Completeness of scattering states of the Dirac Hamiltonian with a step potential

${\bf{A}}{\bf{1}}:\ {\bf{for}}\ {\boldsymbol{E}}\gt {{\boldsymbol{V}}}_{{\bf{0}}}+{\boldsymbol{m}}$ ($E={E}_{p}={E}_{q}+{V}_{0}$ and therefore $p=\sqrt{{E}^{2}-{m}^{2}},q=\sqrt{{(E-{V}_{0})}^{2}-{m}^{2}}$), the left-incident wave function reads as

$$\begin{eqnarray}&&{\psi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{E}}[\theta (-z)\{u(p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}+{Ru}(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}\}+\theta (z){Tu}(q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 12 }$$

with the reflection and transmission coefficients

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}-\sqrt{\tfrac{E-{V}_{0}-m}{E-m}}}{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}+\sqrt{\tfrac{E-{V}_{0}-m}{E-m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}+\sqrt{\tfrac{E-{V}_{0}-m}{E-m}}}\end{eqnarray} \tag{ 13 }$$

determined by the continuity condition at z = 0. Here and in the followings, the dependence on the spin s is and will be suppressed in the wave functions. The continuity of the current defined as ${j}_{z}=\bar{\psi }{\gamma }^{3}\psi ={\psi }^{\dagger }{\alpha }_{z}\psi$ at z = 0 reads as

$$\begin{eqnarray}&&\displaystyle \frac{p}{m}-| R{| }^{2}\displaystyle \frac{p}{m}=| T{| }^{2}\displaystyle \frac{q}{m},\end{eqnarray} \tag{ 14 }$$

which expresses the conservation of the probability

$$\begin{eqnarray}&&{P}_{r}=| R{| }^{2},\quad {P}_{t}=\displaystyle \frac{q}{p}| T{| }^{2},\quad {P}_{r}+{P}_{t}=1.\end{eqnarray} \tag{ 15 }$$

${\bf{A}}{\bf{2}}:\,{\bf{for}}\,{{\boldsymbol{V}}}_{{\bf{0}}}+{\boldsymbol{m}}\gt {\boldsymbol{E}}\gt {{\boldsymbol{V}}}_{{\bf{0}}}-{\boldsymbol{m}}(\gt {\boldsymbol{m}})$, no oscillating solution exists for the region $z\gt 0$ and we have ($q={\rm{i}}\kappa ^{\prime} ={\rm{i}}\sqrt{{m}^{2}-{(E-{V}_{0})}^{2}}$)

$$\begin{eqnarray}&&{\psi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{E}}[\theta (-z)\{u(p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}+{Ru}(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}\}+\theta (z){Tu}({\rm{i}}\kappa ^{\prime} ,s){{\rm{e}}}^{-\kappa ^{\prime} z}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 16 }$$

for $E\gt {V}_{0}$ and

$$\begin{eqnarray}&&{\psi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{E}}[\theta (-z)\{u(p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}+{Ru}(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}\}+\theta (z){T}{\rm{i}}{\sigma }_{z}v(-{\rm{i}}\kappa ^{\prime} ,s){{\rm{e}}}^{-\kappa ^{\prime} z}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 17 }$$

for $E\lt {V}_{0}$, with the reflection and transmission coefficients

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}-{\rm{i}}\sqrt{\tfrac{m-(E-{V}_{0})}{E-m}}}{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}+{\rm{i}}\sqrt{\tfrac{m-(E-{V}_{0})}{E-m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{E-{V}_{0}+m}{E+m}}+{\rm{i}}\sqrt{\tfrac{m-(E-{V}_{0})}{E-m}}},\end{eqnarray} \tag{ 18 }$$

resulting in a total reflection

$$\begin{eqnarray}&&{P}_{r}=| R{| }^{2}=1,\quad {P}_{t}=0,\end{eqnarray} \tag{ 19 }$$

for the current in the region $z\gt 0$ vanishes identically.

${\bf{A}}{\bf{3}}:\ {\bf{for}}\ {{\boldsymbol{V}}}_{{\bf{0}}}-{\boldsymbol{m}}\gt {\boldsymbol{E}}\gt {\boldsymbol{m}}$, the solution in the region $z\gt 0$ is expressed in terms of v ($E={E}_{p}=-{E}_{q}+{V}_{0}$ or $q=\sqrt{{({V}_{0}-E)}^{2}-{m}^{2}}$)

$$\begin{eqnarray}&&{\psi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{E}}[\theta (-z)\{u(p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}+{Ru}(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}\}+\theta (z)T{\sigma }_{z}v(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 20 }$$

with

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{{V}_{0}-E-m}{E+m}}-\sqrt{\tfrac{{V}_{0}-E+m}{E-m}}}{\sqrt{\tfrac{{V}_{0}-E-m}{E+m}}+\sqrt{\tfrac{{V}_{0}-E+m}{E-m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{{V}_{0}-E-m}{E+m}}+\sqrt{\tfrac{{V}_{0}-E+m}{E-m}}}.\end{eqnarray} \tag{ 21 }$$

The continuity of current implies the conservation of probability

$$\begin{eqnarray}&&\displaystyle \frac{p}{m}-| R{| }^{2}\displaystyle \frac{p}{m}=| T{| }^{2}\displaystyle \frac{q}{m}\quad \to \quad {P}_{r}+{P}_{t}=| R{| }^{2}+\displaystyle \frac{q}{p}| T{| }^{2}=1.\end{eqnarray} \tag{ 22 }$$

Recall that the ordinary conservation law of probability is satisfied, but a finite and non-vanishing transmission probability survives even at the infinite-potential limit ${V}_{0}\to \infty$

$$\begin{eqnarray}&&{P}_{t}=\displaystyle \frac{q}{p}| T{| }^{2}\longrightarrow \displaystyle \frac{2\sqrt{{E}^{2}-{m}^{2}}}{E+\sqrt{{E}^{2}-{m}^{2}}}\ne 0.\end{eqnarray} \tag{ 23 }$$

This phenomenon is known as the Klein tunneling [7]. Notice that no left-incident particle is allowed for $m\gt E\gt -m$.

${\bf{A}}{\bf{4}}:\ {\bf{for}}\ {\boldsymbol{-m}}\gt {\boldsymbol{E}}$, the eigenstates are given by the negative-frequency solutions ($E=-| E| =-{E}_{p}=-{E}_{q}+{V}_{0}$)

$$\begin{eqnarray}&&{\psi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{| E| }}[\theta (-z)\{v(p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}+{Rv}(-p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}\}+\theta (z){Tv}(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}]{{\rm{e}}}^{{\rm{i}}| E| t},\end{eqnarray} \tag{ 24 }$$

with

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{| E| +{V}_{0}+m}{| E| +m}}-\sqrt{\tfrac{| E| +{V}_{0}-m}{| E| -m}}}{\sqrt{\tfrac{| E| +{V}_{0}+m}{| E| +m}}+\sqrt{\tfrac{| E| +{V}_{0}-m}{| E| -m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{| E| +{V}_{0}+m}{| E| +m}}+\sqrt{\tfrac{| E| +{V}_{0}-m}{| E| -m}}}.\end{eqnarray} \tag{ 25 }$$

The current continuity and the conservation of probability are the same as in (22).

${\bf{B}}{\bf{1}}:\ {\bf{for}}\ {\boldsymbol{E}}\gt {{\boldsymbol{V}}}_{{\bf{0}}}+{\boldsymbol{m}}$, $E={E}_{q}+{V}_{0}={E}_{p}$ (or $p=\sqrt{{E}^{2}-{m}^{2}},q=\sqrt{{(E-{V}_{0})}^{2}-{m}^{2}}$) and the right-incident wave function is given by

$$\begin{eqnarray}&&{\phi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{E-{V}_{0}}}[\theta (-z){Tu}(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}+\theta (z)\{u(-q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}+{Ru}(q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}\}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 26 }$$

with

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{E+m}{E-{V}_{0}+m}}-\sqrt{\tfrac{E-m}{E-{V}_{0}-m}}}{\sqrt{\tfrac{E+m}{E-{V}_{0}+m}}+\sqrt{\tfrac{E-m}{E-{V}_{0}-m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{E+m}{E-{V}_{0}+m}}+\sqrt{\tfrac{E-m}{E-{V}_{0}-m}}}.\end{eqnarray} \tag{ 27 }$$

The continuity of current and the conservation of probability read as

$$\begin{eqnarray}&&-| T{| }^{2}\displaystyle \frac{p}{m}=-\displaystyle \frac{q}{m}+| R{| }^{2}\displaystyle \frac{q}{m},\quad {P}_{r}+{P}_{t}=| R{| }^{2}+\displaystyle \frac{p}{q}| T{| }^{2}=1.\end{eqnarray} \tag{ 28 }$$

Observe that no right-incident particle is allowed for ${V}_{0}+m\gt E\gt {V}_{0}-m$.

${\bf{B}}{\bf{2}}:\ {\bf{for}}\ {{\boldsymbol{V}}}_{{\bf{0}}}-{\boldsymbol{m}}\gt {\boldsymbol{E}}\gt {\boldsymbol{m}}$ ($E=-{E}_{q}+{V}_{0}={E}_{p}$ or $q=\sqrt{{({V}_{0}-E)}^{2}-{m}^{2}}$), the solution may be written as

$$\begin{eqnarray}&&{\phi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{{V}_{0}-E}}[\theta (-z)T{\sigma }_{z}u(-p,s){{\rm{e}}}^{-{\rm{i}}{p}{z}}+\theta (z)\{v(-q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}+{Rv}(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}\}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 29 }$$

with

$$\begin{eqnarray}&&R=-\displaystyle \frac{\sqrt{\tfrac{E+m}{{V}_{0}-E-m}}-\sqrt{\tfrac{E-m}{{V}_{0}-E+m}}}{\sqrt{\tfrac{E+m}{{V}_{0}-E-m}}+\sqrt{\tfrac{E-m}{{V}_{0}-E+m}}},\quad T=-\displaystyle \frac{2}{\sqrt{\tfrac{E+m}{{V}_{0}-E-m}}+\sqrt{\tfrac{E-m}{{V}_{0}-E+m}}}.\end{eqnarray} \tag{ 30 }$$

The continuity of current and the conservation of probability are the same as in (28).

