# Vacuum decay and fermion total reflection by the Klein step

H. Nakazato<EMAIL_ADDRESS>M. Ochiai
<EMAIL_ADDRESS>Department of Physics, Waseda University, 169-8555
Tokyo, Japan

###### Abstract

The so-called Klein tunneling is re-examined within the framework of quantum
field theory, but from a different point of view on the asymptotic states. We
treat it as a one-dimensional scattering process of a fermion incident to a
step potential and introduce asymptotic operators as appropriate $t=\pm\infty$
limits of the field operator responsible for the process. For the so-called
Klein energy range, two asymptotic vacua naturally emerge which are defined as
states annihilated by the asymptotic annihilation operators. They are related
by a similarity transformation, which entails a vacuum decay and yields a
vacuum decay constant. When a fermion with incident energy in the Klein region
is injected to the step, it is shown to be reflected with probability one,
accompanied by fermion–anti-fermion pairs that are vacuum decay products.

##  I  Introduction

In non-relativistic quantum mechanics, a step potential which rises abruptly
to a finite value at the origin and keeps its value to infinity plays the same
role as a rigid wall in classical mechanics, because the reflection
probability of the stationary scattering problem for the step becomes one when
the incident energy is less than the height of the step. This is because the
wave function under the step is essentially a real function and the
corresponding probability current vanishes. This situation changes
dramatically when one considers relativistic cases. Because the relativistic
dispersion relation between energy and momentum allows solutions with, not
only positive, but also negative energy, an oscillating solution can exist
under the step if the value of the incident energy falls in a particular range
called the Klein region, named after a paradoxical phenomenon for a
relativistic electron described by the Dirac equation [1]. Since the
stationary solutions are understood to represent stationary flows of
probability currents, the phenomenon implies the existence of a non-vanishing
probability current under the step, which may be interpreted as a transmitted
or tunneling current.

Even though the implication seems physically counter-intuitive and, at the
same time, suggestive when one combines it with the idea of Dirac sea, which
may be lifted to produce electron-positron pairs when the applied potential is
strong enough, it has generally been believed that the resolution of this
problem could be sought only within the framework of quantum field theory.
This is because the notion of particles and anti-particles is defined in
quantum field theory, while in a single-particle quantum mechanics, one can
just deal with positive- and negative-energy solutions with no clear
association to particle and anti-particle. Even if the lack of the negative-
energy solution might be interpreted as a presence of anti-particle, such an
interpretation is only justified on the basis of quantum field theory.

In quantum field theory, we calculate the scattering (S) matrix which
describes the transition between two asymptotic states prepared at remote past
$t=-\infty$ and future $t=\infty$ in terms of field operators and therefore it
is crucial that one can define asymptotic states that allow a particle
picture. When the potential is localized in space, the last condition seems to
be satisfied on physical ground because particles are present only in regions
far away from the potential at $t=\pm\infty$. This, on the contrary, means
that for a potential that does not vanish at spatial infinity, like the step
potential, it is not clear whether one can properly define an S matrix like in
the usual cases, even if the potential is supposed to be switched on and off
adiabatically.

One may argue that such a difficulty is mainly due to a mathematical
idealization and over-simplification of a physical setting, however, at the
same time, the problem has attracted researchers for almost ninety years
because its resolution is not only sought from an academic interest but also
is hoped to bring us with some insight into the physical mechanism of pair
creation of particle and anti-particle from the vacuum. In order to somehow
bypass the difficulty, Nikishov proposed to interchange the role of time and
spatial coordinates and to utilize stationary scattering solutions that have a
single plane wave in the spatially asymptotic regions to construct the Green
function [2]. A similar idea is introduced to quantize the field variable of
the scattering problem off the step potential, where solutions of the Dirac
equation that satisfy a particular boundary condition, i.e., those composed of
a single plane wave in regions outside of or under the step, are used to
introduce creation/annihilation operators to discuss vacuum decay and pair
creations [3, 4, 5, 6, 7]. Notice that the dynamics, that is, the time
development of the scattering process under a somewhat smeared potential (the
Sauter potential [8]) has been analyzed numerically within the framework of
quantum field theory [9], where the field operator initially expanded in terms
of solutions of the free Dirac equation is numerically simulated to see the
effect of an incident fermion on the pair creation due to the Pauli exclusion
principle. See also [10] for an introductory review.

In this paper, we shall exclusively consider a relativistic fermion incident
to a step potential with its energy lying in the Klein region and examine the
scattering process within the framework of quantum field theory, but from a
different point of view on the asymptotic states. The strategy adopted here
may be considered more straightforward and to follow a naive physical
expectation. After a brief review on the so-called Klein tunneling in Sec. II
and the relevant stationary scattering solutions of the Dirac equation with
the usual boundary condition in Sec. III, we introduce a field variable and
expand it in terms of the stationary scattering solutions in Sec. IV. It
should be stressed that the fact that they form a complete orthonormal set is
directly shown [11], so that the operators introduced as the expansion
coefficients of the field variable are guaranteed to satisfy the standard
anti-commutation relations. Then we examine the asymptotic limits
$t\to\pm\infty$ of the field operator and find that there are four non-trivial
limits existing, by which the asymptotic “in” and “out” operators are defined.
It is shown that they are related through a similarity transformation, i.e., a
Bogoliubov-like transformation. These asymptotic operators are used to define
two asymptotic vacua in Sec. V. Their overlap can be explicitly evaluated to
yield a vacuum decay constant. Under such an unstable vacuum, an incident
particle with energy in the Klein region is shown to be completely reflected
by the step, accompanying pairs of particle and anti-particle produced from
the unstable vacuum in Sec. VI. The final section VII is devoted to a summary
and discussions and Appendices A and B are added to fix the notation and to
supply additional details and information.

##  II  Klein tunneling in relativistic quantum mechanics

A scattering process of a Dirac fermion with mass $m$ by a one-dimensional
step potential

$\displaystyle V(z)=\theta(z)V_{0},\quad V_{0}>2m,$ (1)

is described by the Dirac Hamiltonian ($\hbar=c=1$)

$\displaystyle H=H_{0}+V(z)=-i\alpha_{z}\partial_{z}+\beta m+V(z)$ (2)

and its stationary solution is easily obtained. Here $\theta(z)$ is the
Heaviside step function and $\alpha_{z}$ and $\beta$ are two of the Dirac
matrices. See Appendix A for the notation adopted here. We are mainly
interested in the stationary scattering problem when the incident energy $E$
lies in the so-called Klein region, $m<E<V_{0}-m$. The stationary solution
which is nothing but an eigenfunction of $H$ belonging to the energy
eigenvalue $E$ is fixed once the boundary conditions and a continuity
condition at $z=0$ have been imposed on the eigenfunction (see Fig. 1). We
shall here treat the problem as an essentially one-dimensional scattering
problem from the beginning and the trivial dependence on the other coordinates
$x$ and $y$ will be ignored completely.

Figure 1: Scattering states are categorized according to whether an incident
flux is coming from the left ($\psi$) or right ($\phi$) of the step potential
and to the range of their energies.

Within the framework of (single-particle) relativistic quantum mechanics, when
a particle with momentum $p>0$ and spin $s$ is incident from the left of the
step, the stationary solution $\psi_{s}^{(E)}(z,t)$ with energy
$E=\sqrt{p^{2}+m^{2}}\equiv E_{p}$ is expressed, in terms of the positive
frequency solution $u(\pm p,s)e^{-iE_{p}t\pm ipz}$ for $z<0$ (left of the
step) and the negative frequency solution $v(q,s)e^{iE_{q}t-iqz}$ for $z>0$
(under the step), as ($E=E_{p}=V_{0}-E_{q},\,E_{q}=\sqrt{q^{2}+m^{2}},\,q>0$)

$\displaystyle\psi_{s}^{(E)}(z,t)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{m}{E}}\left[\theta(-z)\left\\{u(p,s)e^{ipz}+Ru(-p,s)e^{-ipz}\right\\}+\theta(z)T_{\to}\sigma_{z}v(q,s)e^{-iqz}\right]e^{-iEt},$
(3)

with

$\displaystyle
R=\frac{\sqrt{\frac{V_{0}-E-m}{E+m}}-\sqrt{\frac{V_{0}-E+m}{E-m}}}{\sqrt{\frac{V_{0}-E-m}{E+m}}+\sqrt{\frac{V_{0}-E+m}{E-m}}},\quad
T_{\to}=\frac{2}{\sqrt{\frac{V_{0}-E-m}{E+m}}+\sqrt{\frac{V_{0}-E+m}{E-m}}}.$
(4)

(Appendix A summarizes the details of spinors.) The continuity of the current
defined by $j_{z}=\bar{\psi}\gamma^{3}\psi=\psi^{\dagger}\alpha_{z}\psi$ at
$z=0$ implies the consevation of probability

$\displaystyle\frac{p}{m}-|R|^{2}\frac{p}{m}=|T_{\to}|^{2}\frac{q}{m}\quad\longrightarrow\quad
P_{r}+P_{t}=|R|^{2}+\frac{q}{p}|T_{\to}|^{2}=1.$ (5)

Recall that this is a normal conservation law of probability, but a finite and
non-vanishing transmission probability, which does not vanish even at the
infinite-potential limit $V_{0}\to\infty$, survives

$\displaystyle
P_{t}=\frac{q}{p}|T_{\to}|^{2}\quad\xrightarrow{V_{0}\to\infty}\quad\frac{2p}{E_{p}+p}=\frac{2\sqrt{E^{2}-m^{2}}}{E+\sqrt{E^{2}-m^{2}}}\neq
0.$ (6)

This phenomenon is known as the Klein tunneling [4].

Owing to the presence of negative energy solutions to the Dirac equation, we
can set up, not only the left-incident scattering problem, but also the right-
incident problem, because even if the energy is below the potential height,
there exist oscillating solutions of the equation. In fact, we obtain the
right-incident solution $\phi_{s}^{(E)}(z,t)$ with a negative incident
momentum $-q<0$ and spin $s$ ($E=V_{0}-E_{q}=E_{p},\,-p<0$)

$\displaystyle\phi_{s}^{(E)}(z,t)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{m}{V_{0}-E}}\left[\theta(-z)T_{\leftarrow}\sigma_{z}u(-p,s)e^{-ipz}+\theta(z)\left\\{v(-q,s)e^{iqz}+Rv(q,s)e^{-iqz}\right\\}\right]e^{-iEt},$
(7)

where

$\displaystyle
R=-\frac{\sqrt{\frac{E+m}{V_{0}-E-m}}-\sqrt{\frac{E-m}{V_{0}-E+m}}}{\sqrt{\frac{E+m}{V_{0}-E-m}}+\sqrt{\frac{E-m}{V_{0}-E+m}}},\quad
T_{\leftarrow}=-\frac{2}{\sqrt{\frac{E+m}{V_{0}-E-m}}+\sqrt{\frac{E-m}{V_{0}-E+m}}}.$
(8)

The continuity of the current results in the conservation of probability

$\displaystyle-\frac{p}{m}|T_{\leftarrow}|^{2}=-\frac{q}{m}+\frac{q}{m}|R|^{2},$
(9)

which is essentially the same as (5). The reflection coefficient $R$ turns out
to be the same as in the left-incident case, which is nothing but a
realization of reciprocity in quantum mechanics.

It would be worth mentioning that a choice of $v(-q,s)e^{iqz}$, instead of
$v(q,s)e^{-iqz}$, in (3) results in a negative factor in front of the
transmission coefficient squared in (5) and the reflection probability becomes
greater than one, which had been known as the Klein paradox [4]. It is to be
observed that in such a case, the sign of the current (and also the group
velocity) in the potential region ($z>0$) becomes negative, which is not
considered to satisfy the boundary condition for the left-incident scattering
problem where only a positive current (transmitted current) is admissible for
$z>0$.

##  III  Scattering states

The stationary solutions for the left- and right-incident scattering problems
are given by the eigenstates of the Hamiltonian $H$ (2) and are characterized
by their boundary condition, i.e., left-incident ($\psi$) or right-incident
($\phi$), and their eigenvalue $E$. These eigenfunctions form a complete
orthonormal set. Here only those cases where the potential step is higher than
$2m$, $V_{0}>2m$, are considered.

First, the orthogonality of the scattering states follows from the general
argument for hermitian Hamiltonians, that is, eigenfunctions belonging to
different eigenvalues are orthogonal to each other. Two $\psi$’s with
different energies are orthogonal and normalized as

$\displaystyle\int_{-\infty}^{\infty}dz\psi_{s}^{(E)\dagger}(z)\psi_{s^{\prime}}^{(E^{\prime})}(z)=\theta(EE^{\prime})\delta(p-p^{\prime})\delta_{s,s^{\prime}},$
(10)

where $|E|=E_{p},\,|E^{\prime}|=E_{p^{\prime}}$. Similarly, we should have

$\displaystyle\int_{-\infty}^{\infty}dz\phi_{s}^{(E)\dagger}(z)\phi_{s^{\prime}}^{(E^{\prime})}(z)=\theta\left[(E-V_{0})(E^{\prime}-V_{0})\right]\delta(q-q^{\prime})\delta_{s,s^{\prime}},$
(11)

where $|E-V_{0}|=E_{q},\,|E^{\prime}-V_{0}|=E_{q^{\prime}}$, and

$\displaystyle\int_{-\infty}^{\infty}dz\psi_{s}^{(E)\dagger}(z)\phi_{s^{\prime}}^{(E^{\prime})}(z)=0.$
(12)

Since the last orthogonality relation does not follow from the general
argument for both $\psi_{s}^{(E)}$ and $\phi_{s^{\prime}}^{(E^{\prime})}$ can
happen to belong to the same energy $E=E^{\prime}$, its validity has to be
examined separately (see Appendix B). The orthonormality conditions imply that
the following form of completeness relation holds

$\displaystyle\sum_{s,r=\pm}\int_{0}^{\infty}dp\psi_{s}^{(rE_{p})}(z)\psi_{s}^{(rE_{p})\dagger}(z^{\prime})+\sum_{s,r=\pm}\int_{0}^{\infty}dq\phi_{s}^{(V_{0}+rE_{q})}(z)\phi_{s}^{(V_{0}+rE_{q})\dagger}(z^{\prime})=\delta(z-z^{\prime})\mbox{1}\mbox{l}_{4\times
4},$ (13)

where $\mbox{1}\mbox{l}_{4\times 4}$ is the unit matrix acting in the spinor
space. Notice that the fact that they actually constitute a complete
orthonormal set is shown in a straightforward way, i.e., the relation (13) has
been shown explicitly in [11], though the proof itself first appeared more
than forty years ago [12].

##  IV  Quantized field

In order to discuss the scattering process under the Hamiltonian (2) within
the framework of quantum field theory, we introduce a field operator
$\Psi(z,t)$ and set up the equal-time anti-commutation relations between
$\Psi$s and $\Psi^{\dagger}$s, only non-vanishing ones of which are

$\displaystyle\\{\Psi_{\alpha}(z,t),\,\Psi_{\beta}^{\dagger}(z^{\prime},t)\\}=\delta_{\alpha,\beta}\delta(z-z^{\prime}).$
(14)

The field operator $\Psi$ is then expanded in terms of a complete orthonormal
set and creation and annihilation operators are introduced as the expansion
coefficients. Since we are interested in the scattering process described by
the Hamiltonian (2), it would be natural to choose its eigenfunctions that
satisfy the boundary conditions for scattering processes, that is, the left-
incident ($\psi$) and right-incident ($\phi$) scattering states shall be
chosen as the basis functions.

We introduce operators $b_{l(r)},\,d_{l(r)}^{\dagger}$ as expansion
coefficients of $\Psi$ expanded in terms of the scattering states $\psi(\phi)$

$\begin{split}\Psi(z,t)&=\int_{0}^{\infty}dp\sum_{s}\left[b_{l}(p,s)\psi_{s}^{(E_{p})}(z,t)+d_{l}^{\dagger}(p,s)\psi_{s}^{(-E_{p})}(z,t)\right]\\\
&\quad+\int_{0}^{\infty}dq\sum_{s}\left[b_{r}(q,s)\phi_{s}^{(V_{0}+E_{q})}(z,t)+d_{r}^{\dagger}(q,s)\phi_{s}^{(V_{0}-E_{q})}(z,t)\right].\end{split}$
(15)

The orthonormality of scattering eigenstates $\psi_{s}^{(E)}$ and
$\phi_{s}^{(E)}$ guarantees the usual canonical anti-commutation relations
between creation/annihilation operators. Indeed operators are expressed as

$\displaystyle b_{l}(p,s)$
$\displaystyle=\int_{-\infty}^{\infty}dz\psi_{s}^{(E_{p})\dagger}(z,t)\Psi(z,t),$
$\displaystyle b_{r}(q,s)$
$\displaystyle=\int_{-\infty}^{\infty}dz\phi_{s}^{(V_{0}+E_{q})\dagger}(z,t)\Psi(z,t),$
(16) $\displaystyle d_{l}^{\dagger}(p,s)$
$\displaystyle=\int_{-\infty}^{\infty}dz\psi_{s}^{(-E_{p})\dagger}(z,t)\Psi(z,t),$
$\displaystyle d_{r}^{\dagger}(q,s)$
$\displaystyle=\int_{-\infty}^{\infty}dz\phi_{s}^{(V_{0}-E_{q})\dagger}(z,t)\Psi(z,t)$
(17)

and since the right-hand sides becomes all time-independent if $\Psi$
satisfies the same equation of motion as $\psi_{s}^{(E)}$ and
$\phi_{s}^{(E)}$, the equal-time anti-commutation relations (14) and the
orthonormality of $\psi_{s}^{(E)}$ and $\phi_{s}^{(E)}$ (10)–(12) are enough
to show that the operators satisfy the standard anti-commutation relations.

