questions-of-semiconductors-II.md

questions-of-semiconductors-II.md

Questions

The definition of complex $R$ and point group $\mathrm{\textbf{R}}$

In the first half of Lecture II, the complex $R$, not necessarily a group $\mathrm{\textbf{R}}$, is introduced as such.

Consider only the purely translational symmetry operations of an infinite crystal. These operations form a group denoted $\mathrm{\textbf{T}}$, where $\mathrm{\textbf{T}}$ is a subgroup of the full space group $\mathrm{\textbf{G}}$ of the crystal. Denote by $R$, the set of all symmetry operations of $\mathrm{\textbf{G}}$ that involve pure rotations only (proper or improper) as well as rotations accompanied by a translation not in $\mathrm{\textbf{T}}$. $R$ is a subset of $\mathrm{\textbf{G}}$ called a "complex" whose elements are denoted by $\alpha$, $\beta$, $\gamma$, ... since in general $R$ is not a group.

Consider the case wherein $R$ is the set of all pure rotational operations (proper and improper) of $\mathrm{\textbf{G}}$ plus the glide reflection denoted $m$ and shown in the previous slide. $R$ is not a group since $mm$ is a pure translation which is not a member of $R$.

Let $ C = {\alpha, \beta, \gamma, ...}$ be the complex consisting of all elements of $\mathrm{\textbf{G}}$ not in $\mathrm{\textbf{T}}$. Unlike the complex $R$, the translation operations that accompany rotations as members of $C$, can belong in $\mathrm{\textbf{T}}$.

It took me too long to understand this graphically. I went over the slides, the book, and the notes I had when I was taking Math Methods. I finally realized that in the nonsymmorphic example above, both of $R$ and $C$ include the glide reflection $m$ (glide + reflect). $C$ includes a lot more, actually counts a multiple of what $R$ has, by (the number of elements in $\mathrm{\textbf{T}}$) - 1. They obviously also belong in $\mathrm{\textbf{G}}$, because of closure between the elements in $R$ and in $\mathrm{\textbf{T}}$.

And here comes a sudden change. $R$ used to be ${\alpha, \beta, \gamma, ...}$, but now a rather new group is defined, $\mathrm{\textbf{R}} = {\alpha_r, \beta_r, \gamma_r}$. $R$ was not a group before, but at least all the elements that belong in complex $R$ belong in $\mathrm{\textbf{G}}$, but now $\mathrm{\textbf{R}}$ is a group, the point group of a nonsymmorphic group, and contains some elements not part of $\mathrm{\textbf{G}}$, like the reflection part of the glide reflection $m$. Of course, for a symmorphic group $\mathrm{\textbf{G}}$, $R = \mathrm{\textbf{R}}$.

On the other hand, the mapping of the factor group $\mathrm{\textbf{G/T}}$ of the space group G to the constructed sub-group $\mathrm{\textbf{R}} = {\alpha_r, \beta_r, \gamma_r}$ of a symmorphic space group is one-to-one and is called an "isomorphism".

$\mathrm{\textbf{R}} = {\alpha_r, \beta_r, \gamma_r}$ are not necessarily elements of the nonsymmorphic space group and so $\mathrm{\textbf{R}}$ is not a subgroup of $\mathrm{\textbf{G}}$. If $\alpha = \alpha_t \alpha_r$ is a glide or screw, then $\alpha_r$ (improper or proper, respectively) is not an element in $\mathrm{\textbf{G}}$ and $\alpha_t$ is not an element of $\mathrm{\textbf{T}}$. $\mathrm{\textbf{R}}$ is still referred to as the point group of the space group $\mathrm{\textbf{G}}$, since it contains all the information on the rotational symmetries of the space group. However, it is not a sub-group of $\mathrm{\textbf{G}}$ and contains operations that are not elements of $\mathrm{\textbf{G}}$.

(For the diamond crystal) Let $\mathrm{\textbf{G}}$ be the space group of diamond and $\mathrm{\textbf{T}}$ be the translational symmetry operations. $\mathrm{\textbf{T}}$ is the same for zincblende and diamond lattices. Let $C$ be the complex of operations not in $\mathrm{\textbf{T}}$; $C$ is different for the zincblende and diamond lattices. In particular the operation $T(1/4,1/4,1/4) i$ is in $C$ for the diamond lattice, but not in the zincblende lattice. This additional operation is in the form of $\alpha = \alpha_t \alpha_r$. So $\mathrm{\textbf{R}}$ for the diamond lattice (which is not a subgroup because $i$ is not in $\mathrm{\textbf{G}}$, i.e., $\mathrm{\textbf{G}}$ is nonsymmorphic) includes all the rotations of the zincblende lattice plus $i$.

The factor group $\mathrm{\textbf{G/T}}$ of the diamond lattice is isomorphic to $\mathrm{\textbf{R}}$ which has in it all the elements of $\mathrm{\textbf{T_d}}$, but with the addition of $i$. Recall that $\mathrm{\textbf{R}}$ is still referred to as the point group of the nonsymmorphic space group of the diamond lattice because it contains all the information on the rotational symmetries of the space group; however, it contains the element $i$ which is not in the space group.