${\bf{B}}{\bf{3}}:\ {\bf{for}}{\boldsymbol{m}}\gt {\boldsymbol{E}}\gt -{\boldsymbol{m}}$, no oscillating solution exists for the region $z\lt 0$ and we have ($p={\rm{i}}\kappa ={\rm{i}}\sqrt{{m}^{2}-{E}^{2}}$)

$$\begin{eqnarray}&&{\phi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{{V}_{0}-E}}[\theta (-z)T{\sigma }_{z}u(-{\rm{i}}\kappa ,s){{\rm{e}}}^{\kappa z}+\theta (z)\{v(-q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}+{Rv}(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}\}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 31 }$$

for $E\gt 0$ and

$$\begin{eqnarray}&&{\phi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{{V}_{0}-E}}[-{\rm{i}}\theta (-z){Tv}({\rm{i}}\kappa ,s){{\rm{e}}}^{\kappa z}+\theta (z)\{v(-q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}+{Rv}(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}\}]{{\rm{e}}}^{-{\rm{i}}{E}{t}},\end{eqnarray} \tag{ 32 }$$

for $E\lt 0$, with

$$\begin{eqnarray}&&R=-\displaystyle \frac{\sqrt{\tfrac{m+E}{{V}_{0}-E-m}}-{\rm{i}}\sqrt{\tfrac{m-E}{{V}_{0}-E+m}}}{\sqrt{\tfrac{m+E}{{V}_{0}-E-m}}+{\rm{i}}\sqrt{\tfrac{m-E}{{V}_{0}-E+m}}},\quad T=-\displaystyle \frac{2}{\sqrt{\tfrac{m+E}{{V}_{0}-E-m}}+{\rm{i}}\sqrt{\tfrac{m-E}{{V}_{0}-E+m}}}.\end{eqnarray} \tag{ 33 }$$

In this case, all particles incident from the right are reflected

$$\begin{eqnarray}&&{P}_{t}=0,\quad {P}_{r}=| R{| }^{2}=1.\end{eqnarray} \tag{ 34 }$$

${\bf{B}}{\bf{4}}:\ {\bf{for}}\ {\boldsymbol{-m}}\gt {\boldsymbol{E}}$, $E=-{E}_{p}=-{E}_{q}+{V}_{0}=-| E|$, and we have (with positive p and q)

$$\begin{eqnarray}&&{\phi }^{(E)}(z,t)=\displaystyle \frac{1}{\sqrt{2\pi }}\sqrt{\displaystyle \frac{m}{{V}_{0}-E}}[\theta (-z){Tv}(-p,s){{\rm{e}}}^{{\rm{i}}{p}{z}}+\theta (z)\{v(-q,s){{\rm{e}}}^{{\rm{i}}{q}{z}}+{Rv}(q,s){{\rm{e}}}^{-{\rm{i}}{q}{z}}\}]{{\rm{e}}}^{{\rm{i}}| E| t},\end{eqnarray} \tag{ 35 }$$

with

$$\begin{eqnarray}&&R=\displaystyle \frac{\sqrt{\tfrac{| E| +m}{| E| +{V}_{0}+m}}-\sqrt{\tfrac{| E| -m}{| E| +{V}_{0}-m}}}{\sqrt{\tfrac{| E| +m}{| E| +{V}_{0}+m}}+\sqrt{\tfrac{| E| -m}{| E| +{V}_{0}-m}}},\quad T=\displaystyle \frac{2}{\sqrt{\tfrac{| E| +m}{| E| +{V}_{0}+m}}+\sqrt{\tfrac{| E| -m}{| E| +{V}_{0}-m}}}.\end{eqnarray} \tag{ 36 }$$

In this case, the same current continuity relation and probability conservation as in (28) follow.

As described in Introduction, though it is generally taken for granted that eigenfunctions of a Hamiltonian constitute a complete orthonormal set, which can be shown explicitly when it has only a finite number of discrete eigenstates, to prove it for a Hamiltonian endowed with a continuous spectrum can be another, nontrivial task. In the present case, we need to show that the left-hand side of (40) is diagonal in both the coordinate and spinor spaces, which would make the proof more involved than the non-relativistic cases [4].

Since the step potential takes different values depending on the sign of the coordinate z, we need to consider three cases, that is, $z,z^{\prime} \lt 0$, $0\lt z,z^{\prime}$ and $z\lt 0\lt z^{\prime}$, separately to prove (40).

${\bf{Case}}\,{\bf{1}}:\ {\boldsymbol{z}},{\boldsymbol{z}}^{\prime} \lt {\bf{0}}$

When both z and $z^{\prime}$ are negative, i.e., on the left of the potential, the left-hand side of the completeness relation (40) is explicitly written down as

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}}+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}]{[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}^{\prime} }+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[v(p){{\rm{e}}}^{-{\rm{i}}{p}{z}}+{R}_{p}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}}]{[v(p){{\rm{e}}}^{-{\rm{i}}{p}{z}^{\prime} }+{R}_{p}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}{[{T}_{q}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}{\sigma }_{z}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}{[{T}_{q}{\sigma }_{z}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}^{\sqrt{{V}_{0}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}{\sigma }_{z}u(-{\rm{i}}\kappa ){{\rm{e}}}^{\kappa z}{[{T}_{q}{\sigma }_{z}u(-{\rm{i}}\kappa ){{\rm{e}}}^{\kappa z^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{V}_{0}^{2}-{m}^{2}}}^{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}v({\rm{i}}\kappa ){{\rm{e}}}^{\kappa z}{[{T}_{q}v({\rm{i}}\kappa ){{\rm{e}}}^{\kappa z^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}}{[{T}_{q}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}^{\prime} }]}^{\dagger },\end{eqnarray} \tag{ 41 }$$

where it is understood that the summation over spin s has to be taken (though not explicitly written down) and the subscripts p and q are used to distinguish quantities associated with the left-incident (p) case and the right-incident (q) case (with the same energy). It is to be noticed that the reflection amplitude for the left-incident case R_p can be expressed as

$$\begin{eqnarray}&&{R}_{p}=\displaystyle \frac{\sqrt{\tfrac{{E}_{q}+m}{{E}_{p}+m}}-\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}}{\sqrt{\tfrac{{E}_{q}+m}{{E}_{p}+m}}+\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}}\end{eqnarray} \tag{ 42 }$$

for all $p\geqslant 0$ by a proper analytical continuation from a large $p\geqslant \sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}$ (or ${E}_{p}\geqslant {V}_{0}+m$). Similarly the transmission amplitude for the right-incident case T_q can be understood as a proper analytic continuation of

$$\begin{eqnarray}&&{T}_{q}=\displaystyle \frac{2}{\sqrt{\tfrac{{E}_{p}+m}{{E}_{q}+m}}+\sqrt{\tfrac{{E}_{p}-m}{{E}_{q}-m}}}.\end{eqnarray} \tag{ 43 }$$

After the change of variables from q to p (or κ when ${V}_{0}-m\geqslant {E}_{q}\geqslant {V}_{0}+m$), the above expression is simplified as

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\{(u(p)u{(p)}^{\dagger }+| {R}_{p}{| }^{2}v(-p)v{(-p)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{p}(z-z^{\prime} )}+(v(p)v{(p)}^{\dagger }+| {R}_{p}{| }^{2}u(-p)u{(-p)}^{\dagger }){{\rm{e}}}^{-{\rm{i}}{p}(z-z^{\prime} )}\\ &&\quad +\,({R}_{p}^{* }u(p)u{(-p)}^{\dagger }+{R}_{p}v(-p)v{(p)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{p}(z+z^{\prime} )}+({R}_{p}u(-p)u{(p)}^{\dagger }+{R}_{p}^{* }v(p)v{(-p)}^{\dagger }){{\rm{e}}}^{-{\rm{i}}{p}(z+z^{\prime} )}\}\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\displaystyle \frac{p}{q}| {T}_{q}{| }^{2}u(-p)u{(-p)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}(z-z^{\prime} )}\\ &&\quad +\,{\displaystyle \int }_{0}^{m}\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{m}{{E}_{(\kappa )}}\left(\displaystyle \frac{\kappa }{q}| {T}_{q}{| }^{2}u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }+\displaystyle \frac{\kappa }{q}| {T}_{q}{| }^{2}v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }\right){{\rm{e}}}^{\kappa (z+z^{\prime} )}\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\displaystyle \frac{p}{q}| {T}_{q}{| }^{2}v(-p)v{(-p)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}(z-z^{\prime} )},\end{eqnarray} \tag{ 44 }$$

where ${E}_{(\kappa )}={E}_{\pm }{\rm{i}}\kappa =\sqrt{{m}^{2}-{\kappa }^{2}}$ and the fact that ${\sigma }_{z}$ is commutable with both spinors u and v has been used. We have to pay due attention to the fact that, even though not explicitly written, the values of q (and therefore those of E_q) are different when it appears in association with the spinor u or v, which, of course, applies also to R_p and T_q in the above. (This is the reason why apparently the same quantities have not been put together in the last but one line.) Observe that there are terms of functions of the difference $z-z^{\prime}$ and the sum $z+z^{\prime}$ and they are apparently independent of each other.

