The Klein Paradox

In 1929 Oskar Klein set up the most innocent scattering problem imaginable — a Dirac electron running into an electrostatic step $V(x)=V_0\theta(x)$ — and pulled out something that looked impossible. Make the step tall enough, taller than $E+mc^2$ , and the barrier does not reflect the electron the way a mountain reflects a ball. A real, propagating wave leaks through it, undiminished, no matter how high you build the wall. Raise $V_0$ to infinity and the transmission does not vanish. For a physicist trained on the Schrödinger step — where a wall above the particle's energy kills transmission exponentially — this is scandalous. You are ready for this because you already command the free Dirac equation, its plane-wave spinors, and the fact that it carries both positive- and negative-energy solutions. The Klein step is where those negative-energy solutions stop being a bookkeeping nuisance and start doing physics. The resolution turns on one sign — the sign of the transmitted momentum — and getting it right is the entire lesson. Choose it by the phase velocity and you reproduce Klein's headline "reflection greater than one"; choose it by the group velocity, as physical transmission demands, and the anomaly becomes an ordinary, conserved current flowing into the step. That current is carried by a positron. We will solve the Dirac step exactly, pin down the three kinematic regimes, and prove that the "paradox" is really the vacuum handing you a particle-antiparticle pair at the boundary. The single-particle picture cracks; quantum field theory walks in through the crack.

⚡ Key Takeaways

The Klein paradox is the persistence of relativistic transmission through an electrostatic step even when $V_0>E+mc^2$ — where a nonrelativistic barrier would reflect totally.
It happens because inside a supercritical step the propagating channel is the negative-energy branch of the Dirac equation: the incident electron's energy lines up with the far-side antiparticle continuum.
Everything hinges on the sign of the transmitted momentum $p'$ . Fixing it by the group velocity (so the transmitted wave carries charge to the right) gives the flux ratio $\kappa>0$ , and then $R<1$ , $T>0$ , with $R+T=1$ — an ordinary conserved current, not a violation.
The famous " $R>1$ , $T<0$ " statement comes from the wrong momentum sign ( $\kappa<0$ ); it still obeys $R+T=1$ but mislabels a positron current as negative electron transmission.
The transmitted mode is a positron entering the step; the step is a static source of $e^+e^-$ pairs. This is why single-particle relativistic QM must be promoted to a field theory.
Concrete descendants: Schwinger pair production in a strong field, spontaneous positron emission by supercritical nuclei, and perfect Klein tunnelling in graphene.
KEY IDEA: the Klein step transmits because a supercritical potential opens a negative-energy (antiparticle) channel; matching currents correctly gives $R+T=1$ with $T>0$ carried by a positron — the vacuum, not a single tunnelling particle, supplies the transmitted charge.

💡 The Core Insight

A supercritical step $V_0>E+mc^2$ opens a propagating negative-energy channel inside the barrier; the transmitted wave is a right-moving positron, and once its current is signed by the group velocity the scattering obeys $R+T=1$ with both coefficients positive — the "paradox" is pair creation, not a broken conservation law.

The physical picture

Nonrelativistically, a step above the particle's energy is a dead end. The wave inside turns evanescent, $\psi\sim e^{-\kappa x}$ , and the transmission dies exponentially as the wall grows. The article Potential Step and Reflection works this out in full: for $E<V_0$ the Schrödinger step gives total reflection, $R=1$ , $T=0$ , with only an exponentially decaying tail inside. That is the intuition the Dirac equation is about to overturn. Relativistically there is a second continuum sitting a gap $2mc^2$ below the first. A positive electrostatic step lifts both continua on the far side of the boundary. Nudge $V_0$ past $E-mc^2$ and the transmitted electron energy $E-V_0$ enters the mass gap — the wave goes evanescent, exactly like the Schrödinger case, and you get total reflection. But keep pushing, past $V_0=E+mc^2$ , and the far-side negative-energy continuum rises far enough that the incident energy $E$ now lines up inside it. A brand-new travelling wave becomes available — on the lower branch. The wall has become a window.

Figure

The two continua and the supercritical window — as V₀ rises the far-side negative-energy sea lifts until E lands inside it

The transmitted object is best read as a positron leaving the step toward $+\infty$ — equivalently, a negative-energy electron whose group velocity points right even though its wavevector points left. Negative-Energy Solutions and the Interpretive Crisis is where these lower-branch states are catalogued and tied to antiparticles; The Dirac Sea and Antiparticles gives the hole-theory reading, and Charge Conjugation and the Positron supplies the map from a negative-energy electron solution to a positive-energy positron. The Klein step is where all three cash out at once: an electron reflects, a positron is driven into the barrier, and the "extra" current is pair current.

Figure

Reflection and transmission versus step height — the Dirac step keeps transmitting where the nonrelativistic step (dashed) has died to zero

The core equations

The whole problem is one first-order matching at $x=0$ . We keep $\hbar=c=1$ and measure energies in units of $mc^2$ , restoring factors only in the worked examples. The Dirac Hamiltonian with a scalar-potential step reads as follows; it is the same operator developed in The Dirac Equation, now with a piecewise-constant $V(x)$ added to the energy.

\big[\,\boldsymbol{\alpha}\cdot\mathbf p + \beta m + V(x)\,\big]\psi = E\,\psi,\qquad V(x)=V_0\,\theta(x) \qquad (1)

where: $\boldsymbol\alpha,\beta$ are the Dirac matrices ( $\alpha_i=\gamma^0\gamma^i,\ \beta=\gamma^0$ ); $V_0$ is the electrostatic step, coupling like the time component $eA^0$ ; $\theta(x)$ is the Heaviside step; $E$ is the conserved total energy of the stationary state. Because $V$ shifts the energy, in region II the free dispersion holds with $E\to E-V_0$ . On the two sides the momentum is fixed by the relativistic dispersion. In region I ( $x<0$ ) the incident electron is an ordinary positive-energy state; in region II ( $x>0$ ) the available branch depends on where $E-V_0$ falls relative to the mass gap.

E=\sqrt{p^2+m^2}\quad(x<0),\qquad (E-V_0)^2 = p'^2+m^2\quad(x>0) \qquad (2)

where: $p$ is the incident/reflected momentum, $p'$ the transmitted momentum. The inside relation is quadratic in $p'$ ; which root we take is the crux of the paradox and is fixed physically below, not by algebra. The scattering state is built from a right-moving incident spinor, a left-moving reflected spinor in region I, and a single transmitted spinor in region II. For a spin-up electron the relevant components are the first and third of the four-spinor.

\psi_{\mathrm I}=\begin{pmatrix}1\\[2pt]0\\[2pt]\tfrac{p}{E+m}\\[2pt]0\end{pmatrix}e^{ipx}+r\begin{pmatrix}1\\[2pt]0\\[2pt]\tfrac{-p}{E+m}\\[2pt]0\end{pmatrix}e^{-ipx},\qquad \psi_{\mathrm{II}}=t\begin{pmatrix}1\\[2pt]0\\[2pt]\tfrac{p'}{E-V_0+m}\\[2pt]0\end{pmatrix}e^{ip'x} \qquad (3)

where: the lower-to-upper ratio $p/(E+m)$ is the standard positive-energy spinor structure from Dirac Spinors and Plane-Wave Solutions; inside the step it uses $E-V_0$ , which becomes the antiparticle ratio when $E-V_0<-m$ . The reflection and transmission amplitudes $r,t$ are the unknowns. Continuity of $\psi$ at $x=0$ (the Dirac equation is first order, so no derivative condition is imposed) collapses to two scalar equations whose solution is formally identical to the nonrelativistic step — but with a relativistic flux ratio $\kappa$ .

\kappa \equiv \frac{p'}{p}\,\frac{E+m}{E-V_0+m},\qquad r=\frac{1-\kappa}{1+\kappa},\qquad t=\frac{2}{1+\kappa} \qquad (4)

where: $\kappa$ is the ratio of transmitted to incident current amplitude. Everything relativistic hides in $\kappa$ : its sign, set by both $p'$ and the denominator $E-V_0+m$ , decides whether the step behaves ordinarily or "paradoxically." The measured coefficients are current ratios. Reflection is $R=|r|^2$ ; transmission carries the flux factor $\kappa$ , and current continuity forces them to sum to one.

R=\left(\frac{1-\kappa}{1+\kappa}\right)^2,\qquad T=\kappa\left(\frac{2}{1+\kappa}\right)^2,\qquad R+T=1 \qquad (5)

where: $R$ and $T$ are reflection and transmission probabilities. With $\kappa>0$ both lie in $[0,1]$ ; the physics question is whether the supercritical regime forces $\kappa$ positive or negative — and that is exactly the sign question we now settle.

Derivation ASolving the Dirac step: from four-spinor matching to R+T=1

Our goal is to reduce the full first-order matching problem to the two-line system in equation (4) and confirm the current identity $R+T=1$ , with every sign kept honest. The strategy is to match the two active spinor components at $x=0$ , define the flux ratio $\kappa$ , and read off $r,t$ ; the subtlety we are watching for is that $\kappa$ 's numerator and denominator can each change sign in the supercritical regime, so we carry them separately rather than lumping them into $|\kappa|$ .

Step 1. Impose continuity of the upper component of $\psi$ at $x=0$ . The incident and reflected upper entries are both $1$ , and the transmitted upper entry is $1$ , giving

1+r=t.

Why: the first-order Dirac equation only requires $\psi$ itself to be continuous — integrating (1) across the jump gives no delta unless $\psi$ is discontinuous.

Step 2. Impose continuity of the (third) lower component. Dividing through by the incident ratio $p/(E+m)$ and using the definition of $\kappa$ ,

1-r=\kappa\,t.

Why: the lower spinor entry carries the momentum-to-energy ratio, and the ratio of the region-II ratio to the region-I ratio is precisely $\kappa$ .

Step 3. Solve the two-line system. Adding and subtracting,

t=\frac{2}{1+\kappa},\qquad r=\frac{1-\kappa}{1+\kappa}.

Why: this is algebraically identical to the Schrödinger step; all the relativity has been pushed into the single number $\kappa$ .

Step 4. Build the current coefficients. The probability current $j^x=\psi^\dagger\alpha^x\psi$ evaluated on each plane wave gives an incident flux $\propto p/(E+m)$ , so the transmitted-to-incident flux ratio is exactly $\kappa$ :

R=|r|^2,\qquad T=\kappa\,|t|^2.

Why: $T$ must weight $|t|^2$ by how much current the transmitted mode actually carries, which is the flux ratio $\kappa$ .

Step 5. Check conservation. Substituting,

R+T=\frac{(1-\kappa)^2}{(1+\kappa)^2}+\frac{4\kappa}{(1+\kappa)^2}=\frac{(1+\kappa)^2}{(1+\kappa)^2}=1.

Why: current continuity in a stationary state ( $\partial_x j^x=0$ ) demands this, and the algebra confirms it identically in $\kappa$ — the conservation law never breaks. *Verified: reproduces the standard 1D Dirac step of Bjorken–Drell §5.3 and Greiner Ch. 8; as $V_0\to0$ , $\kappa\to1$ , $R\to0$ , $T\to1$ , and a numerical check gives $R+T=1$ to machine precision for every regime.*

Derivation BThe sign of p′: group velocity, not phase velocity, and why κ>0

The single decision that makes or breaks the Klein analysis is which root of $p'^2=(E-V_0)^2-m^2$ to keep. Our goal is to show that the physical requirement — that the transmitted wave carry charge to the right — forces $p'<0$ in the supercritical regime, and that this makes $\kappa$ positive, so $R<1$ and $T>0$ . The thing to watch is that on the negative-energy branch phase velocity and group velocity point in opposite directions; trusting the phase velocity is the classic mistake that manufactures " $R>1$ ."

Step 1. Write the transmitted group velocity from the region-II dispersion $E-V_0=\sqrt{p'^2+m^2}$ on the branch that applies, differentiating:

v_g=\frac{\partial E}{\partial p'}=\frac{p'}{E-V_0}.

Why: the group velocity is the speed at which the wave packet — and hence probability/charge — actually moves.

Step 2. In the supercritical regime $V_0>E+m$ , the inside energy sits on the lower branch, $E-V_0<-m<0$ . For the transmitted wave to move charge rightward we need $v_g>0$ , which with $E-V_0<0$ forces

p'<0.

Why: a negative denominator and a positive required $v_g$ can only be reconciled by a negative numerator $p'$ — the wavevector points left while the current points right.

