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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)?

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  • $\begingroup$ Hermicity and renaming dummy indices should do the trick. $\endgroup$ Oct 12 '21 at 15:48
  • $\begingroup$ @RogerVadim Could you please give it a try. I tried it by renaming $q_1\to k$, I get: $\sum_{k,q}(k^2 -(k+q)^2)c_{k}^\dagger c_{k+q}e^{iqr}=\sum_{k,q}(-q^2 -2kq)c_{k}^\dagger c_{k+q}e^{iqr}$. After that I am stuck. I would be very thankful if you could help me a little $\endgroup$ Oct 12 '21 at 15:51
  • $\begingroup$ Sorry, but it is too tedious to do it just for fun :) But I did it in the past - it surely works. $\endgroup$ Oct 12 '21 at 15:55
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    $\begingroup$ What's the problem? You have $2kq+q^2 = 2q(k+q/2)$; the minus and the factor $2$ will cancel the prefactors of the LHS. BTW: Your definition of $J_q$ seems wrong, since it should depend on $q$, but you sum over all $q$... $\endgroup$ Oct 12 '21 at 16:24
  • $\begingroup$ @Jakob thank you so much for answer. Yes you are right, there was an extra summation over $q$. I have removed it. $\endgroup$ Oct 12 '21 at 17:57
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After a help from @Jakob, I solved it: $$ \sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}} $$ Change dummy variable $q_1\to k$ $$ \sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\sum_{k,q}\bigg\{k^2 -k^2 -q^2-2\mathbf{k}\cdot \mathbf{q} \bigg\}c^\dagger_\mathbf{k}c_\mathbf{k+q} e^{i\mathbf{q}\cdot\mathbf{r}} $$ $$ \mathbf{J_q} = \frac{\hbar}{m}\sum_{k}\big\{ \frac{\mathbf{q}}{2}+\mathbf{k} \big\}c^\dagger_\mathbf{k}c_\mathbf{k+q} $$

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