2.3.10 · Physics › Modern Physics
Intuition Bada picture (WHY yeh matter karta hai)
Ek quantum particle ko do infinitely ऊnchi walls ke beech trap karo. Jaise ek guitar string dono siron par clamped hoti hai, waisi hi sirf kuch standing waves perfectly fit hoti hain. Har allowed wave = ek allowed energy. Toh confinement → quantisation . Yeh sabse simple model hai jo dikhata hai kyun atoms, quantum dots, aur nuclei mein energy levels discrete hote hain.
Definition Infinite square well
Mass m ka ek particle 0 < x < L ke andar freely move karta hai, lekin infinite potential ki walls se trap hota hai:
V ( x ) = { 0 ∞ 0 < x < L otherwise
Particle ==jahan V = ∞ ho wahan exist nahi kar sakta==, isliye wavefunction ψ ( x ) = 0 bahar aur walls par hoti hai.
KYA chahiye: allowed wavefunctions ψ ( x ) aur energies E .
KAISE : box ke andar time-independent Schrödinger equation (TISE) ko boundary conditions ke saath solve karo.
Rearrange karo:
d x 2 d 2 ψ = − ℏ 2 2 m E ψ = − k 2 ψ , k 2 ≡ ℏ 2 2 m E
k define kyun kiya? Kyunki andar E > 0 hai, coefficient negative hai, jo oscillatory solutions deta hai. k naam rakhne se algebra clean rehti hai — aur k wave number nikalta hai.
Toh wave number quantised hai:
k n = L nπ
Intuition Quantisation kyun aata hai
x = L par wall ek node force karti hai. Sirf wahi waves survive karti hain jinki half-wavelengths poori integer times fit hoti hain: L = n ⋅ 2 λ n . Walls referee hain; woh har woh k reject karte hain jo special nahi hai.
Intuition Zero-point energy
Sabse kam energy E 1 = 8 m L 2 h 2 = 0 hai. Ek trapped quantum particle kabhi perfectly rest mein nahi ho sakta . KYU: ψ = const boundary conditions ki node requirement violate karega, aur flat wave matlab Δ x tiny ⟹ Δ p huge (uncertainty). Confinement energy maangta hai.
Intuition Nodes aur shape
ψ n mein n − 1 interior nodes (zeros) aur n antinodes hote hain. Zyada n = zyada wiggles = zyada curvature = zyada kinetic energy (ψ ′′ term bada hota hai).
Worked example Example 1 — 0.10 nm box mein electron (atomic scale)
Electron ke liye E 1 nikalo, L = 1.0 × 1 0 − 10 m.
E 1 = 8 ( 9.11 × 1 0 − 31 ) ( 1.0 × 1 0 − 10 ) 2 ( 6.63 × 1 0 − 34 ) 2
Yeh numbers kyun? m = m e , n = 1 ground state.
Numerator = 4.40 × 1 0 − 67 . Denominator = 8 × 9.11 × 1 0 − 31 × 1.0 × 1 0 − 20 = 7.29 × 1 0 − 50 .
E 1 = 6.03 × 1 0 − 18 J ≈ 37.6 eV
eV kyun? 1.6 × 1 0 − 19 se divide karo. Yeh atomic binding energies jaisa hai — model sahi scale capture karta hai.
Worked example Example 2 — Energy gap
1 → 2
E 2 − E 1 = ( 4 − 1 ) E 1 = 3 E 1 = 3 ( 37.6 ) = 113 eV
Yeh step kyun? E n ∝ n 2 , toh gap mein 2 2 − 1 2 = 3 use hota hai. Aisa jump λ = h c /Δ E ≈ 11 nm (UV) ka photon emit karega.
Common mistake "Energy levels equally spaced hote hain."
Kyun sahi lagta hai: evenly spaced ladders (jaise SHM) common hain. Fix: yahan E n ∝ n 2 hai, toh spacings badhti hain: E 2 − E 1 = 3 E 1 , E 3 − E 2 = 5 E 1 . n aur n + 1 ke beech spacing ( 2 n + 1 ) E 1 hai.
Common mistake "Cosine term bhi rakho."
