1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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 p-bands in fcc crystal, Ashcroft - Mermin book

  1. Sep 28, 2012 #1

    tzo

    User Avatar

    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.

    1. The problem statement, all variables and given/known data
    In problem 2(c) it says:
    For a face-centered Bravais lattice with only nearest-neighbor [itex]\gamma_{ij}[/itex] appreciable, show that the energy bands are given by the roots of

    [itex]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}[/itex]

    where

    [itex]
    \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}).
    [/itex]

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

    [itex]4 \gamma_{1}\cos\frac{1}{2}k_{x}a\cos\frac{1}{2}k_{y}a[/itex]

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

    2. Relevant equations
    http://dileepma.tripod.com/cha10.pdf

    3. The attempt at a solution
    (1,2)-entry[itex] = \beta_{xy}+\widetilde{\gamma}_{xy}(\vec{k})\\[/itex]
    [tex]
    =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})]\\
    [/tex][itex]
    = 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
    [/itex]

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

    Thanks.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Tight-Binding p-bands in fcc crystal, Ashcroft - Mermin book
  1. Band Theory (Replies: 0)

  2. Tight-binding model (Replies: 0)

  3. Atoms in a crystal (Replies: 0)

Loading...