Hydrogen atom — solving in spherical coordinates: Question bank

2.3.12 · D5 · Physics › Modern Physics › Hydrogen atom — solving in spherical coordinates

Shuru karne se pehle, symbols ka ek one-line reminder, taaki kuch bhi unexplained na lage:

Definition Symbol refresher

$r$ = proton se electron ki distance; $θ$ = polar angle (top pole se kitna door); $ϕ$ = azimuthal angle (kitna ghumke, jaise longitude).
$n$ = principal quantum number ( $1, 2, 3, \dots$ ) — energy set karta hai.
$ℓ$ = orbital angular-momentum quantum number ( $0, 1, \dots, n - 1$ ) — shape / total angular momentum set karta hai.
$m_{ℓ}$ = magnetic quantum number ( $- ℓ, \dots, + ℓ$ ) — angular momentum ka ek chosen axis ki taraf tilt set karta hai.
$μ$ = reduced mass $= \frac{m _{e} m _{p}}{m _{e} + m _{p}} \approx m_{e}$ — wo effective mass jo two-body (electron + hilta proton) problem ko one-body mein badal deta hai.
$e$ = elementary charge, electron par electric charge ki size (aur, opposite sign ke saath, proton par); ye Coulomb attraction ki strength set karta hai.
$ε_{0}$ = vacuum permittivity, wo constant jo Coulomb law mein electric force ki strength fix karta hai (ye $4 π ε_{0}$ mein appear hota hai).
$ℏ$ = reduced Planck constant $= h /2 π$ , angular momentum ka fundamental quantum; ye "how-many-wiggles" quantum numbers ko physical energies aur momenta mein convert karta hai.
$V (r) = - \frac{e ^{2}}{4 π ε _{0} r}$ = Coulomb potential; "central" ka matlab hai ye sirf $r$ par depend karta hai.
$E$ = state ki total energy (ek given wavefunction ke liye ek single number); $E_{n}$ = level $n$ ki specific allowed energy, jo $E_{n} = - 13.6 eV / n^{2}$ se milti hai. Negative $E$ matlab bound state.
$V_{eff} (r) = V (r) + \frac{ℏ ^{2} ℓ ( ℓ + 1 )}{2 μ r ^{2}}$ = effective radial potential: asli Coulomb pull plus centrifugal barrier, taaki radial motion is combined potential mein ek 1D particle jaisi lage.
$ψ (r, θ, ϕ)$ = full wavefunction. Hum ise product $ψ = R (r) Y (θ, ϕ)$ ke roop mein likhte hain, jahan $R (r)$ radial part hai (sirf distance par depend karta hai) aur $Y (θ, ϕ)$ angular part hai (ek spherical harmonic, sirf direction par depend karta hai).
$u (r) = r R (r)$ = ek rescaled radial function jo radial equation ko 1D Schrödinger equation jaisa banata hai.

Figure — Hydrogen atom — solving in spherical coordinates

Teesri picture, Figure 3, kuch states ke liye radial function $u (r)$ sketch karti hai taaki tum uske nodes count kar sako — $r = 0$ aur $r = \infty$ ke beech ke zero-crossings. Ye Edge cases section mein node-counting trap ke peeche ka visual hai.

True or false — justify karo

Wavefunction ka single-valued hona hi

m_{ℓ}

ko integer hone par majboor karta hai.

True.

e^{i m_{ℓ} ϕ}

par

Φ (ϕ + 2 π) = Φ (ϕ)

require karne se

e^{i m_{ℓ} 2 π} = 1

hona zaroori hai, jo sirf integer

m_{ℓ}

ke liye hi sach hai — force ki koi physics ki zaroorat nahi. Figure 2 dekho: sirf whole-number windings hi start par wapas aati hain.

Energy ka quantization Coulomb force ki wajah se hota hai.

False.

