Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    Thanks.
     
    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:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook