“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Variational Principle by Hand — A parameter-free trial on a different system

Exercises

Quantum Chemistry I › Unit 2 · The variational game › The Variational Principle by Hand › all problems

Practice Problem 4 of 8

A parameter-free trial on a different system

Problem

Particle in a box of length

L

. Use the parameter-free trial

ψ (x) = x (L - x)

, which satisfies the boundary conditions. Compute

⟨ H ⟩

(in units of

ℏ^{2} / m L^{2}

) and compare with the exact

π^{2} /2 \approx 4.9348

Solution

Predict before reading on. Nothing to differentiate — no parameter. So what is the variational principle still doing, and what would you do to improve the estimate?

With $\hat{H} = - \frac{ℏ ^{2}}{2 m} d^{2} / d x^{2}$ and $- ψ^{''} = 2$ , form the Rayleigh quotient $⟨ H ⟩ = \frac{ℏ ^{2}}{2 m} ⟨ ψ ∣ - ψ^{''} ⟩ / ⟨ ψ ∣ ψ ⟩$ :

⟨ ψ ∣ - ψ^{''} ⟩ = \int_{0}^{L} 2 x (L - x) d x = \frac{L ^{3}}{3}, ⟨ ψ ∣ ψ ⟩ = \int_{0}^{L} x^{2} (L - x)^{2} d x = \frac{L ^{5}}{30}

⟨ H ⟩ = \frac{ℏ ^{2}}{2 m} \cdot \frac{L ^{3} /3}{L ^{5} /30} = \frac{5 ℏ ^{2}}{m L ^{2}}

$5$ versus exact $4.9348$ — only 1.3% high, with no parameter at all. The principle still guarantees the upper bound; to improve it you add parameters, e.g. $x (L - x) (1 + c x (L - x))$ , and minimize over them.