27 e q eit 2q fermis golden then gives the rate with

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ton from one laser beam an atom goes into an excited state (but only virtually since there is strong frequency detuning) and then gets de-excited by a photon from the other beam. By treating optical fields as classical, one can obtain effective Hamiltonian describing interaction of atoms with the laser fields V0 † −iωt (3.27) (ρ e + ρ† q eiωt ) − 2q Fermi’s golden then gives the rate with which excitations are created in the system (this is linear response theory and applies only for exciting a relatively small number of atoms) Veff = W = V02 S (q, ω ) (3.28) What is being measured in experiments is the number of atoms excited into a state with finite momentum as a function of wavevector and frequency differences of the two laser beams (see figs 3.1 and 3.2). To apply formulas (3.25), (3.27) to the BEC we write ρ† in equation (3.26) q using creation and annihilation operators of the Bogoliubov quasiparticles. The 1 /2 leading term in N0 1 ρ† = √ (b0 b† + b† b−q ) = q q 0 V N0 V 1/2 † (uq − vq )(αq − α...
View Full Document

This document was uploaded on 02/27/2014 for the course PHYS 284 at Harvard.

Ask a homework question - tutors are online