# I Total angular momentum for antiparallel electrons

1. Jan 5, 2017

### goodphy

Hello.

Let's have two electrons with same orbital quantum number li and these electrons are in antiparallel; one electron has magnetic quantum number mi = a and and other electron has mi = -a (but we don't know which one has ml = a as we're in coupled representation to talk about total angular momentum).

In this system, we know that total magnetic quantum number m is zero as $$m = \sum\limits_i {{m_i} = 0} .$$ m is allowed to be only 0 so total orbital quantum number l = 0 (l = 0 allows only m = 0). This is exact the same problem about total angular momentum for a closed shell in an atom.

So...can I simply conclude that total angular momentum (and also spin) is always zero for antiparallel electrons (antiparallel-spin electrons)?

2. Jan 6, 2017

### Staff: Mentor

This is not correct. There are other orbital quantum numbers l that also allow m to be zero. Your argument, if correct, would imply that only $s$ orbitals can ever be occupied in an atom, which is obviously false.

No. See above.

3. Jan 7, 2017

### goodphy

Hello. Thanks for replying my question.

I know that magnetic quantum number m can be zero for other orbital quantum number l. In my question, I meant that l = 0 is the only case which allows "only" m = 0.

I had been confused why total orbital quantum number l for a CS (Closed Subshell) is always zero. If l is non-zero for CS, total magnetic quantum number m for CS can be zero and also non-zero value, in principle as m = l, l -1, ... , -l. However, for CS, m is only allowed to be zero! That's why I thought l = 0 for CS.

If I was wrong, could you please tell me the exact reason why l = 0 for CS?

4. Jan 7, 2017

### Staff: Mentor

Yes, but that's the wrong constraint. The constraint is that m = 0 must be possible, not that it must be the only possibility.

5. Jan 7, 2017

### Staff: Mentor

Can you give a reference?

6. Jan 7, 2017

### goodphy

What...?
Reference? I've seen that L = S = 0 for CS (Closed Subshell) many times in various sources. I think it is commonly said, although I'm not fully understanding the exact reason of this. You can easily find this fact by googling.

7. Jan 7, 2017

### goodphy

Hello. I'm sorry why you said m = 0 must be "possible" for CS. I think the fact is that m = 0 is only one possible case for CS.

8. Jan 7, 2017

### Staff: Mentor

I'm looking for a specific reference because I want to see what, specifically, you mean by "L = S = 0 for Closed Subshell" (by "specifically" I mean "mathematically"), and how you think that constraint is derived. If you don't have a single specific reference handy, then post a quick sketch of how you yourself would derive that constraint from first principles. (This is an "I" level thread so you should have sufficient background to do that.)

9. Jan 7, 2017

### Staff: Mentor

@goodphy you might also want to consider this statement of mine from post #2:

$s$ orbitals have $l = 0$, but other orbitals (e.g., $p$ orbitals in an atom like neon--a good test case to consider) do not.

10. Jan 7, 2017

### goodphy

Hello.

These are a reference I can give you.

1. http://quantummechanics.ucsd.edu/ph130a/130_notes/node36.html

2. https://www.researchgate.net/post/Why_are_total_orbital_quantum_number_l_and_total_magnetic_quantum_number_m_zero_for_closed_subshell [Broken]

It looks that you think total orbital quantum number l for CS can be non-zero. Do you really think that?

By the way, the quantum numbers what I meant are "total" quantum numbers for a collection of electrons (So I'm talking about addition of individual angular momentum of electrons), not quantum numbers for individual electrons. The constraint that total magnetic quantum number m = 0 for CS is obvious; a sum of individual magnetic quantum numbers mi of all electrons in CS is zero.

I'm worrying that we're talking about different tings.

Last edited by a moderator: May 8, 2017
11. Jan 7, 2017

### blue_leaf77

@goodphy, I would like to clarify, so in the problem that you described
Is the orbital fully occupied by these two electrons (hence implying that it's an $s$ orbital), or not all occupied?
If the latter applies, then you cannot immediately conclude that $L=0$.

12. Jan 8, 2017

### Staff: Mentor

Yes, these are very helpful.

