Quantum numbers n, l, mₗ, mₛ

WHAT are the four quantum numbers?

Definition The four quantum numbers

Symbol	Name	Allowed values	Physically controls
$n$	Principal quantum number	$1,2,3,\dots$	energy & overall size of orbital
$l$	Azimuthal (orbital angular momentum)	$0,1,2,\dots,(n-1)$	shape of orbital, magnitude of angular momentum
$m_l$	Magnetic quantum number	$-l,\dots,0,\dots,+l$	orientation of orbital in space ( $z$ -component of $\vec L$ )
$m_s$	Spin magnetic quantum number	$+\tfrac12,\,-\tfrac12$	intrinsic spin direction of the electron

The first three ( $n,l,m_l$ ) come from solving Schrödinger's equation for a 3D problem in spherical coordinates $(r,\theta,\phi)$ . The fourth ( $m_s$ ) is extra — it does not appear in the basic Schrödinger equation; it was forced on us by experiment (Stern–Gerlach) and later derived properly by Dirac's relativistic equation.

WHY do quantum numbers exist? (Derivation from scratch)

Step 1 — Separate the wavefunction. We write the 3D wavefunction as a product: $\psi(r,\theta,\phi)=R(r)\,\Theta(\theta)\,\Phi(\phi)$ Why this step? The Coulomb potential $V=-\dfrac{ke^2}{r}$ depends only on $r$ , so the equation separates cleanly into a radial part and an angular part. Each part gives its own quantum number.

Step 2 — The $\phi$ equation gives $m_l$ . The azimuthal part obeys $\frac{d^2\Phi}{d\phi^2}=-m_l^2\,\Phi \quad\Rightarrow\quad \Phi(\phi)=e^{im_l\phi}.$ Why this step? For $\Phi$ to be a single-valued physical wave we need $\Phi(\phi+2\pi)=\Phi(\phi)$ , i.e. $e^{im_l 2\pi}=1$ . This is only true if $m_l$ is an integer. That is where the integers come from — a self-consistency (boundary) condition.

Step 3 — The $\theta$ equation gives $l$ . Solving the polar equation (Legendre's equation) gives finite, non-blowing-up solutions only when the separation constant equals $l(l+1)$ with $l=0,1,2,\dots$ , and only when $|m_l|\le l$ . Why this step? If $l$ were not an integer or if $|m_l|>l$ , the solution diverges at the poles ( $\theta=0,\pi$ ) — unphysical. So $l$ caps $m_l$ .

Step 4 — The $r$ equation gives $n$ . The radial equation has well-behaved (square-integrable) solutions only when $n=1,2,3,\dots$ and $l\le n-1$ . Why this step? The boundary condition is now "the wave must die out as $r\to\infty$ ." This bound-state condition quantizes the energy.

WHY is $L_z$ quantized but never equal to $|\vec L|$ ? Because $m_l\le l < \sqrt{l(l+1)}$ . The vector $\vec L$ can never point fully along $z$ — if it did, $L_x=L_y=0$ exactly, violating the uncertainty principle for angular momentum. This is "space quantization": $\vec L$ is allowed only at discrete tilt angles.

HOW to use them — counting states

The number of allowed $m_l$ values for a given $l$ is $2l+1$ (from $-l$ to $+l$ ). Including spin ( $\times 2$ ), the number of electrons a shell holds is $\sum_{l=0}^{n-1}2(2l+1)=2n^2.$ Why? Each distinct $(n,l,m_l,m_s)$ is one quantum state, and (Pauli) no two electrons share all four.

Common mistake Steel-manned common mistakes

Mistake A: " $m_l$ runs from $0$ to $l$ ." Why it feels right: counting often starts at 0, and $m_l$ looks like a count. Fix: $m_l$ is the projection of a vector on an axis, so it can be negative: $m_l=-l,\dots,+l$ , giving $2l+1$ values.

Mistake B: "Angular momentum is $l\hbar$ (Bohr's old rule)." Why it feels right: Bohr's model gave $L=n\hbar$ , and it's simpler. Fix: Schrödinger gives $|\vec L|=\sqrt{l(l+1)}\,\hbar$ , not $l\hbar$ . For $l=0$ , $|\vec L|=0$ (no orbital angular momentum at all — an s-electron has a spherical, non-rotating cloud).

Mistake C: "Spin means the electron physically spins like a top." Why it feels right: the name and the $\hbar$ units suggest rotation. Fix: Spin is an intrinsic property with no classical rotation; a literal spinning surface would exceed light speed. $m_s$ has only two values, $\pm\tfrac12$ , with no spatial direction to "spin around."

