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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?

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    $\begingroup$ Not sure if I understand you, but are you summing the states of two different electrons? $l \leq n$ for a single electron. $\endgroup$
    – FGSUZ
    Commented Jan 16, 2019 at 15:57
  • $\begingroup$ 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. $\endgroup$
    – Emilio Pisanty
    Commented Jan 16, 2019 at 17:16
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    $\begingroup$ 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 $n$. $\endgroup$
    – IvanaGyro
    Commented Jan 16, 2019 at 17:19
  • $\begingroup$ @Iven If you define n for multielectron states as the sum of single electron n, then total l < total n. $\endgroup$
    – my2cts
    Commented Jan 16, 2019 at 17:55
  • $\begingroup$ @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? $\endgroup$
    – Emilio Pisanty
    Commented Jan 16, 2019 at 18:14

3 Answers 3

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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.

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  • $\begingroup$ 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. $\endgroup$
    – IvanaGyro
    Commented Jan 16, 2019 at 18:34
  • $\begingroup$ 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. $\endgroup$
    – IvanaGyro
    Commented Jan 16, 2019 at 18:40
  • $\begingroup$ @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. $\endgroup$
    – Emilio Pisanty
    Commented Jan 16, 2019 at 19:07
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The angular momentum quantum number $l$ describes the angular momentum of the electron. Basically it is the rotational kinetic energy of the electron. the principal quantum number $n$ describes the total energy of the electron potential + kinetic so the angular quantum number $l$ must be less than $n$.

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The ${\cal l}<n$ rule is valid per orbital. Comparing the total orbital moment $L$ with the single electron n is not meaningful.

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