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Longitudinal free motion turns Landau levels into bands

Physical scope

The exact absolutely continuous spectrum of the same ideal spinless point-particle model on R^3 with a constant perpendicular field and free motion in the field direction. No finite sample, material, or empirical claim is included.

Example

Assume the Axiom of Choice, Countable Choice, and Dependent Choice. For the adopted spinless model on $L^{2} (R^{3}; C)$ , take mass $m > 0$ kilograms, charge $q \neq 0$ coulombs, and the constant field vector $B = (0, 0, B)$ teslas with signed component $B \neq 0$ . Let $A (x, y, z) = (0, B x, 0)$ in tesla-metres, so $\nabla \times A = B$ ; take zero scalar potential and free motion along $z$ . The closed Hamiltonian has bands $E_{n} + \frac{ℏ^{2} k_{z}^{2}}{2 m}, E_{n} = \frac{ℏ ∣ q B ∣}{m} (n + \frac{1}{2}), n \in N_{0},$ where $k_{z}$ is a wave number in inverse metres and each $E_{n}$ is an energy in joules. Its spectrum is the absolutely continuous half-line $[\frac{ℏ ∣ q B ∣}{2 m}, \infty) .$

Example references

Exact planar Landau spectrum in the linear gauge

Closed-form diagonalization with free longitudinal motion

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 adopted spinless Landau model with $m > 0$ , $q B \neq 0$ , $A = (0, B x, 0)$ , $B = (0, 0, B)$ , zero scalar potential, and an additional free coordinate $z$ .

[A1]

The $z$ direction has no potential or boundary constraint and contributes the nonnegative kinetic form $ℏ^{2} ∥ \partial_{z} u ∥_{2}^{2} / (2 m)$ ; the transverse model is the one used in Exact planar Landau spectrum in the linear gauge.

[A2]

The Axiom of Choice and Countable Choice are hypotheses of the closed-form representation and spectral suppliers; Dependent Choice is required by the transverse 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 with free longitudinal motion).

[F1]

Under the three-dimensional transform $U_{3}$ , the closed form is $\sum_{n = 0}^{\infty} \int_{R^{2}} (E_{n} + \frac{ℏ^{2} k_{z}^{2}}{2 m}) ∣ f_{n} (k_{y}, k_{z}) ∣^{2} d k_{y} d k_{z},$ with $E_{n} = ℏ ∣ q B ∣ (n + \frac{1}{2}) / m$ (Closed-form diagonalization with free longitudinal motion).

[F2]

Write $E_{r e f} = 1 J$ for the postulate's reference energy. Apply the closed-form representation theorem to the dimensionless closed form ${\overline{q}}_{B}^{(3)} / E_{r e f}$ . On the closed form domain, its shifted form norm satisfies $({\overline{q}}_{B}^{(3)} / E_{r e f}) [F] + ∥ F ∥_{2}^{2} = E_{r e f}^{- 1} ({\overline{q}}_{B}^{(3)} [F] + E_{r e f} ∥ F ∥_{2}^{2})$ , so normalization preserves the form domain, density, and closedness. Its represented operator $\hat{H}$ has operator domain consisting of form-domain vectors $F$ for which there is a coefficient-space vector $G$ satisfying $({\overline{q}}_{B}^{(3)} / E_{r e f}) (F, V) = ⟨ G, V ⟩$ for every form-domain vector $V$ . The physical Hamiltonian is $H = E_{r e f} \hat{H}$ on this same domain; multiplication by the positive $E_{r e f}$ does not change it. The indexed $E_{0}$ below denotes the lowest Landau energy, not $E_{r e f}$ . The theorem's exact Choice and Countable Choice hypotheses are in force (Unbounded vector potential and effective Pauli model, Semibounded closed-form representation and perturbation).

[F3]

A self-adjoint operator's scalar spectral measures are obtained by its spectral projection-valued measure (Spectral theorem for unbounded self-adjoint operators (PVM form)).

[F4]

Tonelli's theorem gives $w_{n} (k_{z}) = \int_{R} ∣ f_{n} (k_{y}, k_{z}) ∣^{2} d k_{y} \in L^{1} (R)$ for each $n$ , with $\sum_{n} ∥ w_{n} ∥_{1} = ∥ F ∥^{2}$ (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F5]

On each half-line $k_{z} > 0$ and $k_{z} < 0$ , the change of variables $E = E_{n} + c k_{z}^{2}$ , $c = ℏ^{2} / (2 m) > 0$ , has Jacobian $∣ d k_{z} / d E ∣ = 1 / (2 \sqrt{c (E - E_{n})})$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

