Introduction to Quantum Mechanics · Problem 7.54

Time-Independent Perturbation Theory — David J. Griffiths

Physics — Quantum Mechanics · expectation_value_perturbation

Problem

Suppose you want to calculate the expectation value of some observable Ω\Omega, in the nnth energy eigenstate of a system that is perturbed by H′H': ⟨Ω⟩=⟨ψn∣Ω^∣ψn⟩.\langle\Omega\rangle = \left\langle\psi_n\left|\hat{\Omega}\right|\psi_n\right\rangle. Replacing ψn\psi_n by its perturbation expansion, Equation 7.5, ⟨Ω⟩=⟨ψn0∣Ω^∣ψn0⟩+λ[⟨ψn1∣Ω^∣ψn0⟩+⟨ψn0∣Ω^∣ψn1⟩]+λ2(⋯ )+⋯ .\langle\Omega\rangle = \left\langle\psi_n^0\left|\hat{\Omega}\right|\psi_n^0\right\rangle + \lambda\left[\left\langle\psi_n^1\left|\hat{\Omega}\right|\psi_n^0\right\rangle + \left\langle\psi_n^0\left|\hat{\Omega}\right|\psi_n^1\right\rangle\right] + \lambda^2(\cdots) + \cdots. The first-order correction to ⟨Ω⟩\langle\Omega\rangle is therefore ⟨Ω⟩1=2 Re[⟨ψn0∣Ω^∣ψn1⟩],\langle\Omega\rangle^1 = 2\,\text{Re}\left[\left\langle\psi_n^0\left|\hat{\Omega}\right|\psi_n^1\right\rangle\right], or, using Equation 7.13, ⟨Ω⟩1=2 Re∑m≠n⟨ψn0∣Ω^∣ψm0⟩⟨ψm0∣H′∣ψn0⟩En0−Em0(7.118)\langle\Omega\rangle^1 = 2\,\text{Re}\sum_{m \neq n}\frac{\left\langle\psi_n^0\left|\hat{\Omega}\right|\psi_m^0\right\rangle\left\langle\psi_m^0\left|H'\right|\psi_n^0\right\rangle}{E_n^0 - E_m^0} \qquad (7.118) (assuming the unperturbed energies are nondegenerate, or that we are using the "good" basis states). (a) Suppose Ω=H′\Omega = H' (the perturbation itself). What does Equation 7.118 tell us in this case? Explain (carefully) why this is consistent with Equation 7.15. (b) Consider a particle of charge qq (maybe an electron in a hydrogen atom, or a pith ball connected to a spring), that is placed in a weak electric field EextE_{\text{ext}} pointing in the xx direction, so that H′=−qEextx.H' = -qE_{\text{ext}}x. The field will induce an electric dipole moment, pe=q⟨x⟩p_e = q\langle x\rangle, in the "atom." The expectation value of pep_e is proportional to the applied field, and the proportionality factor is called the polarizability, α\alpha. Show that α=−2q2∑m≠n∣⟨ψn0∣x∣ψm0⟩∣2En0−Em0.(7.119)\alpha = -2q^2\sum_{m \neq n}\frac{\left|\left\langle\psi_n^0\left|x\right|\psi_m^0\right\rangle\right|^2}{E_n^0 - E_m^0}. \qquad (7.119) Find the polarizability of the ground state of a one-dimensional harmonic oscillator. Compare the classical answer. (c) Now imagine a particle of mass mm in a one-dimensional harmonic oscillator with a small anharmonic perturbation H′=−16κx3.(7.120)H' = -\frac{1}{6}\kappa x^3. \qquad (7.120) Find ⟨x⟩\langle x\rangle (to first order), in the nnth energy eigenstate. *Answer:* (n+12)ℏκ/(2m2ω3)\left(n + \frac{1}{2}\right)\hbar\kappa/\left(2m^2\omega^3\right). *Comment:* As the temperature increases, higher-energy states are populated, and the particles move farther (on average) from their equilibrium positions; that's why most solids expand with rising temperature.

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