A2.13 B2.21

Foundations of Quantum Mechanics | Part II, 2002

(i) A Hamiltonian H0H_{0} has energy eigenvalues ErE_{r} and corresponding non-degenerate eigenstates ∣r⟩|r\rangle. Show that under a small change in the Hamiltonian δH\delta H,

δ∣r⟩=∑s≠r⟨s∣δH∣r⟩Er−Es∣s⟩,\delta|r\rangle=\sum_{s \neq r} \frac{\langle s|\delta H| r\rangle}{E_{r}-E_{s}}|s\rangle,

and derive the related formula for the change in the energy eigenvalue ErE_{r} to first and second order in δH\delta H.

(ii) The Hamiltonian for a particle moving in one dimension is H=H0+λH′H=H_{0}+\lambda H^{\prime}, where H0=p2/2m+V(x),H′=p/mH_{0}=p^{2} / 2 m+V(x), H^{\prime}=p / m and λ\lambda is small. Show that

iℏ[H0,x]=H′\frac{i}{\hbar}\left[H_{0}, x\right]=H^{\prime}

and hence that

δEr=−λ2iℏ⟨r∣H′x∣r⟩=λ2iℏ⟨r∣xH′∣r⟩\delta E_{r}=-\lambda^{2} \frac{i}{\hbar}\left\langle r\left|H^{\prime} x\right| r\right\rangle=\lambda^{2} \frac{i}{\hbar}\left\langle r\left|x H^{\prime}\right| r\right\rangle

to second order in λ\lambda.

Deduce that δEr\delta E_{r} is independent of the particular state ∣r⟩|r\rangle and explain why this change in energy is exact to all orders in λ\lambda.

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