2
$\begingroup$

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 $q$?
$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.