1
$\begingroup$

The coupled mode space NEGF method for computing transistor characteristics involves expanding the electronic wavefunction in a mode space basis

$$\Psi(x,y,z) = \sum_n\phi_n(x)\xi_n(y,z;x)$$

where $\xi_n(y,z;x)$ is the nth mode at position x (the x dimension being the direction of transport).

The method involves computing coupling between modes along the transport direction. E.g. coupling coefficients like

$$b_{mn}(x) = \langle\xi_m|\frac{\partial}{\partial x}|\xi_n\rangle$$

The numerical algorithm will divide the transistor into $X$ slices $\Delta x$ apart, and compute the modes on each slice. The issue is my numerical solver will sometimes compute $\xi_n$ as the nth eigenvalue, and sometimes compute $-\xi_n$, so phases might not be consistent across each slice. This changing phase will result in a large $\frac{\partial}{\partial x}|\xi_n\rangle$

My question is: Do the phases need to be consistent across each slice? If they do, how do NEGF algorithms ensure they are consistent?

$\endgroup$
  • $\begingroup$ The phases do not matter here, in the calculation of the first derivative they are naturally factored out. Moreover, think about using $\Psi(x,y,z) = \sum_n\phi_n(x)\xi_n(y,z)$ as your ansatz -- note the missing $x$ in $\xi$. That makes thing a lot easier. $\endgroup$ – davidhigh Jan 15 '18 at 13:44
  • $\begingroup$ Hmm, I include the $x$ in order to track how the modes change with $x$. E.g. $\frac{\partial}{\partial x}\xi_n(y,z;x) = \frac{\xi_n(y,z;x+\Delta x)-\xi_n(y,z;x)}{\Delta x}$, which is only 0 if the modes are the same at each $x$. $\endgroup$ – DJames Jan 15 '18 at 18:24
  • $\begingroup$ The "modes" are eigenfunctions of your Hamiltonian, I suppose? If so, you can make the ansatz $\Psi(x,y,z) = \sum_n c_n \xi_n(y,z;x)$. i.e. an expansion in your modes. I don't get, however, why you should have two $x$'es. That complicates everything. Can you please explain? $\endgroup$ – davidhigh Jan 15 '18 at 22:04
  • $\begingroup$ You can find a detailed description of the method in this paper: arxiv.org/ftp/cond-mat/papers/0403/0403739.pdf . In this case, the modes aren't eigenfunctions of the full Hamiltonian. Instead they would be eigenfunctions of the 2D Hamiltonian defined at each position $x$ along the device $\endgroup$ – DJames Jan 16 '18 at 13:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.