Step 3. Insert the signs into $\kappa=\dfrac{p'}{p}\dfrac{E+m}{E-V_0+m}$ . Here $p'<0$ and $E-V_0+m<0$ (since $V_0>E+m$ ), so the two negatives cancel:

\kappa=\underbrace{\frac{p'}{p}}_{<0}\cdot\underbrace{\frac{E+m}{E-V_0+m}}_{<0}>0.

Why: the numerator sign and the denominator sign flip together in the supercritical window, leaving $\kappa$ net positive.

Step 4. With $\kappa>0$ finite, equation (5) gives $0<R<1$ and $0<T<1$ with $R+T=1$ : genuine partial transmission. Had we kept the phase-velocity root $p'>0$ , only the denominator would be negative, giving $\kappa<0$ , $R=\big(\tfrac{1-\kappa}{1+\kappa}\big)^2>1$ , and $T<0$ — the textbook "paradox," which still satisfies $R+T=1$ but mislabels the physics. Why: both sign choices conserve current; only the group-velocity choice assigns the transmitted current to a right-moving object, which is what a scattering experiment measures. *Verified: matches Calogeracos & Dombey, Contemp. Phys. 40, 313 (1999), who identify the group-velocity prescription as the resolution; a numerical evaluation at $E=1.2\,mc^2$ , $V_0=3\,mc^2$ gives $\kappa=+6.205$ , $R=0.522$ , $T=0.478$ , $R+T=1$ .*

Derivation CCharge bookkeeping: the transmitted current is a positron

Having fixed the signs, our goal is to show what the transmitted current physically is, and to confirm that no conservation law is stressed — the Dirac density is positive-definite throughout. The route is to evaluate the current on the transmitted mode and reinterpret the negative-energy electron as a positron via charge conjugation. Watch that, unlike the Klein–Gordon case, the Dirac density never goes negative; only particle number is at stake.

Step 1. The Dirac density $\rho=\psi^\dagger\psi\ge0$ is positive-definite, as established in The Dirac Equation; the pathology that afflicts the scalar theory (see Klein-Gordon Probability Current and Negative Densities) is absent. So probability is never the problem here. Why: any worry about "negative probability" is a Klein–Gordon disease, not a Dirac one; the Dirac paradox is about particle content, not probability sign.

Step 2. The transmitted mode is a negative-energy electron solution moving with $v_g>0$ . By the Feynman–Stückelberg reading and the explicit map in Charge Conjugation and the Positron, a right-moving negative-energy electron is a right-moving positron of positive energy. Why: charge conjugation sends a negative-energy electron solution to a positive-energy positron solution; the transmitted object is physically a positron.

Step 3. Track the charge. The incident electron current splits into a reflected electron current and a transmitted positron current; with $\kappa>0$ this is simply $R+T=1$ with $T>0$ . In the alternative " $T<0$ " convention the same fact appears as $R=1+|T|$ , "more electron-current back than in," the deficit made up by the emitted positron. Why: whichever sign convention one adopts, the conserved quantity is charge, and the ledger balances once the positron is counted.

Step 4. Read the field-theory meaning. A static step steep on the Compton scale and taller than $2mc^2$ makes the local vacuum unstable: it emits $e^+e^-$ pairs, the electron reflecting and the positron transmitting. This is the sharp-step limit of the Schwinger effect. Why: the "extra" current the single-particle picture could not source is supplied by pair creation — the crack through which quantum field theory enters. *Verified: consistent with the hole-theory account (Greiner Ch. 8) and the QFT pair-creation picture; the positive-definiteness of $\rho$ and the identity $R+T=1$ hold in every regime by direct evaluation.*

Limiting cases

With $mc^2<E$ fixed, three regimes appear, bounded by $V_0=E-mc^2$ and $V_0=E+mc^2$ .

Case	Result	Meaning
Weak step, $V_0<E-mc^2$	$\kappa>0$ , $0<R<1$ , $0<T<1$	Ordinary relativistic transmission: a real inside wave on the positive-energy branch. No anomaly.
Gap step, $E-mc^2<V_0<E+mc^2$	$R=1$ , $T=0$	Inside energy $E-V_0$ lies in the mass gap $(-mc^2,mc^2)$ ; the transmitted wave is evanescent — exactly the nonrelativistic Potential Step and Reflection behaviour.
Klein step, $V_0>E+mc^2$	$\kappa>0$ , $R<1$ , $T>0$ (group-velocity signing)	A negative-energy channel reopens; the transmitted wave is a positron. $R+T=1$ holds; the naive phase-velocity signing instead reports $R>1$ , $T<0$ .
$V_0\to\infty$	$T\to$ finite, not $0$	Transmission does not die as the wall grows — the starkest face of the paradox, impossible nonrelativistically.

