Consider a composite lattice such as silicon or graphene (with sublattice A and B). We may call the sublattice tight-binding Bloch sums $\left| k, \alpha \right>$ with $\alpha$ being the sublattice index. This all works fine until I need to express things in the position space. Lets say I have an operator $M$ with matrix elements: $$ M_{k_1, k_2}^{\alpha \beta} \equiv \left< k_1, \alpha \middle| M \middle| k_2, \beta \right> $$ so the $M$ can be written as $$ M = \sum_{k_1, k_2, \alpha, \beta}\left| k_1, \alpha \right> M_{k_1, k_2}^{\alpha \beta} \left< k_2, \beta \right| $$ But now I need to express things in the position space. I tried to insert 2 sets of basis: $$ M = \sum_{r, r', k_1, k_2, \alpha, \beta} \left| r \right> \left< r \middle| k_1, \alpha \right> M_{k_1, k_2}^{\alpha \beta} \left< k_2, \beta \middle| r' \right> \left< r' \right| $$ But this seems problematic because the projection $\left< r \middle| k_1, \alpha \right>$ has different dimensions on two sides, because I think $k$ is supposed to transform into $r$ leaving $\alpha$ as an additional degree of freedom. This may be indicating that we need something like $\left| r, \alpha \right>$ to preserve the sublattice degree of freedom. But what does this state even mean?
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$\begingroup$ What do you mean by "$\langle r| k_1, \alpha\rangle$ has different dimensions on two sides"? And also what does it mean that "$k$ is supposed to transform into $r$"?. The point is that the $r$ index should cover all lattice sites, this means you either interpret $r$ as a lattice site and so there are $NN_\alpha$ of those (with $N$ the number of unit cells and $N_\alpha$ the number of sites in a unit cell) or you give it the structure $(R,\alpha)$ with $R$ spanning just the unit cells and $\alpha$ being the internal degree of freedom. Whatever you choose, be consistent and it will make sense. $\endgroup$– Karim ChahineCommented Jul 8, 2023 at 15:36
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$\begingroup$ What I mean by "having different dimensions on two sides" is that since in Fourier transforms, $k$ and $r$ transform into each other. So if now I add an additional index $\alpha$ it should add an additional degree of freedom. This statement is obviously flawed but I can't see why. $\endgroup$– Frank WangCommented Jul 8, 2023 at 15:43
1 Answer
I would start by noting that, in real space, the number of degrees of freedom you have is $N_c\times N_b$ with $N_c$ the number of unit cells and $N_b$ the number of basis elements for each unit cell. This number of degrees of freedom must be independent of the choice of representation (momentum vs real-space).
When you express the real-space basis as the set of vectors $\{|r\rangle\}$ you must be careful in defining what $r$ is. We know that there must be $N_c\times N_b$ elements of that basis and, since you have indexed the momentum basis as $\{|k, \alpha\rangle\}$, it's best to be consistent and index the real-space basis as $\{|R, \alpha\rangle\}$, where now $R$ indexes the unit cells and $\alpha$ indexes the basis sites. This way the number of degrees of freedom is indeed correct.
So what would $\langle R,\alpha|k, \beta \rangle$ be? Let's first note that, since we have chosen the momentum basis to be indexed as $\{|k, \alpha\rangle\}$, it means that there is a number $N_c$ of $k$ values and so the FT links $k$ with $R$! The FT would be defined as:
\begin{equation} |k, \alpha\rangle = \frac{1}{N_c}\sum_R e^{ikR}|R,\alpha\rangle \end{equation} This is like doing a FT for every sub-lattice. So the elements $\langle R,\alpha|k, \beta \rangle$ would be: \begin{equation} \langle R,\alpha|k, \beta \rangle = \frac{1}{N_c}e^{ikR}\delta_{\alpha,\beta} \end{equation}
With these things, I find it useful to remember how many degrees of freedom you have and how the FT can be defined. Note that you could also bunch $R$ and $\alpha$ together to make up a single index $r$, but then the momentum basis should also be labelled by some index $k$ which runs over the same amount of degrees of freedom, i.e. $N_c\times N_b$.