Since the potential extending to $z=\infty$ is present at all times, it is not
clear whether the standard construction of asymptotic states at $t=\pm\infty$
is consistently applied, especially when one considers the scattering process
occurring in the Klein energy range. We can think, however, that the field
operator $\Psi$ contains all information of scattering and endeavor to extract
asymptotic information by considering appropriate limits. We consider
exclusively the case where the Klein tunneling occurs, i.e., the left-incident
energy $E>m$ is below the potential height, $E=E_{p}<V_{0}-m$. Consider, for
example, the following limit

$\displaystyle\lim_{t\to-\infty}\int_{-\infty}^{\infty}dzu_{p,s}^{\dagger}(z,t)\Psi(z,t),$
(18)

where

$\displaystyle
u_{p,s}(z,t)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{m}{E_{p}}}u(p,s)e^{ipz-
iE_{p}t}\quad(p>0)$ (19)

is nothing but the (normalized) positive-energy solution of the free Dirac
equation. The integration over $z$ in (18) results in one of the following
forms, for those terms with energy lying in the range $m<E<V_{0}-m$,

$\displaystyle\int_{-\infty}^{0}dze^{-ipz+iE_{p}t}e^{ip^{\prime}z-iE_{p^{\prime}}t}=\frac{-i}{p^{\prime}-p-i\epsilon}e^{-i(E_{p^{\prime}}-E_{p})t}\rightarrow
2\pi\delta(p^{\prime}-p),$ (20)
$\displaystyle\int_{-\infty}^{0}dze^{-ipz+iE_{p}t}e^{-ip^{\prime}z-iE_{p^{\prime}}t}=\frac{i}{p^{\prime}+p}e^{-i(E_{p^{\prime}}-E_{p})t}\rightarrow
0,$ (21) $\displaystyle\int_{0}^{\infty}dze^{-ipz+iE_{p}t}e^{\pm
iq^{\prime}z-i(V_{0}-E_{q^{\prime}})t}=\frac{-i}{\pm
q^{\prime}-p}e^{-i(V_{0}-E_{q^{\prime}}-E_{p})t}\rightarrow 0,$ (22)

in the $t\to-\infty$ limit, owing to the Riemann–Lebesgue lemma. The lemma
also implies that the other terms in (18) all disappear in the limit because
their energy does not match $E_{p}$ and oscillates indefinitely. The limit
(18) can be interpreted as the asymptotic annihilation operator for in-coming
particle $b_{\mathrm{in}}(p,s)$ associated with the positive-energy solution
and we find

$\displaystyle b_{\mathrm{in}}(p,s)=b_{l}(p,s).$ (23)

Observe that the above limit (18) just extracts the coefficient of
$u_{p,s}(z,t)$ in $\Psi$. The field operator $\Psi$ also contains other
solutions of the Dirac equation, $u_{-p,s}(z,t)$ and

$\displaystyle v_{\pm
q,s}(z,t)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{m}{E_{q}}}v(\pm q,s)e^{\mp
iqz-i(V_{0}-E_{q})t}\quad(q>0),$ (24)

through the scattering wave functions $\psi$ and $\phi$ in the Klein energy
range. It would be natural to define the asymptotic creation operator for in-
coming anti-particle $d_{\mathrm{in}}^{\dagger}(q,s)$ by the limit

$\displaystyle
d_{\mathrm{in}}^{\dagger}(q,s)=\lim_{t\to-\infty}\int_{-\infty}^{\infty}dzv_{-q,s}^{\dagger}(z,t)\Psi(z,t),$
(25)

which results in the relation

$\displaystyle d_{\mathrm{in}}^{\dagger}(q,s)=d_{r}^{\dagger}(q,s).$ (26)

We understand that the limits (18) and (25) but with opposite momenta $p\to-p$
and $-q\to q$ give nothing

$\displaystyle\lim_{t\to-\infty}\int_{-\infty}^{\infty}dzu_{-p,s}^{\dagger}(z,t)\Psi(z,t)=\lim_{t\to-\infty}\int_{-\infty}^{\infty}dzv_{q,s}^{\dagger}(z,t)\Psi(z,t)=0,$
(27)

as expected from the directions of in-coming particle and anti-particle in the
Klein region. These quantities become non-vanishing, instead in the other
asymptotic limit, i.e., at $t=\infty$, and they shall be denoted as
$b_{\mathrm{out}}$ and $d_{\mathrm{out}}^{\dagger}$, i.e., the particle–anti-
particle interpretation for out-going states. We evaluate the limit to obtain,
for example,

$\displaystyle
b_{\mathrm{out}}(p,s)=\lim_{t\to\infty}\int_{-\infty}^{\infty}dzu_{-p,s}^{\dagger}(z,t)\Psi(z,t)=R(p)b_{l}(p,s)-\sqrt{\frac{E_{q}}{E_{p}}}T(p)d_{r}^{\dagger}(q,\tilde{s}),$
(28)

where $E_{p}=V_{0}-E_{q}$, $R$ and $T=T_{\to}$ are reflection and transmission
coefficients appearing in the left-incident scattering eigenfunction $\psi$
and the spin $\tilde{s}$ stands for the spinor
$\tilde{\xi}(\tilde{s})=\sigma_{z}\xi(s)$. In deriving (28), the following
relations

$\displaystyle\delta(p-p^{\prime})=\frac{p}{E_{p}}\frac{E_{q}}{q}\delta(q-q^{\prime}),\quad
T_{\leftarrow}(q)=-\frac{q}{p}T_{\to}(p)$ (29)

have been used. It is interesting to see that this operator $b_{\mathrm{out}}$
can be expressed as a similarity transformation of $b_{l}$ multiplied by $R$

$\displaystyle
b_{\mathrm{out}}(p,s)=R(p)e^{B^{\dagger}}b_{l}(p,s)e^{-B^{\dagger}},\quad
B^{\dagger}=\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,s)d_{r}^{\dagger}(q,\tilde{s})$
(30)

and satisfies, together with its hermitian conjugate, the standard anti-
commutation relation

$\begin{split}\\{b_{\mathrm{out}}(p,s),\,b_{\mathrm{out}}^{\dagger}(p^{\prime},s^{\prime})\\}&=R^{2}(p)\delta(p-p^{\prime})\delta_{s,s^{\prime}}+\frac{E_{q}}{E_{p}}T^{2}(p)\delta(q-q^{\prime})\delta_{\tilde{s},\tilde{s}^{\prime}}\\\
&=\left[R^{2}(p)+\frac{q}{p}T^{2}(p)\right]\delta(p-p^{\prime})\delta_{s,s^{\prime}}=\delta(p-p^{\prime})\delta_{s,s^{\prime}},\end{split}$
(31)

where the last equality follows from the fact that the quantity in the square
parentheses is unity, which is nothing but the current conservation condition.
Another limit at $t=\infty$ worth evaluating corresponds to the anti-particle
creation operator $d_{\mathrm{out}}^{\dagger}$ and yields

$\displaystyle
d_{\mathrm{out}}^{\dagger}(q,s)=\lim_{t\to\infty}\int_{-\infty}^{\infty}dzv_{q,s}^{\dagger}(z,t)\Psi(z,t)=R(p)d_{r}^{\dagger}(q,s)+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}T(p)b_{l}(p,\tilde{s}),$
(32)

where use has made of $R(p)\equiv R_{\leftarrow}(p)=R_{\to}(q)$ (the
reciprocity). This operator is written as

$\displaystyle
d_{\mathrm{out}}^{\dagger}(q,s)=R(p)e^{-C}d_{r}^{\dagger}(q,s)e^{C},$ (33)

where the operator $C$ turns out to be the same as $B$

$\displaystyle
C=\sum_{s}\int\limits_{E_{q}<V_{0}-m}dq\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}d_{r}(q,s)b_{l}(p,\tilde{s})=\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}d_{r}(q,\tilde{s})b_{l}(p,s)=B,$
(34)

and satisfies

$\displaystyle\\{d_{\mathrm{out}}(q,s),\,d_{\mathrm{out}}^{\dagger}(q^{\prime},s^{\prime})\\}=R^{2}(p)\delta(q-q^{\prime})\delta_{s,s^{\prime}}+\left(\frac{q}{p}\right)^{2}\frac{E_{p}}{E_{q}}T^{2}(p)\delta(p-p^{\prime})\delta_{\tilde{s},\tilde{s}^{\prime}}=\delta(q-q^{\prime})\delta_{s,s^{\prime}}.$
(35)

##  V  Asymptotic vacua

We are now in a position to define asymptotic vacuum states. We define an
asymptotic vacuum state at $t=-\infty$, called “in” vacuum, as the state that
is annihilated by the asymptotic operators $b_{\mathrm{in}}$ and
$d_{\mathrm{in}}$

$\displaystyle
b_{\mathrm{in}}(p,s)\ket{0}_{\mathrm{in}}=d_{\mathrm{in}}(q,s)\ket{0}_{\mathrm{in}}=0.$
(36)

The “in” vacuum $\ket{0}_{\mathrm{in}}$ is nothing but the normalized state
$\ket{0}$ that is annihilated by the operators $b_{l}$ and $d_{r}$

$\displaystyle\ket{0}_{\mathrm{in}}=\ket{0},\quad
b_{l}(p,s)\ket{0}=d_{r}(q,s)\ket{0}=0,\quad\braket{0}{0}=1.$ (37)

We also introduce another asymptotic vacuum state at $t=\infty$, called “out”
vacuum. They shall be defined as the state that is annihilated by the
operators $b_{\mathrm{out}}$ and $d_{\mathrm{out}}$