First, all terms that are functions of $z-z^{\prime}$ are collected to yield

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\left\{\left(u(p)u{(p)}^{\dagger }+| {R}_{p}{| }^{2}v(-p)v{(-p)}^{\dagger }+\displaystyle \frac{p}{q}| {T}_{q}{| }^{2}v(-p)v{(-p)}^{\dagger }\right){{\rm{e}}}^{{\rm{i}}{p}(z-z^{\prime} )}\right.\\ &&\quad +\,\left.\left(v(p)v{(p)}^{\dagger }+| {R}_{p}{| }^{2}u(-p)u{(-p)}^{\dagger }+\displaystyle \frac{p}{q}| {T}_{q}{| }^{2}u(-p)u{(-p)}^{\dagger }\right){{\rm{e}}}^{-{\rm{i}}{p}(z-z^{\prime} )}\right\}.\end{eqnarray} \tag{ 45 }$$

We note that the transmission probability for the right-incident case $\tfrac{p}{q}| {T}_{q}{| }^{2}$ is the same as that for the left-incident case $\tfrac{q}{p}| {T}_{p}{| }^{2}$, which can be explicitly shown by direct calculation, stating that the reciprocal relation in quantum mechanics also holds in this case. Finally, the conservation of probability $| {R}_{p}{| }^{2}+\tfrac{q}{p}| {T}_{p}{| }^{2}=1$ greatly simplifies the above to reach

$$\begin{eqnarray}&&{\displaystyle \int }_{-\infty }^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}(u(p)u{(p)}^{\dagger }+v(-p)v{(-p)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{p}(z-z^{\prime} )}\\ &&\quad =\,{\displaystyle \int }_{-\infty }^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\left(\displaystyle \frac{{E}_{p}{\gamma }^{0}-p{\gamma }^{3}+m}{2m}{\gamma }^{0}+\displaystyle \frac{{E}_{p}{\gamma }^{0}+p{\gamma }^{3}-m}{2m}{\gamma }^{0}\right){{\rm{e}}}^{{\rm{i}}{p}(z-z^{\prime} )}\\ &&\quad =\,\delta (z-z^{\prime} ),\end{eqnarray} \tag{ 46 }$$

where at the first equality the spin sum has explicitly been taken and the unit matrix ${{\mathbb{1}}}_{4\times 4}$ is suppressed on the right-most hand.

Second, the remaining terms that are functions of $z+z^{\prime}$ are shown to cancel to each other. We understand that the contributions coming from the p-integral

$$\begin{eqnarray}&&{\int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\{({R}_{p}^{* }u(p)u{(-p)}^{\dagger }+{R}_{p}v(-p)v{(p)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{p}(z+z^{\prime} )}+({R}_{p}u(-p)u{(p)}^{\dagger }+{R}_{p}^{* }v(p)v{(-p)}^{\dagger }){{\rm{e}}}^{-{\rm{i}}{p}(z+z^{\prime} )}\}\end{eqnarray} \tag{ 47 }$$

can be evaluated on the complex p-plane, because the combination $z+z^{\prime}$ in the exponents is negative definite, the reflection amplitude decays at least like $1/| p|$ for large $| p|$, assuring the convergence of the integrands at $| p| \to \infty$ and the integrands have no singularities other than the several cuts on the real and imaginary p-axis. The integrand proportional to ${{\rm{e}}}^{{\rm{i}}{p}(z+z^{\prime} )}$ can be evaluated on the negative imaginary axis $p=-{\rm{i}}\kappa$, while the other one proportional to ${{\rm{e}}}^{-{\rm{i}}{p}(z+z^{\prime} )}$ is to be evaluated on the positive imaginary axis $p={\rm{i}}\kappa$, where $\kappa \geqslant 0$ in both cases. On the negative imaginary p-axis, we set $p=-{\rm{i}}\kappa$ and

$$\begin{eqnarray}&&u(p)u{(-p)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}(z+z^{\prime} )}=u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }{{\rm{e}}}^{\kappa (z+z^{\prime} )},\quad v(-p)v{(p)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}(z+z^{\prime} )}=v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }{{\rm{e}}}^{\kappa (z+z^{\prime} )},\end{eqnarray} \tag{ 48 }$$

while on the positive imaginary p-axis $p={\rm{i}}\kappa$, we have

$$\begin{eqnarray}&&u(-p)u{(p)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}(z+z^{\prime} )}=u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }{{\rm{e}}}^{\kappa (z+z^{\prime} )},\quad v(p)v{(-p)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}(z+z^{\prime} )}=v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }{{\rm{e}}}^{\kappa (z+z^{\prime} )}.\end{eqnarray} \tag{ 49 }$$

For small $\kappa \leqslant m$, E_p is a real number smaller than m, $0\leqslant {E}_{p}\leqslant m$, and the reflection amplitude associated with spinor u is analytically continued as

$$\begin{eqnarray}&&{R}_{p}^{* }\to {\left.\displaystyle \frac{\sqrt{\tfrac{{E}_{p}-{V}_{0}+m}{{E}_{p}+m}}-\sqrt{\tfrac{{E}_{p}-{V}_{0}-m}{{E}_{p}-m}}}{\sqrt{\tfrac{{E}_{p}-{V}_{0}+m}{{E}_{p}+m}}+\sqrt{\tfrac{{E}_{p}-{V}_{0}-m}{{E}_{p}-m}}}\right|}_{p=-{\rm{i}}\kappa }{\left.=\displaystyle \frac{-{\rm{i}}\sqrt{\tfrac{{V}_{0}-m-E}{E+m}}-\sqrt{\tfrac{{V}_{0}+m-E}{m-E}}}{-{\rm{i}}\sqrt{\tfrac{{V}_{0}-m-E}{E+m}}+\sqrt{\tfrac{{V}_{0}+m-E}{m-E}}}\right|}_{E={E}_{-{\rm{i}}\kappa }}\\ &&\quad {R}_{p}\to {\left.\displaystyle \frac{\sqrt{\tfrac{{E}_{p}-{V}_{0}+m}{{E}_{p}+m}}-\sqrt{\tfrac{{E}_{p}-{V}_{0}-m}{{E}_{p}-m}}}{\sqrt{\tfrac{{E}_{p}-{V}_{0}+m}{{E}_{p}+m}}+\sqrt{\tfrac{{E}_{p}-{V}_{0}-m}{{E}_{p}-m}}}\right|}_{p={\rm{i}}\kappa }{\left.=\displaystyle \frac{{\rm{i}}\sqrt{\tfrac{{V}_{0}-m-E}{E+m}}-\sqrt{\tfrac{{V}_{0}+m-E}{m-E}}}{{\rm{i}}\sqrt{\tfrac{{V}_{0}-m-E}{E+m}}+\sqrt{\tfrac{{V}_{0}+m-E}{m-E}}}\right|}_{E={E}_{{\rm{i}}\kappa }},\end{eqnarray} \tag{ 50 }$$

because the purely imaginary quantity $\sqrt{{E}_{p}-{V}_{0}+m}$ has the same phase as that of p on the imaginary axis, for

$$\begin{eqnarray}&&\sqrt{{E}_{p}-{V}_{0}+m}=\displaystyle \frac{\sqrt{{E}_{p}^{2}-{({V}_{0}-m)}^{2}}}{\sqrt{{E}_{p}+{V}_{0}-m}}=\displaystyle \frac{\sqrt{p-{p}_{0}}\sqrt{p+{p}_{0}}}{\sqrt{{E}_{p}+{V}_{0}-m}},\end{eqnarray} \tag{ 51 }$$

where ${p}_{0}=\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}$ is a real number and located on the real axis. Similarly, for the reflection amplitude associated with spinor v, we obtain

$$\begin{eqnarray}&&{\left.{R}_{p}\to \displaystyle \frac{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}-\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{{E}_{p}-m}}}{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}+\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{{E}_{p}-m}}}\right|}_{p=-{\rm{i}}\kappa }{\left.=\displaystyle \frac{\sqrt{\tfrac{E+{V}_{0}+m}{E+m}}-{\rm{i}}\sqrt{\tfrac{E+{V}_{0}-m}{m-E}}}{\sqrt{\tfrac{E+{V}_{0}+m}{E+m}}+{\rm{i}}\sqrt{\tfrac{E+{V}_{0}-m}{m-E}}}\right|}_{E={E}_{-{\rm{i}}\kappa }}\\ &&\quad {\left.{R}_{p}^{* }\to \displaystyle \frac{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}-\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{{E}_{p}-m}}}{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}+\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{{E}_{p}-m}}}\right|}_{p={\rm{i}}\kappa }{\left.=\displaystyle \frac{\sqrt{\tfrac{E+{V}_{0}+m}{E+m}}+{\rm{i}}\sqrt{\tfrac{E+{V}_{0}-m}{m-E}}}{\sqrt{\tfrac{E+{V}_{0}+m}{E+m}}-{\rm{i}}\sqrt{\tfrac{E+{V}_{0}-m}{m-E}}}\right|}_{E={E}_{{\rm{i}}\kappa }}.\end{eqnarray} \tag{ 52 }$$

All contributions coming from small κ on the imaginary axis are put together to yield

$$\begin{eqnarray}&&{\int }_{0}^{m}\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{m}{{E}_{(\kappa )}}\left\{\left(\displaystyle \frac{\kappa }{q}| {T}_{q}{| }^{2}+{\rm{i}}({R}_{p}-{R}_{p}^{* })\right)u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }+\left(\displaystyle \frac{\kappa }{q}| {T}_{q}{| }^{2}-{\rm{i}}({R}_{p}-{R}_{p}^{* })\right)v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }\right\}{{\rm{e}}}^{\kappa (z+z^{\prime} )}.\end{eqnarray} \tag{ 53 }$$

The coefficient of spinor u vanishes, because

$$\begin{eqnarray}&&\displaystyle \frac{\kappa }{q}{\left|-\displaystyle \frac{2}{\sqrt{\tfrac{m+E}{{V}_{0}-E-m}}+{\rm{i}}\sqrt{\tfrac{m-E}{{V}_{0}-E+m}}}\right|}_{E={E}_{(\kappa )}}^{2}-{\left.2{\mathfrak{I}}\displaystyle \frac{{\rm{i}}\sqrt{\tfrac{{V}_{0}-E-m}{m+E}}-\sqrt{\tfrac{{V}_{0}-E+m}{m-E}}}{{\rm{i}}\sqrt{\tfrac{{V}_{0}-E-m}{m+E}}+\sqrt{\tfrac{{V}_{0}-E+m}{m-E}}}\right|}_{E={E}_{(\kappa )}}\\ &&\quad =\,\displaystyle \frac{4\kappa }{q}{\left(\displaystyle \frac{1}{\tfrac{m+E}{{V}_{0}-E-m}+\tfrac{m-E}{{V}_{0}-E+m}}-\displaystyle \frac{1}{\tfrac{{\kappa }^{2}}{{q}^{2}}(\tfrac{{V}_{0}-E-m}{m+E}+\tfrac{{V}_{0}-E+m}{m-E})}\right)}_{E={E}_{(\kappa )}}=0,\end{eqnarray} \tag{ 54 }$$

where we have used the relations ${\kappa }^{2}={m}^{2}-{E}^{2}$ and ${q}^{2}={({V}_{0}-E)}^{2}-{m}^{2}$ with $E={E}_{(\kappa )}$. Similarly, the coefficient of spinor v, since in this case ${q}^{2}={({V}_{0}+E)}^{2}-{m}^{2}$ with $E={E}_{(\kappa )}$, also vanishes

$$\begin{eqnarray}&&\displaystyle \frac{\kappa }{q}{\left|-\displaystyle \frac{2}{\sqrt{\tfrac{m-E}{{V}_{0}+E-m}}+{\rm{i}}\sqrt{\tfrac{m+E}{{V}_{0}+E+m}}}\right|}_{E={E}_{(\kappa )}}^{2}+{\left.2{\mathfrak{I}}\displaystyle \frac{\sqrt{\tfrac{{V}_{0}+E+m}{m+E}}-{\rm{i}}\sqrt{\tfrac{{V}_{0}+E-m}{m-E}}}{\sqrt{\tfrac{{V}_{0}+E+m}{m+E}}+{\rm{i}}\sqrt{\tfrac{{V}_{0}+E-m}{m-E}}}\right|}_{E={E}_{(\kappa )}}\\ &&\quad =\,\displaystyle \frac{4\kappa }{q}{\left(\displaystyle \frac{1}{\tfrac{m-E}{{V}_{0}+E-m}+\tfrac{m+E}{{V}_{0}+E+m}}-\displaystyle \frac{1}{\tfrac{{\kappa }^{2}}{{q}^{2}}(\tfrac{{V}_{0}+E+m}{m+E}+\tfrac{{V}_{0}+E-m}{m-E})}\right)}_{E={E}_{(\kappa )}}=0.\end{eqnarray} \tag{ 55 }$$

Thus what is remained is the contributions arising from large $\kappa \geqslant m$, where E_p becomes purely imaginary. First we note that the function E_p gains the same phase as that of p on the imaginary axis, i.e., ${E}_{\pm }{\rm{i}}\kappa =\pm {\rm{i}}\sqrt{{\kappa }^{2}-{m}^{2}}\equiv \pm {\rm{i}}{{ \mathcal E }}_{\kappa }$ for $\kappa \geqslant m$. From the direct calculations, we understand that the reflection amplitude for spinor u is replaced by the following complex numbers on the imaginary axis

$$\begin{eqnarray}&&{R}_{p}\to \displaystyle \frac{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}-\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}+\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}={R}_{u}({\rm{i}}{{ \mathcal E }}_{\kappa }),\quad {R}_{p}^{* }\to \displaystyle \frac{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}-\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}+\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}={R}_{u}(-{\rm{i}}{{ \mathcal E }}_{\kappa }),\end{eqnarray} \tag{ 56 }$$

while for spinor v,

$$\begin{eqnarray}&&{R}_{p}\to \displaystyle \frac{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}-\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}+\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}={R}_{v}(-{\rm{i}}{{ \mathcal E }}_{\kappa }),\quad {R}_{p}^{* }\to \displaystyle \frac{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}-\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}+\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}={R}_{v}({\rm{i}}{{ \mathcal E }}_{\kappa }).\end{eqnarray} \tag{ 57 }$$

Observe that the following equalities hold

$$\begin{eqnarray}&&{R}_{v}({\rm{i}}{{ \mathcal E }}_{\kappa })=-{R}_{u}(-{\rm{i}}{{ \mathcal E }}_{\kappa }),\quad {R}_{v}(-{\rm{i}}{{ \mathcal E }}_{\kappa })=-{R}_{u}({\rm{i}}{{ \mathcal E }}_{\kappa }).\end{eqnarray} \tag{ 58 }$$

The spinor parts for u and v are explicitly worked out, after summing over spin degrees freedom, to be

$$\begin{eqnarray}u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger } & = & \displaystyle \frac{\sqrt{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}\sqrt{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}{2m}\left(\begin{array}{c}1\\ \tfrac{-{\rm{i}}{\sigma }_{z}\kappa }{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}\end{array}\right)\left(1,\displaystyle \frac{{\rm{i}}{\sigma }_{z}\kappa }{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}\right)\\ & = & \displaystyle \frac{1}{2m}\left(\begin{array}{lc}\kappa & {\rm{i}}{\sigma }_{z}(-{\rm{i}}{{ \mathcal E }}_{\kappa }+m)\\ -{\rm{i}}{\sigma }_{z}({\rm{i}}{{ \mathcal E }}_{\kappa }+m) & \kappa \end{array}\right),\\ v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger } & = & \displaystyle \frac{\sqrt{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}\sqrt{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}{2m}\left(\begin{array}{c}\tfrac{{\rm{i}}{\sigma }_{z}\kappa }{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}\\ 1\end{array}\right)\left(\displaystyle \frac{-{\rm{i}}{\sigma }_{z}\kappa }{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m},1\right)\\ & = & \displaystyle \frac{1}{2m}\left(\begin{array}{lc}\kappa & {\rm{i}}{\sigma }_{z}(-{\rm{i}}{{ \mathcal E }}_{\kappa }+m)\\ -{\rm{i}}{\sigma }_{z}({\rm{i}}{{ \mathcal E }}_{\kappa }+m) & \kappa \end{array}\right),\end{eqnarray} \tag{ 59 }$$

where the relation $-{{ \mathcal E }}_{\kappa }^{2}={m}^{2}-{\kappa }^{2}$ has been used, showing explicitly that they are actually the same. The remaining integrations over κ are now written as

$$\begin{eqnarray}&&-{\rm{i}}{\displaystyle \int }_{m}^{\infty }\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }}[{R}_{u}(-{\rm{i}}{{ \mathcal E }}_{\kappa })u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }+{R}_{v}(-{\rm{i}}{{ \mathcal E }}_{\kappa })v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }]{{\rm{e}}}^{\kappa (z+z^{\prime} )}\\ &&\quad +\,{\rm{i}}{\displaystyle \int }_{m}^{\infty }\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{m}{{\rm{i}}{{ \mathcal E }}_{\kappa }}[{R}_{u}({\rm{i}}{{ \mathcal E }}_{\kappa })u(-{\rm{i}}\kappa )u{(-{\rm{i}}\kappa )}^{\dagger }+{R}_{v}({\rm{i}}{{ \mathcal E }}_{\kappa })v({\rm{i}}\kappa )v{({\rm{i}}\kappa )}^{\dagger }]{{\rm{e}}}^{\kappa (z+z^{\prime} )},\end{eqnarray} \tag{ 60 }$$

which vanishes owing to the above relations (58) and (59). This completes the proof of the completeness relation (40) for $z,z^{\prime} \lt 0$.

${\bf{Case}}\,{\bf{2}}:\,{\boldsymbol{z}},{\boldsymbol{z}}^{\prime} \gt {\bf{0}}$

As in the previous case, the left-hand side of (40) is explicitly written down

$$\begin{eqnarray}&&{\displaystyle \int }_{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}}{[{T}_{p}u(q){{\rm{e}}}^{{\rm{i}}{p}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{V}_{0}^{2}-{m}^{2}}}^{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}u({\rm{i}}\kappa ^{\prime} ){{\rm{e}}}^{-\kappa ^{\prime} z}{[{T}_{p}u({\rm{i}}\kappa ^{\prime} ){{\rm{e}}}^{-\kappa ^{\prime} z^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}^{\sqrt{{V}_{0}^{2}-{m}^{2}}}\displaystyle \frac{m}{{E}_{p}}{T}_{p}{\rm{i}}{\sigma }_{z}v(-{\rm{i}}\kappa ^{\prime} ){{\rm{e}}}^{-\kappa ^{\prime} z}{[{T}_{p}{\rm{i}}{\sigma }_{z}v(-{\rm{i}}\kappa ^{\prime} ){{\rm{e}}}^{-\kappa ^{\prime} z^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}{\sigma }_{z}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}}{[{T}_{p}{\sigma }_{z}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}}{[{T}_{p}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}[u(-q){{\rm{e}}}^{-{\rm{i}}{q}{z}}+{R}_{q}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}}]{[u(-q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }+{R}_{q}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}[v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}}+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}}]{[v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger },\end{eqnarray} \tag{ 61 }$$

where $\kappa ^{\prime} \geqslant 0$ is defined, when ${E}_{q}\leqslant m$, as ${E}_{q}^{2}={m}^{2}-\kappa {{\prime} }^{2}$. Change of variables $p\to q$ (or $\kappa ^{\prime}$) reduces this to

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}\left\{\left(\left(\displaystyle \frac{q}{p}| {T}_{p}{| }^{2}+| {R}_{q}{| }^{2}\right)u(q)u{(q)}^{\dagger }+v(-q)v{(-q)}^{\dagger }\right){{\rm{e}}}^{{\rm{i}}{q}(z-z^{\prime} )}\right.\\ &&\quad +\,\left(\left(\displaystyle \frac{q}{p}| {T}_{p}{| }^{2}+| {R}_{q}{| }^{2}\right)v(q)v{(q)}^{\dagger }+u(-q)u{(-q)}^{\dagger }\right){{\rm{e}}}^{-{\rm{i}}{q}(z-z^{\prime} )}\\ &&\quad +\,({R}_{q}^{* }u(-q)u{(q)}^{\dagger }+{R}_{q}v(q)v{(-q)}^{\dagger }){{\rm{e}}}^{-{\rm{i}}{q}(z+z^{\prime} )}+({R}_{q}u(q)u{(-q)}^{\dagger }+{R}_{q}^{* }v(-q)v{(q)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{q}(z+z^{\prime} )}\phantom{\rule[-0000]{}{4.0ex}}\}\\ &&\quad +\,{\displaystyle \int }_{0}^{m}\displaystyle \frac{{\rm{d}}\kappa ^{\prime} }{2\pi }\displaystyle \frac{m}{{E}_{p}}\left(\displaystyle \frac{\kappa ^{\prime} }{p}| {T}_{p}{| }^{2}u({\rm{i}}\kappa ^{\prime} )u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }+\displaystyle \frac{\kappa ^{\prime} }{p}| {T}_{p}{| }^{2}v(-{\rm{i}}\kappa ^{\prime} )v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }\right){{\rm{e}}}^{-\kappa ^{\prime} (z+z^{\prime} )}.\end{eqnarray} \tag{ 62 }$$

