why we include the total angular momentum and the z component of angular momentum in our quantum number, but we dont have a quantum number for total spin, only the z component of spin?

Question

osprey · Answer

I can't answer that directly, but I can say bravo to you for asking it.

festinger · Answer

I don't know what specific system you are referring to, but sometimes the spin quantum number $s$ is omitted for electrons because it is implicitly assumed that the number (\s=1/2\), and the fact is *obvious* from the $m_s=\pm\frac{1}{2}$. 

Also, for clarity between the quantum numbers I used and the momentum values you refer to:
$$S=\sqrt{s(s+1)}\hbar$$$$S_{z}=m_s\hbar$$

osprey · Answer

clarity is a rare commodity in science/physics/maths methinks !
(indeed, from my headaches, I get the impression that it's more about "obfuscation" and card tricks, and being "symbolically neat and tidy" ... but then that's just my ((bruised)) "thinking") 
one book I got recently actually did say that a "gauge" was a maths device to make the differential algebra less "ugly" was, I think, the word the author used. Helps make it "canonical" and "symmetrical" in terms of the way the symbols/squiggles are arranged, or some such.

festinger · Answer

@osprey I don't think this question has anything to do with gauge, gauge theory is usually applied to field theory, but in this case no fields are used. But what you say is true. Gauges are a way to simplify the math or to make the math look for "symmetrical".

caozeyuan · Answer

@festinger @osprey
but I still dont get the idea of why are we using both l and ml as quantum number, but only use ms and ignore s?

festinger · Answer

Because for the $l$ quantum number, it very well depends on the shell you are in! If we just list down $m_{l}=0$, it isn't obvious whether we're looking at the $s$ shell, where $l=0$ or the $p$ shell, where $l=1$ but $m_{l}$ could also be 0.

But all electrons have $s=\frac{1}{2}$

caozeyuan · Answer

makes senses, thx!