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 $q$?

1 Answer 1


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 $$ h(k+\frac{q}{2}) - h(k-\frac{q}{2}) = h(k) + \frac{q}{2}\partial_k h(k) + \frac{q^2}{8}\partial^2_k h(k) - h(k) + \frac{q}{2}\partial_k h(k) - \frac{q^2}{8}\partial^2_k h(k) + O(q^3) = q \partial_k h(k) + O(q^3)$$


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