Intuition The big pictureThe Schrödinger equation H ^ ψ = E ψ \hat{H}\psi = E\psi H ^ ψ = E ψ exactly for only a handful of systems (particle in a box, harmonic oscillator, hydrogen atom). For everything else — helium, molecules, anharmonic vibrations — we need approximate methods. Two giants do this job:
Variational principle : guess a trial wavefunction, and the energy you compute is ALWAYS too high. So lower it as far as you can — you're squeezing toward the truth from above.Perturbation theory : start from a problem you CAN solve, then add a small "extra bit" of the Hamiltonian and correct the answer step by step.
Any trial wavefunction is a "mixture" of the true energy eigenstates. The ground state is the lowest possible energy. If your guess accidentally contains a bit of excited states, those drag the average energy up , never down. So the average can never sink below the true ground state. That's the whole idea.
Definition Variational energy (Rayleigh quotient)For any normalizable trial function ϕ \phi ϕ E [ ϕ ] = ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ ⟨ ϕ ∣ ϕ ⟩ . E[\phi] = \frac{\langle \phi | \hat{H} | \phi\rangle}{\langle \phi | \phi\rangle}. E [ ϕ ] = ⟨ ϕ ∣ ϕ ⟩ ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ . variational theorem states E [ ϕ ] ≥ E 0 E[\phi] \ge E_0 E [ ϕ ] ≥ E 0 E 0 E_0 E 0 ϕ \phi ϕ
So strategy: put parameters α \alpha α E ( α ) E(\alpha) E ( α ) minimize :
∂ E ( α ) ∂ α = 0. \frac{\partial E(\alpha)}{\partial \alpha} = 0. ∂ α ∂ E ( α ) = 0.
Worked example Worked example 1 — Particle in a box, polynomial trialBox 0 ≤ x ≤ a 0\le x\le a 0 ≤ x ≤ a H ^ = − ℏ 2 2 m d 2 d x 2 \hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} H ^ = − 2 m ℏ 2 d x 2 d 2 E 0 = π 2 ℏ 2 2 m a 2 ≈ 9.8696 ℏ 2 2 m a 2 E_0 = \frac{\pi^2\hbar^2}{2ma^2}\approx \frac{9.8696\,\hbar^2}{2ma^2} E 0 = 2 m a 2 π 2 ℏ 2 ≈ 2 m a 2 9.8696 ℏ 2 ϕ = x ( a − x ) \phi = x(a-x) ϕ = x ( a − x ) Why this trial? It satisfies the boundary conditions (ϕ ( 0 ) = ϕ ( a ) = 0 \phi(0)=\phi(a)=0 ϕ ( 0 ) = ϕ ( a ) = 0
Numerator: ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ = − ℏ 2 2 m ∫ 0 a x ( a − x ) d 2 d x 2 [ x ( a − x ) ] d x \langle\phi|\hat H|\phi\rangle = -\frac{\hbar^2}{2m}\int_0^a x(a-x)\frac{d^2}{dx^2}[x(a-x)]\,dx ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ = − 2 m ℏ 2 ∫ 0 a x ( a − x ) d x 2 d 2 [ x ( a − x )] d x d 2 d x 2 [ a x − x 2 ] = − 2 \frac{d^2}{dx^2}[ax-x^2] = -2 d x 2 d 2 [ a x − x 2 ] = − 2 = − ℏ 2 2 m ∫ 0 a x ( a − x ) ( − 2 ) d x = ℏ 2 m ⋅ a 3 6 = ℏ 2 a 3 6 m . = -\frac{\hbar^2}{2m}\int_0^a x(a-x)(-2)\,dx = \frac{\hbar^2}{m}\cdot\frac{a^3}{6} = \frac{\hbar^2 a^3}{6m}. = − 2 m ℏ 2 ∫ 0 a x ( a − x ) ( − 2 ) d x = m ℏ 2 ⋅ 6 a 3 = 6 m ℏ 2 a 3 . ∫ 0 a x 2 ( a − x ) 2 d x = a 5 30 . \int_0^a x^2(a-x)^2 dx = \frac{a^5}{30}. ∫ 0 a x 2 ( a − x ) 2 d x = 30 a 5 . E [ ϕ ] = ℏ 2 a 3 / 6 m a 5 / 30 = 5 ℏ 2 m a 2 = 10 ℏ 2 2 m a 2 . E[\phi] = \frac{\hbar^2 a^3/6m}{a^5/30} = \frac{5\hbar^2}{ma^2} = \frac{10\,\hbar^2}{2ma^2}. 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: 10 > 9.8696 10 > 9.8696 10 > 9.8696 1.3% with a parameter-free guess.