ℓ

aur

m_{ℓ}

ka quantization boundary conditions se aata hai aur kisi bhi central potential ke liye hota hai; sirf specific formula

E_{n} \propto 1/ n^{2}

Coulomb ke

1/ r

shape ka use karta hai.

ℓ = 0

wale state mein electron ki koi angular kinetic energy nahi hoti.

True.

ℓ = 0

ke saath centrifugal term

ℏ^{2} ℓ (ℓ + 1) / (2 μ r^{2})

vanish ho jaata hai, isliye saari kinetic energy radial hoti hai. Figure 1 mein ye flat-bottom Coulomb curve hai jisme koi barrier nahi; isiliye

s

-states ka

r = 0

par nonzero density ho sakta hai.

Same

n

but alag

ℓ

wale do states ki hydrogen mein energy hamesha alag hoti hai.

False. Pure hydrogen mein unki energy same hoti hai —

1/ r

potential mein ek hidden symmetry hoti hai jo energy ko sirf

n

par depend banati hai. (Ye "accidental" degeneracy multi-electron atoms mein toot jaati hai.)

Separation constant ko

ℓ (ℓ + 1)

likhna zaroori tha; koi aur naam galat physics deta.

False. Constant bas ek number hai; hum ise

λ

bhi keh sakte the. Ise

ℓ (ℓ + 1)

likhna ek label ki choice hai jo Legendre solutions ko neatly terminate karati hai, lekin physics identical hai.

2 ℓ + 1

number ek given

ℓ

ke liye angular momentum ke allowed orientations count karta hai.

True.

m_{ℓ}

integer steps mein

- ℓ

+ ℓ

tak jaata hai, jo

2 ℓ + 1

values deta hai — har ek

z

-axis par

\hat{L}

ka alag projection

m_{ℓ} ℏ

Kyunki

ψ = R (r) Y (θ, ϕ)

factorize hota hai, electron ki radial aur angular behaviour physically independent motions hain.

Partly false. Factorization ek mathematical convenience hai jo valid hai kyunki

V

sirf

r

par depend karta hai; radial equation mein abhi bhi

ℓ

centrifugal barrier ke through hai (Figure 1), isliye angular quantum number radial problem mein feed back karta hai.

Energy levels

E_{n}

negative hain kyunki humne ek odd sign convention choose kiya.

False. Ye negative hain kyunki bound states free-electron energy se neeche hote hain (jo

r \to \infty

par zero set hai). Negative ka matlab hai "electron ko chiinne mein energy lagegi."

Radial function par boundary conditions

u (0) = 0

aur

u (r \to \infty) = 0

hain.

True.

u = r R

origin par vanish hona chahiye (warna

R \sim 1/ r

wahaan blow up kar deta) aur infinity par decay karna chahiye taaki state normalizable ho. Ye do conditions exactly wahi hain jo energy ko discrete values lene par majboor karti hain.

Error dhundho

n = 2

ke liye, allowed values hain

ℓ = 0, 1, 2

Error:

ℓ

n - 1

par cap hota hai, isliye

n = 2

ke liye sirf

ℓ = 0, 1

. Radial (Laguerre) solution terminate hona fail karta hai — ye

r \to \infty

par blow up karta hai —

ℓ \geq n

ke liye.

"Centrifugal barrier

ℏ^{2} ℓ (ℓ + 1) / (2 μ r^{2})

field mein stored potential energy hai."

Error: ye angular motion ki kinetic energy hai, algebraically ek effective 1D potential

V_{eff}

mein repackage ki gayi hai (Figure 1 mein rising red/green curves). Real field energy sirf Coulomb term

- e^{2} / (4 π ε_{0} r)

hai.

"Hum

m_{e}

ko reduced mass

μ

se replace karte hain kyunki electron hamari soch se bhaari hai."

Error:

μ

proton ke bhi common centre of mass ke around hilne ka account karta hai. Ye ek two-body problem ko one-body mein convert karta hai;

μ < m_{e}

thoda sa, ye electron ki apni mass ki correction nahi hai.

