# Plasmons : doubts about the derivation of the Heisenberg equation of electrons' density

I'm studying plasmons from "Haken-Quantum Field Theory of Solids", and i need some help in the calculation of the equation of motion of eletrons' density $$$$\hat{\rho}_{\overrightarrow{q}} = \frac{1}{\sqrt{V}} \sum_{\overrightarrow{k}} \hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} \label{eq:rhoaadag}$$$$ where $$\hat{a}$$ and $$\hat{a}^{\dagger}$$ are fermionic operator \begin{align} \left\lbrace\hat{a}_{\overrightarrow{k}},\hat{a}^{\dagger}_{\overrightarrow{k'}}\right\rbrace = \delta_{\overrightarrow{k'},\overrightarrow{k}} \\ \left\lbrace\hat{a}_{\overrightarrow{k}},\hat{a}_{\overrightarrow{k'}}\right\rbrace = \left\lbrace\hat{a}^{\dagger}_{\overrightarrow{k}},\hat{a}^{\dagger}_{\overrightarrow{k'}}\right\rbrace = 0 \quad \forall \overrightarrow{k},\overrightarrow{k}' \label{eq:anticomm} \end{align} The book starts from the Heisneberg equation for $$\hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}}\hat{a}_{\overrightarrow{k}}$$ $$$$i\hbar\frac{d}{dt}\left(\hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} \right) = \left[\hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}},\hat{H}\right] \label{eq:eqmotopre} \tag{1}$$$$ where the Hamiltonian operator is \begin{aligned} \hat{H} = \int d\overrightarrow{r} \hat{\psi}^{\dagger}\left(\overrightarrow{r}\right) \left(-\frac{\hbar^2}{2 m^*} \nabla^2 \right) \hat{\psi} \left(\overrightarrow{r}\right) + \\ \frac{1}{2}\int\int d\overrightarrow{r} d\overrightarrow{r}' \hat{\psi}^{\dagger}\left(\overrightarrow{r}'\right) \hat{\psi}^{\dagger}\left(\overrightarrow{r}\right) \frac{e^2}{|\overrightarrow{r}'-\overrightarrow{r}|} \hat{\psi}\left(\overrightarrow{r}\right) \hat{\psi}\left(\overrightarrow{r}'\right) \end{aligned} \label{eq:hsecquant} with $$$$\hat{\psi}\left(\overrightarrow{r}\right)=\frac{1}{\sqrt{V}} \sum_{\overrightarrow{k}} \hat{a}_{\overrightarrow{k}} \exp\left[i\overrightarrow{k}\cdot\overrightarrow{r}\right] \label{eq:annichila}$$$$ $$$$\hat{\psi}^{\dagger}\left(\overrightarrow{r}\right)=\frac{1}{\sqrt{V}} \sum_{\overrightarrow{k}} \hat{a}^{\dagger}_{\overrightarrow{k}} \exp\left[- i\overrightarrow{k}\cdot\overrightarrow{r}\right] \label{eq:crea}$$$$ The first doubt is on the Hamiltonian operator rewritten in terms of creation and annihilation operator. According to my calculations, the Hamiltonian operator is $$$$\hat{H}=\sum_{\overrightarrow{k}} E_{\overrightarrow{k}}\, \hat{a}^{\dagger}_{\overrightarrow{k}} \hat{a}_{\overrightarrow{k}} + \frac{1}{2} \sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2}\, \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_3} \hat{a}_{\overrightarrow{k}_4} \label{eq:hamampiezze}$$$$ where $$$$v_{q'} = \frac{4\pi e^2}{V q'^2};\quad \overrightarrow{q}'=\overrightarrow{k}_1 - \overrightarrow{k}_4 \label{eq:vq}$$$$ while the book states that $$\overrightarrow{q}'=\frac{1}{2}\left(\overrightarrow{k}_1 + \overrightarrow{k}_3 - \overrightarrow{k}_2 - \overrightarrow{k}_4\right)$$ I ignore this thing, so i calculate the commutator into \ref{eq:eqmotopre}, I find \begin{align*} i\hbar\frac{d}{dt}\left(\hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} \right) = \left(E_{\overrightarrow{k}} - E_{\overrightarrow{k}+\overrightarrow{q}}\right) \hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} + %prima somma ps & \frac{1}{2} \sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} \delta_{\overrightarrow{k}_2,\overrightarrow{k}_3} %%%%argomento ps \left[\hat{a}^{\dagger}_{\overrightarrow{k} + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}}, \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_4}\right] \label{eq:primo} \\ %ss &-\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg ss \hat{a}^{\dagger}_{\overrightarrow{k}_1 + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}_3} \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \label{eq:secondo} \\ %ts &+\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg ts \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_3 - \overrightarrow{q} } \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \label{eq:terzo} \\ %qs &-\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg qs \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_3} \hat{a}^{\dagger}_{\overrightarrow{k}_2 + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}_4} \label{eq:quarto} \\ %quintas &+\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg quintas \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_3} \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4 - \overrightarrow{q}} \label{eq:quinto} \end{align*} the sum with the commutator $$\left[\hat{a}^{\dagger}_{\overrightarrow{k} + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}}, \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_4}\right]$$ is equal to zero.

