# Current operator, why is this form valid up to second order in $q$?

Context

In Bernevig´s textbook Topological insulators and topological superconductors, an approximate form of the current operator in momentum space is derived. It is said to be valid to second order in momentum, but I don´t understand why.

They start with a generic lattice Hamiltonian with translational invariance, so that it can be written in momentum space:
$$H=\sum_{ij} c^\dagger_{i}h_{ij}c_{j}, \quad H=\sum_{\mathbf{k}}c_{\mathbf{k}}^\dagger h_\mathbf{k} c_{\mathbf{k}}.\tag{1}$$ Here, $$i,j$$ correspond to the lattice sites and $$\mathbf{k}$$ to momentum.

To obtain the current, they use the continuity equation in momentum space: $$\dot{\rho}(\mathbf{x})+\nabla\cdot \mathbf{J}(\mathbf{x})=0 \implies \dot{\rho}_\mathbf{q}-i\mathbf{q}\cdot\mathbf{J}_\mathbf{q}=0\tag{2} .$$

Next, they use the density operator in momentum space and the Heisenberg equation of motion to write: $$\rho_\mathbf{q}=\frac{1}{\sqrt{N}}\sum_\mathbf{k}c^\dagger_{\mathbf{k+q}}c_\mathbf{k}\implies -i\mathbf{q}\cdot\mathbf{J}_\mathbf{q}=-\dot{\rho}_\mathbf{q}=i[\rho_\mathbf{q},H], \tag{3}$$ which after some algebra becomes $$\mathbf{q}\cdot \mathbf{J}_\mathbf{q}=\frac{1}{\sqrt{N}}\sum_\mathbf{k} (h_{\mathbf{k}+\mathbf{q}}-h_\mathbf{k})c_{\mathbf{k}+\mathbf{q}}^\dagger c_\mathbf{k}. \tag{4}$$

Core of the question

After using $$h_{\mathbf{k}+\mathbf{q}}-h_\mathbf{k} \approx \mathbf{q}\cdot \partial_\mathbf{k} h_\mathbf{k}$$, it is possible to identify the current operator by comparing both sides of equation (4) as: $$\mathbf{J}_\mathbf{q}=\frac{1}{\sqrt{N}}\sum_\mathbf{k} \frac{h_\mathbf{k}}{\partial \mathbf{k}} c^\dagger_{\mathbf{k}+\mathbf{q}}c_\mathbf{k}$$

I understand everything up to this point. Clearly, this form of the current operator is valid to first order in $$\mathbf{q}$$. However, they say that by shifting the $$\mathbf{k}\rightarrow \mathbf{k}-\mathbf{q}/2$$ the approximation is valid to second order:

We can make a better approximation, valid to second order in $$\mathbf{q}$$, by shifting $$\mathbf{k}\rightarrow \mathbf{k}-\mathbf{q}/2$$: $$\mathbf{q}\cdot \mathbf{J}_\mathbf{q}=\frac{1}{\sqrt{N}}\sum_\mathbf{k}(h_{\mathbf{k}+\mathbf{q}/2}-h_{\mathbf{k}-\mathbf{q}/2})c^\dagger_{\mathbf{k}+\mathbf{q}/2}c_{\mathbf{k}+\mathbf{q}/2}\\=\frac{1}{\sqrt{N}}\sum_\mathbf{k}\left(\frac{\partial h_\mathbf{k}}{\partial \mathbf{k}}\cdot \mathbf{q} \right)c^\dagger_{\mathbf{k}+\mathbf{q}/2}c_{\mathbf{k}+\mathbf{q}/2}+\mathcal{O}(q^2)\tag{3.8}$$ The linear term $$q$$ is important to get right in some cases, and hence the shift performed is very important. The current operator at small $$q$$ is, hence, $$\mathbf{J}_\mathbf{q}=\frac{1}{\sqrt{N}}\sum_\mathbf{k} c^\dagger_{\mathbf{k}+\mathbf{q}/2}\frac{\partial h_\mathbf{k}}{\partial \mathbf{k}}c_{\mathbf{k}+\mathbf{q}/2}\tag{3.9}$$

I don't understand:

• Why shifting the momentum makes $$J_\mathbf{q}$$ valid to second order?
• By looking at the second line of (3.8) and equation (3.9), is equation (3.9) then valid only to first order in $$q$$?

this is due to the Taylor expansion. Consider the function $$F(x)$$ with some Taylor expansion about $$x_0$$, then while $$F(x_0+\delta x) = F(x_0) + \delta x \partial_x F(x_0)$$ to first order in $$\delta x$$, if we want to look at the difference of $$F$$ at points around $$x_0$$ we have $$F(x_0+\delta x) - F(x_0-\delta x) = F(x_0) + 2\delta x \partial_x F(x_0)$$ which is true to second order in $$\delta x$$ because that order cancelled out between the two contributions.
Similarly $$h(k+\frac{q}{2}) - h(k-\frac{q}{2}) = h(k) + \frac{q}{2}\partial_k h(k) + \frac{q^2}{8}\partial^2_k h(k) - h(k) + \frac{q}{2}\partial_k h(k) - \frac{q^2}{8}\partial^2_k h(k) + O(q^3) = q \partial_k h(k) + O(q^3)$$