# Identity when Diagonalising Single-Particle Hamiltonian

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?

• Checkout this link for 1D case, extension to multiple dimensions is straightforward. – Sunyam Feb 13 at 20:30
• Thanks! Using the link, I see how this generalises to cubic lattices, but how can that be modified for arbitrary lattices? – FunctionalDefect Feb 14 at 3:33
• 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. – Sunyam Feb 14 at 3:48