Relativistic Landau Levels

A uniform magnetic field turns the transverse Dirac eigenproblem into a harmonic-oscillator problem with a spin-dependent offset. The exact energies are square roots of the Pauli magnetic eigenvalues. The lowest Landau level has one internal state per nonzero energy branch, while each higher level has two. Constructing the full spinors is essential: the spin labels used to diagonalize the squared Hamiltonian are generally not spin eigenvalues of the complete Dirac states.

Required background. Dirac Dynamics in Electromagnetic Fields supplies the squared Hamiltonian, and Landau Levels owns the orbital oscillator and guiding-center construction. Helpful background. The Pauli Equation and Foldy–Wouthuysen Expansion give the low-energy checks.

Uniform-field Dirac Hamiltonian

Take a nonzero signed charge $q$, a constant field $\mathbf B=B\hat{\mathbf z}$ with $B>0$, and $\Phi=0$. Choose a $z$-independent gauge with $A_z=0$, so the conserved longitudinal kinetic momentum equals $p_z$. Initially let $m>0$; the massless limit will be taken separately. With $\boldsymbol\pi=\mathbf p-q\mathbf A$,

$$H=c\boldsymbol\alpha\cdot\boldsymbol\pi+\beta mc^2, \qquad [\pi_x,\pi_y]=iq\hbar B.$$

There is no anomalous Pauli interaction in this Hamiltonian. The magnetic field is prescribed and static; photon emission and radiative energy shifts are outside the model. On infinite space the longitudinal momentum $p_z$ is continuous; box normalization can be used when normalizing individual modes.

Let $r_q=\operatorname{sgn}(q)$ and introduce

$$a=\frac{\pi_x+i r_q\pi_y}{\sqrt{2|q|\hbar B}}, \qquad [a,a^\dagger]=1.$$

Then

$$\pi_x^2+\pi_y^2 =|q|\hbar B(2a^\dagger a+1).$$

The magnetic length is $\ell_B=\sqrt{\hbar/(|q|B)}$. Choose an oscillator basis $a|n,\zeta\rangle=\sqrt n\,|n-1,\zeta\rangle$, where $\zeta$ denotes the independent orbital degeneracy label. Its coordinate realization depends on gauge; the energies do not.

Squared spectrum and the Landau index

In the Dirac basis,

$$H^2=m^2c^4+c^2\boldsymbol\pi^2 -q\hbar Bc^2\Sigma_z.$$

Use a two-component seed $\phi_{n,s}=|n,\zeta\rangle\chi_s$ with $\sigma_z\chi_s=s\chi_s$, $s=\pm1$, and fix the spin phases by $\chi_+=(1,0)^{\mathsf T}$, $\chi_-=(0,1)^{\mathsf T}$. The associated squared energy is

$$E^2=m^2c^4+c^2p_z^2 +|q|\hbar Bc^2(2n+1-r_qs).$$

Define the nonnegative Landau index

$$N=n+\frac{1-r_qs}{2}.$$

Writing the positive energy magnitude as $\mathcal E_N$, the two branches are

$$\begin{aligned} \mathcal E_N(p_z) &=\sqrt{m^2c^4+c^2p_z^2+2|q|\hbar Bc^2N},\\ E_\pm&=\pm\mathcal E_N. \end{aligned}$$

For $N=0$, the only seed is $n=0$, $s=r_q$. For each $N\geq1$, there are two:

Seed	Oscillator index	Spin label
Aligned with the signed charge	$n=N$	$s=r_q$
Opposite label	$n=N-1$	$s=-r_q$

Thus the internal multiplicity on each energy branch is $d_N=2-\delta_{N0}$, before counting orbital degeneracy. For an electron, the lowest seed has spin along $-\hat{\mathbf z}$. The orbital zero-point term cancels its minimal spin magnetic energy.

Squaring only supplies candidate energies and seed spaces. The first-order equation still determines the relation between the upper and lower components.

Constructing normalized Dirac modes

Define $M=mc^2$ and $T=\boldsymbol\sigma\cdot\boldsymbol\pi$. On the $N$th seed space,

$$T^2=f_N,\qquad f_N=p_z^2+2|q|\hbar BN.$$

For any unit-normalized seed $\phi$ in this space, construct

$$\Psi_+[\phi] =C_N \begin{pmatrix} \phi\\[2pt] \dfrac{cT\phi}{\mathcal E_N+M} \end{pmatrix},$$

with $C_N=\sqrt{(\mathcal E_N+M)/(2\mathcal E_N)}$, and

$$\Psi_-[\phi] =C_N \begin{pmatrix} -\dfrac{cT\phi}{\mathcal E_N+M}\\[2pt] \phi \end{pmatrix}.$$

Substitution into $H=\left(\begin{smallmatrix}M&cT\\cT&-M\end{smallmatrix}\right)$ gives $H\Psi_\pm=\pm\mathcal E_N\Psi_\pm$. The identity $c^2f_N=(\mathcal E_N-M)(\mathcal E_N+M)$ proves the normalization. Orthogonal seeds produce orthogonal states within a branch, and the two branches are mutually orthogonal. This is a direct construction of all states in each invariant subspace, not just a count of roots of $H^2$.

For $N\geq1$, order the seed basis as $(\phi_{N,r_q},\phi_{N-1,-r_q})$. With the ladder phases fixed above,

$$T_N= \begin{pmatrix} r_qp_z & t_N\\ t_N & -r_qp_z \end{pmatrix},$$

where $t_N=\sqrt{2|q|\hbar BN}$. For example, starting with the first seed makes the lower component proportional to $r_qp_z\phi_{N,r_q}+t_N\phi_{N-1,-r_q}$. It therefore contains the opposite spin label whenever $N\geq1$. In general $[H,\Sigma_z]\ne0$. Calling the full mode “spin up” simply because its upper seed has $s=+1$ can obscure this mixing.

