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)?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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