Derivation of the Heisenberg equation for electron density

  • #1

Main Question or Discussion Point

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
\begin{equation}
\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}
\end{equation}
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}}##
\begin{equation}
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{4}
\end{equation}
where the Hamiltonian operator is
\begin{equation}
\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}
\end{equation}
with
\begin{equation}
\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}
\end{equation}
\begin{equation}
\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}
\end{equation}
The first doubt is on the Hamiltonian operator rewritten in terms of creation and annihilation operator.
According to my calculations, the Hamiltonian operator is
\begin{equation}
\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}
\end{equation}
where
\begin{equation}
v_{q'} = \frac{4\pi e^2}{V q'^2};\quad \overrightarrow{q}'=\overrightarrow{k}_1 - \overrightarrow{k}_4
\label{eq:vq}
\end{equation}
while the book states that
$$
\overrightarrow{q}'=\overrightarrow{k}_1 + \overrightarrow{k}_3 - \overrightarrow{k}_2 - \overrightarrow{k}_4
$$
I ignore this thing, so i calculate the commutator in the Heisenberg equation
\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 second to last sum, i exchange \[\overrightarrow{k}_1\] with ##\overrightarrow{k}_2## and $\overrightarrow{k}_3$ with $\overrightarrow{k}_4$, while, in the last 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
 

Answers and Replies

  • #2
I solved it. if someone wants the solution, i will post it
 
  • #3
MathematicalPhysicist
Gold Member
4,223
174
Please post for future generations of posters... I might be interested in the future, but not in the near 5-10 years. :-)
 
  • Like
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