The reciprocal relation $\tfrac{q}{p}| {T}_{p}{| }^{2}=\tfrac{p}{q}| {T}_{q}{| }^{2}$ also holds true in this case, which can be proven by direct calculation, and the probability conservation $| {R}_{q}{| }^{2}+\tfrac{p}{q}| {T}_{q}{| }^{2}=1$ brings about the delta function $\delta (z-z^{\prime} )$ from those terms that depend on the difference $z-z^{\prime}$ in the above (62). The remaining integrations over q are evaluated on the imaginary q-axis, where spinors become $u({\rm{i}}\kappa ^{\prime} )u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }$ and $v(-{\rm{i}}\kappa ^{\prime} )v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }$ and the reflection amplitude R_q for spinor u is represented as

$$\begin{eqnarray}&&{R}_{q}=\displaystyle \frac{\sqrt{\tfrac{{E}_{q}+{V}_{0}+m}{{E}_{q}+m}}-\sqrt{\tfrac{{E}_{q}+{V}_{0}-m}{{E}_{q}-m}}}{\sqrt{\tfrac{{E}_{q}+{V}_{0}+m}{{E}_{q}+m}}+\sqrt{\tfrac{{E}_{q}+{V}_{0}-m}{{E}_{q}-m}}}\to \displaystyle \frac{\sqrt{\tfrac{{E}_{q}+{V}_{0}+m}{{E}_{q}+m}}+{\rm{i}}\sqrt{\tfrac{{E}_{q}+{V}_{0}-m}{m-{E}_{q}}}}{\sqrt{\tfrac{{E}_{q}+{V}_{0}+m}{{E}_{q}+m}}-{\rm{i}}\sqrt{\tfrac{{E}_{q}+{V}_{0}-m}{m-{E}_{q}}}}\quad {\rm{for}}\ \kappa ^{\prime} \leqslant m,{E}_{q}=\sqrt{{m}^{2}-\kappa {{\prime} }^{2}}\\ &&\quad \to \displaystyle \frac{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +m}}-\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} -m}}}{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +m}}+\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} +{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} -m}}}={R}_{u}({\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }})\quad {\rm{for}}\ \kappa ^{\prime} \geqslant m,{{ \mathcal E }}_{{\kappa }^{{\prime} }}=\sqrt{\kappa {{\prime} }^{2}-{m}^{2}},\end{eqnarray} \tag{ 63 }$$

and for v,

$$\begin{eqnarray}&&{R}_{q}=\displaystyle \frac{\sqrt{\tfrac{{E}_{q}-{V}_{0}+m}{{E}_{q}+m}}-\sqrt{\tfrac{{E}_{q}-{V}_{0}-m}{{E}_{q}-m}}}{\sqrt{\tfrac{{E}_{q}-{V}_{0}+m}{{E}_{q}+m}}+\sqrt{\tfrac{{E}_{q}-{V}_{0}-m}{{E}_{q}-m}}}\to \displaystyle \frac{-{\rm{i}}\sqrt{\tfrac{{V}_{0}-{E}_{q}-m}{{E}_{q}+m}}-\sqrt{\tfrac{{V}_{0}-{E}_{q}+m}{m-{E}_{q}}}}{-{\rm{i}}\sqrt{\tfrac{{V}_{0}-{E}_{q}-m}{{E}_{q}+m}}+\sqrt{\tfrac{{V}_{0}-{E}_{q}+m}{m-{E}_{q}}}}\quad {\rm{for}}\ \kappa ^{\prime} \leqslant m,{E}_{q}=\sqrt{{m}^{2}-\kappa {{\prime} }^{2}}\\ &&\quad \to \displaystyle \frac{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}+m}}-\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-m}}}{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}+m}}+\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }}-m}}}={R}_{v}(-{\rm{i}}{{ \mathcal E }}_{{\kappa }^{{\prime} }})\quad {\rm{for}}\ \kappa ^{\prime} \geqslant m,{{ \mathcal E }}_{{\kappa }^{{\prime} }}=\sqrt{\kappa {{\prime} }^{2}-{m}^{2}}\end{eqnarray} \tag{ 64 }$$

and ${R}_{q}^{* }$'s as the complex conjugates of the corresponding ones. Thus the contributions of terms that are functions of $z+z^{\prime}$ are

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{m}\displaystyle \frac{{\rm{d}}\kappa ^{\prime} }{2\pi }\displaystyle \frac{m}{{E}_{q}}\left\{\left(\displaystyle \frac{\kappa ^{\prime} }{p}| {T}_{p}{| }^{2}+{\rm{i}}({R}_{q}-{R}_{q}^{* })\right)u({\rm{i}}\kappa ^{\prime} )u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }\right.\\ &&\quad +\,\left.\left(\displaystyle \frac{\kappa ^{\prime} }{p}| {T}_{p}{| }^{2}-{\rm{i}}({R}_{q}-{R}_{q}^{* })\right)v(-{\rm{i}}\kappa ^{\prime} )v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }\right\}{{\rm{e}}}^{-\kappa ^{\prime} (z+z^{\prime} )}\\ &&\quad -{\rm{i}}{\displaystyle \int }_{m}^{\infty }\displaystyle \frac{{\rm{d}}\kappa ^{\prime} }{2\pi }\displaystyle \frac{m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} }[{R}_{u}(-{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} )u({\rm{i}}\kappa ^{\prime} )u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }+{R}_{v}(-{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} )v(-{\rm{i}}\kappa ^{\prime} )v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }]{{\rm{e}}}^{-\kappa ^{\prime} (z+z^{\prime} )}\\ &&\quad +\,{\rm{i}}{\displaystyle \int }_{m}^{\infty }\displaystyle \frac{{\rm{d}}\kappa ^{\prime} }{2\pi }\displaystyle \frac{m}{{\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} }[{R}_{u}({\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} )u({\rm{i}}\kappa ^{\prime} )u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }+{R}_{v}({\rm{i}}{{ \mathcal E }}_{\kappa }^{\prime} )v(-{\rm{i}}\kappa ^{\prime} )v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }]{{\rm{e}}}^{-\kappa ^{\prime} (z+z^{\prime} )},\end{eqnarray} \tag{ 65 }$$

which can be shown to vanish on the basis of the similar arguments as before. This completes the proof for $z,z^{\prime} \gt 0$.

${\bf{Case}}\,{\bf{3}}:\ {\boldsymbol{z}}\lt {\bf{0}}\lt {\boldsymbol{z}}^{\prime}$

If the left-hand side of (40) is explicitly written down in this case, there are eight (actually ten) different terms existing, corresponding to eight different cases ${\bf{A}}{\bf{1}}\mbox{--}{\bf{A}}{\bf{4}}$ and ${\bf{B}}{\bf{1}}\mbox{--}{\bf{B}}{\bf{4}}$. It can be shown, however, that these terms are categorized into four groups, according to their energy eigenvalues, (i) $E\geqslant m$, (ii) $m\gt E\gt 0$, (iii) $0\gt E\gt -m$ and (iv) $-m\geqslant E$.

(i) $E\geqslant m$: Consider first the contributions arising from those wave functions that are belonging to eigenvalues (energies) greater than ${V}_{0}+m$. They are expressed as

$$\begin{eqnarray}&&{\displaystyle \int }_{E={E}_{p}\geqslant {V}_{0}+m}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}}+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}]{[{T}_{p}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}{[u{(-q)}^{-{\rm{i}}{q}{z}^{\prime} }+{R}_{q}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad =\,{\displaystyle \int }_{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\phantom{\rule[-0000]{}{4.0ex}}\{{T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }+{R}_{p}{T}_{p}^{* }u(-p)u{(q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }\\ &&\quad +\,\left.\displaystyle \frac{p}{q}({T}_{q}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }+{T}_{q}{R}_{q}^{* }u(-p)u{(q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} })\right\}.\end{eqnarray} \tag{ 66 }$$

Observe that in this energy range the reflection and transmission amplitudes are all real-valued and, moreover, the following relations hold

$$\begin{eqnarray}&&\displaystyle \frac{p}{q}{T}_{q}=\displaystyle \frac{2}{\left(\tfrac{q}{p}\right)\left(\sqrt{\tfrac{{E}_{p}+m}{{E}_{q}+m}}+\sqrt{\tfrac{{E}_{p}-m}{{E}_{q}-m}}\right)}=\displaystyle \frac{2}{\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}+\sqrt{\tfrac{{E}_{q}+m}{{E}_{p}+m}}}={T}_{p}\end{eqnarray} \tag{ 67 }$$

and

$$\begin{eqnarray}&&\displaystyle \frac{p}{q}{R}_{q}^{* }{T}_{q}=\displaystyle \frac{2\left(\tfrac{q}{p}\right)\left(\sqrt{\tfrac{{E}_{p}+m}{{E}_{q}+m}}-\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}\right)}{{\left(\tfrac{q}{p}\right)}^{2}{\left(\sqrt{\tfrac{{E}_{p}+m}{{E}_{q}+m}}+\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}\right)}^{2}}=\displaystyle \frac{2\left(\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}-\sqrt{\tfrac{{E}_{q}+m}{{E}_{p}+m}}\right)}{{\left(\sqrt{\tfrac{{E}_{q}-m}{{E}_{p}-m}}+\sqrt{\tfrac{{E}_{q}+m}{{E}_{p}+m}}\right)}^{2}}=-{R}_{p}{T}_{p}^{* }.\end{eqnarray} \tag{ 68 }$$

Thus the above expression (66) is simplified to

$$\begin{eqnarray}&&{\int }_{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}({T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }+{T}_{p}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }).\end{eqnarray} \tag{ 69 }$$

