Sorry the title is not precise; wasn't sure how to make it so (this is perhaps a straightforward question).

The following is an identity I see quite often when reading lecture notes about diagonalising single-particle, 1D (tight-binding) Hamiltonians (via changing to $k$-space): $$ \sum_j e^{i(k-k')ja} = N\delta_{k,k'} $$ where as usual $a$ is the lattice constant, $N$ is the number of atoms and $j$ is the site index. Example use (second to third line from equation 3).

First, I was wondering how to prove this (or at least why it makes physical sense). Second, I was wondering if this is just always true for any dimension. i.e. is the following identity true? Does it depend on the lattice? $$ \sum_n e^{i(\vec k-\vec k')\cdot \vec R_n} = N\delta_{\vec k, \vec k'} $$ where now $\vec R_n$ is the lattice vector for site $n$. How would I prove/understand this?

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    $\begingroup$ Checkout this link for 1D case, extension to multiple dimensions is straightforward. $\endgroup$
    – Sunyam
    Commented Feb 13, 2019 at 20:30
  • $\begingroup$ Thanks! Using the link, I see how this generalises to cubic lattices, but how can that be modified for arbitrary lattices? $\endgroup$ Commented Feb 14, 2019 at 3:33
  • $\begingroup$ Do you mean D-dimensional lattices with non-orthogonal basis vectors? I guess reciprocal space vectors are defined as co-orthogonal to original basis. Hence sums can be factored and hence you can proove. $\endgroup$
    – Sunyam
    Commented Feb 14, 2019 at 3:48


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