A semispectral approach for the efficient calculation of scattering matrices in quasi-1D quantum systems and transmission coefficients for the Landauer formula

Abstract

This paper proposes the semispectral method for the calculation of the scattering matrices for the charge carriers in quasi-one-dimensional quantum systems described by the Schrodinger equation. An efficient and accurate calculation method is achieved by adding of reference solutions at selected energy points to the eigenfunction expansion of the Green’s function. A numerical simulation of the quantum wire with the square cross-section in transverse electric field was performed by different methods. The example problem of a quantum wire in a transverse electric field is used to compare the semispectral method with alternative approaches. We find that the semispectral method reliably converges and is significantly faster than the direct solution while the eigenfunction expansion approach has convergence issues. These results allow us to propose the semispectral method as a universal and efficient approach to calculation of the transmission coefficients for the Landauer formula as well as other scattering-based entities.

Appendix

To obtain formula (13), that links the DN-map operator and the scattering matrix, consider formula (5) with the new notations

$$\begin{aligned} a_n^1&= a_n^I e^{i\kappa _n X_I };\;\;\;b_n^1 =b_n^I e^{-i\kappa _n X_I }; \nonumber \\ a_n^3&= a_n^{III} e^{i\kappa _n X_{III} };\;\;\;b_n^3 =b_n^{III} e^{-i\kappa _n X_{III} }. \nonumber \\ \left( {\begin{array}{l} b_n^1 \\ a_n^3 \\ \end{array}} \right)&= \sum _m {\left[ {\left( {{\begin{array}{ll} {S_{n,m}^{I,I} }&{} {S_{n,m}^{I,III} } \\ {S_{n,m}^{III,I} }&{} {S_{n,m}^{III,III} } \\ \end{array} }} \right) \left( {\begin{array}{l} a_m^1 \\ b_m^3 \\ \end{array}} \right) } \right] } \end{aligned}$$

(32)

And consider two representations for the normal derivative of the wavefunction on the interfaces $\Gamma _{I}$ and $\Gamma _{III}$. The first one is based on (5)

$$\begin{aligned} \frac{\partial \psi _n^I }{\partial n}&= -i\kappa _n a_n^1 +i\kappa _n b_n^1 \nonumber \\ \frac{\partial \psi _n^{III} }{\partial n}&= i\kappa _n a_n^3 -i\kappa _n b_n^3 \end{aligned}$$

(33)

The second is based on the definition of the DN-map operator (9)

$$\begin{aligned} \frac{\partial \psi _n^I }{\partial n}&= \sum _m {DN_{n,m}^{I,I} (a_m^1 +b_m^1 )} +\sum _m {DN_{n,m}^{I,III} (a_m^3 +b_m^3 )} \nonumber \\ \frac{\partial \psi _n^{III} }{\partial n}\!&= \!\sum _m {DN_{n,m}^{III,I} (a_m^1 +b_m^1 )} \!+\!\sum _m {DN_{n,m}^{III,III} (a_m^3 \!+\!b_m^3 )} \nonumber \\ \end{aligned}$$

(34)

Here $\psi _n^r$ is a projection of the wavefunction $\psi ^{r}$ onto the n’s channel.

Since the wavefunction and its first derivative should be continuous, both expressions (33) and (34) are equal.

$$\begin{aligned} -i\kappa _n a_n^1 +i\kappa _n b_n^1&= \sum _m {DN_{n,m}^{I,I} (a_m^1 +b_m^1 )}\nonumber \\&+\sum _m {DN_{n,m}^{I,III} (a_m^3 +b_m^3 )} \nonumber \\ i\kappa _n a_n^3 -i\kappa _n b_n^3&= \sum _m {DN_{n,m}^{III,I} (a_m^1 +b_m^1 )}\nonumber \\&+\sum _m {DN_{n,m}^{III,III} (a_m^3 +b_m^3 )} \end{aligned}$$

(35)

After some reordering and switch to the operator form (33) turns into

$$\begin{aligned}&(DN^{I,I}+iK^{I,I})a^{1}+(DN^{I,I}-iK^{I,I})b^{1}\nonumber \\&\quad +\,\,DN^{I,III}a^{3}+DN^{I,III}b^{3}=0\nonumber \\&DN^{III,I}a^{1}+DN^{III,I}b^{1}+(DN^{III,III}-iK^{III,III})a^{3}\nonumber \\&\quad +\,\,(DN^{III,III}+iK^{III,III})b^{3}=0 \end{aligned}$$

(36)

These two expressions can be combined into one, using matrix notation

$$\begin{aligned}&\left( {{\begin{array}{ll} {iK^{I,I}-DN^{I,I}}&{} {-DN^{I,III}} \\ {-DN^{III,I}}&{} {iK^{III,III}-DN^{III,III}} \\ \end{array} }} \right) \left( {\begin{array}{l} b^{1} \\ a^{3} \\ \end{array}} \right) \nonumber \\&\quad =\left( {{\begin{array}{ll} {iK^{I,I}+DN^{I,I}}&{} {DN^{I,III}} \\ {DN^{III,I}}&{} {iK^{III,III}+DN^{III,III}} \\ \end{array} }} \right) \left( {\begin{array}{l} a^{1} \\ b^{3} \\ \end{array}} \right) \nonumber \\ \end{aligned}$$

(37)

The elements of the operator K with differing upper indices are zero, so K$^{\mathrm{I,III}}$ and K$^{\mathrm{III,I}}$ are the zero operators. This fact allows us to note (37) in the following form

$$\begin{aligned} \left( {iK-DN} \right) \left( {\begin{array}{l} b^{1} \\ a^{3} \\ \end{array}} \right) =\left( {iK+DN} \right) \left( {\begin{array}{l} a^{1} \\ b^{3} \\ \end{array}} \right) \end{aligned}$$

(38)

And finally,

$$\begin{aligned} \left( {\begin{array}{l} b^{1} \\ a^{3} \\ \end{array}} \right)&= \left( {iK-DN} \right) ^{-1}\left( {iK+DN} \right) \left( {\begin{array}{l} a^{1} \\ b^{3} \\ \end{array}} \right) \end{aligned}$$

(39)

$$\begin{aligned} S&= (iK-DN)^{-1}(iK+DN) \end{aligned}$$

(40)