Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Derivation of the Heisenberg equation for electron density

  1. Jun 7, 2018 #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
     
  2. jcsd
  3. Jun 12, 2018 #2
    I solved it. if someone wants the solution, i will post it
     
  4. Jun 12, 2018 #3

    MathematicalPhysicist

    User Avatar
    Gold Member

    Please post for future generations of posters... I might be interested in the future, but not in the near 5-10 years. :-)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted