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