"Separated equation ke dono sides ek constant ke barabar hain kyunki humne unhe haath se barabar set kiya."

Error: ye necessarily constant ke barabar hain: sirf

r

ki function sirf angles ki function ke barabar hai variables ki saari values ke liye, jo possible nahi jab tak dono same fixed number na hon.

m_{ℓ}

ki range

0, 1, 2, \dots, ℓ

hai — ye

ℓ + 1

values hain."

Error:

m_{ℓ}

mein negative values bhi shaamil hain:

- ℓ, \dots, 0, \dots, + ℓ

, jo

2 ℓ + 1

values deta hai,

ℓ + 1

nahi.

"Laplacian ke angular terms hydrogen atom ke liye unique physics hain."

Error: spherical-coordinate Laplacian ek pure geometry fact hai chain rule se; ye 3D mein kisi bhi problem ke liye identical hai, hydrogen ho ya na ho.

"Kyunki

ℓ = 0

ka matlab koi angular momentum nahi,

Y_{0}^{0}

solution ko

θ

aur

ϕ

par depend karna chahiye."

Error:

Y_{0}^{0}

ek constant hai (spherically symmetric) — bilkul koi angular dependence nahi. Ye exactly wahi hai jo "zero angular momentum" dikhta hai.

"Kisi state ke radial nodes ki sankhya bas

n

hai."

Error: radial part

u (r)

mein exactly

n - ℓ - 1

nodes (zero-crossings strictly

0

aur

\infty

ke beech) hote hain,

n

nahi. Figure 3 dekho.

Why questions

Hum radial equation mein

R (r)

u (r) = r R (r)

kyun switch karte hain?

Kyunki ye first-derivative term hata deta hai aur radial equation ko jaane-pahchane 1D Schrödinger form

- \frac{ℏ ^{2}}{2 μ} u^{''} + V_{eff} u = E u

mein badal deta hai (Figure 1 mein drawn potential ke saath), clean boundary conditions

u (0) = 0

aur

u (\infty) = 0

ke saath — jinhein analyse karna hum pehle se jaante hain.

Φ (ϕ)

ϕ \to ϕ + 2 π

ke baad apni value par wapas kyun aana chahiye?

Kyunki

ϕ

aur

ϕ + 2 π

space mein same physical point label karte hain; wavefunction ko use ek definite value assign karni chahiye, isliye single-valuedness ek physical requirement hai, math ki nicety nahi. Figure 2 wrap dikhata hai: sirf integer windings hi band hoti hain.

Level

n

ki degeneracy exactly

n^{2}

kyun hai (spin ignore karte hue)?

ℓ = 0

n - 1

tak

2 ℓ + 1

ka sum pehle

n

odd numbers ka sum hai, jo

n^{2}

ke barabar hota hai. Har state

E_{n}

share karta hai kyunki hydrogen mein energy

ℓ

aur

m_{ℓ}

ko ignore karti hai.

Bada

ℓ

electron ko nucleus se door kyun dhakelta hai?

Bada

ℓ

origin ke paas centrifugal barrier

ℏ^{2} ℓ (ℓ + 1) / (2 μ r^{2})

ko stronger banata hai, wahaan

V_{eff}

ko uthata hai. Figure 1 mein,

ℓ = 0

curve (

- \infty

tak dive karta hai) ko

ℓ = 1, 2

curves (origin ke paas ek wall) se compare karo: wall wavefunction ko bahar ki taraf repel karti hai, jaise ek tez-ghoomta skater center ki taraf kheenche jaane ka resist karta hai.

Humne spherical coordinates kyun choose kiye rather than PDE

x, y, z

mein solve karne ke?

Kyunki potential sirf

r

par depend karta hai; coordinate system ko symmetry se match karne se single 3D PDE separate ho jaati hai teen independent one-variable ODEs mein, ek intractable problem ko teen solvable ones mein badal deta hai.

