Tight binding in cubic crystals

[torehan]

\phiHi 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

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


\beta_{xx}= \beta_{yy}=\beta_{zz}=\beta_

and \beta_{xy}=0

So to calculate the \beta_{xx} ;

\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})

\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})

\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})


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 \beta_{xy} must be zero as a concequence of cubic symmetry.

\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})
Is it a concequence of orthanormality of atomic orbitals or what?

Is there anyone who could help me understanding integrals?

Thanks.

 

[kanato]

Yes, your \beta_{xy} = 0 is a consequence of orthonormality of your orbitals. Notice your integrand has xy in it. When you do your integral over dx the function \Delta U will be even, and the function x 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 \beta_xx and \beta_yy. The key point is that it's *cubic* symmetry. What does that mean? How does x relate to y and z?

 

[torehan]

Think about what cubic symmetry means for integrals like \beta_{xx} and \beta_{yy]. The key point is that it's *cubic* symmetry. What does that mean? How does x relate to y and z?

It smells like \pi / 2 rotational symmetry for each axes but.. still not clear.

 

[kanato]

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 \beta_{xx} and rotate the coordinate system used in the integral. You should see it becomes exactly the integral in \beta_{yy}

 

[torehan]

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;
\gamma_{ij}(R)=- \int d\vec{r} \psi^{*}_{i}(\vec{r})\psi_{j}(\vec{r}-\vec{R}) \Delta U(\vec{r})
must be diagonal.
I made some calculations for \gamma_{xy} in attached the document.

Is it diagonal for the same reason?