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In a helium atom, the total spin of the electrons is 0, suggesting that the total spin quantum number S is the sum of the ms quantum numbers (1/2-1/2=0). However, many sources say that the total spin quantum number S is the sum of the s quantum numbers, so in helium S should be 1 (s is always 1/2 for electrons).

In much the same way, I've seen some sources say that L is the sum of the ml quantum numbers and others say that L is the sum of the l quantum numbers.

So, which is correct?

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Both $L$ and $S$ are quantum angular momenta.

Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM.

In QM, angular momentum is complete if you give two numbers: the "angular momentum" and its "third component" $M$. So you give the pairs $|l,\ m_l\rangle$ and $|s, \ m_s\rangle$

So, any angular momentum is defined with those two numbers. But those numbers behave very differently.

The third component is additive. You add $m$'s without problem:

Total $m$ = $m_1+m_2$

On the other hand, $l$'s behave very differently. When we add two angular momenta, the total angular momentum (let's call it $J$), is different: it can go from $|L-S|$ to $L+S$

And the thing is that both things are called "sum", that's the reason of the confusion.

When you add two angular momenta (vectors) $\vec{L}+\vec{S}=\vec{J}$ That's okay, but this is the same as saying

$$|l,\ m_l\rangle \ + \ |s,\ m_s\rangle \ = \ |j,\ m_j\rangle $$

And the total $m$ adds normally: $m_j=m_l+m_s$, okay.

But $j$ is different. $j$ can go from $|l-s|$ to $|l+s|$ in unit steps.


In all angular momenta, as you know, $m$ can go from $-l$ to $+l$. And also $m_s$ from $-s$ to $+s$; and $m_j$ from $-j$ to $+j$.

So, if you have $s=½$, you can have $m_s=\pm ½$

But if you add two spins, $s_1+s_2$, the resulting $j$ can be 0 or 1.

For $j=0$, $m_j$ can only be 0 as well. But for $j=1$, $m$ can have three values: $m_j={-1,0,1}$.

So, "summing two spins" can mean that total spin is "J=1" but "M=0".

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  • $\begingroup$ Brilliant, thank you very much. $\endgroup$ – Ali Chopping Jan 9 at 12:13
  • $\begingroup$ while in general this is an excellent answer, may I object to the line $|l, m_l\rangle + |s, m_s⟩ = |j, m_j\rangle$? This makes it seem as if $l$ and $s$ are referring to the same part of Hilbert space, and that $|j, m_j\rangle$ is something like a superposition of them, while this is not correct. The correct formatting will be $|l, m_l \rangle \otimes |s, m_s\rangle = |j, m_j\rangle$ $\endgroup$ – yu-v Jan 9 at 13:53

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