# Diamagnetic current in the BdG superconductor

Recently, I am following the paper arXiv:1610.01803 to study superfluid weight. In this paper the diamagnetic current is given by:

$$j^D_{\mu}(\vec{q})=\sum_{\vec{k},\sigma}\partial_{\mu}\partial_{\nu}H(\vec{k},\sigma)c^{\dagger}_{\vec{k},\sigma}c_{\vec{k}+\vec{q},\sigma}A_{\mu}(\vec{q})=T_{\mu\nu}A_{\nu}$$

But I met problems when I try to transform this into the Nambu field: $$\langle T_{\mu\nu}\rangle=\sum_{\vec{k}}\langle c^{\dagger}_{\vec{k}\uparrow}\partial_{\mu}\partial_{\nu}H_{\uparrow}(\vec{k})c_{\vec{k}\uparrow}+c^{\dagger}_{\vec{k}\downarrow}\partial_{\mu}\partial_{\nu}H_{\downarrow}(\vec{k})c_{\vec{k}\downarrow}\rangle\\ =\sum_{\vec{k}}\langle c^{\dagger}_{\vec{k}\uparrow}\partial_{\mu}\partial_{\nu}H_{\uparrow}(\vec{k})c_{\vec{k}\uparrow}-c_{\vec{-k}\downarrow}\partial_{\mu}\partial_{\nu}H^{*}_{\downarrow}(\vec{-k})c^{\dagger}_{\vec{-k}\downarrow}+\partial_{\mu}\partial_{\nu}H^{*}_{\downarrow}(\vec{-k})\rangle$$

while in arXiv:1610.01803, authors seems dropped the last term. In their paper, they obtained: $$\langle T_{\mu\nu}\rangle=\sum_{\vec{k}}\langle\psi^{\dagger}_{\vec{k}}\partial_{\mu}\partial_{\nu}H_{BdG}(\vec{k})\psi_{\vec{k}}\rangle$$, and $$\psi_{\vec{k}}=(c_{\vec{k},\uparrow},c^{\dagger}_{-\vec{k},\uparrow})^T$$ is the Nambu field.

I understand why we can drop similar term in the BdG Hamiltonian. But why can we still drop this in the diamagnetic current operators?

Thanks!