$\displaystyle b_{\mathrm{out}}(p,s)\ket{0}_{\mathrm{out}}$
$\displaystyle=R(p)\left(b_{l}(p,s)-\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}d_{r}^{\dagger}(q,\tilde{s})\right)\ket{0}_{\mathrm{out}}=0,$
(38) $\displaystyle d_{\mathrm{out}}(q,\tilde{s})\ket{0}_{\mathrm{out}}$
$\displaystyle=R(p)\left(d_{r}(q,s)+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,\tilde{s})\right)\ket{0}_{\mathrm{out}}=0.$
(39)

It is not difficult to see that the “out” vacuum is explicitly constructed as

$\displaystyle\ket{0}_{\mathrm{out}}=\prod_{p,s}R(p)\left(1+\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,s)d_{r}^{\dagger}(q,\tilde{s})\right)\ket{0},$
(40)

where $L$ is the (infinite) size of the physical system and is formally equal
to $2\pi\delta(p-p)$. To see that it is annihilated by $b_{\mathrm{out}}(p,s)$
and $d_{\mathrm{out}}(q,\tilde{s})$, we only need to confirm that $\forall
p>0$ ($E_{p}=V_{0}-E_{q}$) and $\forall s$, the relations

$\displaystyle\left(b_{l}(p,s)-\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}d_{r}^{\dagger}(q,\tilde{s})\right)\left(1+\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,s)d_{r}^{\dagger}(q,\tilde{s})\right)\ket{0}$
$\displaystyle=0,$ (41)
$\displaystyle\left(d_{r}(q,s)+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,\tilde{s})\right)\left(1+\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,s)d_{r}^{\dagger}(q,\tilde{s})\right)\ket{0}$
$\displaystyle=0$ (42)

hold. The “out” vacuum $\ket{0}_{\mathrm{out}}$ is normalized to unity, as can
be shown explicitly

$\begin{split}{}_{\mathrm{out}}\\!\braket{0}{0}_{\mathrm{out}}&=\bra{0}\prod_{p,s}R^{2}(p)\left(1+\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}d_{r}(q,\tilde{s})b_{l}(p,s)\right)\left(1+\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{l}^{\dagger}(p,s)d_{r}^{\dagger}(q,\tilde{s})\right)\ket{0}\\\
&=\prod_{p,s}R^{2}(p)\left(1+\frac{q}{p}\frac{T^{2}(p)}{R^{2}(p)}\right)=1,\end{split}$
(43)

where the last equality follows from the current conservation condition
$R^{2}(p)+\frac{q}{p}T^{2}(p)=1$. It may be interesting to see that the “out”
vacuum state can be expressed as
$\ket{0}_{\mathrm{out}}=\mathcal{N}e^{B^{\dagger}}\ket{0}$, where the operator
$B^{\dagger}$ is the same as in (30) and the normalization constant is
formally given by $\mathcal{N}=\prod_{p,s}R(p)$. Notice that the meaning of an
infinite product over continuous variable $p$ is unclear and its precise
meaning shall be given in the following.

We are interested in the transition amplitudes to find specific asymptotic
states at $t=\infty$ when the initial state is prepared at $t=-\infty$. It is
thus convenient to express “in” states in terms of “out” states. We can, for
example, invert the relation (28) and (32) to express “in” operators in terms
of “out” operators

$\displaystyle\begin{pmatrix}b_{\mathrm{in}}(p,s)\\\
d_{\mathrm{in}}^{\dagger}(q,\tilde{s})\end{pmatrix}=\begin{pmatrix}R(p)&\sqrt{\frac{E_{q}}{E_{p}}}T(p)\\\
-\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}T(p)&R(p)\end{pmatrix}\begin{pmatrix}b_{\mathrm{out}}(p,s)\\\
d_{\mathrm{out}}^{\dagger}(q,\tilde{s})\end{pmatrix}$ (44)

and we also have, instead of (40),

$\displaystyle\ket{0}_{\mathrm{in}}=\prod_{p,s}R^{-1}(p)\left(1-\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{\mathrm{in}}^{\dagger}(p,s)d_{\mathrm{in}}^{\dagger}(q,\tilde{s})\right)\ket{0}_{\mathrm{out}}.$
(45)

It is not difficult to see that the last expression is actually symmetric
between “in” and “out” for it is rewritten, in terms of “out” operators, as

$\displaystyle\ket{0}_{\mathrm{in}}=\prod_{p,s}R(p)\left(1-\frac{2\pi}{L}\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{\mathrm{out}}^{\dagger}(p,s)d_{\mathrm{out}}^{\dagger}(q,\tilde{s})\right)\ket{0}_{\mathrm{out}}.$
(46)

Notice that the expression (46), which is formal and somewhat ambiguous, can
be written as

$\displaystyle\ket{0}_{\mathrm{in}}=\mathcal{N}e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}},$
(47)

where

$\displaystyle
B_{\mathrm{out}}^{\dagger}=\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{\mathrm{out}}^{\dagger}(p,s)d_{\mathrm{out}}^{\dagger}(q,\tilde{s}).$
(48)

It is easily seen that the state
$e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}$ is actually
annihilated by $b_{\mathrm{in}}$ and $d_{\mathrm{in}}$

$\displaystyle b_{\mathrm{in}}(p,s)\ket{0}_{\mathrm{in}}$
$\displaystyle\propto\left(R(p)b_{\mathrm{out}}(p,s)+\sqrt{\frac{E_{q}}{E_{p}}}T(p)d_{\mathrm{out}}^{\dagger}(q,\tilde{s})\right)e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}=e^{-B_{\mathrm{out}}^{\dagger}}R(p)b_{\mathrm{out}}(p,s)\ket{0}_{\mathrm{out}}=0,$
(49) $\displaystyle d_{\mathrm{in}}(q,\tilde{s})\ket{0}_{\mathrm{in}}$
$\displaystyle\propto\left(R(p)d_{\mathrm{out}}(q,\tilde{s})-\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}T(p)b_{\mathrm{out}}^{\dagger}(p,s)\right)e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}=e^{-B_{\mathrm{out}}^{\dagger}}R(p)d_{\mathrm{out}}(q,\tilde{s})\ket{0}_{\mathrm{out}}=0.$
(50)

The constant $\mathcal{N}$ is calculated from the normalization condition
$\mathcal{N}^{-2}={}_{\mathrm{out}}\\!\braket{0}{e^{-B_{\mathrm{out}}}e^{-B_{\mathrm{out}}\dagger}}{0}_{\mathrm{out}}$.

In order to estimate $\mathcal{N}$, consider a functional

$\displaystyle F[a]\equiv\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}}{0},\quad
B[a]=\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\,a(p,s)d(q,\tilde{s})b(p,s),$ (51)

where the index “out” has been suppressed for notational simplicity. A
functional derivative of $F$ brings down operators $db$ and
$b^{\dagger}d^{\dagger}$

$\displaystyle\frac{\delta}{\delta
a(p,s)}F[a]=-\braket{0}{e^{-B[a]}\left(d(q,\tilde{s})b(p,s)+b^{\dagger}(p,s)d^{\dagger}(q,\tilde{s})\right)e^{-B^{\dagger}[a]}}{0}.$
(52)

These operators can be moved to, say, just the right of $\bra{0}$

$\begin{split}&-\bra{0}\left[d(q,\tilde{s})b(p,s)+\biggl{(}b^{\dagger}(p,s)-a(p,s)d(q,\tilde{s})\biggr{)}\left(d^{\dagger}(q,\tilde{s})+\frac{q}{p}\frac{E_{p}}{E_{q}}a(p,s)b(p,s)\right)\right]e^{-B[a]}e^{-B^{\dagger}[a]}\ket{0}\\\
&=a(p,s)\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}}{0}-\braket{0}{d(q,\tilde{s})b(p,s)e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\left(1-\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)\right).\end{split}$
(53)

The last matrix element satisfies the relations

$\begin{split}&\braket{0}{d(q,\tilde{s})b(p,s)e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\\\
&=\bra{0}e^{-B[a]}e^{-B^{\dagger}[a]}\left(d(q,\tilde{s})+\frac{q}{p}\frac{E_{p}}{E_{q}}a(p,s)b^{\dagger}(p,s)\right)\biggl{(}b(p,s)-a(p,s)d^{\dagger}(q,\tilde{s})\biggr{)}\ket{0}\\\
&=-a(p,s)\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}-\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}b^{\dagger}(p,s)d^{\dagger}(q,\tilde{s})}{0}\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)\\\
&=-a(p,s)\left(1-\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)\right)\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}\\\
&\quad+\braket{0}{d(q,\tilde{s})b(p,s)e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\left(\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)\right)^{2},\end{split}$
(54)

which results in

$\displaystyle\braket{0}{d(q,\tilde{s})b(p,s)e^{-B[a]}e^{-B^{\dagger}[a]}}{0}=-\frac{a(p,s)}{1+\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)}\braket{0}{e^{-B[a]}e^{-B^{\dagger}[a]}}{0}\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}.$
(55)

