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\heading{E6070: Landauer formula for 1D conductance}
\auname{Eran Avraham}
{\bf The problem:}
\Dn
Consider 1D conductor that has transmission coefficient $g$.
The conductor is connected to 1D leads that have
chemical potentials $\mu_a$ and $\mu_b$.
Assume ${\mu_a=\mu}$ and ${\mu_b=\mu+eV}$, where $V$ is the bias.
(1) Write the expression for the current $I$ as an integral
over the occupation function $f(\epsilon)$.
(2) For small bias write the relation as $I=GV$
and obtain an expression for $G$.
Write explicit results for zero temperature Fermi occupation
(Landauer formula)
and for high temperature Boltzman occupation.
(3) Find expressions for $I(V)$ in the case of
arbitrary (possibly large) bias,
for zero temperature Fermi occupation
and for high temperature Boltzmann occupation.
Assume that $g$ is independent of energy
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\textbf{The solution:}
(1) In general for $1D$ we have:
\[I_{a,n}=\frac{e}{L}v_{a,n}\]
Where $v_{a,n}$ is the speed of the electron in the cannel n.
\[I_a=\sum_{n=0}^{\infty} f_a(E_n-\mu_a)I_{a,n}=\int_0^\infty f_a(E-\mu_a)\frac{LdE}{2\pi v_a}(\frac{ev_a}{L})=\frac{e}{2\pi}\int_0^\infty f_a(E-\mu_a)dE\]
Multiplying by the transition coefficient between $a$ and $b$ - $g$ we get the total current:
\[I=g(I_b-I_a)=\frac{e}{2\pi}\int_0^\infty g(f_b(E-\mu_b)- f_a(E-\mu_a))dE\]
(2) We now make use of the fact that the bias is small.
\[I=\frac{e}{2\pi}\int_0^\infty g(f_b(E-\mu-eV)- f_a(E-\mu))dE\approx -\frac{eg}{2\pi}\int_0^\infty f'(E-\mu)eVdE=-\frac{e^2g}{2\pi}V(f(\infty)-f(0))=\frac{e^2g}{2\pi}f(0)V\]
In the fermi case the occupation function is $f=\frac{1}{e^{\beta(E-\mu)}+1}$ therefore:
\[f(0)=1\implies G=\frac{e^2g}{2\pi}\]
Which is the landauer formula.
For $T>0$ we have the Boltzman occupation function $f=e^{-\beta(E-\mu)}$
\[f(0)=e^{\beta\mu}\implies G=\frac{e^2g}{2\pi}e^{\beta\mu}\]
(3) In the case of an arbitrary bias we have:
\[I=\frac{e}{2\pi}\int_0^\infty g(f_b(E-\mu_b)- f_a(E-\mu_a))dE\]
Using the fact that in $T=0$ the fermi a occupation function becomes a step function we get:
\[I(V)=\frac{e}{2\pi}g(\mu_b-\mu_a)=\frac{e^2g}{2\pi}V\]
And in the Boltzman case $T>0$:
\[I(V)=\frac{eg}{2\pi}\int_0^\infty(e^{-\beta(E-\mu_b)}-e^{-\beta(E-\mu_a)})dE=\frac{eg}{2\pi\beta}(e^{\beta\mu_b}-e^{\beta\mu_a})=\frac{eg}{2\pi\beta}e^{\beta\mu}(e^{\beta eV}-1)\]
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