# Current density operator derivation in Fourier space

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)?

• Hermicity and renaming dummy indices should do the trick. Oct 12 '21 at 15:48
• @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 Oct 12 '21 at 15:51
• Sorry, but it is too tedious to do it just for fun :) But I did it in the past - it surely works. Oct 12 '21 at 15:55
• 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 $q$... Oct 12 '21 at 16:24
• @Jakob thank you so much for answer. Yes you are right, there was an extra summation over $q$. I have removed it. Oct 12 '21 at 17:57

## 1 Answer

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 $$q_1\to k$$ $$\sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\sum_{k,q}\bigg\{k^2 -k^2 -q^2-2\mathbf{k}\cdot \mathbf{q} \bigg\}c^\dagger_\mathbf{k}c_\mathbf{k+q} e^{i\mathbf{q}\cdot\mathbf{r}}$$ $$\mathbf{J_q} = \frac{\hbar}{m}\sum_{k}\big\{ \frac{\mathbf{q}}{2}+\mathbf{k} \big\}c^\dagger_\mathbf{k}c_\mathbf{k+q}$$