Tight binding in cubic crystals

[torehan]

$$\phi$$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

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


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

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?