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Homework Help: Solid State II

  1. Mar 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Given some dispersion relation for the tight binding approximation in 2D:
    e(k_x,k_y) = -2t_1[cos(k_x*a)+cos(k_y*a)]-4t_2[cos(k_x*a)cos(k_y*a)]

    Show that the density of states has a logarithmic singularity for some choice of parameters t_i.
    2. Relevant equations
    g(e)de=g integral dS/(2*pi)^3 *1/|grad e(K)|

    3. The attempt at a solution
    When it says logaritmic singularity is it referring to von hove singularity?
    First I expanded the cosines.

    -2t_1[2-k_x^2*a^2/2 -k_y^2*a^2/2] -4t_2[1+k_x^2*a^2 k_y^2*a^2

    Then what do I do?
  2. jcsd
  3. Mar 29, 2008 #2
    I don't think the expression you've got there for the density of states in the one you want --- it's for 3D, where as you want the 2D case. Also, are you sure that you actually have to show there is a *logarithmic* singularity? Because I think the proof for that is *really* quite long. Otherwise, just show that the gradient of e(k) has a zero on contour lines with finite extent --- it will have zeros at the maxima and minima, but the contour lines there have zero length.
  4. Mar 29, 2008 #3
    What formula do I need for the 2D case?
    There is another one in ashcroft
    g_n(e) = g integral (d^3k/(2*pi)^3 delta (e - e_n(k))

    Is that the one I want?
  5. Mar 29, 2008 #4
    All of those assume a 3D situation. You want:

    [tex]g_n(E) = \int \frac{d\mathbf{k}}{(2\pi)^d} \delta\left(E - E_n(\mathbf{k})\right)[/tex]

    where we've used the fact that for most of the things we care about, k-space is discrete with a volume per k-state of [tex]1/(2\pi a)^d[/tex] (a is the crystal size, d is the dimension of the space). The factor of [tex]a^d[/tex] is cancelled by dividing by the volume of the entire crystal.
  6. Mar 29, 2008 #5
    "Otherwise, just show that the gradient of e(k) has a zero on contour lines with finite extent --- it will have zeros at the maxima and minima, but the contour lines there have zero length."

    SO given

    e(k_x,k_y) = -2t_1[\cos(k_x*a)+\cos(k_y*a)]-4t_2[\cos(k_x*a)\cos(k_y*a)]
    Should I expand?

    Do I take the gradient with respect to k or (k_x and k_y)
    Last edited: Mar 29, 2008
  7. Mar 30, 2008 #6
    If I recall correctly, you want to consider the gradient of E in the direction *perpendicular* to the iso-energy contour, since that is what appears in the denominator (once you expand out the delta function). But, since the gradient along the contour is going to be zero, you can just look for critical points of E --- i.e. [tex]dE/dk_x = dE/dk_y = 0[/tex]. After finding these, identify the fact that some of them are neither minima or maxima --- i.e. saddle-points. These should then lie on contour lines of finite extent.

    However, to show that the divergence is logarithmic, rather than say, 1/x, would involve rather a lot more work. If this is for an undergraduate course (well, it would be undergrad in the UK, I don't know about elsewhere), then it's unlikely that the question wants you to actually show the nature of the divergence.
  8. Mar 30, 2008 #7
    After Taylor expanding E(k) I get:


    e(k_x,k_y) = a^2(-t_1+4t_2)(k_{x}^2 +k_{y}^2) +4(t_1+t_2)
    \frac{dE}{dk_x}=0 \rightarrow 4t_2=t_1
    \frac{dE}{dk_y}=0 \rightarrow 4t_2=t_1

    Does that suffice to show the log sing for a choice of parameters t_i?

    Now if I want to obtain the density of states how do I do that?
  9. Mar 30, 2008 #8
    Wait, why did you Taylor expand? There shouldn't be a need to...

    And I would personally think that is sufficient, though of course you should ask your course supervisor or TA or whatever you have at hand.

    The density of states is going to be infinite at a divergence...?
  10. Mar 30, 2008 #9
    so if
    \sin(k_x a)(2t_1 a+ 4t_22 a \cos(k_y a)) =0
    \sin(k_y a)(2t_1 a+ 4t_22 a \cos(k_x a)) =0
    Also given t_1 > 0
    When are these neither min and max? I can't remember anything about multivariable calculus.

    a is non-zero value

    If t2 is zero then that reduces to the tight binding approx.

    is sin(kxa) zero or non zero?
    If kx zero it reduces to t1=-2cos(kya)

    if ky zero then t1 =-2cos(kxa)
    Last edited: Mar 30, 2008
  11. Mar 30, 2008 #10
    The equations are simultaneous equations, so just solve them with the usual methods. Given that you've already got them in factors, you should be able to read off the 4 possibilities. To tell whether they are maxima or minima, you can go via looking at the 2nd derivatives: http://en.wikipedia.org/wiki/Hessian_matrix#Second_derivative_test
  12. Mar 31, 2008 #11
    Okay done. Now I have to evaluate this numerically:

    g_n(E) = \int \frac{d\mathbf{k}}{(2\pi)^d} \delta\left(E - E_n(\mathbf{k})\right)

    Both ti's that I obtained were maximum. Which t's should I use in my dispersion relation?
  13. Apr 1, 2008 #12
    What are my limits of integration?
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