Free particle Hamiltonian is $H_0 = \int d\mathbf{r} \frac{\hbar^2}{2m}(\nabla\Psi^\dagger)(\nabla\Psi)$. The Fourier transform representation of $\Psi^\dagger(\mathbf{r})$ is $$ \Psi^\dagger(\mathbf{r}) = \sum_ke^{-i\mathbf{k}\cdot \mathbf{r}} c_\mathbf{k}^\dagger $$ So, $$ \boxed{H_0 = \frac{\hbar^2}{2m}\sum_k k^2 c_\mathbf{k}^\dagger c_\mathbf{k}} $$ And the density operator $\rho(\mathbf{r})=\Psi^\dagger(\mathbf{r})\Psi(\mathbf{r})$ is $$ \rho(\mathbf{r})=\sum_{q_1q_2}e^{-i\mathbf{q_1}\cdot\mathbf{r}}e^{i\mathbf{q_2}\cdot\mathbf{r}}c^\dagger_\mathbf{q_1}c_\mathbf{q_2} $$
$$ \boxed{\rho(\mathbf{r})=\sum_{q_1q}e^{i\mathbf{q}\cdot\mathbf{r}}c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q}} $$ where $\mathbf{q=q_2-q_1}$. The current density operator is calculated by $$ \nabla\cdot \mathbf{J(r)} = -\frac{i}{\hbar}[H_0,\rho(\mathbf{r})] $$ $$ \nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2[ c_\mathbf{k}^\dagger c_\mathbf{k},c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q}]e^{i\mathbf{q}\cdot\mathbf{r}} $$ $$ \nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2\bigg\{ c_\mathbf{k}^\dagger[ c_\mathbf{k},c^\dagger_\mathbf{q_1}]c_\mathbf{q_1+q}+c^\dagger_\mathbf{q_1}[ c_\mathbf{k}^\dagger ,c_\mathbf{q_1+q}]c_\mathbf{k}\bigg\}e^{i\mathbf{q}\cdot\mathbf{r}} $$ $$ \nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2\bigg\{ c_\mathbf{k}^\dagger c_\mathbf{q_1+q} [ \delta_\mathbf{{k,q_1}}] -c^\dagger_\mathbf{q_1}c_\mathbf{k} [ \delta_\mathbf{{k,q_1+q}}] \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}} $$ $$ \nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q,q_1}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{k,q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}} $$ If we include the Fourier transform of LHS as $f(x)=\sum_q e^{ikx}f_k$ $$ \boxed{\sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}} \tag{1} $$ I know the result for $J_q$ should is $$ \mathbf{J_q} = \frac{\hbar}{m}\sum_{k}(\mathbf{k}+\frac{\mathbf{q}}{2}) c_\mathbf{k}^\dagger c_\mathbf{k+q} \tag{2} $$ Question:
How to reach at Eq (2) from Eq (1)?
