Periodic Schrödinger Operators

An infinite periodic Schrödinger operator has continuous spectral bands even though each fixed-quasimomentum problem has a discrete spectrum. The connection is a direct integral of compact-cell operators. Bloch’s Theorem develops the translation-character theorem and its physical meaning; this page constructs the one-dimensional transform, specifies its domains, and derives the discriminant criterion for band energies.

Required background. Momentum on an Interval explains quasiperiodic domains and their periodic gauge. Unbounded Spectral Theorem supplies the spectral interpretation of unitary decompositions.

The Bloch transform is a unitary change of representation

Let $V\in L^\infty(\mathbb R)$ be real and $L$-periodic, $L>0$. On $L^2(\mathbb R)$ define

$$H=-\alpha\frac{d^2}{dx^2}+V(x),\qquad D(H)=H^2(\mathbb R),\qquad\alpha>0.$$

The bounded real perturbation preserves self-adjointness. For initially compactly supported smooth $\psi$, set

$$(\mathcal B\psi)(k,x)=\sqrt{\frac{L}{2\pi}} \sum_{n\in\mathbb Z}e^{-iknL}\psi(x+nL),$$

where $0\leq x<L$ and $-\pi/L\leq k<\pi/L$. For each fixed $x$, this is the Fourier series of the sequence $\psi(x+nL)$. Parseval followed by integration in $x$ gives

$$\int_{-\pi/L}^{\pi/L}\!\int_0^L |\mathcal B\psi(k,x)|^2\,dx\,dk =\|\psi\|^2.$$

The inverse Fourier-series map also reconstructs every square-integrable fibre field, so $\mathcal B$ extends to a unitary map onto the direct-integral Hilbert space. For general $L^2$ vectors the series is interpreted in this norm, without assuming pointwise convergence.

Relabelling the summation index gives the quasiperiodicity

$$\mathcal B\psi(k,x+L)=e^{ikL}\mathcal B\psi(k,x).$$

Fibre domains carry both boundary conditions

The transformed Hamiltonian is

$$\mathcal B H\mathcal B^{-1} =\int_{[-\pi/L,\pi/L)}^{\oplus}H(k)\,dk,$$

where $H(k)=-\alpha d^2/dx^2+V$ on

$$D(H(k))=\{u\in H^2(0,L): \ u(L)=e^{ikL}u(0),\quad u'(L)=e^{ikL}u'(0)\}.$$

The direct-integral operator domain consists of measurable fibre vectors in these domains almost everywhere with $\int\|H(k)u(k)\|^2dk<\infty$. Fibrewise membership alone is not enough.

Writing $u(x)=e^{ikx}v(x)$ transfers all fibres to the common periodic domain and gives

$$\widetilde H(k)=\alpha\left(-i\frac d{dx}+k\right)^2+V(x), \qquad D(\widetilde H(k))=H^2_{\mathrm{per}}(0,L).$$

The compact Sobolev embedding on the cell implies compact resolvent for every fibre. Its eigenvalues, ordered with multiplicity, are continuous functions $E_j(k)$. They tend to infinity uniformly in $k$ by comparison with the bounded-potential free fibres.

The full spectrum is therefore the union of their ranges over the Brillouin circle. A fixed fibre eigenvector is not a normalized whole-line eigenvector: a field supported at one quasimomentum has zero norm in the direct integral. For this scalar one-dimensional periodic problem, the Floquet spectral theorem gives purely absolutely continuous spectrum. Compactness of each fibre does not imply compactness of the direct-integral resolvent.

The discriminant identifies bands and gaps

For real energy $E$, propagate the two-component vector $(u,u')$ through one period of

$$-\alpha u''+Vu=Eu.$$

Let $M(E)$ denote its monodromy matrix. Constancy of the Wronskian gives $\det M(E)=1$; real $V$ makes $M(E)$ real. Define $D(E)=\operatorname{tr}M(E)$. A quasiperiodic solution exists precisely when $e^{ikL}$ is an eigenvalue of $M(E)$, so

$$e^{2ikL}-D(E)e^{ikL}+1=0, \qquad D(E)=2\cos(kL).$$

Combining this equation with the direct-integral spectrum yields

$$\sigma(H)=\{E\in\mathbb R:|D(E)|\leq2\}.$$

Band endpoints have $D(E)=2$ or $-2$, corresponding to periodic or antiperiodic cell conditions. A gap occurs where $|D(E)|>2$; the two Floquet multipliers are then real reciprocal numbers off the unit circle. Gaps may close, so periodicity alone does not imply a nonzero gap between every pair of adjacent bands.

The two-by-two discriminant construction relies on a scalar second-order equation in one dimension. Higher-dimensional Bloch fibres remain useful, but their spectra are not classified by this scalar trace criterion.

Exercise: band folding need not create gaps

For a constant potential $V_0$, compute the discriminant and the full spectrum. Compare with the fibre energies.

Solution

For $E\geq V_0$, the wavenumber is $q=\sqrt{(E-V_0)/\alpha}$ and $D(E)=2\cos(qL)$. For $E<V_0$, it becomes $2\cosh(L\sqrt{(V_0-E)/\alpha})>2$. Hence $\sigma(H)=[V_0,\infty)$. The fibres have eigenvalues $V_0+\alpha(k+2\pi n/L)^2$, $n\in\mathbb Z$. Folding these parabolas into the chosen quasimomentum cell introduces crossings but no spectral gaps. An arbitrary period assigned to a constant potential does not create a physical scattering mechanism.

References

• Kuchment, Peter, “An overview of periodic elliptic operators,” Bulletin of the American Mathematical Society 53, 343–414 (2016), arXiv:1510.00971, Sections 1, 4, and 6 — Floquet transforms, the Hill discriminant, and absolute continuity.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 7 and 9 — Fourier methods and one-dimensional self-adjoint differential operators.