Difference between revisions of "Bloch's Theorem"

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<math>\vert exp(\overrightarrow{k}j\overrightarrow{a})\vert ^2 = exp (i \overrightarrow{k}j\overrightarrow{a}) exp(-\overrightarrow{k}j\overrightarrow{a})\,\!</math>  
<math>\vert exp(i \overrightarrow{k}j\overrightarrow{a})\vert ^2 = exp (i \overrightarrow{k}j\overrightarrow{a}) exp(-\overrightarrow{k}j\overrightarrow{a})\,\!</math>  




= <math>(cos(\overrightarrow{k}j\overrightarrow{a}))^2 +(sin(\overrightarrow{k}j\overrightarrow{a}))^2\,\!</math>
            = <math>(cos(\overrightarrow{k}j\overrightarrow{a}))^2 +(sin(\overrightarrow{k}j\overrightarrow{a}))^2\,\!</math>


=1
            =1






The phase factor takes into account that there has been a translation.
The phase factor takes into account that there has been a translation.

Revision as of 15:11, 20 May 2009

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Bloch’s Theorem

As polyenes get longer and longer you could calculate and combine the wavefunctions for all the bonds, but this is not very efficient. Instead it is more powerful to consider the periodicity of repeated units.


Note: Polyacetylene is made from acetylene but it does not contain triple bonds. One can concentrate on the repeat unit cell (with cell length a) and the interactions with its neighbors. For example if you move over 2a units from point r you should find the same electron density at that point ja as point r. Because of the translation symmetry electron density <math>\overrightarrow{r}\,\!</math> in cell j (j equiv integer) must equal the electron density at point <math>\overrightarrow{r}\,\!</math> in the origin cell.

<math>\vert \Psi (\overrightarrow{r} + j \overrightarrow{a}) \vert^2 = \vert \Psi (\overrightarrow{r} ) \vert ^2\,\!</math>


<math>\Psi (\overrightarrow{r} + j\overrightarrow{a}) = exp (i \overrightarrow{k}j\overrightarrow{A}) \Psi (\overrightarrow{r})\,\!</math>

Where

<math>exp (i \overrightarrow{k}j\overrightarrow{a})\,\!</math> is the phase factor

The square of the wave function at r plus ja is the same as square of the wave function at r.

reminder

<math>exp (i \overrightarrow{k}j\overrightarrow{a})= cos(\overrightarrow{k}j\overrightarrow{a})+i sin(\overrightarrow{k}j\overrightarrow{a})\,\!</math>


<math>\vert exp(i \overrightarrow{k}j\overrightarrow{a})\vert ^2 = exp (i \overrightarrow{k}j\overrightarrow{a}) exp(-\overrightarrow{k}j\overrightarrow{a})\,\!</math>


           = <math>(cos(\overrightarrow{k}j\overrightarrow{a}))^2 +(sin(\overrightarrow{k}j\overrightarrow{a}))^2\,\!</math>
           =1


The phase factor takes into account that there has been a translation.