No. But it's easier to explain why now that you have given good references for what you mean by "total orbital quantum number", and also given your own explanation:

This is correct, but you apparently haven't fully understood its implications. Let's look at your original description of the scenario you were asking about:

Notice the key thing: you asked about two electrons. But unless the subshell is an $s$ orbital, i.e., $l = 0$, a closed subshell does not contain two electrons. It contains more. For example, the $2p$ subshell in neon, which is a closed subshell, contains six electrons, not two. It is true that the total quantum number l for all six of those $2p$ electrons is zero. But it is not true that the total quantum number l for just two of them, taken individually, is zero. The $p$ orbitals are $l = 1$ states, not $l = 0$ states, so if we take two electrons in the same $p$ orbital (there are three such orbitals at each energy level), they will have $l = 1$ and $m = 0$ (whereas two electrons in an $s$ orbital, which is a closed subshell with only two electrons, will have $l = 0$ and $m = 0$).

So the answer to the question you asked in the OP...

...is no, because two antiparallel electrons don't have to be in a state with total $l = 0$; if that were true, the only orbitals that could ever be filled in atoms would be $s$ orbitals, as I said in post #2. But the answer to the different question, "will the total angular momentum of a closed subshell always be zero", is yes. You were apparently thinking that those two questions were the same. They're not.

Last edited: Jan 8, 2017
13. Jan 8, 2017

### goodphy

Oh Yes!! You're touching the heart of my confusion!

I thought that CS (Closed Subshell) and the situation that all electrons are paired are equivalent. "electrons are paired" here means that if there is an electron with an individual magnetic QN (Quantum Number) mli = a and individual spin projection QN msi = b, there is another electron with mli = - a, msi = - b (of course, all electrons have same individual orbital QN li and individual spin QN si).

You said that the situation that paired electrons are in a partially filled subshell is not equivalent to CS. Could you tell me why these are different? And for CS, why total orbital QN l and total spin QN s must be zero? These are really long-lived questions in my mind...

14. Jan 8, 2017

### goodphy

PeterDonis indicates same things as you mentioned. I would be happy if you follow his last post and my last reply to him and give me some comments:)

15. Jan 8, 2017

### blue_leaf77

I will leave the question before the one I quoted above to Peter since he has been the one who first helped you in that direction.
Firstly, Pauli principle demands that for a closed subshell, there should be as many spin-up electrons as spin-down ones. This result in zero magnetic quantum number. Since in CS there is only one possibility of rearranging the electrons, the magnetic quantum number cannot be different from zero. That's the first part of the proof. Next we also need to prove that the angular momentum QN that corresponds to that $m=0$ can only be $l=0$. Let's assume that $l \neq 0$, then $m = l, l-1, ..., -l+1,-l$. Since the energy levels of a free atom is degenerate in $m$, there will be other values of $m$ different from zero that can be the state of this fully occupied atom (because the energies of these $m$'s are all the same). But this contradicts the requirement from the first part that $m$ can only be zero. So the assumption that $l \neq 0$ must be wrong.

16. Jan 8, 2017

### goodphy

Thanks for giving me some answer.

Let me summarize your comment.
So, the fact that the total orbital QN (Quantum Number) l for CS (Closed Subshell) is always zero stems from a requirement that total magnetic QN m is zero for CS; no other value than 0 is allowed for m (a sum of individual magnetic QN mi of electrons in CS is zero). l of 0 and any positive integers all allows m = 0 in principle (m = l, l - 1, ..., - l), however, l = 0 is the only case which only allows m = 0; no other value than 0 is allowed for m at l = 0. So, l for CS must be zero. The same logic is followed to prove that total spin QN for CS is also zero.

Is my summary correct?

By the way, I think you miswrite this; spin and magnetic quantum numbers are different and independent entities. I think you tried to talk about electron pairs with opposite mi.

17. Jan 8, 2017

### blue_leaf77

Yes.
On a second thought I think my second sentence in the one you quote indeed does not automatically result from the first sentence there. May be I should have added that as a result that all magnetic QN have been claimed by electrons in CS, the total magnetic QN is zero.

18. Jan 8, 2017

### Staff: Mentor

It is. But "all electrons are paired" is not equivalent to "all pairs of electrons have $l = 0$ by themselves". An electron pair that is in a $p$ orbital, for example, is in a state where the pair has $l = 1$. That's the definition of a $p$ orbital. (Similarly, in a $d$ orbital each electron pair has $l = 2$; in an $f$ orbital each pair has $l = 3$; etc.) Only the set of all electrons in a complete subshell (six of them for the $p$ subshell; 10 for $d$; 13 for $f$, etc.) has $l = 0$.