We thus understand that $F[a]$ satisfies

$\displaystyle\frac{\delta}{\delta
a(p,s)}F[a]=\frac{2a(p,s)}{1+\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)}\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}F[a]=\frac{2\frac{q}{p}\frac{E_{p}}{E_{q}}a(p,s)}{1+\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)}\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}F[a],$
(56)

the solution of which with the condition $F[0]=1$ reads as

$\displaystyle
F[a]=\exp\left[\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\ln\left(1+\frac{q}{p}\frac{E_{p}}{E_{q}}a^{2}(p,s)\right)\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}\right].$
(57)

This concludes that the normalization constant $\mathcal{N}$, which is
reproduced by the functional $F[a]$ once $a(p,s)$ is replaced with
$\sqrt{E_{q}/E_{p}}\,T(p)/R(p)$, is given by [13]

$\begin{split}\mathcal{N}&=\exp\left[-\frac{1}{2}\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\ln\left(1+\frac{q}{p}\frac{T^{2}(p)}{R^{2}(p)}\right)\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}\right]\\\
&=\exp\left[\frac{1}{2}\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\ln
R^{2}(p)\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}\right].\end{split}$ (58)

Since the matrix element in the exponent diverges
$\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}=\delta(p-p)=\infty$, this constant
$\mathcal{N}$ formally vanishes, implying that the two vacua,
$\ket{0}_{\mathrm{in}}$ and $\ket{0}_{\mathrm{out}}$, are orthogonal and have
no overlaps, for
$\mathcal{N}={}_{\mathrm{out}}\\!\braket{0}{0}_{\mathrm{in}}$. Observe,
however, that if we replace the delta function in momentum space with that in
energy space, by $\delta(p-p)=\delta(E_{p}-E_{p})\,p/E_{p}$, and recall that
the last delta function is related to the total time $T_{\mathrm{total}}$ by
$\delta(E_{p}-E_{p})=T_{\mathrm{total}}/(2\pi)$, we may interpret that the
“in” vacuum decays with a decay rate $\gamma$

$\displaystyle\gamma=-\frac{1}{\pi}\int\limits_{E_{p}<V_{0}-m}dp\frac{p}{E_{p}}\ln
R^{2}(p)=-\frac{1}{\pi}\int_{m}^{V_{0}-m}dE_{p}\ln
R^{2}(p),\quad\mathcal{N}^{2}=\left|{}_{\mathrm{in}}\\!\braket{0}{0}_{\mathrm{out}}\right|^{2}=e^{-\gamma
T_{\mathrm{total}}}.$ (59)

The above expressions (46)–(48) explicitly show that the “in” vacuum has
nonvanishing overlaps with states of out-going pairs of particle and anti-
particle, which implies that it is not stable in the Klein region. In order to
evaluate explicitly such probabilities of finding out-going pairs of particle
and anti-particle, it is instructive and helpful first to calculate the
following generating functional

$\displaystyle
G[a]=\bra{0}\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\,a(p,s)d(q,\tilde{s})b(p,s)\right]\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b^{\dagger}(p,s)d^{\dagger}(q,\tilde{s})\right]\ket{0},$
(60)

where all operators and states are understood as “out” ones though the index
“out” shall be suppressed for simplicity. Following the similar procedure as
above, it is not difficult to show that $G[a]$ satisfies the functional
differential equation

$\displaystyle\frac{\delta}{\delta
a(p,s)}G[a]=\frac{\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}\braket{0}{d(q,\tilde{s})d^{\dagger}(q,\tilde{s})}{0}G[a]=\frac{\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}G[a].$
(61)

The solution with the condition $G[0]=1$ reads as

$\displaystyle
G[a]=\exp\left[\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\ln\left(1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)\right)\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}\right].$
(62)

Successive functional derivatives of $G$ evaluated at $a=0$ generate
amplitudes of finding out-going particle–anti-particle pairs in the “in”
vacuum. For example, the first derivative yields the amplitude finding a
single pair

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{d_{\mathrm{out}}(q,\tilde{s})b_{\mathrm{out}}(p,s)}{0}_{\mathrm{in}}=\mathcal{N}\left.\frac{\delta}{\delta
a(p,s)}G[a]\right|_{a=0}=\mathcal{N}\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p,s)}{0}_{\mathrm{out}}.$
(63)

The second derivative, on the other hand, yields the amplitude of the form

$\begin{split}{}_{\mathrm{out}}\\!\braket{0}{(db)_{1}(db)_{2}}{0}_{\mathrm{in}}&=\mathcal{N}\left.\frac{\delta^{2}}{\delta
a_{1}\delta a_{2}}G[a]\right|_{a=0}\\\
&=\mathcal{N}\left.\left(-\frac{\beta_{1}^{2}\delta(p_{1}-p_{2})\delta_{s_{1},s_{2}}}{(1+\beta_{1}a_{1})^{2}}+\frac{\beta_{1}\beta_{2}}{(1+\beta_{1}a_{1})(1+\beta_{2}a_{2})}{}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})}{0}_{\mathrm{out}}\right){}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})}{0}_{\mathrm{out}}\right|_{a=0},\end{split}$
(64)

where we have set
$\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}=\beta$ and introduced
a shorthand notation,
$(db)_{1}=d_{\mathrm{out}}(q_{1},\tilde{s}_{1})b_{\mathrm{out}}(p_{1},s_{1})$
etc.. This amplitude vanishes when $p_{1}=p_{2},\,s_{1}=s_{2}$ as it should
be, for
${}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})}{0}_{\mathrm{out}}=\delta(p-p)$.
We thus may write

$\displaystyle\frac{\delta^{2}}{\delta a_{1}\delta
a_{2}}G[a]=\begin{cases}\dfrac{\beta_{1}\,{}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})}{0}_{\mathrm{out}}}{1+\beta_{1}a_{1}}\dfrac{\beta_{2}\,{}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})}{0}_{\mathrm{out}}}{1+\beta_{2}a_{2}}G[a]&\text{for
$1\neq 2$}\\\ 0&\text{otherwise}\end{cases},$ (65)

which can be generalized to higher-order derivatives and we understand that
the amplitude of finding $n$ pairs is given by

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{(db)_{1}(db)_{2}\cdots(db)_{n}}{0}_{\mathrm{in}}=\mathcal{N}\prod_{k=1}^{n}\beta_{k}\,{}_{\mathrm{out}}\\!\braket{0}{(bb^{\dagger})_{k}}{0}_{\mathrm{out}}=\mathcal{N}\prod_{k=1}^{n}\frac{q_{k}}{p_{k}}\sqrt{\frac{E_{p_{k}}}{E_{q_{k}}}}\frac{T(p_{k})}{R(p_{k})}{}_{\mathrm{out}}\\!\braket{0}{b(p_{k},s_{k})b^{\dagger}(p_{k},s_{k})}{0}_{\mathrm{out}},$
(66)

where all momenta $p_{k}$ are different, for otherwise it vanishes.

In order to properly normalize the multi-pair states, so that their inner
products are just given by the product of delta functions, each state has to
be devided by
$\sqrt{\frac{q}{p}\frac{E_{p}}{E_{q}}{}_{\mathrm{out}}\\!\braket{0}{b(p,s)b^{\dagger}(p,s)}{0}_{\mathrm{out}}}$
because

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{d_{\mathrm{out}}(q,\tilde{s})b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p^{\prime},s^{\prime})d_{\mathrm{out}}^{\dagger}(q^{\prime},\tilde{s}^{\prime})}{0}_{\mathrm{out}}=\frac{q}{p}\frac{E_{p}}{E_{q}}{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p,s)}{0}_{\mathrm{out}}\delta(p-p^{\prime})\delta_{s,s^{\prime}}.$
(67)

The probability of finding an out-going single pair thus reads as

$\displaystyle
P_{1}=\mathcal{N}^{2}\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\frac{q}{p}\frac{T^{2}(p)}{R^{2}(p)}{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p,s)}{0}_{\mathrm{out}}.$
(68)

Generally speaking, however, it is not easy to write down the probability of
finding $n$ pairs of out-going particle and anti-particle, denoted as $P_{n}$,
because a naive expectation for $P_{n}$,

$\displaystyle\mathcal{N}^{2}\frac{1}{n!}\left(\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\frac{q}{p}\frac{T^{2}(p)}{R^{2}(p)}{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p,s)}{0}_{\mathrm{out}}\right)^{n},\quad
n=0,1,2,\cdots,$ (69)

is not precise and has to be corrected by properly subtracting all possible
coincident contributions where at least two pairs share the same momentum. The
explicit form of $P_{n}$ could instead be read from the power expansion of the
generating functional $G[a]$ (62).