Worked example Worked example 2 — Adding a parameterTry ϕ = x ( a − x ) + β x 2 ( a − x ) 2 \phi = x(a-x) + \beta\, x^2(a-x)^2 ϕ = x ( a − x ) + β x 2 ( a − x ) 2 β \beta β E E E 9.8696 9.8696 9.8696 Why this step? More flexibility = a richer mixture = you can cancel out more of the unwanted excited-state contamination, so the upper bound tightens. This is the engine of quantum chemistry (basis sets = many parameters).
If a problem is almost one you can solve, write the Hamiltonian as a solved part H ^ ( 0 ) \hat H^{(0)} H ^ ( 0 ) λ H ^ ′ \lambda\hat H' λ H ^ ′ λ \lambda λ
H ^ = H ^ ( 0 ) + λ H ^ ′ , H ^ ( 0 ) ψ n ( 0 ) = E n ( 0 ) ψ n ( 0 ) . \hat H = \hat H^{(0)} + \lambda \hat H',\qquad \hat H^{(0)}\psi_n^{(0)} = E_n^{(0)}\psi_n^{(0)}. H ^ = H ^ ( 0 ) + λ H ^ ′ , H ^ ( 0 ) ψ n ( 0 ) = E n ( 0 ) ψ n ( 0 ) . E n = E n ( 0 ) + λ E n ( 1 ) + λ 2 E n ( 2 ) + ⋯ , ψ n = ψ n ( 0 ) + λ ψ n ( 1 ) + ⋯ E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots,\qquad \psi_n = \psi_n^{(0)} + \lambda\psi_n^{(1)} + \cdots 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 + smallx 2 x^2 x 2 H ^ ′ = 1 2 k ′ x 2 \hat H' = \tfrac12 k' x^2 H ^ ′ = 2 1 k ′ x 2 n n n E n ( 1 ) = 1 2 k ′ ⟨ n ∣ x 2 ∣ n ⟩ = 1 2 k ′ ⋅ ℏ 2 m ω ( 2 n + 1 ) . E_n^{(1)} = \tfrac12 k'\langle n|x^2|n\rangle = \tfrac12 k'\cdot \frac{\hbar}{2m\omega}(2n+1). E n ( 1 ) = 2 1 k ′ ⟨ n ∣ x 2 ∣ n ⟩ = 2 1 k ′ ⋅ 2 mω ℏ ( 2 n + 1 ) . Why this step? Standard result ⟨ n ∣ x 2 ∣ n ⟩ = ℏ 2 m ω ( 2 n + 1 ) \langle n|x^2|n\rangle = \frac{\hbar}{2m\omega}(2n+1) ⟨ n ∣ x 2 ∣ n ⟩ = 2 mω ℏ ( 2 n + 1 ) rescales k k k
Common mistake Steel-manned misconceptions(a) "A lower variational energy means a more accurate wavefunction." Feels right because energy is the thing we minimize. Why it's tempting: energy converges fast. Fix: energy is stationary (2nd-order accurate) near the truth, so wavefunctions can still be poor while energy looks great. Properties other than energy can be badly off.
(b) "Variational energy could come out below E 0 E_0 E 0 Why tempting: approximations sometimes undershoot. Fix: the theorem forbids it — E [ ϕ ] ≥ E 0 E[\phi]\ge E_0 E [ ϕ ] ≥ E 0
(c) Using ordinary (non-degenerate) E n ( 2 ) E_n^{(2)} E n ( 2 ) Why tempting: the formula is right there. Fix: if E n ( 0 ) = E m ( 0 ) E_n^{(0)}=E_m^{(0)} E n ( 0 ) = E m ( 0 ) degenerate perturbation theory (diagonalize H ^ ′ \hat H' H ^ ′
Recall Explain it to a 12-year-old (Feynman)Imagine the true answer is the lowest point in a valley you can't see. The variational method: you toss balls (guesses) into the valley; every ball lands somewhere above or at the bottom, never below it — so the lowest ball you find is your best estimate of the bottom, and adding adjustable knobs lets the ball roll lower. The perturbation method: you already know a nearby easy hill perfectly, and the real hill is almost the same with a tiny extra bump; you just add small corrections — first the average bump height, then a finer fix — to get the real height.
"VAULT" — V ariational gives an A bove-bound (U pper L imit) that you T ighten. And for perturbation: "FIRST is the AVERAGE, SECOND always SINKS the ground."
State the variational theorem inequality. For any normalizable trial
ϕ \phi ϕ :
E [ ϕ ] = ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ ⟨ ϕ ∣ ϕ ⟩ ≥ E 0 E[\phi]=\dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi|\phi\rangle}\ge E_0 E [ ϕ ] = ⟨ ϕ ∣ ϕ ⟩ ⟨ ϕ ∣ H ^ ∣ ϕ ⟩ ≥ E 0 , equality iff
ϕ \phi ϕ is the true ground state.
