= Solution
\b[Uniform system and dispersion.] A <plane wave> reduces the differential <Hamiltonian> to a two-by-two <matrix> acting on its <spinor>:
$$
H(\mathbf k)=
\begin{pmatrix}
V&\hbar c(k_x-ik_y)\\
\hbar c(k_x+ik_y)&-V
\end{pmatrix}
=\mathbf d(\mathbf k)\cdot\boldsymbol\sigma,
\qquad
\mathbf d=(\hbar ck_x,\hbar ck_y,V).
$$
The Pauli anticommutation relations imply $H(\mathbf k)^2=|\mathbf d|^2I$. Choosing a <spinor> in one of its <eigenspaces> therefore gives
$$
\boxed{E_\pm(k)=\pm\sqrt{\hbar^2c^2k^2+V^2},\qquad k=|\mathbf k|.}
$$
An arbitrary constant <spinor> multiplying a <plane wave> is not necessarily an <energy eigenstate>; it must be an eigenvector of this <matrix>. For example, for the upper branch and $\varepsilon=|\mathbf d|$, one can use
$$
u_+(\mathbf k)=
\frac1{\sqrt{2\varepsilon(\varepsilon+V)}}
\begin{pmatrix}\varepsilon+V\\\hbar c(k_x+ik_y)\end{pmatrix},
$$
away from the coordinate singularity $\varepsilon+V=0$, where another normalized eigenvector chart is used.
For the usual $c>0$ convention, the massless branches are $E_\pm=\pm\hbar ck$. For nonzero $V$, the two branches have a gap $2|V|$ and are quadratic near $k=0$:
$$
E_\pm\simeq\pm|V|\pm\frac{\hbar^2c^2k^2}{2|V|}.
$$
\b[Changing the sign of $V$ does not change these energies.] It reverses the out-of-plane <spin> <polarization> of a given band at the origin. The plots for equal positive and negative mass magnitudes therefore have identical dispersion curves, despite their different <spinors>.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2016/paper-1-dirac-bands.png]
{title=Original calculated massive and massless two-dimensional Dirac dispersions; equal and opposite masses share energies but reverse the band spin at zero momentum}
\b[<Spin> direction in the massless case.] For $k>0$ put $k_x+ik_y=ke^{i\varphi}$. The normalized eigenspinors can be chosen as
$$
u_s(\mathbf k)=\frac1{\sqrt2}
\begin{pmatrix}1\\s e^{i\varphi}\end{pmatrix},
\qquad E_s=s\hbar ck,\qquad s=\pm1.
$$
Direct multiplication gives $\langle\sigma_x\rangle=s\cos\varphi$, $\langle\sigma_y\rangle=s\sin\varphi$ and $\langle\sigma_z\rangle=0$. Since the <Pauli matrices> represent physical <spin> in this question,
$$
\boxed{\langle\mathbf S\rangle_s
=s\frac{\hbar}{2}\frac{(k_x,k_y,0)}{k}.}
$$
The <spin> is \b[parallel to <momentum> on the positive-energy branch and antiparallel on the negative-energy branch]. Thus the branch-independent statement is collinearity; the printed parallel claim needs the positive-branch qualification. At $k=0,V=0$, the <Hamiltonian> is zero and <spin> is not fixed by a <wavevector> direction.
\b[Mass-domain-wall <eigenstate>.] Assume $c>0$ for the stated normalizable wall state and write
$$
F(y)=\exp\!\left[-\frac1{\hbar c}\int_0^y V(s)\,\mathrm ds\right],
\qquad u_-=\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
Then $F'=-VF/(\hbar c)$, $\sigma_xu_-=-u_-$, and $(i\sigma_y+\sigma_z)u_-=0$. Applying the full operator gives
$$
H\{e^{ik_xx}F(y)u_-\}
=e^{ik_xx}F(y)
\{\hbar ck_x\sigma_x+iV(y)\sigma_y+V(y)\sigma_z\}u_-
=-\hbar ck_x\,e^{ik_xx}F(y)u_-.
$$
The transverse derivative and mass terms cancel pointwise. Hence \b[the specified state is an <eigenstate> with $E=-\hbar ck_x$].
For a step mass $V(y)=V_+>0$ above the interface and $V(y)=V_-<0$ below it,
$$
F(y)=\begin{cases}
e^{-y/\xi_+},&y>0,\\
e^{y/\xi_-},&y<0,
\end{cases}
\qquad
\boxed{\xi_+=\frac{\hbar c}{V_+},\qquad
\xi_-=\frac{\hbar c}{|V_-|}.}
$$
The <Localization length at a step Dirac mass wall> sets the two exponential tails. The amplitude is continuous, peaks at the interface and decays on both sides, with generally unequal decay lengths. The <probability density> decays with lengths $\xi_\pm/2$. A normalized transverse function is $f(y)=\sqrt{2/(\xi_++\xi_-)}F(y)$ when multiplying the unit <spinor> $u_-$. The longitudinal <plane wave> can independently be box-normalized. A step in $V$ allows a step in $F'$; the first-order Dirac equation does not impose continuity of that derivative.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2016/paper-1-dirac-interface.png]
{title=Original mass-wall solution: asymmetric exponential confinement, the chiral energy branch and rigid longitudinal translation of a localized packet}
\b[Propagation.] The interface dispersion is the straight line
$$
\boxed{E(k_x)=-\hbar ck_x,\qquad
v_x=\frac1{\hbar}\frac{\mathrm dE}{\mathrm dk_x}=-c.}
$$
Every component on this branch travels in the negative $x$ direction. A packet has the form
$$
\Psi(x,y,t)=f(y)u_-\int A(k_x)e^{ik_x(x+ct)}\,\mathrm dk_x
=f(y)u_-\,g(x+ct).
$$
This <dispersionless chiral interface wavepacket> therefore translates rigidly with speed $c$ in that direction, without dispersion in this exactly linear model. It remains confined in $y$; there is no free transverse <group velocity>, because the bound-state family is not labelled by a transverse travelling-wave <momentum>.
The <continuity equation> gives the current components $j_x=c\Psi^\dagger\sigma_x\Psi=-c|\Psi|^2$ and $j_y=c\Psi^\dagger\sigma_y\Psi=0$. This explicitly confirms absence of <probability> transport away from the interface. For the step-mass case the profile is normalizable because its tails have positive decay constants and $c>0$. A general sign-changing profile also requires $F\in L^2(\mathbb R)$; nonzero asymptotic masses suffice, whereas signs alone do not guarantee confinement if their integrated magnitudes grow too slowly. If $c<0$, the supplied exponential instead grows at infinity; a normalizable state requires the opposite <spinor> and envelope sign. Reversing the wall orientation also reverses the propagation chirality.
Back to article page
