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Exact planar Landau spectrum in the linear gauge

Physical scope

Exact spectral prediction of the adopted spinless nonrelativistic point-particle model on the infinite plane, for a constant nonzero perpendicular field in the unbounded linear Landau gauge. It makes no finite-sample degeneracy count, boundary claim, or empirical assertion.

Statement

Assume the Axiom of Choice, Countable Choice, and Dependent Choice, and the adopted spinless minimal-coupling form model. Let $m > 0$ be mass in kilograms, $q \neq 0$ charge in coulombs, $B \neq 0$ the constant signed perpendicular field component in teslas, and $ℏ > 0$ in joule-seconds. Let $A : R^{2} \to R^{2}$ be $A (x, y) = (0, B x)$ in tesla-metres; then $B_{⊥} = \partial_{x} A_{y} - \partial_{y} A_{x} = B$ is its scalar perpendicular field. Take wavefunctions in $L^{2} (R^{2}; C)$ . Set $V = 0$ and define the Hamiltonian by the closed form of the core expression $q_{B} [u] = \frac{1}{2 m} (∥ - i ℏ \partial_{x} u ∥_{2}^{2} + ∥ (- i ℏ \partial_{y} - q B x) u ∥_{2}^{2}), u \in C_{c}^{\infty} (R^{2}; C) .$ The resulting Friedrichs Hamiltonian is unitarily equivalent, under partial Fourier transformation in $y$ followed by $x = ξ + ℏ k / (q B)$ , to the constant family of oscillators $h (k) = - \frac{ℏ^{2}}{2 m} \partial_{ξ}^{2} + \frac{(q B ξ)^{2}}{2 m}, k \in R,$ with its form-closed realization. Its spectrum is ${\frac{ℏ ∣ q B ∣}{m} (n + \frac{1}{2}) : n \in N_{0}} .$ Every listed eigenvalue has infinite multiplicity, and every spectral measure is purely atomic on this set; there is no continuous spectral measure. This is a conditional model deduction, not a claim that a material realizes the ideal infinite-plane Hamiltonian.

Statement references

Unbounded vector potential and effective Pauli model

Closed-form diagonalization in Landau gauge

Semibounded closed-form representation and perturbation

Oscillator scaling and exact realization

The Axiom of Choice

The Axiom of Countable Choice ( ${A C}_{ω}$ )

The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain

Facts & Assumptions

Given: the Axiom of Choice, Countable Choice, Dependent Choice, the spinless model postulate, $m > 0$ , $q B \neq 0$ , $ℏ > 0$ , $A = (0, B x)$ , $V = 0$ , and the core form in the Statement. Use SI units.

[A1]

Adopt the spinless minimal-coupling assignment and the stated Friedrichs-form prescription (Unbounded vector potential and effective Pauli model).

[A2]

The Axiom of Choice and Countable Choice are hypotheses of the closed-form representation and unbounded spectral suppliers. Dependent Choice is required by the Landau form supplier (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, Closed-form diagonalization in Landau gauge).

[F1]

The closed magnetic form is unitarily equivalent to the diagonal form $\sum_{n} E_{n} ∥ f_{n} ∥_{2}^{2}$ on coefficient families, with $E_{n} = ℏ ∣ q B ∣ (n + \frac{1}{2}) / m$ , under the explicitly normalized partial Fourier and translation map (Closed-form diagonalization in Landau gauge).

[F2]

The closed-form representation theorem applies to a dimensionless closed form. Write $E_{r e f} = 1 J$ for the postulate's reference energy and apply the theorem to $q / E_{r e f}$ . On the closed form domain, $(q / E_{r e f}) [F] + ∥ F ∥_{2}^{2} = E_{r e f}^{- 1} (q [F] + E_{r e f} ∥ F ∥_{2}^{2})$ , so normalization preserves its form domain, density, and closedness. The theorem's operator domain consists of form-domain vectors $F$ for which there is a coefficient-space vector $G$ satisfying $(q / E_{r e f}) (F, V) = ⟨ G, V ⟩$ for every form-domain vector $V$ . If $\hat{H}$ represents this normalized form, then the model Hamiltonian is $H = E_{r e f} \hat{H}$ on the same operator domain, so its spectral values are multiplied by $E_{r e f}$ . Here $E_{r e f}$ is distinct from the indexed Landau level $E_{0}$ used below. The theorem's exact hypotheses include the Axiom of Choice and Countable Choice (Unbounded vector potential and effective Pauli model, Semibounded closed-form representation and perturbation).