##  VI  Scattering process seen as transition from single-particle “in” state

Even though the asymptotic “in” state is shown to decay into pairs of particle
and anti-particle and it is not clear whether to discuss stationary scattering
process within this framework is meaningful, we can explore what happens to a
single-particle state prepared at $t=-\infty$. Such a state would be
interpreted as an initial state of the scattering problem. Consider, for
definiteness, a state that corresponds to a left-incident particle moving
right at $t=-\infty$, which is represented by the state
$b_{\mathrm{in}}^{\dagger}(p,s)\ket{0}_{\mathrm{in}}$. This state is shown to
contain an out-going particle and multi-pairs of particle and anti-particle at
$t=\infty$. We just rewrite the initial state in terms of the “out” operators
and “out” vacuum

$\displaystyle
b_{\mathrm{in}}^{\dagger}(p,s)\ket{0}_{\mathrm{in}}=\left[R(p)b_{\mathrm{out}}^{\dagger}(p,s)+\sqrt{\frac{E_{q}}{E_{p}}}T(p)d_{\mathrm{out}}(q,\tilde{s})\right]\mathcal{N}e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}=\frac{\mathcal{N}}{R(p)}b_{\mathrm{out}}^{\dagger}(p,s)e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}},$
(70)

where the last equality follows if the operators in the parentheses are moved
just next to the vacuum. Since the state contains only those states that
contain one more particles than anti-particles, we just concentrate on the
transition amplitudes to such states.

It is possible to evaluate the following amplitude, which will turn out to be
relevant to the current problem, to obtain

$\displaystyle{}_{\mathrm{out}}\\!\bra{0}\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\,a(p,s)d_{\mathrm{out}}(q,\tilde{s})b_{\mathrm{out}}(p,s)\right]b_{\mathrm{out}}(p^{\prime},s^{\prime})b_{\mathrm{out}}^{\dagger}(p,s)e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}=\frac{\delta(p-p^{\prime})\delta_{s,s^{\prime}}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}G[a],$
(71)

where $G[a]$ is given in (62). If we set
$a(p,s)=\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}$, $G$ becomes
${}_{\mathrm{out}}\\!\braket{0}{e^{-B_{\mathrm{out}}}e^{-B_{\mathrm{out}}^{\dagger}}}{0}_{\mathrm{out}}=\mathcal{N}^{-2}$
and the above relation just implies that

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{e^{-B_{\mathrm{out}}}b_{\mathrm{out}}(p^{\prime},s^{\prime})b_{\mathrm{out}}^{\dagger}(p,s)e^{-B_{\mathrm{out}}^{\dagger}}}{0}_{\mathrm{out}}=\mathcal{N}^{-2}R^{2}(p)\delta(p-p^{\prime})\delta_{s,s^{\prime}}.$
(72)

Observe also that

$\begin{split}&{}_{\mathrm{out}}\\!\bra{0}\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\,a(p,s)d_{\mathrm{out}}(q,\tilde{s})b_{\mathrm{out}}(p,s)\right]d_{\mathrm{out}}^{\dagger}(q^{\prime},\tilde{s}^{\prime})\\\
&=a(p^{\prime},s^{\prime})\frac{q^{\prime}}{p^{\prime}}\frac{E_{p^{\prime}}}{E_{q^{\prime}}}{}_{\mathrm{out}}\\!\bra{0}\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\,a(p,s)d_{\mathrm{out}}(q,\tilde{s})b_{\mathrm{out}}(p,s)\right]b_{\mathrm{out}}(p^{\prime},s^{\prime}),\end{split}$
(73)

implying that the states with anti-particles one less than particles is
related to those with particles one more than anti-particles.

We expect that the initial state with a single particle moving right with
momentum $p$ and spin $s$,
$b_{\mathrm{in}}^{\dagger}(p,s)\ket{0}_{\mathrm{in}}$, corresponds to a final
state with a single particle moving left and multi-pairs of particle and anti-
particle because the former is written as in (70). The probability of finding
one out-going particle with no pairs of particle and anti-particle is simply
proportional to the matrix element squared

$\displaystyle\left|{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p^{\prime},s^{\prime})b_{\mathrm{in}}^{\dagger}(p,s)}{0}_{\mathrm{in}}\right|^{2}=\frac{\mathcal{N}^{2}}{R^{2}(p)}\left(\delta(p-p^{\prime})\delta_{s,s^{\prime}}\right)^{2}.$
(74)

Though the quantity still has to be properly normalized in order to be
interpreted as a probability density, we understand that such a probability
becomes exponentially small, owing to the fact that the “in” vacuum is not
stable and decays out into pairs of particle and anti-particle. If we take
into account of such decay products of particle–anti-particle pairs, the
matrix element squared for a specific final state with a particle and multi-
pairs of particle and anti-particle has to be summed over all degrees of
freedom, i.e., we have to sum over spins and integrate over momenta of the
pairs, after properly normalizing the final state. Specifically, we calculate

$\displaystyle\sum_{n\geq
0}\frac{1}{n!}\prod_{k=1}^{n}\sum_{s_{k}}\int\limits_{E_{p_{k}}<V_{0}-m}\frac{dp_{k}}{\xi_{k}}\left|{}_{\mathrm{out}}\\!\bra{0}(db)_{1}\cdots(db)_{n}b_{\mathrm{out}}(p^{\prime},s^{\prime})\frac{\mathcal{N}}{R(p)}b_{\mathrm{out}}^{\dagger}(p,s)e^{-B_{\mathrm{out}}^{\dagger}}\ket{0}_{\mathrm{out}}\right|^{2},$
(75)

where the shorthand notation
$(db)_{k}=d_{\mathrm{out}}(q_{k},\tilde{s}_{k})b_{\mathrm{out}}(p_{k},s_{k})$
has again been introduced and $\xi_{k}$ is the normalization factor

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{(db)_{k}(b^{\dagger}d^{\dagger})_{l}}{0}_{\mathrm{out}}=\frac{q_{k}}{p_{k}}\frac{E_{p_{k}}}{E_{q_{k}}}{}_{\mathrm{out}}\\!\braket{0}{b_{\mathrm{out}}(p_{k},s_{k})b_{\mathrm{out}}^{\dagger}(p_{k},s_{k})}{0}_{\mathrm{out}}\delta(p_{k}-p_{l})\delta_{s_{k},s_{l}}\equiv\xi_{k}\delta(p_{k}-p_{l})\delta_{s_{k},s_{l}}.$
(76)

We note that the matrix element can be written as an $n$th functional
derivative of (71) evaluated at $a=0$

$\displaystyle{}_{\mathrm{out}}\\!\braket{0}{(db)_{1}\cdots(db)_{n}b_{\mathrm{out}}(p^{\prime},s^{\prime})b_{\mathrm{out}}^{\dagger}(p,s)e^{-B_{\mathrm{out}}^{\dagger}}}{0}_{\mathrm{out}}=\frac{-\delta}{\delta
a_{1}}\cdots\frac{-\delta}{\delta
a_{n}}\left[\frac{\delta(p-p^{\prime})\delta_{s,s^{\prime}}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}G[a]\right]_{a=0},$
(77)

where $a_{k}=a(p_{k},s_{k})$ and therefore the above quantity (75), which is
now written as

$\begin{split}\frac{\mathcal{N}^{2}}{R^{2}(p)}\sum_{n}\frac{1}{n!}\prod_{k=1}^{n}\sum_{s_{k}}\int\limits_{E_{p_{k}}<V_{0}-m}dp_{k}\,{}_{\mathrm{out}}\bra{0}e^{-B_{\mathrm{out}}}b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p^{\prime},s^{\prime})\frac{(b^{\dagger}d^{\dagger})_{n}}{\xi_{n}}\cdots\frac{(b^{\dagger}d^{\dagger})_{1}}{\xi_{1}}\ket{0}_{\mathrm{out}}\\\
\times\frac{-\delta}{\delta a_{1}}\cdots\frac{-\delta}{\delta
a_{n}}\left[\frac{\delta(p-p^{\prime})\delta_{s,s^{\prime}}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}G[a]\right]_{a=0},\end{split}$
(78)

is reduced to

$\displaystyle\frac{\mathcal{N}^{2}}{R^{2}(p)}{}_{\mathrm{out}}\\!\bra{0}e^{-B_{\mathrm{out}}}b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p^{\prime},s^{\prime})\left[\frac{\delta(p-p^{\prime})\delta_{s,s^{\prime}}}{1+\frac{q}{p}\sqrt{\frac{E_{p}}{E_{q}}}\frac{T(p)}{R(p)}a(p,s)}G[a]\right]_{a=-\frac{(b^{\dagger}d^{\dagger})}{\xi}}\ket{0}_{\mathrm{out}}.$
(79)

Since $b_{\mathrm{out}}^{\dagger}$ and $d_{\mathrm{out}}^{\dagger}$ are
fermion operators that satisfy
$(b_{\mathrm{out}}^{\dagger}(p,s))^{2}=0=(d_{\mathrm{out}}^{\dagger}(q,\tilde{s}))^{2}$,
the operator inserted in the denominator gives no contribution and we have

$\displaystyle
G[a]\biggr{|}_{a=-\frac{(b^{\dagger}d^{\dagger})}{\xi}}=\exp\left[-\sum_{s}\int\limits_{E_{p}<V_{0}-m}dp\sqrt{\frac{E_{q}}{E_{p}}}\frac{T(p)}{R(p)}b_{\mathrm{out}}^{\dagger}(p,s)d_{\mathrm{out}}^{\dagger}(q,\tilde{s})\right],$
(80)

which is nothing but $e^{-B_{\mathrm{out}}^{\dagger}}$. The quantity (75)
finally turns out to be

$\displaystyle\frac{\mathcal{N}^{2}}{R^{2}(p)}{}_{\mathrm{out}}\\!\braket{0}{e^{-B_{\mathrm{out}}}b_{\mathrm{out}}(p,s)b_{\mathrm{out}}^{\dagger}(p^{\prime},s^{\prime})e^{-B_{\mathrm{out}}^{\dagger}}}{0}_{\mathrm{out}}\delta(p-p^{\prime})\delta_{s,s^{\prime}}=\delta(p-p^{\prime})\delta_{s,s^{\prime}}\delta(p-p).$
(81)

The last equality follows from the normalization condition (72).

In order to properly reproduce probabilities or “cross section” in scattering
problem, the result (81) has to be devided by the incident flux and the total
scattering time. The incident flux
$j_{\mathrm{inc}}=\psi_{s}^{\dagger}(p)\alpha_{z}\psi_{s}(p)$ is calculated
from the incident wave function

$\displaystyle\psi_{s}(p)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{m}{E_{p}}}u(p,s)$
(82)

and we obtain

$\displaystyle
j_{\mathrm{inc}}=\frac{1}{2\pi}\frac{m}{E_{p}}\frac{E_{p}+m}{2m}\frac{2p}{E_{p}+m}=\frac{1}{2\pi}\frac{p}{E_{p}}.$
(83)

Since the total scattering time $T_{\mathrm{total}}$ is formally written as

$\displaystyle
T_{\mathrm{total}}=2\pi\delta(E_{p}-E_{p})=2\pi\frac{E_{p}}{p}\delta(p-p),$
(84)

the amplitude squared has to be multiplied by

$\displaystyle\frac{1}{T_{\mathrm{total}}\,j_{\mathrm{inc}}}=\frac{1}{\delta(p-p)}$
(85)

to yield the probability density. Notice that the same result could be
attained if the initial state is normalized just by formally dividing it by
${}_{\mathrm{in}}\\!\braket{0}{b_{\mathrm{in}}(p,s)b_{\mathrm{in}}^{\dagger}(p,s)}{0}_{\mathrm{in}}=\delta(p-p)$.

We arrive at a conclusion that when a particle with momentum $p>0$ and spin
$s$ is injected, the probability of finding an out-going particle moving left
asymptotically is just one, when one does not care about the number of
particle–anti-particle pairs produced

$\displaystyle\text{Prob. (particle moving right with $p,s$ $\to$ particle
moving left with arbitrary numbers of pairs)}=1.$ (86)

Since the out-going particle moves left, this is considered to be the
reflection probability, i.e., total reflection and states that this is all
what happens.

##  VII  Summary and discussions

The Klein tunneling process in relativistic quantum mechanics is examined on
the basis of quantum field theory, where the field operator $\Psi$ which is
assumed to have all information on the physical process plays crucial and
central roles. It is constructed on the basis of the solutions of the Dirac
equation that satisfy boundary conditions for stationary scattering problem.
It is stressed that the condition that they constitute a complete orthonormal
set is crucial in order for operators introduced as the expansion coefficients
to satisfy the standard anti-commutation relations. In constrast to most cases
where the completeness of the basis functions has just been assumed on
physical ground, the stationary solutions of the Dirac equation on which the
field quantization of this paper is based have been shown explicitly to
satisfy the completeness condition [11]. This fact guarantees the standard
anti-commutation relations among creation/annihilation operators introduced in
the field operator. Other choices of basis functions, of course, would be
possible, but only when they are shown (not simply assumed) to be complete and
orthonormal.

In this respect, it should be remarked that the orthogonality of the
stationary solutions belonging to the same energy eigenvalue is closely
connected to the choice of proper boundary conditions. As stated above and can
be shown explicitly in simple cases, the usual boundary condition for the
scattering process, that is, the existence of incident and reflected plane
waves on one far side of the interaction region and the transmitted plane wave
on the other far side, entails, in general, to the orthogonality of left-
incident and right-incident solutions. The result can be derived in connection
with a position-independent current formed from these solutions. It should be
observed that the other solutions that satisfy, say, spatially asymptotic
plane wave conditions, that is, solutions that have either out-going or in-
coming plane waves on the same far side of the interaction region, are not
orthogonal to each other. See Appendix B for details.

Next, the asymptotic operators are introduced as the appropriate $t=\pm\infty$
limits of the field operator $\Psi$. At $t=-\infty$, only the right-moving
wave exsisting at the left of the step ($z<0$) and left-moving wave in the
potential region ($z>0$) survive in $\Psi$, which are supposed to represent
incident waves. On the other hand, only the left-moving and right-moving waves
are extracted for $z<0$ and $z>0$, respectively, at $t=\infty$, representing
out-gonig scattered ones. The asymptotic operators are just defined as the
coefficients of these waves in $\Psi$. The asymptotic vacuum states are then
defined to be those annihilated by the asymptotic annihilation operators. Two
sets of asymptotic operators are related by the Bogoliubov-like transformation
in the Klein energy range, on which we have exclusively forcused in this
paper, resulting in an unstable “in” vacuum.

In this way, we arrive at the following view on the Klein tunneling on the
basis of quantum field theory. When the step potential is higher than $2m$,
not only the “in” vacuum is unstable, but also a particle injected into such a
potential is totally reflected and is accompanied with pair-produced particles
moving left and anti-particles moving right. No transmitted particles moving
right exist. This view would have been somewhat expected when the field
operator $\Psi$ is expanded in terms of $b_{\mathrm{out}}$ and
$d_{\mathrm{out}}^{\dagger}$, however, the explicit calculations presented in
the previous sections show consistently that it is indeed the case.

Even though in this paper we have exclusively considered operators relevant to
the Klein region, the framework developed here surely admits presence of other
particle and anti-particle modes, for which both asymptotic vacua are the same
because no mixing between creation and annihilation operators is necessary to
connect different asymptotic operators. Nothing peculiar would happen for such
modes.

##  Appendix A  Dirac spinors

We fix the notations and normalizations of four-component spinors. The
stationary plane-wave solutions for the free Dirac Hamiltonian

$\displaystyle H_{0}=-i\alpha_{z}\partial_{z}+\beta
m,\quad\alpha_{z}=\begin{pmatrix}0&\sigma_{z}\\\
\sigma_{z}&0\end{pmatrix},\quad\beta=\begin{pmatrix}\mbox{1}\mbox{l}&0\\\
0&-\mbox{1}\mbox{l}\end{pmatrix}$ (87)

are

$\displaystyle u(p,s)e^{-ip\cdot
z}=\sqrt{\frac{E_{p}+m}{2m}}\begin{pmatrix}\mbox{1}\mbox{l}\\\
\frac{\sigma_{z}p}{E_{p}+m}\end{pmatrix}\bm{\xi}(s)e^{ipz-iE_{p}t}$ (88)

for a positive frequency $E_{p}=\sqrt{p^{2}+m^{2}}$ and

$\displaystyle v(p,s)e^{ip\cdot
z}=\sqrt{\frac{E_{p}+m}{2m}}\begin{pmatrix}\frac{\sigma_{z}p}{E_{p}+m}\\\
\mbox{1}\mbox{l}\end{pmatrix}\bm{\xi}(s)e^{-ipz+iE_{p}t}$ (89)

for a negative frequency $-E_{p}$, where $\sigma_{z}$ is the third Pauli
matrix, 1l the $2\times 2$ unit matrix and $\bm{\xi}(s)$ a two-component
spinor. They are normalized and form a complete orthonormal set

$\displaystyle\bar{u}(p,s)u(p,s^{\prime})=\delta_{s,s^{\prime}},\quad\bar{v}(p,s)v(p,s^{\prime})=-\delta_{s,s^{\prime}},$
(90) $\displaystyle
u^{\dagger}(p,s)u(p,s^{\prime})=v^{\dagger}(p,s)v(p,s^{\prime})=\frac{E_{p}}{m}\delta_{s,s^{\prime}},$
(91)
$\displaystyle\sum_{s}\left[u(p,s)\bar{u}(p,s)-v(p,s)\bar{v}(p,s)\right]=\mbox{1}\mbox{l}_{4\times
4}.$ (92)

Now, if a constant potential is added to the above Hamiltonian

$\displaystyle H_{0}^{\prime}=H_{0}+V_{0}=-i\alpha_{z}\partial_{z}+\beta
m+V_{0}\quad(-\infty<z<\infty),$ (93)

the corresponding plane-wave solutions are ($E_{q}=\sqrt{q^{2}+m^{2}}$)

$\displaystyle u(q,s)e^{-iq\cdot
z}=\sqrt{\frac{E_{q}+m}{2m}}\begin{pmatrix}\mbox{1}\mbox{l}\\\
\frac{\sigma_{z}q}{E_{q}+m}\end{pmatrix}\bm{\xi}(s)e^{iqz-i(E_{q}+V_{0})t}$
(94)

for a “positive” frequency $E_{q}+V_{0}\geq V_{0}$ and

$\displaystyle v(q,s)e^{iq\cdot
z}=\sqrt{\frac{E_{q}+m}{2m}}\begin{pmatrix}\frac{\sigma_{z}q}{E_{q}+m}\\\
\mbox{1}\mbox{l}\end{pmatrix}\bm{\xi}(s)e^{-iqz-i(-E_{q}+V_{0})t}$ (95)

for a “negative” frequency $-E_{q}+V_{0}\leq V_{0}$. They are again normalized
and form a complete orthonormal set in spinor space, just like in (90)–(92),
with $E_{p}$ replaced with $E_{q}$. Notice that the frequency $-E_{q}+V_{0}$
can be positive for small momenta $|q|<\sqrt{V_{0}^{2}-m^{2}}$, even though it
is to be associated with the spinor $v$. Note also that the above explicit
forms of spinors are not unique since $\sigma_{z}$ is commutable with the
above $H_{0}$ and $H_{0}^{\prime}$ and therefore, e.g., $\sigma_{z}u(p,s)$ can
be used instead of $u(p,s)$.

