5.1.2 · Chemistry › Physical Chemistry (Advanced)
Schrödinger equation H ^ ψ = E ψ sirf kuch hi systems ke liye exactly solve ho sakti hai (particle in a box, harmonic oscillator, hydrogen atom). Baaki sab ke liye — helium, molecules, anharmonic vibrations — hume approximate methods chahiye. Do bade kaam karte hain ye:
Variational principle : ek trial wavefunction guess karo, aur jo energy tum compute karte ho wo HAMESHA zyada hogi. Toh use jitna ho sake kam karo — tum upar se sachchi answer ki taraf squeeze kar rahe ho.
Perturbation theory : ek aisa problem lo jo tum solve KAR SAKTE ho, phir Hamiltonian ka ek chhota sa "extra tukda" add karo aur jawab ko step by step theek karte jao.
Koi bhi trial wavefunction true energy eigenstates ka ek "mixture" hota hai. Ground state sabse lowest possible energy hoti hai. Agar tumhara guess accidentally excited states ka thoda sa hissa le leta hai, toh wo average energy ko upar kheenchte hain, neeche nahi. Isliye average kabhi true ground state se neeche nahi ja sakta. Yahi poori baat hai.
Definition Variational energy (Rayleigh quotient)
Kisi bhi normalizable trial function ϕ ke liye, define karo
E [ ϕ ] = ⟨ ϕ ∣ ϕ ⟩ ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ .
Variational theorem kehta hai ki E [ ϕ ] ≥ E 0 , jahan E 0 true ground-state energy hai. Equality tab hogi jab ϕ sach mein ground state ho.
Toh strategy: apne trial function mein parameters α daalo, E ( α ) compute karo, phir minimize karo:
∂ α ∂ E ( α ) = 0.
Worked example Worked example 1 — Particle in a box, polynomial trial
Box 0 ≤ x ≤ a , H ^ = − 2 m ℏ 2 d x 2 d 2 . Sach mein E 0 = 2 m a 2 π 2 ℏ 2 ≈ 2 m a 2 9.8696 ℏ 2 .
Trial jo dono walls par zero ho: ϕ = x ( a − x ) .
Ye trial kyun? Ye boundary conditions satisfy karta hai (ϕ ( 0 ) = ϕ ( a ) = 0 ) — zaroori hai, warna ye space mein hai hi nahi.
Numerator: ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ = − 2 m ℏ 2 ∫ 0 a x ( a − x ) d x 2 d 2 [ x ( a − x )] d x .
Kyunki d x 2 d 2 [ a x − x 2 ] = − 2 :
= − 2 m ℏ 2 ∫ 0 a x ( a − x ) ( − 2 ) d x = m ℏ 2 ⋅ 6 a 3 = 6 m ℏ 2 a 3 .
Denominator: ∫ 0 a x 2 ( a − x ) 2 d x = 30 a 5 .
E [ ϕ ] = a 5 /30 ℏ 2 a 3 /6 m = m a 2 5 ℏ 2 = 2 m a 2 10 ℏ 2 .
Compare karo: 10 > 9.8696 . Sach se sirf 1.3% upar — aur ye bina kisi parameter ke guess se!
Worked example Worked example 2 — Ek parameter add karna
Try karo ϕ = x ( a − x ) + β x 2 ( a − x ) 2 (agla symmetric correction). β optimize karne se E aur neeche 9.8696 ki taraf aa jaata hai. Ye step kyun? Zyada flexibility = richer mixture = excited-state contamination aur cancel ho sakti hai, toh upper bound tighten hoti hai. Yahi quantum chemistry ka engine hai (basis sets = bahut saare parameters).
Agar ek problem almost wahi hai jo tum solve kar sakte ho, toh Hamiltonian ko ek solved part H ^ ( 0 ) aur ek chhoti si takleef λ H ^ ′ mein likho. Phir energy aur wavefunction ko smallness parameter λ mein power series ke roop mein expand karo — jaise "perturbation kitni badi hai" mein Taylor series.
H ^ = H ^ ( 0 ) + λ H ^ ′ , H ^ ( 0 ) ψ n ( 0 ) = E n ( 0 ) ψ n ( 0 ) .
Expand karo:
E n = E n ( 0 ) + λ E n ( 1 ) + λ 2 E n ( 2 ) + ⋯ , ψ n = ψ n ( 0 ) + λ ψ n ( 1 ) + ⋯
Worked example Worked example 3 — Harmonic oscillator + chhota
x 2 bump
H ^ ′ = 2 1 k ′ x 2 . Level n ke liye first-order correction:
E n ( 1 ) = 2 1 k ′ ⟨ n ∣ x 2 ∣ n ⟩ = 2 1 k ′ ⋅ 2 mω ℏ ( 2 n + 1 ) .