As a matter of fact, it can be confirmed that the contributions coming from the wave functions belonging to energy eigenvalues greater than or equal to m, $E\geqslant m$, are consistently written in this form with the lower limit replaced with p = 0. When $\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}\gt p\geqslant \sqrt{{V}_{0}^{2}-{m}^{2}}$, in order to make the integrand convergent at $| p| \to \infty$, the first term in (69) is defined by

$$\begin{eqnarray}&&{{T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }| }_{q\to -{\rm{i}}\kappa ^{\prime} }={T}_{p}^{* }u(p)u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-\kappa ^{\prime} z^{\prime} },\end{eqnarray} \tag{ 70 }$$

while the second term is defined as

$$\begin{eqnarray}&&{{T}_{p}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }| }_{q\to {\rm{i}}\kappa ^{\prime} }={T}_{p}u(-p)u{({\rm{i}}\kappa ^{\prime} )}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}-\kappa ^{\prime} z^{\prime} }.\end{eqnarray} \tag{ 71 }$$

Notice that here on the left-hand side, the variable q is to be replaced with that specified at vertical bar $-{\rm{i}}\kappa ^{\prime}$ or ${\rm{i}}\kappa ^{\prime}$ in the quantities which are considered as functions of q, while on the right-hand side, conjugate operations are to be taken for functions of specified variables. For example, $u(q){| }_{q\to {\rm{i}}\kappa ^{\prime} }=u({\rm{i}}\kappa ^{\prime} )$ and $u{(q)}^{\dagger }{| }_{q\to {\rm{i}}\kappa ^{\prime} }=u{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }$. Since, in the second term of (69), we analytically continue $q\to {\rm{i}}\kappa ^{\prime}$, the term $\sqrt{{E}_{q}-m}=\sqrt{{E}_{p}-{V}_{0}-m}$ is replaced with ${\rm{i}}\sqrt{{V}_{0}-{E}_{p}+m}$ in the amplitudes T_p, which results in the relation ${T}_{p}={R}_{p}{T}_{p}^{* }$. (This relation also holds when ${V}_{0}\gt E\geqslant {V}_{0}-m$.) The sum of these two terms is just the contribution from the energy range ${V}_{0}\leqslant E\lt {V}_{0}+m$.

In the energy range ${V}_{0}-m\leqslant E\lt {V}_{0}$, we note that $\kappa ^{\prime} =\sqrt{({E}_{p}-{V}_{0}+m)(m-{E}_{p}+{V}_{0})}$ and

$$\begin{eqnarray}u{({\rm{i}}\kappa ^{\prime} )}^{\dagger } & = & \sqrt{\displaystyle \frac{{E}_{p}-{V}_{0}+m}{2m}}\left(1,\displaystyle \frac{-{\rm{i}}{\sigma }_{z}\kappa ^{\prime} }{{E}_{p}-{V}_{0}+m}\right)=\displaystyle \frac{1}{\sqrt{2m}}(\sqrt{{E}_{p}-{V}_{0}+m},-{\rm{i}}{\sigma }_{z}\sqrt{m-{E}_{p}+{V}_{0}})\\ & = & -{\rm{i}}\displaystyle \frac{1}{\sqrt{2m}}({\rm{i}}{\sigma }_{z}\sqrt{{E}_{p}-{V}_{0}+m},\sqrt{m-{E}_{p}+{V}_{0}}){\sigma }_{z}\\ & = & \sqrt{\displaystyle \frac{{V}_{0}-{E}_{p}+m}{2m}}\left(\displaystyle \frac{{\rm{i}}{\sigma }_{z}\kappa ^{\prime} }{{V}_{0}-{E}_{p}+m},1\right){\sigma }_{z}=-{\rm{i}}{v}{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }{\sigma }_{z}.\end{eqnarray} \tag{ 72 }$$

Therefore, the integrand in the parentheses in (69) becomes nothing but

$$\begin{eqnarray}&&(-{\rm{i}}){T}_{p}^{* }u(p)v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }{\sigma }_{z}{{\rm{e}}}^{{\rm{i}}{p}{z}-\kappa ^{\prime} z^{\prime} }+(-{\rm{i}}){R}_{p}{T}_{p}^{* }u(-p)v{(-{\rm{i}}\kappa ^{\prime} )}^{\dagger }{\sigma }_{z}{{\rm{e}}}^{-{\rm{i}}{p}{z}-\kappa ^{\prime} z^{\prime} },\end{eqnarray} \tag{ 73 }$$

which is the contribution from this energy range ${V}_{0}-m\leqslant E\lt {V}_{0}$.

Finally in the energy range $m\leqslant E\lt {V}_{0}-m$, the variable q is a positive number and appears in association with the spinor v as $v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }$ or $v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }$, which implies that the term $\sqrt{E-{V}_{0}\pm m}$ has to be replaced with ${\rm{i}}\sqrt{{V}_{0}-E\mp m}$ in the amplitude T_p and with $-{\rm{i}}\sqrt{{V}_{0}-E\mp m}$ in ${T}_{p}^{* }$ in (69). Note that in either case, q is to be replaced with $-q$. It is straightforward to confirm that the following relations hold, for the left-incident contribution,

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}}+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}]{[{T}_{p}{\sigma }_{z}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad =\,{\left.{\displaystyle \int }_{0}^{\sqrt{({V}_{0}-{m}^{2})-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}}+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}]{[{T}_{p}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\right|}_{\sqrt{{E}_{p}-{V}_{0}\pm m}\to -{\rm{i}}\sqrt{{V}_{0}-{E}_{p}\mp m}},\end{eqnarray} \tag{ 74 }$$

and for the right-incident contribution,

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}\displaystyle \frac{p}{q}{T}_{q}{\sigma }_{z}u(-p){{\rm{e}}}^{-{\rm{i}}{pz}}{[v(-q){{\rm{e}}}^{{\rm{i}}{qz}^{\prime} }+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{qz}^{\prime} }]}^{\dagger }\\ &&\quad =\,{\left.{\displaystyle \int }_{0}^{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{pz}}{[u(-q){{\rm{e}}}^{-{\rm{i}}{qz}^{\prime} }-{R}_{p}u(q){{\rm{e}}}^{{\rm{i}}{qz}^{\prime} }]}^{\dagger }\right|}_{\sqrt{{E}_{p}-{V}_{0}\pm m}\to i\sqrt{{V}_{0}-{E}_{p}\mp m}}.\end{eqnarray} \tag{ 75 }$$

The second terms on the right-hand sides of these relations (i.e., those proportional to the product of reflection and transmission amplitudes) are canceled to each other and finally we arrive at the conclusion that the contributions coming from the energy range $E\geqslant m$ are consistently written down as

$$\begin{eqnarray}&&{\int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}({T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }+{T}_{p}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }).\end{eqnarray} \tag{ 76 }$$

(ii) $m\gt E\gt 0$: A straightforward calculation yields

$$\begin{eqnarray}&&{\displaystyle \int }_{\sqrt{{({V}_{0}-m)}^{2}-{m}^{2}}}^{\sqrt{{V}_{0}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}{\sigma }_{z}u(-{\rm{i}}\kappa ){{\rm{e}}}^{\kappa z}{[v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad =\,{\displaystyle \int }_{0}^{m}\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{\kappa }{q}\displaystyle \frac{m}{{E}_{p}}({T}_{q}{\sigma }_{z}u(-{\rm{i}}\kappa )v{(-q)}^{\dagger }{{\rm{e}}}^{\kappa z-{\rm{i}}{q}{z}^{\prime} }-{T}_{q}^{* }{\sigma }_{z}u(-{\rm{i}}\kappa )v{(q)}^{\dagger }{{\rm{e}}}^{\kappa z+{\rm{i}}{q}{z}^{\prime} })\\ &&\quad =\,{\left.-{\displaystyle \int }_{0}^{{\rm{i}}{m}}\displaystyle \frac{{\rm{d}}({\rm{i}}\kappa )}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }\right|}_{p\to {\rm{i}}\kappa }\\ &&\quad \quad -\,{\left.{\displaystyle \int }_{0}^{-{\rm{i}}{m}}\displaystyle \frac{{\rm{d}}(-{\rm{i}}\kappa )}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }\right|}_{p\to -{\rm{i}}\kappa },\end{eqnarray} \tag{ 77 }$$

where ${T}_{q}{R}_{q}^{* }=-{T}_{q}^{* }$ has been used in the first equality. Actually in the first term of the last line, T_p is evaluated by replacing $\sqrt{{E}_{p}-{V}_{0}\pm m}$ by ${\rm{i}}\sqrt{{V}_{0}-{E}_{p}\mp m}$ in the original one in (13) and it is confirmed that it coincides with $-\tfrac{\kappa }{q}{T}_{q}$. Under these replacements together with $p\to {\rm{i}}\kappa$, we can show that

$$\begin{eqnarray}&&u{(-q)}^{\dagger }=\sqrt{\displaystyle \frac{{E}_{p}-{V}_{0}+m}{2m}}\left(1,-{\sigma }_{z}\sqrt{\displaystyle \frac{{E}_{p}-{V}_{0}-m}{{E}_{p}-{V}_{0}+m}}\right)\\ &&\quad \to \,{\rm{i}}\sqrt{\displaystyle \frac{{V}_{0}-{E}_{p}-m}{2m}}\left(1,-{\sigma }_{z}\sqrt{\displaystyle \frac{{E}_{p}-{V}_{0}-m}{{E}_{p}-{V}_{0}+m}}\right)\\ &&\quad =\,-{\rm{i}}\sqrt{\displaystyle \frac{{V}_{0}-{E}_{p}+m}{2m}}\left(-{\sigma }_{z}\sqrt{\displaystyle \frac{{V}_{0}-{E}_{p}-m}{{V}_{0}-{E}_{p}+m}},1\right){\sigma }_{z}=-{\rm{i}}{v}{(-q)}^{\dagger }{\sigma }_{z}.\end{eqnarray} \tag{ 78 }$$

