Peter Mason- The Effect of the Trapping Potential on the Behaviour of a Fast Rotating Condensate

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Transcript of Peter Mason- The Effect of the Trapping Potential on the Behaviour of a Fast Rotating Condensate

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    The Effect of the Trapping Potential on the

    Behaviour of a Fast Rotating Condensate

    Peter Mason

    Centre de Mathematiques Appliquees, Ecole PolytechniqueParis, France

    Work in collaboration with Amandine Aftalion: Project ANR VoLQuan

    Verona, September 2009

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    Introduction

    Experimental examples of rotating condensates: Harmonic Traps

    [Madison et al. PRL, 2000]

    [Coddington et al. PRL, 2003]

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    Introduction

    More experimental examples of rotating condensates: Harmonic Plus Gaussian traps

    [Bretin et al. PRL, 2000]

    [Ryu et al. PRL, 2007]

    [Weiler et al. Nature, 2008]

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    Outline of the Talk

    1. Fast Rotation in Harmonic Traps

    Minimisation of the Gross-Pitaevskii energy Lowest Landau level (LLL) analysis

    2. Fast Rotation in Toroidal Traps (Annular condensates)

    Harmonic + Quartic and Harmonic + Gaussian traps

    LLL analysis vs. Thomas-Fermi analysis

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    Section 1: Fast Rotation in Harmonic Traps

    The non-dimensional Gross-Pitaevskii energy is

    E[] = [H] +

    g

    2||4 d

    2r

    for Hamiltonian

    H

    =

    1

    22 +

    r2

    2 L

    z=

    1

    2(

    iA)2 + (1

    2)r2

    2

    K.E. P.E. rotation energy Coroilis

    centrifugal

    restoring

    where Lz = i(yxxy) and A = r with = (0, 0, ) and r = (x,y, 0).

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    The Landau Levels

    A common eigenbasis of Lz and H is the Hermite functions

    j,k = er2/2(x + iy)

    j(x iy)k(er2).

    The eigenvalues for Lz are j k, while for H, they are

    Ej,k = 1 + (1 )j + (1 + )k.

    Suppose = 1. These are the Landau levels.

    If 1 then two adjacent levels are separated by 2. However the distancebetween two adjacent states is 1

    1.

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    The Lowest Landau Level

    We are interested in the lowest energy state: the lowest Landau level (LLL). Thisoccurs when k = 0.

    Any function of the LLL is a linear combination of the j,0s and we can write

    (r) = er2/2

    P(u) = er2/2

    nj=1

    (u uj)

    for P an analytic function of u = x + iy and uj the n complex zeros of P.

    Each uj is the position of a vortex!

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    Link to Ginzberg-Landau Problems

    The GP energy can be rewritten by introducing a parameter ,

    = 14g2/5

    If we rescale R = 1r and u(r) = R3/2(Rr) then we get

    E[u] =

    12|u|2 Lz + 1

    22r2|u|2 + 1

    42|u|4

    =

    1

    2|u|2 Lz + 1

    42 |u|2 t

    2

    where t = 0 r2.Thomas-Fermi analysis considers the limit 0. However the LLL considers theasymptotic limit

    1.

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    Minimisation of the GP Energy (1)

    Remember the GP energy is

    E[] = [H] +

    g

    2||4 d

    2r

    for which the wave function minimising E[] is

    H(r) + g

    |(r)

    |2(r) = (r)

    where the chemical potential can be determined from the normalisation condition.

    We can then write the energy in the LLL as

    ELLL = +

    (1 )r2||2 + g

    2||4

    d2r

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    Minimisation of the GP Energy (2)

    The minimisation of

    ELLL = + (1 )r2||2 + g

    2||4 d

    2r

    is equivalent to minimising

    E[] =ELLL[]

    1 = r2||2 +

    2 ||4 d2r

    where = g1.

    Minimisation of ELLL depends only on one parameter : a combination of g and.