Ye step kyun? Standard result ⟨ n ∣ x 2 ∣ n ⟩ = 2 mω ℏ ( 2 n + 1 ) . Ye yahan exact hai kyunki perturbation sirf k ko rescale karti hai, lekin ye machinery ko clearly dikhata hai.
Common mistake Steel-manned misconceptions
(a) "Kam variational energy matlab zyada accurate wavefunction." Sahi lagta hai kyunki energy hi wo cheez hai jo hum minimize karte hain. Kyun tempting lagta hai: energy fast converge hoti hai. Fix: energy truth ke paas stationary hoti hai (2nd-order accurate), toh wavefunctions phir bhi kharab ho sakti hain jabki energy achi lagti hai. Energy ke alawa doosri properties bahut galat ho sakti hain.
(b) "Variational energy E 0 se neeche aa sakti hai agar kismat ho." Kyun tempting lagta hai: approximations kabhi kabhi undershoot karti hain. Fix: theorem ise forbid karta hai — E [ ϕ ] ≥ E 0 rigorously. Agar tumhe chhota number mila, toh arithmetic/normalization mein galti hui hai.
(c) Degenerate levels par ordinary (non-degenerate) E n ( 2 ) use karna. Kyun tempting lagta hai: formula wahan hai. Fix: agar E n ( 0 ) = E m ( 0 ) toh denominator blow up ho jaata hai — tumhe degenerate perturbation theory use KARNI CHAHIYE (pehle degenerate subspace mein H ^ ′ diagonalize karo).
Recall Ise ek 12-saal ke bache ko samjhao (Feynman)
Socho sachi answer ek valley ka sabse nichla point hai jo tum dekh nahi sakte. Variational method: tum balls (guesses) valley mein phenko; har ball kahin na kahin upar ya bottom par hi girti hai, kabhi neeche nahi — toh jo sabse nichli ball tum pate ho wo bottom ka tumhara best estimate hai, aur adjustable knobs add karne se ball aur neeche roll karne lagti hai. Perturbation method: tum ek nearby easy hill ko already perfectly jaante ho, aur asli hill almost wahi hai bas ek chhoti si extra bump ke saath; tum real height paane ke liye bas chhote corrections add karte ho — pehle average bump height, phir ek aur barik fix.
"VAULT" — V ariational ek A bove-bound (U pper L imit) deta hai jise tum T ighten karte ho. Aur perturbation ke liye: "FIRST is the AVERAGE, SECOND always SINKS the ground."
Variational theorem inequality batao. Kisi bhi normalizable trial ϕ ke liye: E [ ϕ ] = ⟨ ϕ ∣ ϕ ⟩ ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ ≥ E 0 , equality tab jab ϕ true ground state ho.
Variational energy hamesha ≥ E 0 kyun hoti hai? Ye eigenenergies ka weighted average ∑ ∣ c n ∣ 2 ∑ ∣ c n ∣ 2 E n hai, aur har E n ≥ E 0 .
First-order energy correction kya hai? E n ( 1 ) = ⟨ ψ n ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ — perturbation ki expectation value unperturbed state mein.
Second-order energy correction (non-degenerate) kya hai? E n ( 2 ) = ∑ m = n E n ( 0 ) − E m ( 0 ) ∣ ⟨ ψ m ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ ∣ 2 .
Ground-state second-order correction hamesha ≤ 0 kyun hoti hai? Numerator ≥ 0 hai aur har denominator E 0 ( 0 ) − E m ( 0 ) < 0 hai.
Particle in a box ke liye trial wavefunction mein kya hona chahiye? Boundary conditions ϕ ( 0 ) = ϕ ( a ) = 0 (aur normalizable/single-valued honi chahiye).
Non-degenerate perturbation theory kab fail hoti hai? Jab unperturbed levels degenerate hon (E n ( 0 ) = E m ( 0 ) ) — denominators diverge ho jaate hain; degenerate PT use karo.
PIB trial ϕ = x ( a − x ) ke liye exact se compare karke kya energy milti hai? E = m a 2 5 ℏ 2 = 2 m a 2 10 ℏ 2 vs exact 2 m a 2 9.8696 ℏ 2 — sirf 1.3% zyada.
Schrodinger equation Hpsi=Epsi
Rayleigh quotient E of phi
Variational theorem E>=E0
Expand phi in eigenstates
Minimize E over parameters
Trial wavefunction with params