[F3]

The one-dimensional oscillator with $ω_{c} = ∣ q B ∣ / m$ has the complete Hermite eigenbasis and eigenvalues $E_{n} = ℏ ω_{c} (n + \frac{1}{2})$ (Oscillator scaling and exact realization).

[F4]

A self-adjoint operator has a unique regular spectral projection-valued measure (Spectral theorem for unbounded self-adjoint operators (PVM form)).

Proof

1.1givenA1A2F1F3

The hypotheses $m, ℏ > 0$ and $q B \neq 0$ make $ω_{c} = ∣ q B ∣ / m > 0$ and the translation $x_{0} (k) = ℏ k / (q B)$ well-defined. Apply [F1]: the unitary $U$ sends the closed form to the coefficient form $q [F] = \sum_{n = 0}^{\infty} E_{n} ∥ f_{n} ∥_{L^{2} (k)}^{2},$ where $F (ξ, k) = \sum_{n} f_{n} (k) ψ_{n} (ξ)$ and $(ψ_{n})$ is the complete oscillator basis from [F3]. In particular, $q [F] \geq E_{0} ∥ F ∥_{2}^{2} > 0$ for nonzero $F$ .

2.1A2F1F2step 1.1constructalgebra

Apply the closed-form representation theorem [F2] to the normalized form $\hat{q} = q / E_{r e f}$ . If $F = (f_{n})$ is in its representing operator domain, there is $G = (g_{n})$ in the coefficient Hilbert space such that $\hat{q} (F, V) = ⟨ G, V ⟩$ for every form-domain vector $V$ . Fix $n$ and take $V$ to have sole nonzero component $V_{n} = (E_{n} / E_{r e f}) f_{n} - g_{n}$ . It lies in the form domain because $f_{n}, g_{n} \in L^{2} (k)$ and $E_{n} / E_{r e f}$ is finite. The representation identity gives $0 = \hat{q} (F, V) - ⟨ G, V ⟩ = ∥ (E_{n} / E_{r e f}) f_{n} - g_{n} ∥_{2}^{2},$ so $g_{n} = (E_{n} / E_{r e f}) f_{n}$ for every $n$ . Since $G$ belongs to the Hilbert space, $\sum_{n} (E_{n} / E_{r e f})^{2} ∥ f_{n} ∥_{2}^{2} < \infty$ , equivalently $\sum_{n} E_{n}^{2} ∥ f_{n} ∥_{2}^{2} < \infty$ because $E_{r e f} > 0$ .

3.1A2F1F2step 2.1algebra

Conversely, suppose $F$ is in the form domain and $\sum_{n} E_{n}^{2} ∥ f_{n} ∥_{2}^{2} < \infty$ . Define $G_{n} = (E_{n} / E_{r e f}) f_{n}$ , which is a vector in the coefficient Hilbert space. For every form-domain vector $V = (v_{n})$ , Cauchy–Schwarz gives absolute convergence and $\hat{q} (F, V) = \sum_{n} (E_{n} / E_{r e f}) ⟨ f_{n}, v_{n} ⟩ = \sum_{n} ⟨ G_{n}, v_{n} ⟩ = ⟨ G, V ⟩ .$ Thus [F2] represents the normalized operator by $(f_{n}) \mapsto ((E_{n} / E_{r e f}) f_{n})$ on the stated domain. The model Hamiltonian is $E_{r e f}$ times this operator and therefore acts by $(f_{n}) \mapsto (E_{n} f_{n})$ on that same domain; uniqueness identifies it with the postulated Hamiltonian.

4.1A2F1F3F4step 3.1algebra∎

For each $n$ , the eigenspace at $E_{n}$ is $L^{2} (R_{k}) \otimes C ψ_{n}$ , which is infinite-dimensional; the energies are strictly increasing and tend to $+ \infty$ . If $z \notin {E_{n} : n \geq 0}$ , then $\inf_{n} ∣ E_{n} - z ∣ > 0$ and $\sup_{n} E_{n} / ∣ E_{n} - z ∣ < \infty$ . The diagonal inverse is therefore bounded and maps into the operator domain, so $z$ is in the resolvent. Each $E_{n}$ is an eigenvalue, and the complete coefficient decomposition together with [F4] shows that all spectral projections are sums of eigenspace projections. Hence every spectral measure is atomic on the displayed set and the spectrum is exactly that set.

Remarks

The linear potential $A_{y} = B x$ is unbounded. The proof works with its closable quadratic form and never applies a bounded-coefficient $H^{2}$ realization. The finite-rectangle degeneracy count in the cited treatments depends on boundary conditions; no exact hard-wall or finite-area count is claimed here.

Exact planar Landau spectrum in the linear gauge

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources