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The Landauer Formula

Landauer Formula is really useful in order to compute current in nanoscale devices. Actually this formula has been included in the opensource code NanoTCAD ViDES, which compute transport in Carbon Nanotubes and Silicon Nanowire Transistors.

Let us consider the following system, composed by a one-dimensional channel (a quantum wire) with length $L$, and two metallic reservoirs with electrochemical potential $\mu_L$ and $\mu_R$ ( $\mu_L>\mu_R$) (Fig. 2).

Figure 2: Sketch of the considered one-dimensional system.

We suppose that only the first one-dimensional subband is occupied, that the electrons in the channel does not suffer any scattering mechanism, i.e. transport in the channel is ballistic, and that the electrons entering the reservoirs contacts are instantaneously in equilibrium with them. For the moment, let the temperature be equal to zero (T=0 K) and the contacts reflectionless, that means that the transmission probability from the contact to contact is unitary. Being a two-dimensional confined system, the dispersion relation is equal to (32), where $\epsilon_i$ is also referred as the cut-off energy of $i$-th 1D subband or transversal mode. The positive current, carried by $k>0$ states in the $i$-subband, reads [4]

I^{>}_i=\frac{q}{L}\sum_{k>0}v_i(k) f(E-\mu_L)
\end{displaymath} (41)

where $v_i$ is the velocity and $f$ is the Fermi-Dirac distribution function. From ([*]) we obtain
\end{displaymath} (42)

and from (4), we can express (43) as
I^{>}_i=\frac{2q}{h}\int_{\epsilon_i}^\infty f(E-\mu_L) dE
\end{displaymath} (43)

where $h$ is the Planck constant. Equivalently, the negative current reads
I^{<}_i=\frac{2q}{h}\int_{\epsilon_i}^\infty f(E-\mu_R) dE
\end{displaymath} (44)

If we now define the function $M(E)$ as
M(E)=\sum_i u(E-\epsilon_i),
\end{displaymath} (45)

where $u(E)$ is the Heavyside function, we can express the total positive current as
$\displaystyle I^>$ $\textstyle =$ $\displaystyle \sum_i I_i^>=\sum_i\frac{2q}{h}\int_{\epsilon_i}
  $\textstyle =$ $\displaystyle \frac{2q}{h}\int_{-\infty}^{+\infty}f(E-\mu_L)M(E)dE$ (46)

Working at $T= 0 K$, the Fermi-Dirac function is a step function, and considering $M(E)$ constant and equal to an integer $M$ between the energy range $[\mu_R,\mu_L]$, the total current reads

I=I^>-I^<=\frac{2 q^2}{h}M\frac{\mu_L-\mu_R}{q}
\end{displaymath} (47)

We note that (49) can be reduced to the form

\end{displaymath} (48)

where $V$ is the applied potential ( $\frac{\mu_L-\mu_R}{q}$) and $G$ is the conductance, that in this case is
G=\frac{2 q^2}{h}M
\end{displaymath} (49)

Since the channel is purely ballistic, the resistance of the channel is equal to zero. This means that all the dissipation of the energy happens in the reservoirs, since a charge propagating from the left transverse the channel conserving the energy, and when it reaches the right reservoir, it looses energy thermalizing in the contact. That's why the inverse of $G$ is often referred as the contact resistance.

If we now relax the hypothesis of reflectionless contacts, we have to consider in (48) $T(E)$, the probability that an electron originated by one electrode reaches the other reservoir propagating through the channel.

Figure 3: Electrons propagating from the left contact undergo elastic scattering and $T$ is the probability of reaching the other reservoir.

If $I_L^>$ is the influx of electrons from the left reservoir, $I_L^<$ is the flux of back-scattered electrons in the left reservoir, and $I_R^>$ is the flux of electrons that has reached the right reservoir (Fig. 3), if we consider $T(E)$ constant and equal to $\tau$ between the energy range $[\mu_R,\mu_L]$, we deal with the following relations

$\displaystyle I_L^>$ $\textstyle =$ $\displaystyle \frac{2 q}{h}M (\mu_L-\mu_R)$ (50)
$\displaystyle I_L^<$ $\textstyle =$ $\displaystyle \frac{2 q}{h}M(1-\tau) (\mu_L-\mu_R)$ (51)
$\displaystyle I_R^>$ $\textstyle =$ $\displaystyle \frac{2 q}{h}M \tau (\mu_L-\mu_R)$ (52)

The total current $I$ is then equal to
I=I_L^>-I_L^<=I_R^>=\frac{2 q^2}{h}M\tau\frac{\mu_L-\mu_R}{q}
\end{displaymath} (53)

and the conductance in case of reflecting contacts can be expressed as
G=\frac{2 q^2}{h}M\tau
\end{displaymath} (54)

that is the Landauer Formula.

We can now study the case in which the temperature is not equal to zero. The total current now reads

f(E-\mu_L)-f(E-\mu_R)\right] M(E)T(E)dE
\end{displaymath} (55)

If $\mu_R=\mu$ and $\mu_L=\mu+\delta\mu$, with $\delta\mu\ll \mu$, (57) becomes

f(E-(\mu+\delta\mu))-f(E-\mu)\right] M(E)T(E)dE
\end{displaymath} (56)

and by means of the Taylor expansion

f(E-(\mu+\delta\mu))-f(E-\mu)=-\delta\mu\frac{\partial f(E,\mu)}{\partial E}
\end{displaymath} (57)

we can express (58) as

I=\frac{2q^2}{h}\left [\int_{-\infty}^{+\infty}-\frac{\partial f(E,\mu)}{\partial E} M(E)T(E)dE \right] \frac{\delta \mu}{q}
\end{displaymath} (58)

that can be reduced to the form expressed in (50), if we define

G(\mu)=\frac{2q^2}{h}\int_{-\infty}^{+\infty}-\frac{\partial f(E,\mu)}{\partial E} M(E)T(E)dE
\end{displaymath} (59)

where $-\frac{\partial f(E,\mu)}{\partial E}$ is often referred as the broadening function.

As a numerical example, we can consider the case of a QPC with reflectionless contacts in which the transversal potential is a harmonic potential [5]. The transversal modes are equal to

E_{tn}=\left(n-\frac{1}{2} \right) \hbar \omega_0.
\end{displaymath} (60)

In Fig. 4 we show the conductance at different temperatures, as a function of the electrochemical potential of the reservoir ($\mu$), supposing a small voltage $\delta \mu$ is applied between the two reservoir, and $\hbar\omega_0=0.1 eV$. As can be noted, as the temperature is increased, the step-like function of the conductance at $T= 0 K$ is smoothed. Such an effect is due to the fact that in (61), the broadening function ``broadens'' over the energy as the temperatures increases (Fig. 5) : Equation (61) can be then seen as an average over a window that increases as the temperature is increased.

Figure: Conductance of a QPC with reflectionless contact at different temperatures, in which the transversal potential has been approximated with a harmonic potential ( $\hbar\omega_0=0.1 eV$).

Figure 5: Broadening function computed at different temperatures. At T=0 K the broadening function is a delta-Dirac function with area equal to one.

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Next: Single Electron Box Up: Quantum point contact and Previous: Quantum point contact and
Fiori Gianluca 2005-11-11