Given a trial wave function ψ = C₁φ₁ + C₂φ₂ and the Hamiltonian matrix elements. ∫ φ₁^* Hφ₁ dv = 0, ∫ φ₁^*… · ABC26PH1294
Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Variation Method · Exam: CSIR NET JUNE 2016 · Marks: 2 · Difficulty: Medium
Given a trial wave function $\psi = C_1\phi_1 + C_2\phi_2$ and the Hamiltonian matrix elements. $\int \phi_1^* H\phi_1\,dv = 0$, $\int \phi_1^* H\phi_2\,dv = 2.5$, $\int \phi_2^* H\phi_2\,dv = 12.0$ the variationally determined ground state energy is
(a)-0.52
(b)-0.50
(c)12.50
(d)12.52
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Secular equation (0 − E)(12 − E) − 2.5² = 0 ⇒ E² − 12E − 6.25 = 0 ⇒ E = (12 − √169)/2 = −0.50 (the lower root).