[tex]\phi[/tex]Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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 x^{2}, y^{2}and z^{2}.

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.

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# Tight binding in cubic crystals

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