19. Jan 9, 2017

### goodphy

Hello. I appreciate your effort to follow all kinds of questions that I have posted. Thank you.

Right now, I'm thinking that the question about "pairing electrons in a PS (Partially filled Subshell)" doesn't make sense at all. We're talking about angular momentum (and spin)-related total quantum numbers; we're talking about additions of angular momentums and spins; It automatically means that we're in coupled representation like |l, m, s, ms, l1, l2, ..., s1, s2, ...>. We can not say anything about mi and msi in this representation. Pairing electrons (grouping two electrons with opposite mi or msi) requires knowledge of mi and msi. However, we don't know anything about them since we're in the coupled representation so we can not do pairing electrons.

In CS (Closed Subshell), at least we know that every electron has its counter partner in the pairing. That's why we know total magnetic QN (Quantum Number) m = 0. But in PS, we don't know which values of mi and msi are there so we don't know m.

However...there is a special case; spins of two electrons. Let's say there are two electrons (they can be even free electrons) and we're only interested in spin. If ms = 0 is observed, we immediately know that they're paired as m = m1 + m2. Since they're paired, ms = 0 is only one possible outcome, so we can say that s = 0. In fact, I'm really not sure a state of s = 1, ms = 0 even exists for two electrons.

I would like to ask you to get some comments about this.

20. Jan 9, 2017

### Staff: Mentor

It depends on what kinds of states you are considering. The states I described--two electrons paired in a single $p$ orbital, for example--are certainly valid states, theoretically speaking, and they have definite values for all three quantum numbers $n$, $l$, and $m$ (in the case of a $2p$ orbital, for example, the values are $n = 2$, $l = 1$, $m = 0$). However, it is also true that the electrons in a partially filled subshell do not have to be in those particular states, and in practical terms they probably won't be; they will probably be in other states which can be expressed as superpositions of the states I described, and which will not in general have definite values of the $l$ and $m$ quantum numbers (though they will probably still have definite values of $n$).

Given the specific observation you describe, yes, you can say that after the observation. But that doesn't mean you can say the joint spin state of the electrons was $s = 0$ before the observation. To say that you would have to know that the electrons were prepared in a particular way so that their spins were entangled.

21. Jan 9, 2017

### blue_leaf77

Yes it exists. For a system comprising two electrons, such state is one of the triplet states. As described above, knowing only the value of $m_s$ is not enough to deduce the complete state of the system.

22. Jan 10, 2017

### Staff: Mentor

I'm a bit late to the game, but could we all agree on using the standard convention that single-electron states are labelled with lowercase letters, $l, m_l, s, m_s$, while many-electron states are labelled with uppercase letters, $L, M_L, S, M_S$? Otherwise, this gets impossible to discuss.

The best way to figure these things out is to write out all possible microstates, i.e., listing all orbitals with proper quantum numbers $l$ and $m_l$, then filling them with electrons of a given spin, respecting the Pauli exclusion principle. Then, to figure out the quantum numbers for the full state, one calculates $M_L = \sum m_l$ and $M_S = \sum m_s$, and this will indicate the possible values of $L$ and $S$. For filled (sub) shells, one obviously gets $M_L=0$ only, so the only possible value of $L$ is 0. Likewise, $M_S=0$, so $S=0$.

If you were asked the question "You have a state with angular momentum $J=1$, what are the possible values of $M_J$?", I hope you would answer $M_J= -1, 0, 1$. So obviously, if $S=1$, there has to be a state with $M_S=0$.

Do the exercise I mentioned above. Take for instance two p electrons, and list all possible states (you should get 15) and find the possible values of L and S. You will see that there are five micro states corresponding to $L=2$, $S=0$, 9 corresponding to $L=1$, $S=1$, and 1 corresponding to $L=0$, $S=0$.

Last edited: Jan 10, 2017
23. Jan 14, 2017

### goodphy

Oh, Yes. You're absolutely right. I was confused.

Yes, for two electrons in a partially filled subshell or different subshells or even free states, we really don't know a definite value of total spin projection quantum number ms for these electrons before its measurement, unlikely to CS (Closed Subshell) where total quantum numbers ms and ml are very obvious; they are only zero, no matter we measure them or not. If we measure ms = 0 for these two electrons, we just can say that ms = 0 is just measured, among multiple possibilities of ms. With ms = 0, the state of two electrons is either (s = 0, ms = 0) or (s = 1, ms = 0), no other things we can say.

Thanks for making my mind clear.

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