⚠️ The transmitted momentum sign is not the direction of the current

The single most common error is to equate the sign of the transmitted wavevector $p'$ with the direction the wave carries charge. On the negative-energy branch inside a supercritical step, the group velocity (hence the physical current) points opposite to the phase velocity. Choosing $p'$ by the phase velocity yields the spurious " $R>1$ , $T<0$ ." Fix $p'$ by demanding positive group velocity and you get $\kappa>0$ , $R<1$ , $T>0$ , and $R+T=1$ — the same conserved current, now physically labelled.

Worked examples

Example 1 — The supercritical Dirac step (a positron gets through). Take a spin-up electron of energy $E=1.2\,mc^2$ incident on a step $V_0=3\,mc^2$ . Since $V_0=3>E+mc^2=2.2$ , this is the Klein regime. Work in units $\hbar=c=m=1$ .

Step 1. Incident momentum: $p=\sqrt{E^2-1}=\sqrt{1.44-1}=\sqrt{0.44}=\mathbf{0.663}$ .

Step 2. Inside energy: $E-V_0=1.2-3=-1.8$ , safely below $-1$ , so a propagating lower-branch wave exists. Its momentum magnitude is $|p'|=\sqrt{(E-V_0)^2-1}=\sqrt{3.24-1}=\sqrt{2.24}=1.497$ . The group-velocity prescription fixes $p'=-1.497$ .

Step 3. Flux ratio: $\kappa=\dfrac{p'}{p}\dfrac{E+1}{E-V_0+1}=\dfrac{-1.497}{0.663}\cdot\dfrac{2.2}{-0.8}=(-2.258)(-2.75)=\mathbf{6.205}$ . Positive, as the two sign flips demand.

Step 4. Coefficients: $R=\left(\dfrac{1-6.205}{1+6.205}\right)^2=\left(\dfrac{-5.205}{7.205}\right)^2=(-0.7224)^2=\mathbf{0.522}$ , and $T=\kappa\left(\dfrac{2}{1+\kappa}\right)^2=6.205\left(\dfrac{2}{7.205}\right)^2=6.205\times0.07705=\mathbf{0.478}$ .

Step 5. Conservation: $R+T=0.522+0.478=\mathbf{1.000}$ , both positive. A real fraction of the incident current — carried by a transmitted positron — enters the "forbidden" region. Transmission does not vanish, yet nothing is over-unity.

Example 2 — The nonrelativistic contrast (total reflection). Put the same under-barrier condition to a Schrödinger particle: energy $E$ below the step $V_0$ , so $E<V_0$ . There is no negative-energy sea, so no second channel.

Step 1. Region-I wavevector $k=\sqrt{2mE}/\hbar$ ; region-II is classically forbidden, wave $\sim e^{-qx}$ with $q=\sqrt{2m(V_0-E)}/\hbar$ real.

Step 2. Matching $\psi$ and $\psi'$ (second-order equation needs both) gives $r=\dfrac{k-iq}{k+iq}$ , so $R=|r|^2=\dfrac{k^2+q^2}{k^2+q^2}=\mathbf{1}$ .

Step 3. With $R=1$ and no propagating inside wave, $T=\mathbf{0}$ : total reflection, exactly as in Potential Step and Reflection. No matter how tall the Schrödinger wall, transmission is strictly zero once $E<V_0$ — the precise opposite of the Dirac step's persistent transmission.

History

The Klein paradox was born from a sign kept honest for the first time only decades after Klein wrote it down, and its interpretation tracks the birth of quantum field theory.

Klein Paradox

posed 1929, resolved via QED 1930s–1990s · The prediction that a Dirac electron transmits through a supercritical electrostatic step, resolved as electron–positron pair creation by the vacuum — a foundational stress test of single-particle relativistic quantum mechanics.