(Remember that under these replacements, q is to be replaced with $-q$ and we get the right exponential factors.) A similar argument is applied to the second term. The contribution is now summarized as

$$\begin{eqnarray}&&{\int }_{-{\rm{i}}{m}}^{0}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}^{* }u(p)u{(q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }+{\int }_{{\rm{i}}{m}}^{0}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}u(-p)u{(-q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} },\end{eqnarray} \tag{ 79 }$$

where the integration contours are taken along the imaginary p-axis. (iii) and (iv) $0\geqslant E$: We first consider the region (iv) where E ≤ −m. The contribution coming from this energy range is

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}[v(p){{\rm{e}}}^{-{\rm{i}}{p}{z}}+{R}_{p}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}}]{[{T}_{p}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad +\,{\displaystyle \int }_{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}^{\infty }\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}v(-p){{\rm{e}}}^{{\rm{i}}{p}{z}}{[v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\\ &&\quad =\,{\displaystyle \int }_{0}^{\infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}({T}_{p}^{* }v(p)v{(q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }+{T}_{p}v(-p)v{(-q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }).\end{eqnarray} \tag{ 80 }$$

Since in the contribution arising from the region (iii) $0\gt E\gt -m$

$$\begin{eqnarray}&&{\int }_{\sqrt{{V}_{0}^{2}-{m}^{2}}}^{\sqrt{{({V}_{0}+m)}^{2}-{m}^{2}}}\displaystyle \frac{{\rm{d}}{q}}{2\pi }\displaystyle \frac{m}{{E}_{q}}{T}_{q}(-{\rm{i}})v({\rm{i}}\kappa ){{\rm{e}}}^{\kappa z}{[v(-q){{\rm{e}}}^{{\rm{i}}{q}{z}^{\prime} }+{R}_{q}v(q){{\rm{e}}}^{-{\rm{i}}{q}{z}^{\prime} }]}^{\dagger }\end{eqnarray} \tag{ 81 }$$

the relation ${T}_{q}{R}_{q}^{* }=-{T}_{q}^{* }$ holds, we can rewrite this as

$$\begin{eqnarray}&&-{\int }_{m}^{0}\displaystyle \frac{{\rm{d}}\kappa }{2\pi }\displaystyle \frac{m}{{E}_{p}}\displaystyle \frac{\kappa }{q}(-{{\rm{i}}{T}}_{q}v({\rm{i}}\kappa )v{(-q)}^{\dagger }{{\rm{e}}}^{\kappa z-{\rm{i}}{q}{z}^{\prime} }+{{\rm{i}}{T}}_{q}^{* }v({\rm{i}}\kappa )v{(q)}^{\dagger }{{\rm{e}}}^{\kappa z+{\rm{i}}{q}{z}^{\prime} }).\end{eqnarray} \tag{ 82 }$$

Furthermore, we confirm that the transmission amplitude in this energy range (25) is analytically continued for ${E}_{p}\lt m$ as

$$\begin{eqnarray}&&{T}_{p}\to \displaystyle \frac{2}{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}+{\rm{i}}\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{m-{E}_{p}}}}=-\displaystyle \frac{\kappa }{q}{T}_{q},\quad {T}_{p}^{* }\to \displaystyle \frac{2}{\sqrt{\tfrac{{E}_{p}+{V}_{0}+m}{{E}_{p}+m}}-{\rm{i}}\sqrt{\tfrac{{E}_{p}+{V}_{0}-m}{m-{E}_{p}}}}=-\displaystyle \frac{\kappa }{q}{T}_{q}^{* },\end{eqnarray} \tag{ 83 }$$

so that the contribution coming from region (iv) is expressed as integrations along imaginary p-axis

$$\begin{eqnarray}&&{\int }_{-{\rm{i}}{m}}^{0}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}v(-p)v{(-q)}^{\dagger }{{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }+{\int }_{{\rm{i}}{m}}^{0}\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}{T}_{p}^{* }v(p)v{(q)}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }.\end{eqnarray} \tag{ 84 }$$

Putting all contributions (i)–(iv) together, we understand that the left-hand side of (40) is conveniently written as

$$\begin{eqnarray}&&{\displaystyle \int }_{-{\rm{i}}{m}\to 0\to \infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}({T}_{p}^{* }u(p)u{(q)}^{\dagger }+{T}_{p}v(-p)v{(-q)}^{\dagger }){{\rm{e}}}^{{\rm{i}}{p}{z}-{\rm{i}}{q}{z}^{\prime} }\\ &&\quad +\,{\displaystyle \int }_{{\rm{i}}{m}\to 0\to \infty }\displaystyle \frac{{\rm{d}}{p}}{2\pi }\displaystyle \frac{m}{{E}_{p}}({T}_{p}u(-p)u{(-q)}^{\dagger }+{T}_{p}^{* }v(p)v{(q)}^{\dagger }){{\rm{e}}}^{-{\rm{i}}{p}{z}+{\rm{i}}{q}{z}^{\prime} }.\end{eqnarray} \tag{ 85 }$$

Observe that the exponential factors make the integrands convergent at $| p| \to \infty$ in the lower-half and upper-half planes for the first and second lines, respectively, if proper branches have been chosen for momentum q. Since singularities only appear on the real and imaginary axises as branch cuts, the integrals are evaluated on the imaginary p-axis, from $-{\rm{i}}{m}$ to $-{\rm{i}}\infty$ for the first integral and from im to ${\rm{i}}\infty$ for the second. (Remember that here we are considering the case where $z\lt 0$ and $z^{\prime} \gt 0$.) We note, however, that the momentum q for spinor u, $q=\sqrt{{({E}_{p}-{V}_{0})}^{2}-{m}^{2}}$, and that for spinor v, $q=\sqrt{{({E}_{p}+{V}_{0})}^{2}-{m}^{2}}$, are different quantities, though expressed with the same symbol q for notational simplicity, and we have to choose a proper phase when they are analytically continued.

Actually, if we put $p=\pm {\rm{i}}\kappa$ ($\kappa \geqslant m$) and thus ${E}_{p}=\pm \sqrt{{\kappa }^{2}-{m}^{2}}\equiv \pm {\rm{i}}{{ \mathcal E }}_{\kappa }$ on the imaginary p-axis, we have, for spinor u,

$$\begin{eqnarray}&&q=\sqrt{{(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0})}^{2}-{m}^{2}}=\sqrt{(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m)(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m)}\equiv {q}_{\pm },\end{eqnarray} \tag{ 86 }$$

which are different from those for spinor v

$$\begin{eqnarray}q & = & \sqrt{{(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0})}^{2}-{m}^{2}}=\sqrt{(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m)(\pm {\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m)}\\ & = & -\sqrt{(\mp {\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m)(\mp {\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m)}=-{q}_{\mp }.\end{eqnarray} \tag{ 87 }$$

We will put together those terms that have the same $(z,z^{\prime} )$-dependence. When analytically continued on the imaginary p-axis, the first term and the last term in (85) have the same $(z,z^{\prime} )$-dependence ${e}^{\kappa z-{{\rm{i}}{q}}_{-}z^{\prime} }$. The first term is evaluated as

$$\begin{eqnarray}&&{\int }_{-{\rm{i}}{m}}^{-{\rm{i}}\infty }\displaystyle \frac{{\rm{d}}(-{\rm{i}}\kappa )}{2\pi }\displaystyle \frac{m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }}{T}_{p}^{* }u(-{\rm{i}}\kappa )u{({q}_{-})}^{\dagger }{{\rm{e}}}^{\kappa z-{{\rm{i}}{q}}_{-}z^{\prime} },\end{eqnarray} \tag{ 88 }$$

while the last term as

$$\begin{eqnarray}&&{\int }_{{\rm{i}}{m}}^{{\rm{i}}\infty }\displaystyle \frac{{\rm{d}}({\rm{i}}\kappa )}{2\pi }\displaystyle \frac{m}{{\rm{i}}{{ \mathcal E }}_{\kappa }}{T}_{p}^{* }v({\rm{i}}\kappa )v{(-{q}_{-})}^{\dagger }{{\rm{e}}}^{\kappa z-{{\rm{i}}{q}}_{-}z^{\prime} }.\end{eqnarray} \tag{ 89 }$$

Observe that the amplitude ${T}_{p}^{* }$ is actually the same quantity for both cases, for

$$\begin{eqnarray}&&{T}_{p}^{* }\,({\rm{for}}\,{\rm{spinor}}\,u)=\displaystyle \frac{2}{\sqrt{\tfrac{-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}+\sqrt{\tfrac{-{{ \mathcal E }}_{\kappa }-{V}_{0}-m}{-{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}}=\displaystyle \frac{2}{\sqrt{\tfrac{{\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m}{{\rm{i}}{{ \mathcal E }}_{\kappa }-m}}+\sqrt{\tfrac{{{ \mathcal E }}_{\kappa }+{V}_{0}+m}{{\rm{i}}{{ \mathcal E }}_{\kappa }+m}}}={T}_{p}^{* }\,({\rm{for}}\,{\rm{spinor}}\,v).\end{eqnarray} \tag{ 90 }$$

After the spin sum, the spinor parts are explicitly written down as

$$\begin{eqnarray}&&{u(-{\rm{i}}\kappa )u{({q}_{-})}^{\dagger }\equiv u(p)u{(q)}^{\dagger }| }_{p=-{\rm{i}}\kappa }\\ \quad =\,\displaystyle \frac{1}{2m}\left(\begin{array}{cc}\sqrt{(-{\rm{i}}{{ \mathcal E }}_{\kappa }+m)(-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m)} & {\sigma }_{z}\sqrt{(-{\rm{i}}{{ \mathcal E }}_{\kappa }+m)(-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m)}\\ {\sigma }_{z}\sqrt{(-{\rm{i}}{{ \mathcal E }}_{\kappa }-m)(-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}+m)} & \sqrt{(-{\rm{i}}{{ \mathcal E }}_{\kappa }-m)(-{\rm{i}}{{ \mathcal E }}_{\kappa }-{V}_{0}-m)}\end{array}\right)\end{eqnarray} \tag{ 91 }$$

and

$$\begin{eqnarray}{v({\rm{i}}\kappa )v{(-{q}_{-})}^{\dagger }\equiv v(p)v{(q)}^{\dagger }| }_{p={\rm{i}}\kappa }=\displaystyle \frac{1}{2m}\left(\begin{array}{cc}\sqrt{({\rm{i}}{{ \mathcal E }}_{\kappa }-m)({\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m)} & {\sigma }_{z}\sqrt{({\rm{i}}{{ \mathcal E }}_{\kappa }-m)({\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m)}\\ {\sigma }_{z}\sqrt{({\rm{i}}{{ \mathcal E }}_{\kappa }+m)({\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}-m)} & \sqrt{({\rm{i}}{{ \mathcal E }}_{\kappa }+m)({\rm{i}}{{ \mathcal E }}_{\kappa }+{V}_{0}+m)}\end{array}\right).\end{eqnarray} \tag{ 92 }$$

We carefully examine the phases of the square roots and understand that these two spinor parts have opposite signs and are canceled to each other. This means that the first term and the last term in (85) are canceled to give vanishing contribution. A similar argument shows that the remaining two terms, the second and the third terms in (85), cancel to each other and we can conclude that (85) vanishes identically. This completes the proof of (40) for the case of $z\lt 0\lt z^{\prime}$, where its right-hand side vanishes.

