314 | Chapter 9 wavelength excitations of any id are sound waves. The density of sound excitations is {exp[bho(k)]1}1. However, in evaluating the number operator n, we want the number " of quasiparticles of wave vector k, which is not the same as the density of sound excitations. Change integration variables to x e(k)/(kBT), set y bl, and d 1 2mkB T 3/2 n 2 F(bl) 4p h "2 p冿瑑�Z 楼 x dx F(y) exy 1 0 (9:163) (9:164)The left-hand side (n) of (9.163) is a ed number. As the temperature T becomes lower, the integral on the right becomes smaller due to the factor of T 3/2. The term F(y) compensates by having y become smaller in magnitude (less negative). Eventually, y bl 0. This condition dees the Bose-Einstein temperature:kB Tk 2/3 h "2 4p2 n 2m F(0) (9:165)For T < Tl, l(T) 0 and one has to replace eqn. (9.161) byn n0 + Z d3 k 1 (2p)3 ebe(k) 1 (9:166)The quantity n0 f0n is the density of particles in the k 0 state, which is usually represented by the fraction f0 of all of the helium atoms. The deition of n0 is 1 2mkB T 3/2 F(0) 4p2 h "2 " # T 3/2 n 1 Tk T 3/2 f0 T) 1 Tk n0 n (9:167) (9:168)(9:169)The previous deition of Tl has been used to write the condensate fraction as a 3 power 2 of the reduced temperature. For the noninteracting boson gas, all atoms are in the condensate at zero temperature [ f0(0) 1]. The condensate is the macroscopic quantum state. At k 0 the particle wavelength extends throughout the id. In the next section we d a very different behavior for the actual 4He superid.9.6.2 Off-Diagonal Long-Range OrderThe above derivation of BEC (Bose-Einstein condensation) treats the particles as noninteracting. That model is a poor approximation to 4He. In the actual liquid, each atom has 6� near neighbors within its range of potential energy. Also, each atom has a large zero-point kinetic energy due to rapid motion within its small space in liquid.