The Klein paradox is one of the few "named results" in physics that is a problem rather than a person's theorem, and its lifetime maps the maturation of relativistic quantum theory. Oskar Klein wrote it down in 1929 as an apparent embarrassment; Fritz Sauter and Friedrich Hund sharpened the strong-field, pair-creating reading in the 1930s and 40s; Julian Schwinger's 1951 vacuum-decay calculation gave it a dynamical cousin; and only in 1999 did Calogeracos and Dombey settle the sign question definitively with the group-velocity prescription. Today it is taught not as a paradox but as the cleanest demonstration that relativistic wave mechanics is the one-particle shadow of a field theory — and, since 2006, as a phenomenon you can measure on a chip of graphene.

⏳ History

1929

— Oskar Klein analyses the Dirac equation on a step potential and finds transmission that does not vanish for tall barriers (Z. Phys. 53, 157)

1931

— Fritz Sauter studies the smooth-step and strong-field versions, sharpening the pair-creation reading (Z. Phys. 69, 742)

1941

— Friedrich Hund connects the effect to spontaneous emission of electron–positron pairs by strong fields

1951

— Julian Schwinger computes the vacuum pair-production rate in a constant field, the dynamical cousin of the Klein step

1999

— Calogeracos and Dombey give the definitive group-velocity resolution and history (Contemp. Phys. 40, 313)

2006

— Katsnelson, Novoselov and Geim identify perfect Klein tunnelling in graphene (Nat. Phys. 2, 620)

Where this lives today

The Klein step is no longer a curiosity — it is the clean idealisation of vacuum instability in strong fields, and it has become experimentally live. In graphene, low-energy carriers obey a two-dimensional massless Dirac equation with an effective speed $v_F\approx10^6\,\mathrm{m/s}$ ; a potential step transmits with probability one at normal incidence regardless of height, because there is no mass gap to suppress the negative-energy channel and pseudospin conservation forbids backscattering. This perfect Klein tunnelling, seen in graphene p–n junctions, frustrates electrostatic confinement of electrons and is now a design constraint for graphene electronics. At the frontier, the same physics drives the search for the Schwinger effect — real $e^+e^-$ pairs pulled from the vacuum by fields approaching the critical value $E_c=m^2c^3/e\hbar\approx1.3\times10^{18}\,\mathrm{V/m}$ . Next-generation ultra-intense lasers and heavy-ion collisions (where transient fields around supercritical nuclear charges $Z\alpha\to1$ can dive a bound level into the negative continuum) aim to catch the vacuum in the act. In every case the bookkeeping is the Klein step's: an incident amplitude, a reflected electron, and a transmitted positron, with charge — never particle number — the conserved quantity.

Going further

References. Griffiths, Introduction to Elementary Particles, 2nd ed., §7 · Sakurai, Advanced Quantum Mechanics, §3.10 · Bjorken & Drell, Relativistic Quantum Mechanics, §5.3 · Greiner, Relativistic Quantum Mechanics: Wave Equations, Ch. 8 · Calogeracos & Dombey, Contemp. Phys. 40, 313 (1999), "History and physics of the Klein paradox" · Katsnelson, Novoselov & Geim, Nat. Phys. 2, 620 (2006).

Provenance. The spinor matching gives $r=(1-\kappa)/(1+\kappa)$ , $t=2/(1+\kappa)$ with $\kappa=(p'/p)(E+m)/(E-V_0+m)$ , reproducing Bjorken–Drell §5.3, Greiner Ch. 8, and Sakurai §3.10. This article adopts the group-velocity signing of $p'$ (Calogeracos & Dombey 1999), which yields the physically labelled $R<1$ , $T>0$ , $R+T=1$ throughout, rather than the historically famous but sign-artefact " $R>1$ , $T<0$ ." Both conventions obey $R+T=1$ ; the difference is only whether the transmitted current is booked as a right-moving positron ( $T>0$ ) or a negative electron transmission ( $T<0$ ). The regime boundaries $V_0=E\pm mc^2$ and the critical field $E_c=m^2c^3/e\hbar\approx1.3\times10^{18}\,\mathrm{V/m}$ are the standard QED values; $Z\alpha\to1$ point-charge supercriticality (near $Z\approx170$ ) is textbook. All coefficients recomputed numerically ( $E=1.2\,mc^2$ , $V_0=3\,mc^2$ : $\kappa=6.205$ , $R=0.522$ , $T=0.478$ ).

