Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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rel="nofollow" href="#ulink_9d39eda4-0863-5d7f-8f8a-60e093e0b882">(49) to obtain the Laplacian:

      (54)image

      (55)image

      This result must be equal to the right-hand side of (53). We therefore get, after dividing both sides by —2n(r, t):

      (56)image

      To prevent any confusion with the azimuthal angle φ we now call χ the Gross-Pitaevskii wave function. The time-independent Gross-Pitaevskii equation then becomes (in the absence of any potential except the wall potentials of the box):

      We look for solutions of the form:

      where eφ is the tangential unit vector (perpendicular both to r and the Oz axis). Consequently, the fluid rotates along the toroidal tube, with a velocity proportional to l. As v is a gradient, its circulation along a closed loop “equivalent to zero” (i.e. which can be contracted continuously to a point) is

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