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

With the physics convention $⟨\Phi |\Psi ⟩=\int {\Phi }^{\ast }\Psi \mathrm{d}\tau$, the adjoint operator is defined byThis definition includes the operator domains and their boundary conditions. If $\hat{A}$ is a Hermitian operator on the common domain, then ${\hat{A}}^{†}\Phi =\hat{A}\Phi$. Conjugate symmetry of the inner product consequently giveswhich is the required integral identity.

The boundary conditions must make the boundary form from integration by parts vanish for every pair of allowed functions. For a Schrödinger kinetic operator in one dimension, this meansSufficient examples are decay of the functions and derivatives at spatial infinity, homogeneous hard-wall conditions $\Phi =\Psi =0$ at finite endpoints, or matching periodic conditions on both functions and their derivatives. The exact boundary form depends on $\hat{A}$; square integrability alone does not replace the domain requirement.

The unperturbed Hamiltonian satisfiesThe supplied eigenfunctions can be chosen as an orthonormal eigenbasis, with $⟨{\Psi }_j^{(0)}|{\Psi }_i^{(0)}⟩={\delta }_{ji}$.

Insert the expansions into the energy eigenvalue equation $\hat{H}{\Psi }_i=E_i{\Psi }_i$. Through first order,The zeroth-order coefficients reproduce part (i). Equating the coefficients of $\lambda$ and moving them to one side givesTerms involving ${\hat{H}}^{(2)}$ first enter at order ${\lambda }^2$ and do not affect this equation of first-order nondegenerate perturbation theory.

Multiply the first-order equation by ${\Psi }_i^{(0)\ast }$ and integrate. The first term vanishes because Hermitian conjugation moves the unperturbed Hamiltonian onto the known bra:Here $E_i^{(0)}$ is real. The remaining equation isUsing wavefunction normalization therefore yieldsThus the first-order energy coefficient is the perturbation's expectation value in the unperturbed quantum state.

For this quantum harmonic oscillator, ${\hat{H}}^{(1)}=x^4$. The Gaussian integral givesThe distinction between the coefficient and the actual first-order shift matters: the energy shift is $3\lambda /4$, so $E_0=1/2+3\lambda /4+O({\lambda }^2)$ in the stated atomic units.

Project the first-order equation onto ${\Psi }_j^{(0)}$, with $j\ne i$. By orthonormality and the Hermitian operator property,ConsequentlyandThe 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 $⟨{\Psi }_i^{(0)}|{\Psi }_i^{(1)}⟩=0$, a permitted first-order phase and wavefunction normalization convention. A component parallel to ${\Psi }_i^{(0)}$ is not determined by the projected equation. Nondegeneracy ensures nonzero denominators; a small energy gap can still make the perturbative expansion inaccurate unless $|\lambda H_{ji}|\ll |E_i^{(0)}-E_j^{(0)}|$.