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3. NANOMEMS PHYSICS: Quantum Wave Phenomena                   117


             and

                 *
               χ ~   ηe − iEt+ ipx  ,                                                                               (107)

             substituting them into (104) and (105), together with (101), and solving the
             resulting system of equations for E, one obtains the result,


                                p 4
                         2
                            2
               E () p =  S  p +     .                                                                     (108)
                               4m  2
             This is the dispersion relation of a superfluid. Expressing the fluid velocity
             in terms of it, we obtain,


               v =  E  =  S  2  +  p  2  .                                                                        (109)
                   p          4m  2

             This equation has a positive minimum, occurring at  p  →  0, and given by
             the constant velocity  S. Since this  velocity is  independent of  momentum,
             E(p)  must  contain  an  energy  gap. An energy gap in its spectrum, thus,  is
             another manifestation of superfluidic behavior.
               The zero-vorticity property of a superfluid is derived from first principles
             as follows. From (103) it may be seen that the wave function for the Bose
             condensate in its lowest energy state is a  one-particle  complex  wave.
             Generalizing this expression to,

               ψ  () =x  ψ  e  iχ ()  ,                                                                           (110)
                               x

             one can express the mass density as  =ρ  m  ψ , where  () x  and the current
                                                            ψ
                                                    2
             are related, as usual, by,
               G     i=
                j =  −  (ψ  * ∇ ψ  −  ψ ∇ ψ  *  ).                                                           (111)
                      2

             It then follows that, inserting (110) into (101) one obtains,
               G       2         =
                j  = = ψ  ∇ χ  =  ρ  ∇ χ ,                                                                (112)
                                m

             which, upon comparison with (94) yields,
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