For $N=0$, $T_0=r_qp_z$ and the invariant Dirac matrix is only

$$H_0= \begin{pmatrix} M&cr_qp_z\\ cr_qp_z&-M \end{pmatrix}.$$

Both nonzero components now carry the same spin label $s=r_q$, so the lowest modes are also $\Sigma_z$ eigenstates. At $p_z=0$ their energies are $\pm mc^2$, independent of $B$ in minimal Dirac theory. Jentschura (2023) gives an explicit coordinate bispinor treatment in the symmetric gauge.

Orbital degeneracy and bulk state counting

The label $\zeta$ has not disappeared. For each independent internal mode, the bulk orbital state density per transverse area is

$$\frac{|q|B}{2\pi\hbar} =\frac{1}{2\pi\ell_B^2}.$$

For a flux-compatible periodic area $A_\perp$, the orbital count is $N_\Phi=|q|BA_\perp/(2\pi\hbar)$, an integer. For a large open sample this is the bulk count, with edge states and boundary corrections treated separately. The derivation belongs to Landau Levels.

At fixed $p_z$, a nonzero energy branch therefore has $d_NN_\Phi$ states. Counting the two energy signs as an additional spin degeneracy would double this number incorrectly. In a quantum field theory, negative-frequency solutions instead enter the antiparticle expansion; a one-particle squared spectrum alone does not specify occupation or vacuum charge.

Nonrelativistic expansion and its scale

If $f_N\ll m^2c^2$,

$$\mathcal E_N =mc^2+\frac{f_N}{2m} -\frac{f_N^2}{8m^3c^2} +O\!\left(\frac{f_N^3}{m^5c^4}\right).$$

The first correction is

$$\mathcal E_N-mc^2 =\frac{p_z^2}{2m} +\frac{|q|\hbar B}{m}N+\cdots ,$$

which is the paired $g=2$ Landau–Zeeman spectrum of the Pauli equation. The quartic correction is a square of the complete magnetic eigenvalue $f_N$. It agrees with the $-F_{\rm P}^2/(8m^3c^2)$ operator on Foldy–Wouthuysen Expansion, not with a spin-independent $\boldsymbol\pi^4$ term alone.

The expansion parameter is

$$\epsilon_N =\frac{p_z^2+2|q|\hbar BN}{m^2c^2}.$$

For a fixed field, sufficiently large $N$ leaves the nonrelativistic regime. A small cyclotron energy at the lowest levels does not justify the expansion uniformly over the infinite oscillator spectrum.

Massless lowest modes

For $m=0$ and nonzero $\mathcal E_N$ the same mode construction is well defined. The spectrum becomes

$$E_\pm =\pm c\sqrt{p_z^2+2|q|\hbar BN}.$$

At $N=0$, the two-component energy-block matrix is $cr_qp_z\tau_x$, where $\tau_x$ interchanges upper and lower Dirac components. In the Dirac basis used here, its eigenvalue $\chi=\pm1$ is also the chirality eigenvalue on this lowest-level subspace. The two chiral dispersions are

$$E_\chi=r_q\chi\,cp_z.$$

They move in opposite longitudinal directions. At fixed nonzero $p_z$, only one lies in the positive-energy branch. At $p_z=0$ both energies vanish: the zero eigenspace has dimension two before orbital degeneracy, and assigning two distinct energy-sign projectors there is not meaningful. The formulas with $1/\mathcal E_N$ must be replaced by direct diagonalization of this zero matrix.

This is a three-dimensional, four-component Dirac problem. A two-dimensional band system requires its own velocity, mass terms, valley content, and physical-spin counting. One cannot import this multiplicity unchanged into graphene or infer a quantum anomaly from the one-particle spectrum alone. A static pure magnetic field also does not produce the electric work or time-dependent mixing required for the usual pair-production processes.

Exercises

The first electron pair. Take $q=-e$. List the oscillator and seed-spin labels for $N=0$ and $N=1$.

Solution

Here $r_q=-1$. The lowest level has $(n,s)=(0,-1)$. The $N=1$ pair is $(1,-1)$ and $(0,+1)$. Each seed yields one state in each energy branch through the full-spinor construction. The orbital flux degeneracy multiplies these internal counts.

Checking the negative branch. Apply the Dirac block matrix to $\Psi_-[\phi]$. Verify both component equations.

Solution

The upper component is $cT\phi[1-M/(\mathcal E_N+M)] =\mathcal E_N cT\phi/(\mathcal E_N+M)$. The lower component is $-[c^2f_N/(\mathcal E_N+M)+M]\phi =-\mathcal E_N\phi$. They equal $-\mathcal E_N$ times the respective components of $\Psi_-$.

Where the expansion fails. At $p_z=0$, express $\epsilon_N$ using $B_*=m^2c^2/(|q|\hbar)$. Does the lowest level acquire a minimal Dirac magnetic shift?

Solution

The parameter is $\epsilon_N=2NB/B_*$. The expansion needs $2NB/B_*\ll1$. For $N=0$ the exact energy magnitude is $mc^2$ at every $B$ within this ideal model, because orbital and minimal spin contributions cancel. An anomalous moment or additional confinement changes the model.

References

• Jentschura, Ulrich D. “Algebraic Approach to Relativistic Landau Levels in the Symmetric Gauge.” Physical Review D 108, 016016 (2023). doi:10.1103/PhysRevD.108.016016. arXiv:2306.01155.
• Landau, Lev D., and Evgeny M. Lifshitz. Quantum Mechanics: Non-Relativistic Theory. 3rd ed. Pergamon Press (1977).
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.