Why is the variational energy always ≥ E 0 \ge E_0 ≥ E 0 It is a weighted average
∑ ∣ c n ∣ 2 E n ∑ ∣ c n ∣ 2 \frac{\sum|c_n|^2E_n}{\sum|c_n|^2} ∑ ∣ c n ∣ 2 ∑ ∣ c n ∣ 2 E n of eigenenergies, and every
E n ≥ E 0 E_n\ge E_0 E n ≥ E 0 .
What is the first-order energy correction? E n ( 1 ) = ⟨ ψ n ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ E_n^{(1)}=\langle\psi_n^{(0)}|\hat H'|\psi_n^{(0)}\rangle E n ( 1 ) = ⟨ ψ n ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ — the expectation value of the perturbation in the unperturbed state.
What is the second-order energy correction (non-degenerate)? E n ( 2 ) = ∑ m ≠ n ∣ ⟨ ψ m ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ ∣ 2 E n ( 0 ) − E m ( 0 ) E_n^{(2)}=\sum_{m\ne n}\dfrac{|\langle\psi_m^{(0)}|\hat H'|\psi_n^{(0)}\rangle|^2}{E_n^{(0)}-E_m^{(0)}} E n ( 2 ) = ∑ m = n E n ( 0 ) − E m ( 0 ) ∣ ⟨ ψ m ( 0 ) ∣ H ^ ′ ∣ ψ n ( 0 ) ⟩ ∣ 2 .
Why is the ground-state second-order correction always ≤ 0 \le 0 ≤ 0 Numerator is
≥ 0 \ge0 ≥ 0 and every denominator
E 0 ( 0 ) − E m ( 0 ) < 0 E_0^{(0)}-E_m^{(0)}<0 E 0 ( 0 ) − E m ( 0 ) < 0 .
What must a trial wavefunction satisfy for a particle in a box? The boundary conditions
ϕ ( 0 ) = ϕ ( a ) = 0 \phi(0)=\phi(a)=0 ϕ ( 0 ) = ϕ ( a ) = 0 (and be normalizable/single-valued).
When does non-degenerate perturbation theory fail? When unperturbed levels are degenerate (
E n ( 0 ) = E m ( 0 ) E_n^{(0)}=E_m^{(0)} E n ( 0 ) = E m ( 0 ) ) — denominators diverge; use degenerate PT.
For PIB trial ϕ = x ( a − x ) \phi=x(a-x) ϕ = x ( a − x ) E = 5 ℏ 2 m a 2 = 10 ℏ 2 2 m a 2 E=\frac{5\hbar^2}{ma^2}=\frac{10\hbar^2}{2ma^2} E = m a 2 5 ℏ 2 = 2 m a 2 10 ℏ 2 vs exact
9.8696 ℏ 2 2 m a 2 \frac{9.8696\hbar^2}{2ma^2} 2 m a 2 9.8696 ℏ 2 — only 1.3% high.
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
Intuition Hinglish mein samjhoDekho, asli quantum problems mein Schrödinger equation exactly solve nahi hoti — sirf box, oscillator, hydrogen jaise chhote cases mein hoti hai. Baaki sab ke liye approximation chahiye, aur do main tareeke hain.
Variational principle bola: tum koi bhi trial wavefunction guess karo, aur usse jo energy niklegi wo hamesha asli ground-state energy se zyada ya barabar hogi — kabhi kam nahi. Iska reason: koi bhi guess asal mein true eigenstates ka mixture hai, aur agar usmein excited states ka thoda hissa ghus gaya, wo average ko upar khichta hai. Isliye strategy simple — trial function mein ek parameter α \alpha α E ( α ) E(\alpha) E ( α ) ∂ E ∂ α = 0 \frac{\partial E}{\partial \alpha}=0 ∂ α ∂ E = 0
Perturbation theory alag style hai: agar problem kisi solved problem ke almost barabar hai, to Hamiltonian ko H ^ ( 0 ) + λ H ^ ′ \hat H^{(0)} + \lambda\hat H' H ^ ( 0 ) + λ H ^ ′ λ \lambda λ E ( 1 ) = ⟨ ψ ( 0 ) ∣ H ^ ′ ∣ ψ ( 0 ) ⟩ E^{(1)}=\langle\psi^{(0)}|\hat H'|\psi^{(0)}\rangle E ( 1 ) = ⟨ ψ ( 0 ) ∣ H ^ ′ ∣ ψ ( 0 ) ⟩ neeche le jaata hai.
Ek important warning: jab energy levels degenerate hon (barabar hon), to normal second-order formula ka denominator phat jaata hai — tab degenerate perturbation theory use karni padti hai. Aur yaad rakho: kam variational energy ka matlab achhi energy hai, par wavefunction phir bhi kharaab ho sakti hai!
