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Tight binding in cubic crystals

  1. Apr 5, 2010 #1
    [tex]\phi[/tex]Hi all,

    I would like to make band structure calculations with tight binding method and I start reading about this method from Ashcroft - Mermin, Chapter 10: The Tight Binding Method and try to solve the problems at the and of the chapter.

    In problem 2
    a. As a consequence of cubic symmetry, show that

    [tex]\gamma_{ij}(R)=- \int d\vec{r} \psi^{*}_{i}(\vec{r})\psi_{j}(\vec{r}-\vec{R}) \Delta U(\vec{r})[/tex]
    [tex]\beta_{ij}= \gamma_{ij}(R=0)[/tex]

    [tex]\beta_{xx}= \beta_{yy}=\beta_{zz}=\beta_ [/tex]

    and [tex]\beta_{xy}=0[/tex]

    So to calculate the [tex]\beta_{xx}[/tex] ;

    [tex]\beta_{xx}= - \int d\vec{r} \psi^{*}_{x}(\vec{r})\psi_{x}(\vec{r}) \Delta U(\vec{r})= - \int d\vec{r} x^{2} \left| \phi \right|^{2} \Delta U(\vec{r})[/tex]

    [tex]\beta_{yy}= - \int d\vec{r} \psi^{*}_{y}(\vec{r})\psi_{y}(\vec{r}) \Delta U(\vec{r})= - \int d\vec{r} y^{2} \left| \phi \right|^{2} \Delta U(\vec{r})[/tex]

    [tex]\beta_{zz}= - \int d\vec{r} \psi^{*}_{z}(\vec{r})\psi_{z}(\vec{r}) \Delta U(\vec{r})= - \int d\vec{r} z^{2} \left| \phi \right|^{2} \Delta U(\vec{r})[/tex]

    These must be equal to but I couldn't see any equalities as all of the integrals has difference in their integrands as x2 , y2 and z2.
    And also [tex]\beta_{xy}[/tex] must be zero as a concequence of cubic symmetry.

    [tex]\beta_{xy}= - \int d\vec{r} \psi^{*}_{x}(\vec{r})\psi_{y}(\vec{r}) \Delta U(\vec{r})= - \int d\vec{r} xy \left| \phi \right|^{2} \Delta U(\vec{r})[/tex]
    Is it a concequence of orthanormality of atomic orbitals or what?

    Is there any one who could help me understanding integrals?

    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2
    Yes, your [tex]\beta_{xy} = 0[/tex] is a consequence of orthonormality of your orbitals. Notice your integrand has [tex]xy[/tex] in it. When you do your integral over dx the function [tex]\Delta U[/tex] will be even, and the function [tex]x[/tex] is odd, so you are integrating the product of an even function with an odd function over all x. This will give you zero.

    Think about what cubic symmetry means for integrals like [tex]\beta_xx[/tex] and [tex]\beta_yy[/tex]. The key point is that it's *cubic* symmetry. What does that mean? How does x relate to y and z?
  4. Apr 5, 2010 #3
    It smells like [tex]\pi / 2[/tex] rotational symmetry for each axes but.. still not clear.
  5. Apr 5, 2010 #4
    Yeah.. what happens to your coordinate system if you rotate by 90 degrees about some axis. If you rotate around z, it will transform x into y and y into -x. So take your [tex]\beta_{xx}[/tex] and rotate the coordinate system used in the integral. You should see it becomes exactly the integral in [tex]\beta_{yy}[/tex]
  6. Apr 5, 2010 #5
    Yes, now it's more clear.
    Thanks for your constructive talk.

    Know another question appers..
    In simple cubic crystal for the first 6 nearest-neighbor;
    [tex]\gamma_{ij}(R)=- \int d\vec{r} \psi^{*}_{i}(\vec{r})\psi_{j}(\vec{r}-\vec{R}) \Delta U(\vec{r})[/tex]
    must be diagonal.
    I made some calculations for [tex]\gamma_{xy}[/tex] in attached the document.

    Is it diagonal for the same reason?

    Attached Files:

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