    We get that

    min[E] =2

    2

    3

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    Minimisation of the GP Energy (3)

    The density and radius of the disk are then

    ||2 = 2R20

    1 r

    2

    R20

    R0 =

    2

    1/4

    This is an inverted parabola! However it is not in the space of minimisation: to

    alleviate this we need to add vortices.

    Note

    The LLL analysis is valid for g(1

    )

    1 and is obtained by balancing K.E. and

    P.E. terms. We have that the vortex size is of the same order as the intervortexdistance (i.e. interaction between vortices becomes important). Vortices neednot be small.

    In contrast a TF analysis neglects the K.E. and finds a balance between the P.E.and interaction energy. We require large g so that vortices are small.

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    Gaussian or Inverted Parabola?

    Is the lattice regular or distorted?

    A regular lattice provides a Gaussian decay. [Ho, PRL 2001].

    However the optimum energy is obtained for the inverted parabola. In this case thelattice is distorted towards the edges and extends to infinity. The mean behaviouris that of an inverted parabola. To get an inverted parabola in the LLL, we requirevortices.

    distorted (lower energy)regular

    So a distorted lattice can change the decay to an inverted parabola and improve

    the energy. [Wanatabe et al. PRL 2004; Cooper et al. PRA 2004; Aftalion et al. PRA 2005].The lattice extends to infinity.

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    Section 2: Fast Rotation in Toroidal Traps

    Introducing an extra term to the harmonic trap removes the possibility of asingularity occurring when 1. There are two types of trap that can create anannular condensate;

    Harmonic Plus Quartic: V(r) = 12r2 + kr4

    Harmonic Plus Gaussian: V(r) = 12r2 + A exp(l2r2)

    for constants k, l and A.

    The expression for the energy is

    1

    2||2 + V(r)||2 + 1

    2g||4 Lz

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    Harmonic Plus Quartic Traps

    V(r) = 12r2 + kr4

    If we take the negative sign (a Mexican hat potential) then the condensate isalways annular (provided < 0), irrespective of the angular rotational velocity.

    [Cozzini et al. PRA, 2006].

    If we take the positive sign then the condensate ground state at = 0 is circular.As increases then an annulus will develop. [Fetter et al. PRA, 2005].

    increasing

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    Thomas-Fermi analysis (large g)

    We can work with the normalisation condition:

    1 =

    R20

    ||2 rdrd

    2c = 1 + 2

    k

    3kg

    2

    1/3

    .

    Thus c is the value of the rotational velocity at which the central hole first appears.

    If > c we can again use the normalisation condition, now integratingR2R1

    togive that

    Area = (R22 R21) =2k

    3kg2

    1/3 constant!Width = R2 R1 =

    (2c 1)2

    k[2

    1 + (2 1)2 (2c 1)2]

    .

    [Fetter et al. PRA, 2005]

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    Numerical Examples

    Examples from Fetter et al. PRA, 2005.

    1. Low interaction strength g = 80 and k = 0.5.

    increasing

    2. High interaction strength g = 1000 and k = 0.5.

    increasing

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    Harmonic Plus Gaussian Traps

    Experimental Examples of Vortices in Toroidal Condensates:

    [Bretin et al. PRL, 2004]

    [Weiler et al. Nature, 2008]

    All these examples use a harmonic plus Gaussian potential trap

    V(r) =

    1

    2r2

    + A exp(l2

    r2

    ).

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    Inner Boundary Existence

    The existence of the inner boundary comes from the effective potential:

    V(r) = A exp(l2r2) + (1 2)r2/2

    Look for the minimum:V

    r= 0 r2 = 1

    l2log

    2Al2

    1 2

    2Al2

    1 2 > 1 + for an inner boundary to exist

    0 2 4 60

    10

    20

    30

    40

    r

    V

    5 0 50

    0.02

    0.04

    0.06

    x

    5 0 50

    0.005

    0.01

    0.015

    0.02

    0.025

    0.03

    x

    0 2 4 60

    10

    20

    30

    40

    r

    V

    g = 100, A = 25, l = 0.03 g = 500, A = 100, l = 0.9

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    Disk Condensate

    Suppose that A, l and are such th