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Bosons on a square lattice

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose an ideal bose gas sees a periodic potential with a period a in both x and y directions. Its eigenstates are altered from the free-particle form. The lowest band has energies
    [itex]\epsilon_\vec{k}=2t(2-cos(k_xa)-cos(k_ya))[/itex]
    where t is an energy scale that depends on the amplitude of the periodic potential, and the wavevectors [itex]\vec{k}[/itex] are restricted to the first Brillouin zone: [itex]\pi/a <k_{x,y}<\pi/a[/itex].

    (i) Show that the density of states [itex]g(\epsilon)[/itex] is zero for [itex]\epsilon>8t[/itex] and [itex]\epsilon<0[/itex]

    (ii) Show that [itex]g(\epsilon) \simeq L^2/4\pi ta^2[/itex] at low energies [itex]0\leq\epsilon\ll 8t[/itex]

    Hint: Sketch the dispersion relation above consider the approximate form of [itex]\epsilon_{\vec{k}}[/itex] at lom wavelengths ([itex]ka \ll 1[/itex]).

    2. Relevant equations



    3. The attempt at a solution

    Number (i) is quite obvious since the cosine can only take values between -1 and 1. Therefore the expression in brackets can only take value between 0 and 4 which shows that there are no states with energies < 0 or energies > 8t.

    (ii) For long wavelengths we can approximate
    [itex] \epsilon_\vec{k}=2t(2-cos(k_xa)-cos(k_ya)) = -2t(k_x^2 a^2+k_y^2a^2) = -2ta^2\vec{k}^2[/itex]
    But how does that help me to get the density of states? I don't know how to begin. Can someone give me an ansatz?


    Cheers, physicus
     
  2. jcsd
  3. Jan 13, 2013 #2

    TSny

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    Check your result for your approximation of [itex] \epsilon_\vec{k}[/itex]. I think you might be off by a sign and a factor of 2.

    To proceed to the density of states, compare your approximate result for [itex] \epsilon_\vec{k}[/itex] with [itex] \epsilon_{\vec{k},free}[/itex] for a free particle of mass m moving in 2 dimensions.
     
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