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In four dimensions, $[\psi ]=3/2$, $[A_{\mu }]=1$, and $[F_{\mu \nu }]=2$. The minimal-coupling operator $\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$ has dimension four, so $[e]=0$ and $e$ is a marginal coupling. The Pauli term $\bar{\psi }[{\gamma }^{\mu },{\gamma }^{\nu }]F_{\mu \nu }\psi$ has dimension five, so $[\lambda ]=-1$ and $\lambda$ is an irrelevant coupling by power counting in quantum field theory.

Separate $x,z$ plane waves in the spinless equation. The transverse operator is $-{\hslash }^2{d}^2/d{y}^2+(p_x+eBy/c{)}^2$, the same shifted oscillator without a Pauli term. Its eigenvalues are $|e|\hslash |B|(2n+1)/c$. Thuswhich is the printed formula for $B>0$. The absence of the spin splitting is the substantive difference from the Dirac result.

The Pauli term couples a fermion bilinear to electromagnetic field strength through $\bar{\psi }{\sigma }^{\mu \nu }{F}_{\mu \nu }\psi$. In four dimensions it is a dimension-five interaction.

In Feynman gauge, use the auxiliary photon propagator $-i{g}_{\mu \nu }/(k^2-{\lambda }^2+i0)$ and $r=\lambda /m>0$. The projected one-loop Feynman parameter integral isHere $x$ is the source's photon parameter $z$.^[ref]

To check the reduction, use $N_2=4z(1-z)m^2$ and ${\Delta }^{\prime }=(1-z{)}^2{m}^2+z{\lambda }^2$. The three-parameter expression has prefactor $e^2/(16{\pi }^2)$ multiplying $N_2/{\Delta }^{\prime }$. At fixed $z$, ${\int }_0^{1-z}du=1-z$, giving $e^2/(4{\pi }^2){\int }_0^{1}z(1-z{)}^2/[(1-z{)}^2+z{r}^2]dz$. Substituting $e^2=4\pi \alpha$ gives exactly the displayed integral. This starts from the projected QED numerator; it is not a derivation of the whole QED calculation from an arbitrarily chosen integral.

The regulator can then be removed by a short independent argument. Write its integrand as $f_r(x)$. For $0\le x<1$, $f_r(x)\to 2x$, andThe dominated convergence theorem therefore gives $I(r)\to 1$ and $F_2^{(1)}(0)=\alpha /(2\pi )$. At $x=1$ the regulated integrand is zero, whereas the limiting function can be assigned value $2$; one point has zero measure and changes no integral. For every $r>0$, $0<I(r)<1$, and decreasing $r$ increases the integral. The visible suppression near $x=1$ narrows as the regulator shrinks; it is not a uniform pointwise approximation at that endpoint.

This auxiliary mass regulates an infrared divergence, not an ultraviolet divergence; the finite-mass prescription alone is not fully gauge-invariant.^[ref]

The ultraviolet problem belongs to the full QED amplitude. Electron self-energy, wave-function renormalization and the charge vertex require consistent counterterms and renormalization conditions. The Ward identity for QED renormalization constants ties $Z_1$ to $Z_2$, rather than allowing the vertex and external electron normalization to be fitted independently. Minimal renormalizable QED has no independent Pauli term to adjust this finite one-loop $F_2(0)$: its dimension-five coefficient is predicted from the mass and electric charge inputs. Adding a free Pauli term would define a different theory with an extra parameter. Finiteness of the reduced integral alone proves neither all-order renormalizability nor finiteness of every QED observable.