= Solution
For $E>0$, put $k=\sqrt{2mE}$ and $q=k/\hbar>0$. Outside the support of the <potential energy>, a <generalized eigenfunction> of the <Hamiltonian operator> has the <plane wave> form
$$
\psi(x)=\begin{cases}A e^{iqx}+B e^{-iqx},&x<-R_0,\\C e^{iqx}+D e^{-iqx},&x>R_0.\end{cases}
$$
Here $R_0$ denotes the support radius, to distinguish it from the <quantum reflection probability> $R$. The <momenta> are $\pm k$ and the <velocities> are $\pm k/m$. Thus $A,D$ describe incoming <scattering channels> from the left and right, and $B,C$ describe outgoing <scattering channels> towards the left and right. These <energy eigenfunctions> are not <square-integrable functions>; normalizable incoming <wave packets> are formed by superposition over <energy>, and their passage through the interaction region gives the physical <quantum scattering> experiment.
For incidence from the left, normalize $A=1,D=0$ and write $B=r_L,C=t_L$. The <quantum reflection amplitude> is $r_L$ and the <transmission amplitude> is $t_L$. The ratios of outgoing to incident <probability currents> define
$$
\boxed{R=|r_L|^2,\qquad T=|t_L|^2.}
$$
No extra <velocity> ratio occurs because both exterior regions have the same <potential energy> and <mass>. The <scattering matrix>, with a fixed choice of channel order, is
$$
\boxed{\begin{pmatrix}B\\C\end{pmatrix}=S(q)\begin{pmatrix}A\\D\end{pmatrix},\qquad S(q)=\begin{pmatrix}r_L&t_R\\t_L&r_R\end{pmatrix}.}
$$
Its first and second columns are the responses to unit incidence from the left and right, respectively. For a reflection-symmetric <potential energy>, these conventions give $r_L=r_R=r$ and $t_L=t_R=t$.
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