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Choose the vector potential $A=(-{\int }_0^{y}B(s)ds,0,0)$, whose curl is $(0,0,B(y))$, and a spin-down wavefunction $\Psi =e^{-ikx}\chi (y)(0,1{)}^{\mathsf{T}}$, independent of $z$. DefineThe covariant $x$ derivative acts as $-i(k+a)$, and ${\sigma }_3$ acts as $-1$. The Pauli equation consequently reduces toThe sign of the plane-wave phase was chosen to match the printed $k+a$ convention; choosing $e^{ikx}$ simply reverses the label $k$.

Set $W=k+a$ and $M={\partial }_y+W$. On the square-integrable operator domain with vanishing boundary terms, $M^{†}=-{\partial }_y+W$, andTaking the inner product with $\chi$ gives $ϵ\Vert \chi {\Vert }_2^2=\Vert M\chi {\Vert }_2^2\ge 0$. Thus $\boxed{E\ge 0}$.

A zero mode solves $M\chi =0$, soSince $a(0)=0$ and $a^{\prime }\ge b_0=e{B}_0/\hslash$, integration on either side of zero gives ${\int }_0^{y}a(s)ds\ge b_0{y}^2/2$. Therefore $|{\chi }_k(y)|\le |C_k|e^{-ky-b_0{y}^2/2}$, proving square integrability for every real $k$. Each can be normalized and supplies a distinct momentum label at zero energy. Hence zero energy is infinitely degenerate. In an infinite $x$ direction the individual plane waves are generalized momentum states; square-integrable superpositions of their normalized transverse zero modes give ordinary transverse states by the Fourier isometry. The stipulated zero $z$ momentum is understood per unit longitudinal length, or with periodic longitudinal normalization.