Second doubt : I exchange the indices in the last two sums (can i do it ?)

In the penultimate sum i exchange $$\overrightarrow{k}_1$$ with $$\overrightarrow{k}_2$$ and $$\overrightarrow{k}_3$$ with $$\overrightarrow{k}_4$$, while, in the las sum, $$\overrightarrow{k}_3$$ with $$\overrightarrow{k}_4$$ and $$\overrightarrow{k}_1$$ with $$\overrightarrow{k}_2$$ \begin{align*} %ss &-\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg ss \hat{a}^{\dagger}_{\overrightarrow{k}_1 + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}_3} \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \label{eq:primosec} \\ %ts &+\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg ts \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_3 - \overrightarrow{q} } \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \label{eq:secondosec} \\ %qs &-\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg qs \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \hat{a}^{\dagger}_{\overrightarrow{k}_1 + \overrightarrow{q}} \hat{a}_{\overrightarrow{k}_3} \label{eq:terzosec} \\ %quintas &+\sum v_{q'} \; \delta_{\overrightarrow{k}_1,\overrightarrow{k}_3+\overrightarrow{k}_4-\overrightarrow{k}_2} %arg quintas \hat{a}^{\dagger}_{\overrightarrow{k}_2} \hat{a}_{\overrightarrow{k}_4} \hat{a}^{\dagger}_{\overrightarrow{k}_1} \hat{a}_{\overrightarrow{k}_3 - \overrightarrow{q}} \label{eq:quartosec} \end{align*} i manipulate them, but i don't find the solution of the book, that is \begin{align*} i\hbar\frac{d}{dt}\left(\hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} \right) = \left(E_{\overrightarrow{k}} - E_{\overrightarrow{k}+\overrightarrow{q}}\right) \hat{a}^{\dagger}_{\overrightarrow{k}+\overrightarrow{q}} \hat{a}_{\overrightarrow{k}} %ss \\ &+\sum_{k', q'} v_{q'} \; %arg ss \left[\left(\hat{a}^{\dagger}_{k + q} \hat{a}_{k + q'} \hat{a}^{\dagger}_{k' + q'} \hat{a}_{k'} \right) - \left(\hat{a}^{\dagger}_{k' + q'} \hat{a}_{k'} \hat{a}^{\dagger}_{k + q- q'} \hat{a}_k\right) \right] \end{align*} Am I on the right way? Can you give me some hint in order to find the solution of the book?

Thank you for who will answer me

• I solved it. if someone wants the solution, i will post it Commented Jun 12, 2018 at 16:14

As regard the first doubt, From the delta Kronecker we have $$\vec{k}_1-\vec{k}_4 = \vec{k}_3-\vec{k}_2$$ so \begin{align*} \vec{q}'&=\vec{k}_1-\vec{k}_4 \\ &=\frac{1}{2}\left(\vec{k}_1 - \vec{k}_4\right) + \frac{1}{2}\left(\vec{k}_1 - \vec{k}_4\right) \\ &=\frac{1}{2}\left(\vec{k}_1 - \vec{k}_4\right) + \frac{1}{2}\left(\vec{k}_3-\vec{k}_2\right) \\ &=\frac{1}{2}\left(\vec{k}_1 - \vec{k}_4 + \vec{k}_3-\vec{k}_2\right) \end{align*}