Incidentally, the current conservation
$\partial_{0}j^{0}+\partial_{z}j^{z}=0$, resulting from the invariance under
the global phase transformation, implies a $z$-independent current
$j^{z}=\psi^{\dagger}\alpha_{z}\psi$ for stationary states and the above
positive- and negative-frequency solutions give both positive currents

$\displaystyle
u^{\dagger}(q,s)\sigma_{z}u(q,s)=\frac{q}{m}=v^{\dagger}(q,s)\sigma_{z}v(q,s)$
(96)

for positive $q>0$, irrespectively of the value of $V_{0}$. In this respect,
both $u(q,s)$ and $v(q,s)$ are considered to express positive flows of current
with positive and negative frequencies.

##  Appendix B  Orthogonality and boundary conditions of stationary solutions

We show here that the orthogonality of stationary solutions of the Dirac
equation that belong to the same energy eigenvelue is closely connected to the
boundary conditions they satisfy and the existence of spatially constant
quantities. First, under the standard boundary conditions for scattering
problems, a left-incident solution $\psi$ is characterized by its asymptotic
behavior

$\displaystyle z\to-\infty:\quad\psi\sim e^{\pm
ipz}\chi_{p}+R_{\leftarrow}\chi_{-p}e^{\mp ipz},\qquad
z\to\infty:\quad\psi\sim T_{\to}\chi_{p^{\prime}}e^{\pm ip^{\prime}z},$ (97)

and a right-incident solution $\phi$ by

$\displaystyle z\to-\infty:\quad\phi\sim T_{\leftarrow}e^{\mp
ipz}\chi_{-p},\qquad z\to\infty:\quad\phi\sim e^{\mp
ip^{\prime}z}\chi_{-p^{\prime}}+R_{\to}\chi_{p^{\prime}}e^{\pm ip^{\prime}z},$
(98)

where $\chi$ represents either spinor $u$ (positive frequency solution) or $v$
(negative frequency solution), upper and lower signs in the exponent are to be
associated with $u$ and $v$, respectively (spin degrees of freedom is
neglected) and $p>0$ and $p^{\prime}>0$ are magnitudes of asymptotic momenta
at $z=-\infty$ and $z=\infty$ where the potential takes constant values. Since
both $\psi$ and $\phi$ satisfy the same Dirac equation with the same energy,
the quantity $\psi^{\dagger}\alpha_{z}\phi$ is shown to be $z$-independent
$\partial_{z}(\psi^{\dagger}\alpha_{z}\phi)=0$, in particular, its values at
$z=\pm\infty$ are the same. Inserting their spatially asymptotic forms (97)
and (98), we have

$\displaystyle\psi^{\dagger}\alpha_{z}\phi\sim\left(e^{\mp
ipz}\chi_{p}^{\dagger}+R_{\leftarrow}^{*}e^{\pm
ipz}\chi_{-p}^{\dagger}\right)\alpha_{z}T_{\leftarrow}e^{\mp
ipz}\chi_{-p}=-\frac{p}{m}R_{\leftarrow}^{*}T_{\leftarrow}$ (99)

at $z=-\infty$ and

$\displaystyle\psi^{\dagger}\alpha_{z}\phi\sim T_{\to}^{*}e^{\mp
ip^{\prime}z}\chi_{p^{\prime}}^{\dagger}\alpha_{z}\left(e^{\mp
ip^{\prime}z}\chi_{-p^{\prime}}+R_{\to}e^{\pm
ip^{\prime}z}\chi_{p^{\prime}}\right)=\frac{p^{\prime}}{m}T_{\to}^{*}R_{\to}$
(100)

at $z=\infty$, resulting in the equality (reciprocity)

$\displaystyle-\frac{p}{m}R_{\leftarrow}^{*}T_{\leftarrow}=\frac{p^{\prime}}{m}T_{\to}^{*}R_{\to}.$
(101)

Here we have used the relation
$\chi_{p}^{\dagger}\alpha_{z}\chi_{p}=\frac{p}{m}$ and
$\chi_{p}^{\dagger}\alpha_{z}\chi_{-p}=0$ that hold for any choice of $\chi$.

Now consider an inner product between $\psi_{1}$ with left-incident momentum
$p_{1}$ and $\phi_{2}$ with right-incident momentum $p_{2}^{\prime}$. If
$p_{1}\neq p_{2}$ where $p_{2}$ is the asymptotic momentum corresponding to
$p_{2}^{\prime}$, two solutions are belonging to different energy eigenvalues
and are thus orthogonal to each other. We thus anticipate that if two momenta
become identical, only singular contributions proportional to delta functions
would appear in the inner product (for any finite contributions are absent if
the momenta are different). Such contributions only appear from the spatially
asymptotic regions $z\sim\infty$ and $z\sim-\infty$ and can be estimated as

$\begin{split}\int_{-\infty}^{\infty}dz\psi_{1}^{\dagger}\phi_{2}&=\text{divergent
parts of}\left[\int_{-\infty}^{z_{-}}dz\left(e^{\mp
ip_{1}z}\chi_{p_{1}}^{\dagger}+R_{\leftarrow}^{*}e^{\pm
ip_{1}z}\chi_{-p_{1}}^{\dagger}\right)T_{\leftarrow}e^{\mp
ip_{2}z}\chi_{-p_{2}}\right.\\\ &\phantom{=\text{divergent parts of
}}\left.+\int_{z_{+}}^{\infty}dz\,T_{\to}^{*}e^{\mp
ip_{1}^{\prime}z}\chi_{p_{1}^{\prime}}^{\dagger}\left(e^{\mp
ip_{2}^{\prime}z}\chi_{-p_{2}^{\prime}}+R_{\to}e^{\pm
ip_{2}^{\prime}z}\chi_{p_{2}^{\prime}}\right)\right]\\\ &=\pi
R_{\leftarrow}^{*}T_{\leftarrow}\frac{E_{p_{1}}}{m}\delta(p_{1}-p_{2})+\pi
T_{\to}^{*}R_{\to}\frac{E_{p_{1}^{\prime}}}{m}\delta(p_{1}^{\prime}-p_{2}^{\prime})=0,\end{split}$
(102)

where $z_{+}>0$ and $z_{-}<0$ are arbitrary large and small but finite
numbers. In this estimation, rapidly oscillating terms give no contributions
owing to the Riemann–Lebesgue lemma and use has been made of
$p_{1}^{\prime}E_{p_{1}}\delta(p_{1}-p_{2})=p_{1}E_{p_{1}^{\prime}}\delta(p_{1}^{\prime}-p_{2}^{\prime})$
which follows from the equality
$E_{p_{1}}-E_{p_{2}}=E_{p_{1}^{\prime}}-E_{p_{2}^{\prime}}$ and of (101). The
orthogonality between left-incident and right-incident degenerate solutions of
the stationary Dirac equation is thus proved.

Consider next another degenerate solution of the same Dirac equation but with
different boundary condition $\tilde{\phi}$ that has only left-moving
component at the far right end $z\sim\infty$. It should behave asymptotically
like

$\displaystyle z\to-\infty:\quad\tilde{\phi}\sim Ae^{\pm ipz}\chi_{p}+Be^{\mp
ipz}\chi_{-p},\qquad z\to\infty:\quad\tilde{\phi}\sim e^{\mp
ip^{\prime}z}\chi_{-p^{\prime}}.$ (103)

Following the same line of thought, a spatially conserved quantity
$\psi^{\dagger}\alpha_{z}\tilde{\phi}$ derives the equality

$\displaystyle\frac{p}{m}(A-R_{\leftarrow}^{*}B)=0.$ (104)

The inner product between $\psi_{1}$ and $\tilde{\phi}_{2}$ is similarly
evaluated as

$\displaystyle\int_{-\infty}^{\infty}dz\psi_{1}^{\dagger}\tilde{\phi}_{2}\propto\pi(A+R_{\leftarrow}^{*}B)\frac{E_{p_{1}}}{m}\delta(p_{1}-p_{2})=2\pi
A\frac{E_{p_{1}}}{m}\delta(p_{1}-p_{2}),$ (105)

which does not vanish because none of $A$ and $B$ are allowed to vanish. A
similar argument is also applied to cases with two solutions that have a
single component in the other asymptotic region $z\to-\infty$ and we can show
that such solutions are not orthogonal to each other. We thus conclude that
the degenerate solutions of the stationary Dirac equation, one with only
right-moving component and the other with only left-moving component in the
same spatial asymptotic region, are never orthogonal to each other.

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