n$. – IvanaGyro Jan 16 at 17:19
• @Iven If you define n for multielectron states as the sum of single electron n, then total l < total n. – my2cts Jan 16 at 17:55
• @my2cts That metric is essentially useless - there is basically no literature that even looks at that sum, let alone uses it for anything useful. Or do you have a relevant counter-example? – Emilio Pisanty Jan 16 at 18:14

Which is wrong, my example or the answer in the textbook?

Your example is wrong. You have two active electrons in the $$p$$ shell, and their total spin must couple either to $$S=0$$ or $$S=1$$, which correspond to singlet ($$2S+1=1$$) or triplet ($$2S+1=3$$) states. The target state you've been given is a doublet state (indicated by the $$2S+1=2$$ superscript), so you've already missed the mark.

More generally, if you want a doublet state (with $$S=1/2$$), then you need an odd number of electrons, since even numbers of electrons always have integer-valued total spin.

This then puts you into trouble, because having $$n=2$$ limits you to having only $$p$$ electrons with $$\ell=1$$ contributing to the orbital angular momentum, and if you have an odd number of such electrons, then you're restricted to an odd-integer value for $$L$$. This then completely eliminates the possibility of any $$2 \ {}^2\mathrm{D}_J$$ state, whatever the $$J$$.

(If any of the above is unfamiliar, then it's almost certainly because of an incomplete preparation in the quantum-mechanical procedure for adding angular momenta. This is a large and complex topic, and you should take it from the ground up.)

As for your more general question,

I want to confirm that whether the total orbital angular momentum quantum number $$L$$ is always less than $$n$$.

No, this is not the case (at least, for excited states). With a half-filled shell, say, on atomic nitrogen, it's perfectly possible to achieve $$\rm F$$ states with $$L=3$$, by taking the parallel configuration for the three individual orbital angular momenta.

• It seems that I had a mistake on my grammer. In my example, I meant one of the electron has $+1/2$ as the magnetic quantum number and the ohter has $-1/2$, so $S=0$. The two electrons spin in anti-parallel. However, your answer really solves my problem, thank you. – IvanaGyro Jan 16 at 18:34
• If the three outer electrons of atomic nitrogen are in the parallel configuration, it seems that $L$ should be $0$, ($1 + 0 + \left(-1\right)$). The information is from the table in the Wikipedia page. – IvanaGyro Jan 16 at 18:40
• @IvenCJ7 No. That's not how the addition of angular momentum works in QM - both of the assertions in your comments above are dead wrong. (Some short heuristics as to why: you're adding the $z$ components of the angular momenta, but you should be doing a vector addition of different vectors whose components don't commute with each other, which is what complicates the problem.) Like I said, this is not a trivial method, and you should do a dedicated reading of the topic on a serious QM textbook. – Emilio Pisanty Jan 16 at 19:07