Verification

1.1A1A2F1F2constructalgebra

Put $c = ℏ^{2} / (2 m)$ , $M_{n} (k_{y}, k_{z}) = E_{n} + c k_{z}^{2}$ , and ${\hat{M}}_{n} = M_{n} / E_{r e f}$ . Apply [F1] to the three-dimensional core form. The normalized closed form has coefficient multiplier ${\hat{M}}_{n}$ ; apply [F2] under the choice hypotheses in [A2]. For a form-domain vector $F$ , its representing operator domain consists exactly of those $F$ for which there is $G \in L^{2}$ with $\sum_{n} \int {\hat{M}}_{n} F_{n} \overline{H_{n}} = \sum_{n} \int G_{n} \overline{H_{n}}$ for every form-domain $H$ . If such $G$ exists, fix $n$ and any bounded box $K \subset R^{2}$ . Since ${\hat{M}}_{n}$ is bounded on $K$ , both $1_{K} {\hat{M}}_{n} F_{n}$ and $1_{K} G_{n}$ are in $L^{2} (K)$ . Choose the test vector with sole component $H_{n} = 1_{K} ({\hat{M}}_{n} F_{n} - G_{n})$ ; it belongs to the form domain because its index and support are bounded. The identity forces $\int_{K} ∣ {\hat{M}}_{n} F_{n} - G_{n} ∣^{2} = 0.$ Exhausting $R^{2}$ by bounded boxes and all $n$ gives $G_{n} = {\hat{M}}_{n} F_{n}$ almost everywhere, so $\hat{M} F \in L^{2}$ . Conversely, let $F \in L^{2}$ satisfy $\hat{M} F \in L^{2}$ . Since ${\hat{M}}_{n} \geq a : = E_{0} / E_{r e f} > 0$ , one has $\sum_{n} \int {\hat{M}}_{n} ∣ F_{n} ∣^{2} \leq a^{- 1} \sum_{n} \int {\hat{M}}_{n}^{2} ∣ F_{n} ∣^{2} < \infty,$ so $F$ belongs to the form domain. For every form-domain $H$ , Cauchy–Schwarz gives $\sum_{n} \int {\hat{M}}_{n} F_{n} \overline{H_{n}} = ⟨ \hat{M} F, H ⟩,$ so [F2] represents the normalized operator by multiplication by ${\hat{M}}_{n}$ , with domain $\sum_{n} \int {\hat{M}}_{n}^{2} ∣ F_{n} ∣^{2} < \infty$ . Since $E_{r e f} > 0$ , this is equivalent to $\sum_{n} \int M_{n}^{2} ∣ F_{n} ∣^{2} < \infty$ . The physical Hamiltonian $H = E_{r e f} \hat{H}$ therefore acts by multiplication by $M_{n}$ on this same domain. This identification follows under [A2].

2.1A2F1F2step 1.1algebra

The essential range of the multiplier $M_{n}$ is $[E_{0}, \infty)$ : every $M_{n} \geq E_{n} \geq E_{0}$ , while the $n = 0$ branch attains every value in that interval as $k_{z}$ varies. To verify the spectrum directly, if $λ \notin [E_{0}, \infty)$ then $\inf_{t \geq E_{0}} ∣ t - λ ∣ > 0$ and $\sup_{t \geq E_{0}} t / ∣ t - λ ∣ < \infty$ . Thus multiplication by $(M - λ)^{- 1}$ is bounded and maps into the operator domain, so $λ$ is in the resolvent. If $λ \geq E_{0}$ , set $k_{0} = \sqrt{(λ - E_{0}) / c}$ . For every $δ > 0$ , a sufficiently small interval around $k_{0}$ (or an interval $0 < k_{z} < \sqrt{δ / c}$ when $λ = E_{0}$ ), together with any bounded positive-measure interval of $k_{y}$ , gives a set of positive finite measure on which $∣ E_{0} + c k_{z}^{2} - λ ∣ < δ$ . Normalized indicator functions of these sets are in the operator domain and have residual norm at most $δ$ ; hence $λ$ is in the spectrum. The spectrum is exactly the stated half-line.

3.1A2F3F4F5step 1.1step 2.1algebra∎

For a Borel set $S \subseteq R$ , define $(P (S) F)_{n} = 1_{{M_{n} \in S}} f_{n}$ . These are orthogonal projections, $P (R) = I$ , and disjoint Borel sets give orthogonal projections whose sum converges strongly, because the squared $L^{2}$ norms of the tails are tails of a convergent nonnegative mass series. Thus $P$ is a projection-valued measure. Its scalar measures are finite Borel measures on $R$ , hence regular. Its spectral integral acts by multiplication by $M_{n}$ on exactly the square-integrability domain of step 1.1; uniqueness in [F3] identifies $P$ as the Hamiltonian's spectral measure. Consequently, for every coefficient vector $F = (f_{n})$ , $⟨ P (S) F, F ⟩ = \sum_{n} \int_{R} 1_{S} (E_{n} + c k_{z}^{2}) w_{n} (k_{z}) d k_{z},$ by [F4]. Each band measure is therefore the pushforward of $w_{n} (k_{z}) d k_{z}$ under $k_{z} \mapsto E_{n} + c k_{z}^{2}$ . Splitting the integral into $k_{z} > 0$ and $k_{z} < 0$ and applying [F5] gives, for $E > E_{n}$ , the density $ρ_{n} (E) = \frac{w_{n} ⁣ (\sqrt{(E - E_{n}) / c}) + w_{n} ⁣ (- \sqrt{(E - E_{n}) / c})}{2 \sqrt{c (E - E_{n})}} .$ The threshold $E_{n}$ has no atom because its preimage is the null set ${0}$ . Therefore each band measure is absolutely continuous, and the countable nonnegative sum of these finite measures remains absolutely continuous. By [F3] these are all scalar spectral measures of the represented Hamiltonian. There are no eigenvalues or singular-continuous spectral measures. The conclusion is conditional on the adopted model assumptions.

Remarks

Each transverse Landau level retains an infinite $L^{2} (k_{y})$ multiplicity. The continuous band parameter comes from free longitudinal momentum $k_{z}$ ; this ideal model makes no finite-sample or material claim.

Longitudinal free motion turns Landau levels into bands

Example

Example references

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources