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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?

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  • $\begingroup$ 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

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