Klein Paradox

A sufficiently high Dirac potential step has propagating states on its far side even when the kinetic energy lies on the negative branch. This is the essential setting of the Klein problem. Selecting the transmitted mode by its outgoing velocity gives nonnegative Dirac reflection and transmission probabilities with $R+T=1$. The surprising transmission is real; a negative probability is not required. A quantum-field interpretation of a critical step additionally needs asymptotic particle states and their occupations.

Required background. The Dirac Hamiltonian as an Operator provides the spectral branches; Dirac Current defines density and flux; Dirac in Electromagnetic Fields fixes the electrostatic coupling.

Helpful background. Pair Creation explains particle counting; Conserved Klein–Gordon Current supplies the signed-current scalar comparison.

The Dirac step and its asymptotic continua

Use $\hbar=c=1$, $m>0$, and a fixed spin channel at normal incidence. A constant unitary choice of its two components gives

$$H=-i\sigma_x\partial_x+m\sigma_z+V(x),$$ $$V(x)=V_0\theta(x),\qquad V_0>0.$$

Here $V_0=q\,\Delta\Phi$ is a potential-energy step, not a voltage. This Hamiltonian is a stationary one-body wave problem in a prescribed source. The finite jump imposes continuity of the spinor at $x=0$; its derivative need not be continuous.

Take an incident energy $E>m$ on the left and define

$$p=\sqrt{E^2-m^2},\qquad a=\frac{p}{E+m},$$ $$\epsilon_R=E-V_0.$$

In a region of constant $V$, the local dispersion is $(E-V)^2=k^2+m^2$. For real $k$ and kinetic energy $\epsilon=E-V$, a useful unnormalized eigenspinor is

$$w(\epsilon,k)= \begin{pmatrix}1\\ k/(\epsilon+m)\end{pmatrix}.$$

Direct multiplication by $k\sigma_x+m\sigma_z$ gives $\epsilon w$. Its density and current obey

$$\rho=w^\dagger w>0,\qquad j=w^\dagger\sigma_xw,\qquad \frac{j}{\rho}=\frac{k}{\epsilon}.$$

The last ratio is also the group velocity $\partial E/\partial k$. On the negative branch, velocity and momentum have opposite signs. The displayed spinor chart is singular at $\epsilon=-m$; threshold values must be obtained by a limit or a nonsingular normalization.

The right-hand region has three distinct possibilities:

Right kinetic energy	Right mode	Boundary condition
$\epsilon_R>m$	Positive-branch propagation	Choose $k_R>0$ for outgoing velocity.
$-m<\epsilon_R<m$	Evanescent mode	Choose $\operatorname{Im}k_R>0$ for decay as $x\to+\infty$.
$\epsilon_R<-m$	Negative-branch propagation	Choose $k_R<0$ for outgoing velocity.

The last interval is the Klein region, $m<E<V_0-m$. It exists only when $V_0>2m$: the left positive-energy continuum then overlaps the right negative-energy continuum at the same conserved $E$.

Matching amplitudes and currents

Suppressing the common factor $e^{-iEt}$, write

$$\psi_L(x)= \begin{pmatrix}1\\a\end{pmatrix}e^{ipx} +r\begin{pmatrix}1\\-a\end{pmatrix}e^{-ipx},$$ $$\psi_R(x)= t\begin{pmatrix}1\\b\end{pmatrix}e^{ik_Rx}, \qquad b=\frac{k_R}{\epsilon_R+m}.$$

For either propagating outgoing branch, $b>0$. Continuity gives $1+r=t$ and $a(1-r)=bt$, hence

$$r=\frac{a-b}{a+b},\qquad t=\frac{2a}{a+b}.$$

The incident, reflected, and transmitted currents are $2a$, $-2a|r|^2$, and $2b|t|^2$. Cross terms in the left current cancel. Therefore

$$R=|r|^2=\frac{(a-b)^2}{(a+b)^2},$$ $$T=\frac{b}{a}|t|^2=\frac{4ab}{(a+b)^2}.$$

Both are nonnegative and $R+T=1$. Squaring $t$ without the current factor $b/a$ would give an incorrect transmission probability.

In the gap, set $k_R=i\kappa$, where $\kappa=\sqrt{m^2-\epsilon_R^2}>0$. Then $b$ is purely imaginary, the decaying mode has zero stationary current, and $|r|=1$. It can have nonzero density near the interface even though there is no transmitted flux at infinity.

For a high step in the Klein region,

$$k_R=-\sqrt{\epsilon_R^2-m^2},\qquad b>1.$$

As $V_0\to\infty$ at fixed $E>m$, $b\to1$ and $T\to4a/(1+a)^2$, a nonzero limit. A semi-infinite nonrelativistic forbidden region would instead support only an evanescent wave. Here the right region contains a propagating branch. A smooth finite-width electric barrier is a different boundary problem and can introduce strong suppression; the abrupt step is not a universal strong-field transmission law.

An exact Klein-region example

In units with $m=1$, choose

$$E=\frac53,\qquad V_0=\frac{35}{12}, \qquad \epsilon_R=-\frac54.$$

Then

$$p=\frac43,\quad a=\frac12,\quad k_R=-\frac34,\quad b=3.$$

The matching coefficients and probabilities are

$$r=-\frac57,\quad t=\frac27,\qquad R=\frac{25}{49},\quad T=\frac{24}{49}.$$

The transmitted momentum is negative but its velocity is $k_R/\epsilon_R=3/5>0$, exactly as required by an outgoing condition at the right boundary.

