[Kwant] particle-hole symmetry in the tutorial sec2.6

[wilson2048]

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