💼 Case: Consider a spinless particle of charge $q=e$ which is incident from the left upon a one-dimensional electrostatic potential defined by
$$ e\Phi(x)=V(x) =\left \{ \begin{matrix} 0 & \text{for} & x<0 \\ V_0 & \text{for} & x>0 \end{matrix} \right . \qquad \text{with} \quad V_0 >0 $$
The general expression previously derived the Klein-Gordon is
$$ [(\hat{\vec p} -\vec V )^2 c^2 +(mc^2 +S)^2 ]\psi (\vec r,t) = (i\hbar \partial _t - V_0 )^2 \psi ( \vec r,t ) $$
From there we set $S=0$ and $q\vec A = \vec V=0$ as per out condition with $V_0=e\Phi(x)=V(x)$
$$ \left ( - \hbar ^2 c^2 \frac{\partial}{\partial x^2 } + m^2 c^ 4 \right ) \psi ( x,t)=\left ( i\hbar \frac{\partial}{\partial t} -V(x) \right )^2 \psi (x,t) $$
where $p_x=-i\hbar \partial _x$
Separating the time time-dependent part by attempting solutions of the form
$$ \psi(x,y)=\phi(x)e^{-iE_pt/\hbar} $$
where $E_p>0$ we can then sub in to our equation to get the stationary equation
$$ \left ( (E_p-V(x))^2 + \hbar^2 c^2 \frac{\partial ^2}{\partial x^2} -m^2 c^4 \right )\phi(x)=0 $$
1️⃣ For the first region $x<0$
$$ \phi_\text{FR}(x)=\phi_i(x) +\phi_r(x) $$
with $\phi_i(x) = Ae^{ipx/\hbar}$ and $\phi_r (x) = B e^{-ipx/\hbar}$ being the incident and reflected components
We can then substitute into into our Klein-Gordon equation with $V(x)=0$ to get
$$ p^2 = \frac{E_p^2 -m^2 c^4}{c^2} $$
as expected
2️⃣ For the second region $x>0$
$$ \phi_\text{SR}(x)=\phi_t(x) = Ce^{ip'x/\hbar} $$
Again substituting into the Klein-Gordon equation with $V(x)=V_0$ gives us
$$ p'^2 = \frac{(E_p-V_0)^2-m^2 c^4}{c^2} $$
🟰 Combining
From the continuity of the wavefunction $\phi (x)$ and its spatial derivative $\partial _x \phi(x)$ at $x=0$
$$ B=\frac{p-p'}{p+p'}A \quad \text{and}\quad C= \frac{2p}{p+p'}A $$
Thus the conserved current is given by
$$ j(x)=-i\hbar c \left [ \psi ^\frac{\partial \psi}{\partial x}-\psi \frac{\partial \psi^}{\partial x} \right ] $$
from which we obtain the incident current and the reflected current
$$ j_i = 2c|A|^2 p \quad j_r=-2c|B|^2 p=-2c|A|^2 p \left | \frac{p-p'}{p+p'} \right |^2 $$
Giving a reflection probability of
$$ R=-\frac{j_r}{j_i}=\left | \frac{p-p'}{p+p'} \right |^2 $$
💎 Conclusion:
If $E_p>V_0 + mc^2$ then $p'$ is real with transmitted current j$_t=2c|C|^2 p'$ which satisfies $j_t=j_i+j_r$ and transmission probability
$$ T=\frac{j_t}{j_i} =\frac{4pp'}{(p+p')^2}=1-R $$
as expected from non-relativistic physics
If $V_0-mc^2 <E_p<V_0+mc^2$ then $p'$ is imaginary. In this case $p'=i|p'|$ to give $\phi_t(x)=Ce^{-|p'|x/\hbar}$ and we have exponential decay in the second region with $j_t=0$
If $V_0>E_P+mc^2$ then $p'$ is real. This is weird since it means the kinetic energy is lower than the barrier yet the wave can go through it. This is the Klein paradox. We can calculate the group velocity in region two to be
$$ \frac{\partial E_p}{\partial p'}=-\frac{c^2 p'}{V_0 - E_p} $$
For it to travel to the right then $p'<0$ which means $R>1$ and $T<0$ with $R+T=1$ still holding, so a larger negative current inside the barrier.
Calculating the density inside the barrier we find
$$ \rho=2(E_0-V_0)|C|^2 $$
which is negative! The way to interpret this is that the beam is either completely reflected or particle-antiparticle pairs appear with the anti-particles going through