# Landauer formula in 1D

1. Apr 3, 2009

### carbon9

Hi all,

I have studied the Landauer's formula from the book "Current at the Nanoscale", but a formula made me confused. In the general case, for a nanoscopic transport, the book gives the formula:

http://img19.imageshack.us/img19/2986/33768834.jpg [Broken]

In the following paragragraph it is told:

"For 1-D case, (current density has no meaning in 1D, so we replace J with I)"

and then it gives

http://img19.imageshack.us/img19/8793/23920600.jpg [Broken]

From these equations I understand that he used

D(E)=1/(2$$\pi$$)

for 1D system (I may be wrong!). But in the previous chapters, he gives another function for D(E) in 1D:

http://img19.imageshack.us/img19/734/75994918.jpg [Broken]

So, what is the point that I'm missing here?

Cheers

Last edited by a moderator: May 4, 2017
2. Apr 6, 2009

### sokrates

The book is being a little bit sloppy.

First off, he didn't use

D(E) = 1/ 2pi --- which makes no sense at all as you figured out.

The correct formula for density of states in a 1D conductor is given by the formula you secondly gave.

The point you are missing (or the author is failing to describe) is :

The current in a 1D conductor where there's only "one mode" can be written as:

$$I = \frac{2e}{h}\int^{ul}_{ur} T ( f^+(E)-f^-(E) )dE$$

and there's no "density of states" term here because it is implicitly included in the "number of modes" term.

A more rigorous way of writing this could've been:

$$I = \frac{2e}{h}\int_{ur}^{ul} T(E) M(E) f^+(E)-f^-(E) dE$$
where

$$M(E) = v_x D(E) / 2L$$

But I guess the reason the book skips these 'details' is that Landauer gave his formula in the following form:

$$I = \frac{2e}{h} M T \int^{ul}_{ur} f^+(E)-f^-(E) dE$$

assuming Tranmission and number of modes are constant within the energy range you are biasing the device.
This becomes the book's formula for M=1.

3. Apr 10, 2009

### carbon9

Thank you very much Sokrates.

Cheers