Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A1 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv
- v b
  - Solution v

i

Solution

The unperturbed Hamiltonian satisfies

\hat{H}^{(0)} Ψ_{i}^{(0)} = E_{i}^{(0)} Ψ_{i}^{(0)} .

The supplied eigenfunctions can be chosen as an orthonormal eigenbasis, with

⟨ Ψ_{j}^{(0)} ∣ Ψ_{i}^{(0)} ⟩ = δ_{ji}

.

ii

Solution

Insert the expansions into the energy eigenvalue equation

\hat{H} Ψ_{i} = E_{i} Ψ_{i}

. Through first order,

\hat{H} Ψ_{i} E_{i} Ψ_{i} = \hat{H}^{(0)} Ψ_{i}^{(0)} + λ (\hat{H}^{(0)} Ψ_{i}^{(1)} + \hat{H}^{(1)} Ψ_{i}^{(0)}) + O (λ^{2}), = E_{i}^{(0)} Ψ_{i}^{(0)} + λ (E_{i}^{(0)} Ψ_{i}^{(1)} + E_{i}^{(1)} Ψ_{i}^{(0)}) + O (λ^{2}) .

The zeroth-order coefficients reproduce part (i). Equating the coefficients of

λ

and moving them to one side gives

(\hat{H}^{(0)} - E_{i}^{(0)}) Ψ_{i}^{(1)} + (\hat{H}^{(1)} - E_{i}^{(1)}) Ψ_{i}^{(0)} = 0.

Terms involving

\hat{H}^{(2)}

first enter at order

λ^{2}

and do not affect this equation of first-order nondegenerate perturbation theory.

iii

Solution

Multiply the first-order equation by

Ψ_{i}^{(0) *}

and integrate. The first term vanishes because Hermitian conjugation moves the unperturbed Hamiltonian onto the known bra:

⟨ Ψ_{i}^{(0)} ∣ (\hat{H}^{(0)} - E_{i}^{(0)}) ∣ Ψ_{i}^{(1)} ⟩ = ⟨ \hat{H}^{(0)} Ψ_{i}^{(0)} ∣ Ψ_{i}^{(1)} ⟩ - E_{i}^{(0)} ⟨ Ψ_{i}^{(0)} ∣ Ψ_{i}^{(1)} ⟩ = 0.

Here

E_{i}^{(0)}

is real. The remaining equation is

⟨ Ψ_{i}^{(0)} ∣ \hat{H}^{(1)} ∣ Ψ_{i}^{(0)} ⟩ - E_{i}^{(1)} ⟨ Ψ_{i}^{(0)} ∣ Ψ_{i}^{(0)} ⟩ = 0.

Using wavefunction normalization therefore yields

E_{i}^{(1)} = ⟨ Ψ_{i}^{(0)} ∣ \hat{H}^{(1)} ∣ Ψ_{i}^{(0)} ⟩ = \int Ψ_{i}^{(0) *} \hat{H}^{(1)} Ψ_{i}^{(0)} d τ .

Thus the first-order energy coefficient is the perturbation's expectation value in the unperturbed quantum state.

iv

Solution

For this quantum harmonic oscillator,

\hat{H}^{(1)} = x^{4}

. The Gaussian integral gives

E_{0}^{(1)} = \frac{1}{π} \int_{- \infty}^{\infty} x^{4} e^{- x^{2}} d x = \frac{1}{π} \frac{3 π}{4} = \frac{3}{4} .

The distinction between the coefficient and the actual first-order shift matters: the energy shift is $3 λ /4$ , so

E_{0} = 1/2 + 3 λ /4 + O (λ^{2})

in the stated atomic units.

v

Solution

Project the first-order equation onto

Ψ_{j}^{(0)}

, with

j \neq = i

. By orthonormality and the Hermitian operator property,

⟨ Ψ_{j}^{(0)} ∣ (\hat{H}^{(0)} - E_{i}^{(0)}) ∣ Ψ_{i}^{(1)} ⟩ ⟨ Ψ_{j}^{(0)} ∣ (\hat{H}^{(1)} - E_{i}^{(1)}) ∣ Ψ_{i}^{(0)} ⟩ = (E_{j}^{(0)} - E_{i}^{(0)}) c_{ij}, = H_{ji} .

Consequently

c_{ij} = \frac{H _{ji}}{E _{i}^{(0)} - E _{j}^{(0)}} = \frac{H _{ij}^{*}}{E _{i}^{(0)} - E _{j}^{(0)}} (j \neq = i),

and

Ψ_{i}^{(1)} = j \neq = i \sum \frac{⟨ Ψ _{j}^{(0)} ∣ H ^ ^{(1)} ∣ Ψ _{i}^{(0)} ⟩}{E _{i}^{(0)} - E _{j}^{(0)}} Ψ_{j}^{(0)} .

The numerator order is important: the question defines

H_{ij}

with

i

on the bra, whereas projecting the correction onto state

j

gives

H_{ji}

. The two coincide only when the relevant matrix element is real.

Excluding an

i

component chooses

⟨ Ψ_{i}^{(0)} ∣ Ψ_{i}^{(1)} ⟩ = 0

, a permitted first-order phase and wavefunction normalization convention. A component parallel to

Ψ_{i}^{(0)}

is not determined by the projected equation. Nondegeneracy ensures nonzero denominators; a small energy gap can still make the perturbative expansion inaccurate unless

∣ λ H_{ji} ∣ ≪ ∣ E_{i}^{(0)} - E_{j}^{(0)} ∣

.

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