ℓ = 0, 1, 2, 3

ko s, p, d, f kyun kaha jaata hai rather than

0, 1, 2, 3

Historical spectroscopy labels (sharp, principal, diffuse, fundamental) stuck ho gaye; ye integer

ℓ

jaisi hi information carry karte hain aur solution ko periodic-table structure se connect karte hain.

Angular momentum magnitude

ℓ (ℓ + 1) ℏ

kyun nikalta hai aur

ℓ ℏ

nahi?

Kyunki

\hat{L}^{2}

(squared magnitude ke liye operator) ki eigenvalue

ℓ (ℓ + 1) ℏ^{2}

hai; square root lene par

ℓ (ℓ + 1) ℏ

milta hai, jo hamesha

ℓ ℏ

se kam se kam utna bada hota hai — ek genuinely quantum feature.

Edge cases

n = 1

hone par saare allowed states kya hain?

Sirf ek:

ℓ = 0

(

n - 1 = 0

par capped) aur

m_{ℓ} = 0

. Ye single

1 s

state

- 13.6

eV par ground state hai.

n \to \infty

hone par energy ka kya hota hai?

E_{n} = - 13.6/ n^{2} \to 0^{-}

: levels neeche se zero ki taraf ek saath aa jaate hain. Exactly zero par electron free (unbound) ho jaata hai — ye ionization threshold hai.

Kya

ℓ = 0, m_{ℓ} = 0

possible hai, aur iska angular part kaisa dikhta hai?

Haan; ye

s

-state hai jiska spherical harmonic

Y_{0}^{0}

ek constant hai, isliye probability cloud bilkul spherical hai — koi preferred direction nahi.

Kya

m_{ℓ}

kabhi

ℓ

se magnitude mein zyada ho sakta hai, jaise

m_{ℓ} = 3

aur

ℓ = 1

Nahi. Associated Legendre solutions finite hote hain tabhi jab

∣ m_{ℓ} ∣ \leq ℓ

; is se aage ye poles

θ = 0, π

par blow up karte hain, isliye aisa state exist nahi karta.

3 s

state mein kitne radial nodes hain, aur

3 d

state mein?

3 s

mein

n = 3, ℓ = 0 \Rightarrow n - ℓ - 1 = 2

nodes hain;

3 d

mein

n = 3, ℓ = 2 \Rightarrow n - ℓ - 1 = 0

nodes hain. Same energy, bahut alag radial shapes — Figure 3 dekho.

Real experiment mein hydrogen level ki

ℓ

-degeneracy kya todta hai?

Ek external magnetic field (Zeeman effect) states ko

m_{ℓ}

ke hisaab se split karta hai, aur pure

1/ r

se departures (multi-electron atoms mein, ya relativistic corrections)

ℓ

ke hisaab se split karte hain — "accidental" symmetry fragile hai.

Sabse chhota possible total angular momentum kya hai, aur kis state mein hota hai?

ℓ = 0

0 \cdot 1 ℏ = 0

milta hai:

s

-states mein exactly zero orbital angular momentum hota hai, true minimum.

Agar proton infinitely heavy hota, to

μ

kaise change hota?

μ = \frac{m _{e} m _{p}}{m _{e} + m _{p}} \to m_{e}

jab

m_{p} \to \infty

; reduced mass plain electron mass ban jaata hai, kyunki immovable proton ka matlab hai sirf electron hilta hai.

Wo boundary conditions kya hain jo actually discrete energies pick out karti hain?

u (0) = 0

(origin par regular) aur

u (r \to \infty) = 0

(normalizable). Sirf special energies par hi ek solution dono ko ek saath satisfy kar sakta hai — wo discreteness hi energy quantization hai.

Recall Ek-line self-test

Upar har answer chhupaao. Har ek ke liye pehle bolo kaun sa rule (single-valuedness, Legendre/Laguerre ka termination, centrifugal barrier, Coulomb-specific energy, boundary conditions $u (0) = u (\infty) = 0$ ) kaam kar raha hai — ye understanding ka asli test hai.