Tight-Binding p-bands in fcc crystal, Ashcroft - Mermin book

[tzo]

Hi all,
I'm trying to solve a problem in Ashcroft - Mermin "Solid State Physics", Chapter 10: The Tight-Binding Method, and need your help.

Homework Statement

In problem 2(c) it says:
For a face-centered Bravais lattice with only nearest-neighbor $\gamma_{ij}$ appreciable, show that the energy bands are given by the roots of

$0 = \begin{vmatrix} \varepsilon(\vec{k})-\varepsilon^{0}(\vec{k})+4\gamma_{0}\cos\frac{1}{2}k_{y}a\cos\frac{1}{2}k_{z}a & -4\gamma_{1}\sin\frac{1}{2}k_{x}a\sin\frac{1}{2}k_{y}a & -4\gamma_{1}\sin\frac{1}{2}k_{x}a\sin\frac{1}{2}k_{z}a \\ -4\gamma_{1}\sin\frac{1}{2}k_{y}a\sin\frac{1}{2}k_{x}a & \varepsilon(\vec{k})-\varepsilon^{0}(\vec{k})+4\gamma_{0}\cos\frac{1}{2}k_{z}a\cos\frac{1}{2}k_{x}a & -4\gamma_{1}\sin\frac{1}{2}k_{y}a\sin\frac{1}{2}k_{z}a\\ -4\gamma_{1}\sin\frac{1}{2}k_{z}a\sin\frac{1}{2}k_{x}a & -4\gamma_{1}\sin\frac{1}{2}k_{z}a\sin\frac{1}{2}k_{y}a & \varepsilon(\vec{k})-\varepsilon^{0}(\vec{k})+4\gamma_{0}\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{y}a \end{vmatrix}$

where

$\varepsilon^{0}(\vec{k})={E}_{p}-\beta-4\gamma_{2}(\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{z}a+\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}{k}_{y}a+\cos\frac{1}{2}k_{y}a\cos\frac{1}{2}k_{z}a),\\ \gamma_{0}=-\int d \vec{r}[{x}^{2}-y(y-\frac{1}{2}a)]\phi(r)\phi(\sqrt{{x}^{2}+(y-\frac{a}{2})^{2}+(z-\frac{a}{2})^{2}})\Delta U(\vec{r}),\\ \gamma_{1}=-\int d \vec{r}(x-\frac{1}{2}a)(y-\frac{1}{2}a)\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+{z}^{2}})\Delta U(\vec{r}),\\ \gamma_{2}=-\int d \vec{r}x(x-\frac{1}{2}a)\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+{z}^{2}})\Delta U(\vec{r}).$

I could derive diagonal terms properly, but off-diagonal terms, for example (1,2)-entry, become :

$4 \gamma_{1}\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{y}a$

Below is my derivation. Could you tell me where I made a mistake?

Homework Equations

http://dileepma.tripod.com/cha10.pdf

The Attempt at a Solution

(1,2)-entry$= \beta_{xy}+\widetilde{\gamma}_{xy}(\vec{k})\\$
$$=0+\sum_{n.n.}e^{i\vec{k}\cdot\vec{R}}[-\int d \vec{r} \psi_{x}^{*}(\vec{r})\psi_{y}(\vec{r}-\vec{R})\Delta U(\vec{r})]\\$$$= e^{i(\frac{1}{2}k_{x}a+\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{x}a-\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{x}a+\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{x}a-\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{x}a+\frac{1}{2}k_{z}a)}(-\int d \vec{r} xy\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+{y}^{2}+(z+\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{x}a-\frac{1}{2}k_{z}a)}(-\int d \vec{r} xy\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+{y}^{2}+(z-\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{x}a+\frac{1}{2}k_{z}a)}(-\int d \vec{r} xy\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+{y}^{2}+(z+\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{x}a-\frac{1}{2}k_{z}a)}(-\int d \vec{r} xy\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+{y}^{2}+(z-\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{y}a+\frac{1}{2}k_{z}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{{x}^{2}+(y+\frac{a}{2})^{2}+(z+\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{y}a-\frac{1}{2}k_{z}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{{x}^{2}+(y+\frac{a}{2})^{2}+(z-\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{y}a+\frac{1}{2}k_{z}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{{x}^{2}+(y-\frac{a}{2})^{2}+(z+\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ + e^{i(-\frac{1}{2}k_{y}a-\frac{1}{2}k_{z}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{{x}^{2}+(y-\frac{a}{2})^{2}+(z-\frac{a}{2})^{2}} )\Delta U(\vec{r}))\\ =e^{i(\frac{1}{2}k_{x}a+\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + e^{i(\frac{1}{2}k_{x}a-\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{(x+\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))---- (*)\\ + e^{i(-\frac{1}{2}k_{x}a+\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r})) ----(*)\\ + e^{i(-\frac{1}{2}k_{x}a-\frac{1}{2}k_{y}a)}(-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\ + 8 \cdot 0\\ =4 \gamma_{1}\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{y}a$

I'm suspecting that two starred terms in second last equation has negative sign.

Thanks.

 

[mark726]

Hello, thank you for reaching out for help. I have looked over your derivation and I believe I have found the mistake.

In the two starred terms in the second to last equation, you have the following integrals:

(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r})) ----(*)\\
+ (-\int d \vec{r} x(y-\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\

The problem is that in the first integral, you have x(y+a/2), but in the second integral, you have x(y-a/2). This is incorrect because the two integrals should have the same sign in front of the y term. Therefore, the correct expression should be:

(-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y+\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r})) ----(*)\\
+ (-\int d \vec{r} x(y+\frac{a}{2})\phi(r)\phi(\sqrt{(x-\frac{a}{2})^{2}+(y-\frac{a}{2})^{2}+z^2} )\Delta U(\vec{r}))\\

This will result in a negative sign in front of the two starred terms, which will then cancel out the negative sign in front of the first starred term, giving you the correct expression of:

4 \gamma_{1}\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{y}a

I hope this helps! Let me know if you have any further questions. Good luck with your problem.