Current density free electron gas

[peterprp]

Hello,
I am studying transport in the free electron gas model and I don't understand how to compute the average of the electron density current.
We are given the hamiltonian
$H=\int \psi^\dagger(r,t)(-\frac{\hbar^2\nabla^2}{2m}+e\vec{E}\cdot\vec{r})\psi(r,t)$
where the $psi$ operator is the solution of the Heisenberg equation for field operators
$[H,\psi]=-i\hbar\frac{d\psi(r,t)}{dt}$,
namely
$\psi(r,t) = \frac{1}{\sqrt{V}}\sum_k e^{i(\vec{k}-e\vec{E}t/\hbar)\cdot\vec{r}} e^{\int_0^t \epsilon(k-iEt'/\hbar)dt'}c_k$
where
$\epsilon(k)=\frac{\hbar^2 k^2}{2m}$
and I need to compute the average of
$J(r,t)=\frac{ie\hbar}{2m}[\psi^\dagger(r,t) \nabla \psi(r,t) - \nabla\psi^\dagger(r,t) \psi(r,t)]$
Now, the result is Drude's one:
$n e^2 \tau/m$
Could someone please give me some hint on how to sketch out this calculation?
I know that I have to take the average over k weighting via the Fermi distribution and I also have to average over collision times, namely
$\int e^{-t/\tau}/\tau (...) dt$
but why does the dependence on position disappear?
Thank you in advance

 

[BigJDubs]

.To compute the average of the electron density current, you need to first calculate the expectation value of $J(r,t)$. This can be done by taking the trace of the density matrix $\rho$ times $J(r,t)$, which is equal to:$$\langle J(r,t)\rangle = \text{tr}(\rho J(r,t)) = \sum_{k,k'}\langle c_k^\dagger c_{k'}\rangle \langle k|J(r,t)|k'\rangle$$ where the $\langle k|J(r,t)|k'\rangle$ is given by:$$\langle k|J(r,t)|k'\rangle = \frac{ie\hbar}{2m} \int d^3 r\left[\psi^\dagger_k(r,t) \nabla \psi_{k'}(r,t) - \nabla\psi^\dagger_k(r,t) \psi_{k'}(r,t)\right] $$Using the expression for $\psi_k(r,t)$, we can rewrite this as:$$ \langle k|J(r,t)|k'\rangle = \frac{ie\hbar}{2mV} \sum_{\vec{q}}e^{i(\vec{k}-\vec{k'}-e\vec{E}t/\hbar)\cdot\vec{q}} \left[\vec{k}-\vec{k'}-e\vec{E}t/\hbar\right] $$Now, to get the average of $J(r,t)$, we need to average over all possible values of $k$ and $k'$ with the Fermi distribution, which gives:$$ \langle J(r,t)\rangle = \frac{ie\hbar}{2mV} \sum_{\vec{q}} \left[f(\vec{k}+\vec{q})-f(\vec{k})\right]\left[\vec{k}-e\vec{