Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 6

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. The opposite-frequency branch is not obtained by replacing ${\omega }_k$ by the same shifted positive frequency in both terms. A globally constant offset produces no electric field, whereas different offsets on opposite sides of a step cannot be removed by a single smooth global phase.

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

The negative-kinetic-energy branch is interpreted as an antiparticle channel. In the overlap $m{c}^2<E<V_0-m{c}^2$, particles can escape to the left and oppositely charged antiparticles to the right. This interval exists only when $V_0>2m{c}^2$. The external electrostatic source supplies the energy for pair creation; the reflected particle flux can exceed the incident one because the outgoing state need not contain only the incident particle.

The conserved quantity is signed charge current, not a positive one-particle probability in both regions. Negative $\mathcal{T}$ denotes rightward transport of opposite charge, so it is consistent with $\mathcal{R}>1$ and $\mathcal{R}+\mathcal{T}=1$. A quantized charged field with particle and antiparticle creation operators gives the appropriate interpretation of this Klein paradox. This bosonic result must not be replaced by a fermionic reflection law or interpreted as a failure of charge conservation.^[ref]