To summarize, ${\rm{Cases}}\,{\rm{1}}\mbox{--}{\rm{3}}$ complete the proof of completeness relation (40).

As is already stated, only non-trivial relation is the orthogonality relation between ${\psi }^{(E)}$ and ${\phi }^{(E^{\prime} )}$ (39). For definiteness, the orthogonality shall be demonstrated here only for the case of $E,E^{\prime} \gt m$. In this case, the inner product between ${\psi }^{(E)}$ and ${\phi }^{(E^{\prime} )}$ is explicitly calculated as (again a trivial factor ${\delta }_{s,s^{\prime} }$ arising from the spinor inner product shall be suppressed)

$$\begin{eqnarray}&&{\displaystyle \int }_{-\infty }^{\infty }{\rm{d}}{z}{\psi }^{(E)}{(z)}^{\dagger }{\phi }^{(E^{\prime} )}(z)\propto {\displaystyle \int }_{-\infty }^{0}{\rm{d}}{z}{[u(p){{\rm{e}}}^{{\rm{i}}{p}{z}}+{R}_{p}u(-p){{\rm{e}}}^{-{\rm{i}}{p}{z}}]}^{\dagger }{T}_{q^{\prime} }u(-p^{\prime} ){{\rm{e}}}^{-{\rm{i}}{p}^{\prime} z}\\ &&\quad +\,{\displaystyle \int }_{0}^{\infty }{\rm{d}}{z}{[{T}_{p}u(q){{\rm{e}}}^{{\rm{i}}{q}{z}}]}^{\dagger }[u(-q^{\prime} ){{\rm{e}}}^{-{\rm{i}}{q}^{\prime} z}+{R}_{q^{\prime} }u(q^{\prime} ){{\rm{e}}}^{{\rm{i}}{q}^{\prime} z}]\\ &&\quad =\,{{\rm{i}}{T}}_{q^{\prime} }\displaystyle \frac{u{(p)}^{\dagger }u(-p^{\prime} )}{p\,+\,p^{\prime} }-{{\rm{i}}{T}}_{q^{\prime} }{R}_{p}\displaystyle \frac{u{(-p)}^{\dagger }u(-p^{\prime} )}{p-p^{\prime} -{\rm{i}}\epsilon }-{{\rm{i}}{T}}_{p}\displaystyle \frac{u{(q)}^{\dagger }u(-q^{\prime} )}{q\,+\,q^{\prime} }-{{\rm{i}}{T}}_{p}{R}_{q^{\prime} }\displaystyle \frac{u{(q)}^{\dagger }u(q^{\prime} )}{q-q^{\prime} -{\rm{i}}\epsilon }.\end{eqnarray} \tag{ 93 }$$

The second and the last terms contain terms proportional to delta functions

$$\begin{eqnarray}&&{T}_{q^{\prime} }{R}_{p}u{(-p)}^{\dagger }u(-p^{\prime} )\pi \delta (p-p^{\prime} )+{T}_{p}{R}_{q^{\prime} }u{(q)}^{\dagger }u(q^{\prime} )\pi \delta (q-q^{\prime} )=\displaystyle \frac{{E}_{p}}{m}\left({T}_{q}{R}_{p}+{T}_{p}{R}_{q}\displaystyle \frac{q}{p}\right)\pi \delta (p-p^{\prime} ),\end{eqnarray} \tag{ 94 }$$

which vanishes because of the relation ${T}_{p}{R}_{q}\tfrac{q}{p}=-{T}_{q}{R}_{p}$. The remaining parts are collected to yield

$$\begin{eqnarray}&&\displaystyle \frac{{{\rm{i}}{T}}_{q^{\prime} }}{{p}^{2}-p{{\prime} }^{2}}\sqrt{\displaystyle \frac{({E}_{p}+m)({E}_{p^{\prime} }+m)}{2m}}\,\phantom{\rule[-0000]{}{3.7ex}}\{p-p^{\prime} \\ &&\quad -\,\left.(p+p^{\prime} ){R}_{p}-((p-p^{\prime} +(p+p^{\prime} ){R}_{p})\displaystyle \frac{{pp}^{\prime} }{({E}_{p}+m)({E}_{p^{\prime} }+m)}\right\}\\ &&\quad -\,\displaystyle \frac{{{\rm{i}}{T}}_{p}}{{q}^{2}-q{{\prime} }^{2}}\sqrt{\displaystyle \frac{({E}_{q}+m)({E}_{q^{\prime} }+m)}{2m}}\,\{q-q^{\prime} +(q+q^{\prime} ){R}_{q^{\prime} }\\ &&\quad -\,\left.(q-q^{\prime} -(q+q^{\prime} ){R}_{q^{\prime} })\displaystyle \frac{{qq}^{\prime} }{({E}_{q}+m)({E}_{q^{\prime} }+m)}\right\}.\end{eqnarray} \tag{ 95 }$$

We note that ${p}^{2}-p{{\prime} }^{2}={E}_{p}^{2}-{E}_{p^{\prime} }^{2}$ and ${q}^{2}-q{{\prime} }^{2}={E}_{q}^{2}-{E}_{q^{\prime} }^{2}$ and furthermore

$$\begin{eqnarray}p-p^{\prime} -(p+p^{\prime} ){R}_{p} & = & p(1-{R}_{p})-p^{\prime} (1+{R}_{p})=\left(\sqrt{\displaystyle \frac{{E}_{q}-m}{{E}_{p}-m}}p-\sqrt{\displaystyle \frac{{E}_{q}+m}{{E}_{p}+m}}p^{\prime} \right){T}_{p},\\ p-p^{\prime} +(p+p^{\prime} ){R}_{p} & = & \left(\sqrt{\displaystyle \frac{{E}_{q}+m}{{E}_{p}+m}}p-\sqrt{\displaystyle \frac{{E}_{q}-m}{{E}_{p}-m}}p^{\prime} \right){T}_{p},\\ q-q^{\prime} +(q+q^{\prime} ){R}_{q^{\prime} } & = & \left(\sqrt{\displaystyle \frac{{E}_{p^{\prime} }+m}{{E}_{q^{\prime} }+m}}q-\sqrt{\displaystyle \frac{{E}_{p^{\prime} }-m}{{E}_{q^{\prime} }-m}}q^{\prime} \right){T}_{q^{\prime} },\\ q-q^{\prime} -(q+q^{\prime} ){R}_{q^{\prime} } & = & \left(\sqrt{\displaystyle \frac{{E}_{p^{\prime} }-m}{{E}_{q^{\prime} }-m}}q-\sqrt{\displaystyle \frac{{E}_{p^{\prime} }+m}{{E}_{q^{\prime} }+m}}q^{\prime} \right){T}_{q^{\prime} }.\end{eqnarray} \tag{ 96 }$$

Substitution of these expressions into (95) yields, after a straightforward calculation,

$$\begin{eqnarray}&&{\rm{i}}\displaystyle \frac{{T}_{p}{T}_{q^{\prime} }}{{E}_{p}^{2}-{E}_{p^{\prime} }^{2}}\displaystyle \frac{{E}_{p}+{E}_{p^{\prime} }}{2m}(\sqrt{({E}_{q}-m)({E}_{p^{\prime} }+m)}-\sqrt{({E}_{q}+m)({E}_{p^{\prime} }-m)})\\ &&\quad -\,{\rm{i}}\displaystyle \frac{{T}_{p}{T}_{q^{\prime} }}{{E}_{q}^{2}-{E}_{q^{\prime} }^{2}}\displaystyle \frac{{E}_{q}+{E}_{q^{\prime} }}{2m}(\sqrt{({E}_{p^{\prime} }+m)({E}_{q}-m)}-\sqrt{({E}_{q}+m)({E}_{p^{\prime} }-m)})\\ &&\quad =\,{\rm{i}}\displaystyle \frac{{T}_{p}{T}_{q^{\prime} }}{2m}(\sqrt{({E}_{q}-m)({E}_{p^{\prime} }+m)}-\sqrt{({E}_{q}+m)({E}_{p^{\prime} }-m)})\left(\displaystyle \frac{1}{{E}_{p}-{E}_{p^{\prime} }}-\displaystyle \frac{1}{{E}_{q}-{E}_{q^{\prime} }}\right),\end{eqnarray} \tag{ 97 }$$

which is zero because ${E}_{q}={E}_{p}-{V}_{0}$ and ${E}_{q^{\prime} }={E}_{p^{\prime} }-{V}_{0}$. Thus the right-hand side of (93) is shown to vanish, which means that the wave function for left-incident case ${\psi }^{(E)}$ and that for right-incident case ${\phi }^{(E^{\prime} )}$ are orthogonal when $E,E^{\prime} \gt m$. The above argument would easily be extended to other energy ranges to show the orthogonality between ${\psi }^{(E)}$ and ${\phi }^{(E^{\prime} )}$ $\forall E,E^{\prime}$.