Mistake D: " $n=1$ can have a p-orbital." Why it feels right: every other shell has p, d… Fix: $l\le n-1$ , so $n=1$ allows only $l=0$ (1s). No 1p exists.

Recall Feynman: explain it to a 12-year-old

Imagine the electron is a little drum skin wrapped around the nucleus. A drum can only make certain notes — you can't get an in-between note. $n$ tells you how big the drum is (big drum = low note = far from nucleus). $l$ tells you the pattern of vibration (calm and round, or with stripes). $m_l$ tells you which way the striped pattern is turned. And $m_s$ is a tiny built-in magnet in the electron that can only point "up" or "down." No two electrons in the same atom are allowed to have the exact same four labels — like assigned seats with a unique seat number.

Recall Active recall — cover the answers

What boundary condition forces $m_l$ to be an integer?
Why is $|\vec L| \ne l\hbar$ ?
Max electrons in shell $n$ ?

What does the principal quantum number

n

physically control?

The energy and overall size of the orbital;

E_n=-13.6/n^2

eV for hydrogen.

What are the allowed values of

l

for a given

n

l=0,1,2,\dots,(n-1)

What are the allowed values of

m_l

for a given

l

m_l=-l,-l+1,\dots,0,\dots,+l

(that is

2l+1

values).

What are the only allowed values of

m_s

+\tfrac12

and

-\tfrac12

Where do

n,l,m_l

come from mathematically?

From boundary/single-valued conditions when solving the Schrödinger equation in spherical coordinates (radial → n, polar → l, azimuthal → m_l).

What boundary condition quantizes

m_l

Single-valuedness:

\Phi(\phi+2\pi)=\Phi(\phi)

, so

e^{i m_l 2\pi}=1

forces

m_l

integer.

Formula for magnitude of orbital angular momentum?

|\vec L|=\sqrt{l(l+1)}\,\hbar

Formula for the z-component of orbital angular momentum?

L_z=m_l\hbar

Why can

\vec L

never point fully along the z-axis?

Because

m_l\le l<\sqrt{l(l+1)}

, so

L_z<|\vec L|

; pointing fully would fix

L_x=L_y=0

, violating angular-momentum uncertainty (space quantization).

How many electrons can shell

n

hold and why?

2n^2

, from

\sum_{l=0}^{n-1}2(2l+1)

, with the factor 2 for spin.

(n,l,m_l,m_s)=(2,2,0,\tfrac12)

allowed?

No —

l

must be

\le n-1=1

, so

l=2

is forbidden.

What is

|\vec L|

for an s-electron (

l=0

Zero, since

\sqrt{0(0+1)}\hbar=0

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, electron ko hum "chhoti ball" maan ke nahi sochte — wo ek standing wave hai, jaise guitar ki taar par bante hue pattern. Jab Schrödinger equation ko hydrogen atom ke liye solve karte hain, to har valid solution ke saath kuch integers attach ho jaate hain. Yahi integers hain quantum numbers: $n$ (size aur energy), $l$ (orbital ka shape), $m_l$ (orbital kis direction me tilt hai), aur $m_s$ (electron ka built-in spin, sirf up ya down).

Sabse important baat — ye integers kahin se "rakhe" nahi gaye, ye boundary conditions se aate hain. Jaise $\Phi(\phi)$ ko $2\pi$ ghoomne ke baad apne aap se match karna padta hai, isliye $m_l$ majboori me integer ban jaata hai. Isi tarah radial wave ko infinity par die out hona padta hai, jisse $n$ quantize hota hai. Rules yaad rakho: $l$ hamesha $n$ se ek kam tak ( $0$ se $n-1$ ), $m_l$ minus $l$ se plus $l$ tak, aur $m_s=\pm\tfrac12$ .

Angular momentum me ek common galti hoti hai: log Bohr wala $L=n\hbar$ use kar lete hain. Lekin sahi formula hai $|\vec L|=\sqrt{l(l+1)}\,\hbar$ . Aur $L_z=m_l\hbar$ kabhi bhi pure $|\vec L|$ ke barabar nahi ho sakta — isliye vector hamesha thoda tilted rehta hai (space quantization). Iska reason uncertainty principle hai.

Practical use: ek shell $n$ me total $2n^2$ electrons aate hain, kyunki har $(n,l,m_l)$ orbital ko spin ki wajah se 2 se multiply karte hain, aur Pauli ke according koi do electron ke chaaron number same nahi ho sakte. Isi se poora periodic table banta hai — yahi 80/20 wali asli cheez hai jo yaad rakhni hai.

WHAT are the four quantum numbers?

WHY do quantum numbers exist? (Derivation from scratch)

HOW to use them — counting states

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

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