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\heading{E7050: The Drude formula} 
\auname{Shay Inbar}


{\bf The problem:}
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\Dn

Consider a ring of length $L$,
with a particle that has the Drude
velocity-velocity correlation function
with a time constant $\tau$.
The temperature is $T$.

\Dn

(a)
Find the conductance of the ring
using the canonical FDT.

\Dn

(b)
What is the conductance if there are $N$ fermions
at zero temperature instead of a single particle.

\Dn

(c)
What is $\tau$, and hence what is the conductance,
if the scattering in the ring is due to
a stochastic segment that has a transmission $g$. 

\Dn\Dn


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{\bf The solution:}
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\Dn

(a) 
The current obeys $I=-\frac{\partial H}{\partial \Phi}$ and so $I$ is the ``generalized force''.
We define
\begin{equation}
	I = -G\dot{\Phi}
\end{equation}

where $G$ is a dissipation term called the ``Drude conductance''.
Using the fluctuation-dissipation theorem we have

\begin{equation}
	G = \frac{1}{2T} \tilde{C}(\omega = 0 )=\frac{1}{2T} \langle I(t) I(0) \rangle
	  = \frac{1}{2T} \left( \frac{e}{L} \right)^2 \langle v(t) v(0) \rangle
\end{equation}

where in the last term we used $I=\frac{e}{L}v$.
Now, the velocity-velocity correlation function is, by the Drude assumption,

\begin{equation}
	\langle v(t) v(0) \rangle = \int_{-\infty}^{\infty} {v_T}^2 e^{-|t|/\tau} dt
								= 2{v_T}^2 \tau 
\end{equation}

We can also express the thermal velocity from the equipartition theorem
\begin{equation}
	\frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}T
	\quad \Longrightarrow \quad \langle v^2 \rangle = \frac{T}{m}
\end{equation}

putting this all together, we get

\begin{equation}
	G = \frac{1}{2T} \left(\frac{e}{L}\right)^2 \cdot 2 \frac{T}{m} \tau
	= \frac{1}{L^2} \frac{e^2}{m} \tau
\end{equation}

Note that this is independent of $T$.

%%%%%%%%%%%%%%%
\Dn

(b)
For a system of $N$ Fermions, the correct result is simply $N$ times the single-particle conductance 
as we show below:

\begin{equation}
	G^{[N]} = \frac{N}{L^2} \frac{e^2}{m} \tau
\end{equation}

The only difference is in the time constant $\tau$, which we assume here to be the same.

This is actually a result of 2 effects that exactly cancel each other out: 
The Fermion nature of the system and the quantum-mechanical correlation function.
A quantum mechanical calculation shows that for Fermions
\begin{equation}
	\tilde{C}^{[N]}(\omega) = \omega \coth \left(\frac{\omega}{2T}\right)\tilde{C}_{E_F}(\omega)
\end{equation}

The conductance is given by FDT
\begin{align*}
	G^{[N]} &= \frac{1}{\omega} \tanh \left(\frac{\omega}{2T}\right) \tilde{C}^{[N]}(\omega) \\
	        &= \frac{1}{\omega} \tanh \left(\frac{\omega}{2T}\right) \cdot \omega \coth \left(\frac{\omega}{2T}\right)\tilde{C}_{E_F}(\omega) \\
	        &= \tilde{C}_{E_F}(\omega) = \frac{N}{L^2} \frac{e^2}{m} \tau_F
\end{align*}	
where at $T=0$ we have $\tau_F = \tau$. 

\Dn

(c)
We take the time constant to be $\tau = \frac{\ell}{v}$ where $\ell$ is the mean free path.
The correlation function would then have an exponential decay $\exp[-v|t|/\ell]$ as usual.
We recognize, however, that after each ``scattering'' event the current is a fraction $g-(1-g) = 2g-1$
of the original (with a combination of clockwise and anticlockwise currents), and the time between adjacent scattering events is
$\frac{L}{v}$, so the correlation function takes the form of a discrete sum

\begin{equation}
	C(t) = e^2\frac{v}{L} \sum_{n=-\infty}^{\infty} (2g-1)^n \delta \left( t-\left( \frac{L}{v}n \right) \right)
\end{equation}

from which we get by integration

\begin{equation}
	\frac{\ell}{L} = \frac{g}{g-1}
\end{equation}

and the conductance in the ring is given by

\begin{equation}
	G = \left(\frac{e^2}{mvL}\right) \frac{\ell}{L} = \left(\frac{g}{g-1}\right) \frac{e^2}{mvL}
\end{equation}

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