Advanced Quantum Mechanics Schwabl · Problem 3.8

Correlation Functions, Scattering, and Response

Physics — Quantum Mechanics · ground-state-energy-bosons-contact-interaction

Problem

*Ground-state energy for bosons with contact interaction* Consider a system of NN identical bosons of mass mm interacting with one another via a two-particle contact potential, H=∑α=1Npi22m+λ∑i<jδ(xi−xj).H = \sum_{\alpha=1}^N \frac{p_i^2}{2m} + \lambda\sum_{i<j}\delta(\mathbf{x}_i - \mathbf{x}_j). In the limit of weak interaction, the Bogoliubov transformation can be used to express the Hamiltonian in the form H=N22Vλ−12∑k′(k22m+nλ−ωk)+∑k′ωkαk†αk.H = \frac{N^2}{2V}\lambda - \frac{1}{2}\sum_\mathbf{k}'\left(\frac{k^2}{2m} + n\lambda - \omega_\mathbf{k}\right) + \sum_\mathbf{k}'\omega_\mathbf{k}\alpha^\dagger_\mathbf{k}\alpha_\mathbf{k}. The ground-state energy E0=N2λ2V−∑k′(k22m+nλ−ωk)E_0 = \frac{N^2\lambda}{2V} - \sum_\mathbf{k}'\left(\frac{k^2}{2m} + n\lambda - \omega_\mathbf{k}\right) diverges at the upper integration limit (ultraviolet divergence). The reason for this is the unphysical form of the contact potential. The divergence is removed by introducing the physical scattering length, which describes the ss-wave scattering by a short-range potential. Show that the scattering amplitude ff of particles in the condensate is given, in first-order perturbation theory, by a:=−f(k1=k2=k3=k4=0)=m4πℏ2λ{1−λV∑k′mk2+O(λ2)}.a := -f(\mathbf{k}_1 = \mathbf{k}_2 = \mathbf{k}_3 = \mathbf{k}_4 = 0) = \frac{m}{4\pi\hbar^2}\lambda\left\{1 - \frac{\lambda}{V}\sum_\mathbf{k}'\frac{m}{k^2} + \mathcal{O}(\lambda^2)\right\}. Eliminate λ\lambda from the expression for the ground-state energy by introducing aa. For small values of a/r0a/r_0, where r0=(N/V)−1/3r_0 = (N/V)^{-1/3} is the mean separation of particles, show that the ground-state energy is given by E0=N2πℏ2anm{1+12815π(ar0)3/2}.E_0 = N\frac{2\pi\hbar^2 an}{m}\left\{1 + \frac{128}{15\sqrt{\pi}}\left(\frac{a}{r_0}\right)^{3/2}\right\}. Calculate from this the chemical potential μ=∂E0∂N\mu = \frac{\partial E_0}{\partial N} and the sound velocity cc c=∂P∂ρ,ρ=mn,P=−∂E0∂V.c = \sqrt{\frac{\partial P}{\partial \rho}}, \quad \rho = mn, \quad P = -\frac{\partial E_0}{\partial V}.

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