Derivation of the Heisenberg equation for electron density

In summary, the conversation is discussing the calculation of the equation of motion for electrons' density in the context of studying plasmons from "Haken-Quantum Field Theory of Solids". The equation of motion is given by the equation for the fermionic operator, which involves creation and annihilation operators and follows the Heisenberg equation. The Hamiltonian operator is also discussed, and there are some doubts and discrepancies between the calculations and the solution in the book. The person is seeking clarification and guidance on how to find the correct solution.
  • #1
GiovanniNunziante
2
1
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
 
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  • #2
I solved it. if someone wants the solution, i will post it
 
  • #3
Please post for future generations of posters... I might be interested in the future, but not in the near 5-10 years. :-)
 
  • Like
Likes GiovanniNunziante

Related to Derivation of the Heisenberg equation for electron density

1. What is the Heisenberg equation for electron density?

The Heisenberg equation for electron density is a fundamental equation in quantum mechanics that describes the time evolution of the electron density in a system. It is named after German physicist Werner Heisenberg, who first derived it in 1925.

2. How is the Heisenberg equation for electron density derived?

The Heisenberg equation for electron density is derived using the Heisenberg uncertainty principle, which states that the position and momentum of a particle cannot be known simultaneously with absolute certainty. By considering the time evolution of the electron density operator, the equation is derived as a result of the commutation relations between position and momentum operators.

3. Why is the Heisenberg equation for electron density important?

The Heisenberg equation for electron density is important because it allows us to understand and predict the behavior of electrons in quantum systems. It is a fundamental equation in quantum mechanics that is used to solve many problems in fields such as chemistry, material science, and condensed matter physics.

4. What are some applications of the Heisenberg equation for electron density?

The Heisenberg equation for electron density has many applications in various fields of science and technology. It is used to study the electronic structure of atoms and molecules, understand the behavior of electrons in materials, and develop new technologies such as quantum computing and nanotechnology.

5. Are there any limitations or assumptions in the Heisenberg equation for electron density?

Like any other equation in physics, the Heisenberg equation for electron density has its limitations and assumptions. One of the main limitations is that it only describes the time evolution of a single electron density, and cannot account for interactions between multiple electrons. Additionally, it is based on the assumption that the system is in a state of equilibrium, which may not always be the case in real-world scenarios.

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