Perturbations of the Hamiltonian

Often, the physics of a system are easily solved in a simpler case H0H_0, on top of which a higher order perturbation is applied. This is the case with spin-orbit coupling and hyperfine coupling in hydrogen, for example. In these cases we can write the Hamiltonian as

H(λ)=H0+λH1=H0+ΔH(λ). \begin{align*} H(\lambda) = H_0 + \lambda H_1 = H_0 + \Delta H(\lambda). \end{align*}

When λ1\lambda \ll 1, we can calculate the change in energy of an eigenstate ψ0\ket{\psi_0} of H0H_0

ΔE=EHE0=ψ0ΔH(λ)ψ0+O(λ2). \begin{align*} \Delta E = E_H - E_0 = \braket{\psi_0 | \Delta H(\lambda) | \psi_0} + O(\lambda^2). \end{align*}

The proof follows from the Hellman-Feynman theorem. Let H(λ)H(\lambda) have some normalized eigenstate ψλ\ket{\psi_\lambda} with eigenvalue EλE_\lambda. From H-F,

dEλdλ=ψλdHλdλψλ=ψλH1ψλ. \begin{align*} \frac{d E_\lambda}{d \lambda} &= \left\langle \psi_\lambda \left| \frac{dH_\lambda}{d\lambda} \right| \psi_\lambda \right\rangle = \braket{\psi_\lambda | H_1 | \psi_\lambda}. \end{align*}

Taylor expanding,

E(λ)=E0+λdEλdλλ=0+O(λ2)E(λ)E0=ψ0ΔHψ0+O(λ2). \begin{align*} E(\lambda) &= E_0 + \lambda \left.\frac{d E_\lambda}{d \lambda}\right | _{\lambda=0} + O(\lambda^2) \\ E(\lambda) - E_0 &= \braket{\psi_0 | \Delta H | \psi_0} + O(\lambda^2). \end{align*}

This is the simplest application of perturbation theory.