22 we can interpret the appearance of the gapless

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: ive modes is given by Ep = p( p + 2n0 U0 ) (3.20) We can define the healing length from 1 2 = n0 U0 mξh (3.21) In the long wavelength limit, qξh << 1, we find sound-like dispersion Eq = vs |q |. Sound velocity 1/2 n0 U0 m vs = (3.22) We can interpret the appearance of the gapless mode as manifestation of spontaneously broken symmetry: this mode arises because the superfluid state spontaneously breaks the U (1) symmetry corresponding to the conservation in the number of of particles. However sound mode by itself does not imply supefluidity. As we know, sound modes exist in room temperature gases. In the short wavelength limit, qξh >> 1, we find free particle dispersion Eq = q 2 /2m. It is natural to ask about the change in the wavefunction (3.8) implied by the Bogoliubov analysis. From the form of the mean-field Hamiltonian (3.12) we expect that it should have coherent superpositions of p, −p pairs. So we expect the wavefunction to be of the form † P |ΨBog = C eαbp=0 + p fp b† b† p p− |vac (3.2...
View Full Document

Ask a homework question - tutors are online