Dear kwant users: I am reading the tutorial `sec2.6` and encounter the disscussion about the particle symmetry.
(Note you can put the following text into the editor which supports markdown for better format, like jupyter.) >From the given code, it seems that we have $S_{ee}=S_{hh}^*$ and >$S_{eh}=-S_{he}^*$: ``` def check_PHS(syst): # Scattering matrix s = kwant.smatrix(syst, energy=0) # Electron to electron block s_ee = s.submatrix((0,0), (0,0)) # Hole to hole block s_hh = s.submatrix((0,1), (0,1)) print('s_ee: \n', np.round(s_ee, 3)) print('s_hh: \n', np.round(s_hh[::-1, ::-1], 3)) print('s_ee - s_hh^*: \n', np.round(s_ee - s_hh[::-1, ::-1].conj(), 3), '\n') # Electron to hole block s_he = s.submatrix((0,1), (0,0)) # Hole to electron block s_eh = s.submatrix((0,0), (0,1)) print('s_he: \n', np.round(s_he, 3)) print('s_eh: \n', np.round(s_eh[::-1, ::-1], 3)) print('s_he + s_eh^*: \n', np.round(s_he + s_eh[::-1, ::-1].conj(), 3)) ``` However, from my naive reasoning I would expect that $S_{ee}=S_{hh}$ and $S_{he}=-S_{eh}$. Here is my reasoning, $$ H \begin{pmatrix}\psi_e \\ \psi_h \end{pmatrix}=E\begin{pmatrix}\psi_e \\ \psi_h \end{pmatrix} \Rightarrow \begin{pmatrix}\psi_e^{(out)} \\ \psi_h^{(out)} \end{pmatrix} = S(E) \begin{pmatrix}\psi_e^{(in)} \\ \psi_h^{(in)} \end{pmatrix} $$ With partilce symmetry $\sigma_y H \sigma_y = -H$, we have: $$ H \sigma_y \begin{pmatrix}\psi_e \\ \psi_h \end{pmatrix}=-E \sigma_y \begin{pmatrix}\psi_e \\ \psi_h \end{pmatrix} \Rightarrow \sigma_y\begin{pmatrix}\psi_e^{(out)} \\ \psi_h^{(out)} \end{pmatrix} = S(-E) \sigma\begin{pmatrix}\psi_e^{(in)} \\ \psi_h^{(in)} \end{pmatrix} $$ which leads to $$ S(E) = \sigma_y S(-E) \sigma_y, $$ giving $S_{ee}(E)=S_{hh}(-E)$ and $S_{he}(E)=-S_{eh}(-E)$. Best regards, Wilson