Visual walkthrough — Quantum numbers n, l, mₗ, mₛ

2.3.13 · D2 · Physics › Modern Physics › Quantum numbers n, l, mₗ, mₛ

---

Step 1 — Ek trapped wave sirf kuch khaas shapes mein vibrate kar sakti hai

KYA. Ek minute ke liye atoms bhool jao. Ek guitar string lo jo dono siron par pakdi hui hai. Use pluck karo. Woh kisi bhi random shape mein nahi hilti — woh ek aisa pattern adopt karti hai jisme bumps ki poori sankhya hoti hai: 1 bump, 2 bumps, 3 bumps. Kabhi $1.5$ bumps nahi.

KYO. Dono siron par pin ki gayi string ko har end par zero hona chahiye. Aadhe bump wali shape string ko wall par zero se upar chhod deti hai — impossible, wall use neeche rok leti hai. Isliye "walls se match karo" ka rule har non-whole-number pattern ko baahar nikal deta hai. Sirf integers bachte hain.

PICTURE. Figure dekho: red curve ek legal 3-bump pattern hai (woh dono walls par zero ko touch karti hai). Faint black curve ek illegal koshish hai — woh wall par band nahi hoti, isliye nature ise forbid karta hai.

---

Step 2 — Wave ko ring mein wrap karo: yahan se $m_l$ paida hota hai

KYA. Atom mein electron ek wave hai jo 3D mein rehti hai, lekin pehle hum sirf yeh dekhte hain ki woh vertical axis ke around kaise wrap hoti hai — woh angle jise hum $\varphi$ (bolo "fye") kehte hain, compass direction jab tum ek circle ke around chalte ho. Guitar string ko ek ring mein bend karo aur wave ko usके around daudao.

KYO. Ring par ends ko pin karne ke liye koi wall nahi hoti. Iski jagah wave ko khud se mila lena hoga: $360°$ ka poora chakkar lagane ke baad tum wahin aate ho jahan se chale the, isliye wahan wave ki height wohi value honi chahiye jo tumne chhalte waqt chodhi thi. Kuch bhi aur matlab hai wave mein ek "cliff" — ek break — jo ek smooth physical wave nahi hoti.

PICTURE. Figure mein red wave loop ke around exactly 2 whole bumps fit karti hai, isliye woh smoothly reconnect ho jaati hai (green dot: start = end). Black wave $2\frac{1}{2}$ bumps fit karti hai aur ek mismatch (ek jump) se takraati hai — forbidden.

Aao us wave ko naam dete hain jo ring ke around daudti hai: $\Phi (\varphi )$ — padho "wave-height as a function of compass angle $\varphi$." Ek wave likhne ka sabse saaf tarika jo ek circle ke around steadily daudti hai:

$\Phi (\varphi )=e^{i{m}_l\varphi }$

Term by term:

• $\varphi$ hai circle ke around tum kitna chal chuke ho ($0$ se $2\pi$ radians = ek lap).
• $m_l$ hai ek lap mein wave kitne full bumps banati hai — yeh ek counter hai.
• $e^{i(\cdots )}$ sirf mathematician ka compact tarika hai "ek wave jo round aur round jaati hai" likhne ka; ek circle par ghoomte dot ki tarah socho. (Exactly is form ke liye dekho Angular Momentum in Quantum Mechanics)

Match-yourself rule kehta hai $\Phi (\varphi +2\pi )=\Phi (\varphi )$. Plug in karo:

$e^{i{m}_l(\varphi +2\pi )}=e^{i{m}_l\varphi }⟹e^{i{m}_{l}2\pi }=1.$

Ek ghoomta dot apne exact start par wapas aata hai sirf tabhi jab usne poore number of turns liye hon. Isliye:

Isliye orbital angular momentum ka $z$-component $L_z=m_l\hslash$ hai: $m_l$ literally count karta hai ki wave axis ke baare mein kitni tezi se circulate karti hai.

---

Step 3 — Wave ko poles ke upar aur neeche chalao: yahan se $l$ paida hota hai

KYA. Ab upar-neeche direction add karo: angle $\theta$ ("theta") jo north pole ($\theta =0$) se neeche south pole ($\theta =\pi$) tak measure hota hai. Wave ab pole se pole tak bhi ripple karti hai. Is piece ko $\Theta (\theta )$ bolo.

