Derive Fermion Particle Density Operator in q-space

I am reading Bernevig's Topological Insulators and Topological Superconductors. In section 3.1.1, it gives an expression for the Fourier transformation of electron density operator, which I cannot derive.

(Notation: $k,q$ for wave vector, $i,j$ for lattice cell.)

I tried to derive this operator from: $\rho_i = c_{i}^{\dagger}c_{i}$ on each lattice cell. Then if I assume $c_j^{} = \frac{1}{\sqrt{N}}\sum_k e^{ikj}c_k^{}$, I get:

• $c_j^{\dagger} = \frac{1}{\sqrt{N}}\sum_k e^{-ikj}c_k^{\dagger}$, simply by conjugating above equation.

Then:

\begin{align} \rho_q &= \frac{1}{\sqrt{N}}\sum_j e^{-iqj}c_{j}^{\dagger}c_{j}\\ &= \frac{1}{\sqrt{N}}\sum_j e^{-iqj} \sum_{k,k'} \frac{1}{\sqrt{N}}e^{-ikj}c_k^{\dagger} \frac{1}{\sqrt{N}} e^{ik'j}c_{k'}\\ &= \frac{1}{\sqrt{N}} \sum_{j,k,k'}\frac{1}{N} e^{i(k'-k-q)j} c_k^{\dagger}c_{k'} \\ &= \frac{1}{\sqrt{N}} \sum_{k,k'} \delta(k'-k-q) c_k^{\dagger}c_{k'}\\ &= \frac{1}{\sqrt{N}} \sum_{k} c_k^{\dagger}c_{k+q} \end{align}
But in my book it simply gives the answer:

$$\rho_q = \frac{1}{\sqrt{N}} \sum_{k} c_{k+q}^{\dagger}c_{k}$$

which is different from mine. I checked and failed to catch my error. I found expression for density operator similar to Bernevig's on other books, but I could not find a derivation for it. Can someone help?

• Your $\rho_q$ is Bernevig's $\rho_{-q}$. – Meng Cheng Apr 14 '16 at 17:02
• Thanks for that. I am new to solid theories and feels a little bit unaccustomed to the slightly different conventions throughout many books. – taper Apr 20 '16 at 1:22