If one instead chooses $k_R=+3/4$, then $b=-3$ and obtains

$$R_{\rm formal}=\frac{49}{25},\qquad T_{\rm signed}=-\frac{24}{25}.$$

That right-hand wave travels toward the interface. The algebra describes a different scattering arrangement with an incoming right-hand current. It is not the prescribed left-incident, right-outgoing one-body problem. These coefficients neither imply a negative Dirac probability density nor directly count pairs created from a vacuum.

Scalar current and quantum-field interpretation

The scalar Klein–Gordon comparison needs separate sign bookkeeping. In a stationary electrostatic region its conserved signed Klein–Gordon density is proportional to $\epsilon|\phi|^2$ and its current to $k|\phi|^2$, with the same positive overall normalization. The physical electric current has an additional factor of the signed charge $q$. In the negative branch, the outgoing choice $k<0$ gives $j/\rho=k/\epsilon>0$ while both the signed density and current are negative.

For a scalar step, matching $\phi$ and $\partial_x\phi$ can therefore give a reflected signed-flux coefficient larger than one and a negative transmitted signed-flux coefficient with an outgoing transmitted group velocity. The Dirac explanation “that sign chose an incoming mode” cannot be transferred to this scalar calculation. Its conserved charge form is not a positive one-particle probability norm. Singular parameter choices of the ideal scalar step also require care, rather than interpreting divergent stationary amplitudes as finite production probabilities.

At the field-theory level, the overlap of asymptotic continua requires distinguishing particle annihilation from antiparticle creation operators in the mode expansions. To compute a pair count, specify the in/out basis, its flux normalization, the initial vacuum or occupied state, and how the source is prepared. The Bogoliubov counting construction then determines mean occupations with the appropriate statistics. For a spatial step these asymptotic channels are organized by the left and right regions; temporal-pulse coefficients cannot be imported without changing that construction.

Consequently, a stationary one-body $R$ or $T$ is not by itself a vacuum production probability. Transmission through an abrupt high step, emission by a supercritical source with a specified preparation, and the vacuum persistence calculation in a sustained electric field are related questions with different boundary and state data. Calogeracos and Dombey discuss the wave-mechanical distinction; Gavrilov and Gitman develop the spatial-step field quantization.

Checks and useful limits

Restoring units, set $M=mc^2$. Then

$$p=\frac{\sqrt{E^2-M^2}}{c},$$ $$a=\frac{cp}{E+M},\qquad b=\frac{cp_R}{\epsilon_R+M}.$$

Plane-wave phases are $e^{ipx/\hbar}$, the current is $c\psi^\dagger\sigma_x\psi$, and the velocity ratio is $c^2p_R/\epsilon_R$. The Klein condition becomes $V_0>2mc^2$ with $mc^2<E<V_0-mc^2$. A constant shift of the potential and of the energy origin leaves $E-V$ and every physical result invariant.

In the massless limit at nonzero asymptotic kinetic energies, the outgoing branches have $a=b=1$, giving $R=0$, $T=1$ at normal incidence in this model. This limit is not a claim of perfect transmission for arbitrary transverse momentum, mass terms, or interfaces that mix additional channels.

Exercises

Current is the diagnostic. Recompute the three currents in the exact example and verify conservation without first using $R+T=1$.

Solution

$j_{\rm in}=1$, $j_{\rm ref}=-25/49$, and $j_{\rm trans}=2(3)(4/49)=24/49$. Thus $j_{\rm in}+j_{\rm ref}=j_{\rm trans}$. The right-hand density is $(1+3^2)(4/49)=40/49$, so $j_{\rm trans}/\rho_{\rm trans}=3/5$.

A gap solution. Take $E=2m$ and $V_0=2m$. Determine $k_R$, $b$, and $R$ for the decaying solution.

Solution

$\epsilon_R=0$, $k_R=im$, $b=i$, and $a=\sqrt3/3$. Therefore $r=(a-i)/(a+i)$ has unit magnitude. The right current is $2\operatorname{Re}(b)|t|^2e^{-2mx}=0$ and $R=1$.

What does the step establish? Does a measured stationary transmission coefficient determine the mean number of pairs produced from an incoming vacuum?

Solution

No. The coefficient specifies a flux ratio for the stated one-body boundary problem. Vacuum pair counting additionally requires quantized-field in/out modes and their relation, a choice of initial state, and source preparation. Those data determine which operator mixing coefficient contributes to the occupation number.

References

• Calogeracos, A., and N. Dombey. “History and Physics of the Klein Paradox.” Contemporary Physics 40, 313–321 (1999). doi:10.1080/001075199181387; author manuscript. Historical and wave-mechanical analysis, including the velocity choice in the step problem.
• Gavrilov, S. P., and D. M. Gitman. “Quantization of Charged Fields in the Presence of Critical Potential Steps.” Physical Review D 93, 045002 (2016). doi:10.1103/PhysRevD.93.045002; author manuscript. Quantized-field channel definitions and vacuum processes for spatial electric steps.
• Greiner, Walter. Relativistic Quantum Mechanics: Wave Equations. 3rd ed. Springer (2000). doi:10.1007/978-3-662-04275-5. Relativistic potential scattering and interpretation of negative-energy branches.