# Quantum numbers in the electron cloud

**URL:** <https://boards.straightdope.com/t/quantum-numbers-in-the-electron-cloud/781818>\
**Category:** Factual Questions\
**Created:** [March 9, 2017, 11:50pm UTC](https://boards.straightdope.com/t/quantum-numbers-in-the-electron-cloud/781818 "2017-03-09T23:50:39Z")\
**Posts on this page:** 3\
**Page:** 1

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**Author:** ![Cardinal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cardinal/32/4000_2.png) [@Cardinal](https://boards.straightdope.com/u/Cardinal)\
**Post date:** [March 9, 2017, 11:50pm UTC](https://boards.straightdope.com/t/quantum-numbers-in-the-electron-cloud/781818/1 "2017-03-09T23:50:39Z")

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I can memorize the possible values for the numbers, but what governs what they can be? Why does l start at zero? Why does ml start at negative l? I know that gives you the number of orbitals in the subshell, but why express it in that way?

Or is the answer, “supermath that you wouldn’t understand”, as I suspect?

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**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\
**Post date:** [March 10, 2017, 12:50am UTC](https://boards.straightdope.com/t/quantum-numbers-in-the-electron-cloud/781818/2 "2017-03-10T00:50:23Z")

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> [@Cardinal](#):
>
> Or is the answer, “supermath that you wouldn’t understand”, as I suspect?

Its not actually that complicated.

At the most simplest, it just has to produce the series 2, 8 , 8 for Bohr’s electron shells.  
The first two electrons can be 1,0,0, + 1/2 and 1,0,0, -1/2. Where n = 1, l and ml MUST be 0.

So there’s your starting points.

The supermath comes in when you ascribe a meaning to l and ml.  
The momentum and the “projection” of the momentum

You will notice n=3 does not have 8. For that you need to involve the presence of spin–orbit interaction and find states that are “eigenvectors of the Hamiltonian” (i.e. each represents a state that does not mix with others over time).

This leaves 2 , 8 and 8 it seems.

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**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\
**Post date:** [March 10, 2017, 9:00am UTC](https://boards.straightdope.com/t/quantum-numbers-in-the-electron-cloud/781818/3 "2017-03-10T09:00:25Z")

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To elaborate a little bit:  
[ul]  
[li] _l_ describes the total amount of angular momentum of the orbital, which is one of those things which can only take on specific possible values, the lowest of which is zero. Because the lowest possible value of angular momentum you can measure is zero, _l_ starts at zero.[sup]1[/sup][/li][li] Similarly, _m[sub]l_[/sub] describes how much of that total angular momentum is in a particular direction, and this also is one of those things which can only take on specific possible values. Since the total amount of angular momentum is _l_, the maximum amount possible in any one direction is _l_. Similarly, the most negative possible amount is -_l_. So _m[sub]l[/sub]_ is between -_l_ and _l_, inclusively.[/li][/ul]  
[sup]1[/sup] Technically, _l_ is associated with the _square_ of the angular momentum, and a squared thing can’t be negative, which is why _l_ starts at zero. For a state with quantum number _l_, the square of the angular momentum is _l_ (_l_+1) ~ _l_[sup]2[/sup], so it makes a certain amount of sense to say that the most angular momentum you can have in any one direction is _l_.