Kyun sahi lagta hai: general solution mein sin aur cos dono hote hain. Fix: boundary condition ψ ( 0 ) = 0 cosine ko khatam kar deti hai (B = 0 ). BCs hamesha turant apply karo.
ψ probability hai."
Fix: ψ amplitude hai; probability density ∣ ψ ∣ 2 hoti hai. Probability nikaalने ke liye ∣ ψ ∣ 2 ko region par integrate karna padta hai.
Common mistake "Ground state
n = 0 par hoti hai."
Kyun sahi lagta hai: counting usually 0 se start hoti hai. Fix: n = 0 se ψ ≡ 0 milta hai — koi particle nahi. Sabse kam physical state n = 1 hai jisme nonzero energy hai.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Socho ek jump rope dono walls se bandhi hai. Jab tum use hilao, toh sirf kuch neat patterns aate hain: ek bada arch, phir do arches, phir teen — kabhi "dedh" arch nahi, kyunki ends hamesha still rehne chahiye. Har neat pattern ek certain shaking speed = ek certain energy maangta hai. Ek tiny particle ek box mein exactly usi rope jaisa behave karta hai: sirf special wave patterns fit hote hain, toh uske paas sirf special energies ho sakti hain. Aur woh kabhi poori tarah se wiggling band nahi kar sakta — sabse calm pattern (ek arch) mein bhi kuch energy hoti hai.
Mnemonic Energy formula yaad rakhо
"n 2 over 8 m L 2 , times h 2 ." Mnemonic: "En square = (nh)² / 8mL²" — kyunki E n = ( nh ) 2 / ( 8 m L 2 ) . Bada box (L ↑ ) → lower, denser levels (electron zyada free hai); bhaari particle → lower levels.
ψ = A sin k x + B cos k x mein B = 0 kyun hota hai?
n = 3 ke liye box ke andar kitne nodes hote hain?
E 3 / E 1 kya hai?
Infinite square well kaunsa potential define karta hai? V = 0 for 0 < x < L aur V = ∞ bahar (walls particle ko trap karti hain).
ψ kaunsi boundary conditions satisfy karta hai?ψ ( 0 ) = ψ ( L ) = 0 (continuity; particle wahan nahi ho sakta jahan V = ∞ ho).
Cosine term kyun vanish ho jaata hai? ψ ( 0 ) = 0 se B = 0 force hota hai, sirf A sin ( k x ) bachta hai.
k par quantisation condition kya hai?sin ( k L ) = 0 ⇒ k n = nπ / L , n = 1 , 2 , 3 , …
Particle in a box ke energy levels? E n = 8 m L 2 n 2 h 2 = 2 m L 2 ℏ 2 n 2 π 2 .
Normalised wavefunctions? Nonzero ground-state energy kyun hoti hai? Confinement + uncertainty: flat/zero wave forbidden hai, toh E 1 = h 2 /8 m L 2 > 0 (zero-point energy).
Level spacings kaisa behave karte hain? Woh badhte hain: E n + 1 − E n = ( 2 n + 1 ) E 1 , equal nahi.
ψ n mein kitne interior nodes hote hain?n − 1 nodes (aur n antinodes).
Probability density aur wavefunction mein kya fark hai? Density ∣ ψ n ∣ 2 hai; probability ke liye ise region par integrate karo.
Bade box L ka energies par kya effect hota hai? E n ∝ 1/ L 2 , toh levels girate hain aur paas aa jaate hain.
Schrödinger Equation (TISE) — woh master equation jo humne solve ki.
Wavefunction and Born Rule — ∣ ψ ∣ 2 ka matlab aur normalisation.
Heisenberg Uncertainty Principle — zero-point energy explain karta hai.
Quantum Harmonic Oscillator — ek aur bound state, lekin equally spaced levels.
Quantum Tunnelling and Finite Well — jab walls finite hoti hain toh kya badalta hai.
Standing Waves on a String — quantisation ka classical analogue.
Quantum Dots — real-world particle-in-a-box (size se colour tune hota hai).
Infinite square well V=0 inside
Time-independent Schrodinger eqn
Define k squared = 2mE/hbar squared
General solution A sin kx + B cos kx
Boundary conditions psi=0 at walls
Energy levels E_n = n^2 h^2 / 8mL^2