KYO. Sphere ke do khaas danger points hain: poles. Wahan, saari compass lines ek single point mein crash karti hain. Agar humara pole-to-pole ripple bahut wild hai, toh iska height exactly poles par infinity tak blow up ho jaata hai — infinite wave unphysical hai. Mathematics (Legendre's equation, from Schrödinger Equation) kehta hai ripple finite rahti hai sirf tabhi jab pole-to-pole pattern ko ek whole number se label kiya jaaye:

$l=0,1,2,3,\dots$

aur — crucially — sirf tabhi jab Step 2 ka ring-count isse exceed na kare:

$|m_l|\le l.$

PICTURE. Figure teen legal pole-to-pole patterns dikhata hai $l$ ke hisaab se stacked: $l=0$ ek smooth round blob hai (koi stripes nahi), $l=1$ mein ek nodal line hai, $l=2$ mein do hain. Red band equator mark karta hai jahan $l=2$ ki extra ripple hai. Har ek ke neeche, ek tally dikhata hai ki $m_l$ ke kitne sideways orientations allow hain: $l=0$ deta hai 1, $l=1$ deta hai 3, $l=2$ deta hai 5 — hamesha $2l+1$.

Term by term, orbital angular momentum ki magnitude jo yahan nikalti hai:

$|\vec{L}|=\sqrt{l(l+1)}\hslash$

• $l$ hai pole-to-pole kitne stripes hain angular wave mein — shape ki "roughness."
• $\sqrt{l(l+1)}$ us stripe-count ko angular-momentum arrow ki actual length mein convert karta hai.
• $\hslash$ ("h-bar") angular momentum ki fixed quantum unit hai — woh sabse chhota coin jo nature use karta hai.

---

Step 4 — $L_z$ kabhi seedha upar kyun point nahi kar sakta (space quantization)

KYA. Step 2 ne sideways spin-count $m_l$ diya (jo $L_z=m_l\hslash$ set karta hai). Step 3 ne total arrow length $|\vec{L}|=\sqrt{l(l+1)}\hslash$ diya. Ab inhe compare karo.

KYO. Sabse bada $m_l$ hota hai $l$ khud. Isliye sabse bada possible upward reach hai $L_z=l\hslash$. Lekin arrow ki length hai $\sqrt{l(l+1)}\hslash$, jo strictly badi hai:

$l\hslash <\sqrt{l(l+1)}\hslash .$

Ek arrow apni khud se zyada lambi length vertical par project nahi kar sakta. Isliye $\vec{L}$ kabhi $z$ ke saath flat nahi let sakta — woh hamesha tilted rahta hai. Agar woh perfectly $z$ ke saath let jaata, toh $L_x$ aur $L_y$ dono exactly zero aur ek saath perfectly known ho jaate, jo uncertainty principle forbid karta hai.

PICTURE. $l=2$ ke liye red arrow ki length $\sqrt{6}\hslash \approx 2.45\hslash$ hai. Iske allowed vertical projections paanch rungs hain $L_z=-2\hslash ,-\hslash ,0,\hslash ,2\hslash$. Top rung bhi ($2\hslash$) tip se chhota hai — isliye arrow ek fixed cone ki taraf point karta hai, kabhi seedha upar nahi. Yahi hai space quantization.

---

Step 5 — Wave ko baahir ki taraf push karo: yahan se $n$ aata hai aur energy fix hoti hai

KYA. Aakhri direction hai radial: wave ki strength kaise badlti hai jab tum nucleus se baahir jaate ho, distance $r$. Is piece ko $R(r)$ bolo.

KYO. Ek electron bound hai — woh infinity tak escape nahi kar sakta. Isliye iska wave $r\to \infty$ par zero tak fade ho jana chahiye; infinity par abhi bhi buzzing wave ek free, un-trapped particle describe karti hai, bound electron nahi. "Door fizzle ho jaao" demand karna ek aur boundary condition hai, aur — same kahani as always — sirf countable set of radial patterns survive karti hain, labeled by

$n=1,2,3,\dots \text{with the extra rule}l\le n-1.$

PICTURE. Figure $n=1,2,3$ ke liye radial wave plot karta hai. Har ek ke radial nodes ki alag sankhya hai (woh points jahan woh zero cross karta hai) — red curve ($n=3$) $n=1$ se do zyada baar zero cross karti hai. Zyada crossings = zyada energy = bada orbital. Dashed envelope dikhata hai ki har curve door jaake zero ho jaati hai (boundary condition).