Builds on: The Dirac Equation · Dirac Spinors and Plane-Wave Solutions · Negative-Energy Solutions and the Interpretive Crisis · Klein-Gordon Probability Current and Negative Densities · Potential Step and Reflection. Leads to: The Dirac Sea and Antiparticles · Charge Conjugation and the Positron · field quantization and the Schwinger effect.

Is the Klein paradox a real contradiction in quantum mechanics?

No. Nothing in the mathematics is inconsistent — current is conserved ( $R+T=1$ ) in every regime. What fails is the interpretation of the Dirac equation as a single-particle theory. The "paradox" is only paradoxical if you insist one electron goes in and demand a single-particle account of the outgoing currents. Once you allow the step to create an electron–positron pair, the bookkeeping is ordinary. The lesson is historical and conceptual: relativistic wave equations must be quantised as fields.

Where exactly does the sign convention "R>1" come from, if it's wrong?

It comes from choosing the transmitted momentum $p'>0$ by matching the phase velocity direction to the wavevector. That choice makes the flux ratio $\kappa<0$ , and then $R=\big(\tfrac{1-\kappa}{1+\kappa}\big)^2>1$ with $T=\kappa|t|^2<0$ . It is not "wrong" arithmetic — it still gives $R+T=1$ — but it mislabels a right-moving positron current as negative electron transmission. Fixing $p'$ by the group velocity ( $p'<0$ here) flips $\kappa$ positive and gives the physically transparent $R<1$ , $T>0$ .

Why doesn't the transmission die as V₀ → ∞, the way it does for a Schrödinger wall?

Because a Schrödinger wall higher than $E$ has no propagating inside channel — the wave is evanescent and $T\to0$ . A Dirac step taller than $E+mc^2$ has a propagating negative-energy channel that never closes as $V_0$ grows; if anything it becomes more available. The transmitted current settles to a finite fraction rather than vanishing. Persistent transmission through an ever-taller wall is the sharpest signature that a genuinely new (antiparticle) degree of freedom is participating.

Is the Dirac density ever negative here, like in Klein–Gordon?

No. The Dirac density $\rho=\psi^\dagger\psi$ is a sum of squared moduli and is positive-definite everywhere, as established in The Dirac Equation. The negative-density disease belongs to the Klein–Gordon equation (see Klein-Gordon Probability Current and Negative Densities). In the Klein paradox the only quantity that behaves unexpectedly is the transmission coefficient under a bad sign choice, and even that is a current, not a probability density. Particle number, not probability, is what the single-particle picture cannot hold onto.

What is the role of the mass gap in the three regimes?

The gap $2mc^2$ between the positive- and negative-energy continua defines the boundaries. If $V_0<E-mc^2$ , the inside energy stays in the upper continuum — ordinary transmission. If $E-mc^2<V_0<E+mc^2$ , the inside energy $E-V_0$ falls inside the gap, where no propagating state exists — the wave is evanescent and you get total reflection, just like nonrelativistic quantum mechanics. Only when $V_0>E+mc^2$ does $E-V_0$ drop below $-mc^2$ into the lower continuum, opening the antiparticle channel and the paradox.

How is the Klein step related to the Schwinger effect?

The Klein step is the static, sharp-boundary idealisation of vacuum pair production. Schwinger's calculation treats a constant electric field, which does work $\sim eE\,\lambda_C$ over a Compton wavelength; when that approaches $2mc^2$ the vacuum decays into pairs at a rate $\propto\exp(-\pi m^2c^3/\hbar eE)$ . The Klein step replaces the smooth field by an infinitely steep potential jump; the condition $V_0>E+mc^2$ is the same "the field can supply $2mc^2$ " statement in a different geometry. Both describe the vacuum emitting an $e^+e^-$ pair when a strong field overlaps the two continua.

Does graphene really realise the Klein paradox, or is it just an analogy?

It realises the same equation, which is what matters. Graphene's low-energy electrons obey a 2D massless Dirac equation, with the sublattice degree of freedom playing the role of spin (pseudospin) and $v_F\approx10^6\,\mathrm{m/s}$ playing the role of $c$ . A gate-defined potential step is a genuine relativistic step for these carriers, and the predicted perfect transmission at normal incidence — no mass gap, pseudospin conservation forbidding backscattering — has been observed. It is not a loose metaphor: it is the massless limit of exactly the physics analysed here, in a tabletop material.