Is "must die at infinity" condition se jo energy nikalti hai woh exactly hydrogen ladder hai (dekho Hydrogen Atom Energy Levels):

$E_n=-\frac{13.6\text{ eV}}{n^2}$

• $n$ hai wave kitni door aur kitni energetic hai — master counter.
• Minus sign batata hai ki electron trapped hai (ise free karne ke liye tumhe energy add karni padegi).
• Denominator mein $n^2$ high shells ko near zero energy par crowded karta hai.

---

Step 6 — Ek extra number jo wave NAHI hai: spin $m_s$

KYA. Teen counts ($n,l,m_l$) wave ko teen tarao se wrap karne se aaye. Lekin experiment (Stern-Gerlach Experiment) ne atoms ki beam ko exactly do mein split kiya, yahan tak ki $l=0$ wale atoms ke liye bhi (koi orbital angular momentum nahi). Ek fourth label ki zaroorat thi: spin, $m_s$.

KYO. Koi bhi amount of spatial wave-wrapping ek $l=0$ electron ko do mein split nahi kar sakti. Two-way split zaroor electron ke andar se hi aani chahiye, koi sideways pointing ki jagah nahi — sirf "up" ya "down":

$m_s=+\frac{1}{2}\text{or}{m}_s=-\frac{1}{2}.$

PICTURE. Figure dikhata hai Stern–Gerlach beam unified enter karti hai aur two red dots ke roop mein nikalti hai — up aur down, beech mein kuch nahi. Yahan ek spinning ball ki koi classical picture nahi hai (ek real spinning electron surface light speed exceed kar leti); spin simply ek intrinsic two-valued tag hai.

---

Ek-picture summary

Ek trapped wave par teen boundary conditions, plus ek intrinsic tag, chaar labels dete hain — aur kisi bhi atom mein do electrons charon share nahi kar sakte (Pauli Exclusion Principle), yahi poora Electron Configuration & Periodic Table banata hai.

Recall Feynman: poora walkthrough simple shabdon mein

Electron ko ek wave socho jo nucleus ke paas phasi hui hai. Use teen directions mein squeeze karo aur har squeeze ek "match yourself" rule demand karta hai. Vertical ke around (compass loop), wave ko ek lap ke baad reconnect karna hoga — isliye bumps ki poori sankhya fit honi chahiye: woh count hai $m_l$, aur woh dono taraf run kar sakta hai, isliye negative ho sakta hai. Pole to pole jaate hue, wave ko do poles par blow up nahi karna chahiye — yeh shape kitni striped ho sakti hai isko limit karta hai, $l$ deta hai, aur $l$ kam se kam $|m_l|$ jitna bada hona chahiye. Baahir ki taraf jaate hue, wave ko door jaake kuch nahi ho jaana chahiye kyunki electron trapped hai — yeh $n$ deta hai aur energy $-13.6/n^2$ eV fix karta hai, is rule ke saath ki $l$ $n$ se ek chhota rukta hai. Arrow ki length ($\sqrt{l(l+1)}\hslash$) ko iske tallest reach ($l\hslash$) se compare karne par pata chalta hai arrow hamesha tilted hai — space quantization. Aakhir mein, experiment ek aur non-wave tag force karta hai, spin, jo sirf up ya down hota hai. Chaar labels; koi bhi do electrons charon match nahi kar sakte. Woh single "no repeats" rule hi poora periodic table hai.

Recall Active recall — answers chhupa lo

Kaun si boundary condition $m_l$ ko uske integer values deti hai? ::: Ring ka khud par band hona, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, jo $e^{i{m}_{l}2\pi }=1$ force karta hai. Kaun si condition $n$ deti hai? ::: Wave ko $r\to \infty$ par die out karna hoga (bound-state condition). $\vec{L}$ hamesha $z$-axis se tilted kyun rehta hai? ::: Kyunki $L_z^{\max }=l\hslash <\sqrt{l(l+1)}\hslash =|\vec{L}|$; projection kabhi full length ke barabar nahi ho sakta. Spin wave-count kyun nahi hai? ::: Yeh $l=0$ atoms ko bhi do mein split karta hai, isliye yeh ek intrinsic two-valued property honi chahiye, spatial wrapping nahi.