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Gauge multiplication preserves H2 and transports the magnetic momentum

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $ℏ > 0$ , $m > 0$ , $q \in R$ , an open interval $I \subseteq R$ , admissible magnetic data $(A, Φ)$ on $R^{3}$ and a real function $χ : R^{3} \times I \to R$ such that:

for every $t \in I$ , $χ (\cdot, t)$ is $C^{2}$ on $R^{3}$ , and the functions $D_{j} χ (\cdot, t)$ , $D_{k} D_{j} χ (\cdot, t)$ are bounded on $R^{3}$ with bounds independent of $t \in I$ ;
for every $x \in R^{3}$ , $t \mapsto χ (x, t)$ is differentiable on $I$ , and $\partial_{t} χ$ is bounded on $R^{3} \times I$ .

Put $θ (x, t) : = \frac{q}{ℏ} χ (x, t)$ and let $G (t) : L^{2} (R^{3}; C) \to L^{2} (R^{3}; C)$ be multiplication by $e^{i θ (\cdot, t)}$ (The complex exponential by its power series). Then for every $t \in I$ :

(a) $G (t)$ is unitary and preserves $W^{1, 2}$ and $W^{2, 2}$ : $G (t) W^{k, 2} (R^{3}; C) = W^{k, 2} (R^{3}; C)$ for $k = 1, 2$ ;

(b) for every $u \in W^{1, 2}$ and every $j$ , $D_{j} (G (t) u) = G (t) D_{j} u + i (D_{j} θ (\cdot, t)) G (t) u as L^{2} classes;$

(c) for every $u \in W^{1, 2}$ the gauge covariance identity holds: $(P_{j} - q (A_{j} + D_{j} χ (\cdot, t))) G (t) u = G (t) (P_{j} - q A_{j}) u;$

(d) $t \mapsto G (t) h$ is continuous on $I$ for every $h \in L^{2}$ , and for every $h \in L^{2}$ and $t \in I$ the difference quotients converge in $L^{2}$ : $\lim_{s \to 0} \frac{1}{s} (G (t + s) h - G (t) h) = \frac{i q}{ℏ} (\partial_{t} χ (\cdot, t)) G (t) h .$

The function $χ$ is an arbitrary function with the stated regularity; no identification with a physical gauge choice is asserted.

Facts & Assumptions

Given: $ℏ, m > 0$ , $q \in R$ , admissible $(A, Φ)$ , a gauge function $χ$ as in the Statement, $θ = q χ / ℏ$ and the multipliers $G (t)$ .

[F1]

Euler's formula and trigonometric parity give $∣ e^{i θ (x, t)} ∣ = 1$ and $e^{- i θ (x, t)} = \overline{e^{i θ (x, t)}}$ for real $θ$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ). For a measurable multiplier $a$ with $∣ a ∣ = 1$ , $M_{a} h : = a h$ satisfies $∥ M_{a} h ∥_{2}^{2} = \int ∣ a ∣^{2} ∣ h ∣^{2} = ∥ h ∥_{2}^{2}$ and $M_{\overline{a}} M_{a} = M_{a} M_{\overline{a}} = I$ , so $M_{a}$ is unitary; pointwise modulus also gives its $L^{p}$ isometry for $1 \leq p \leq \infty$ . In particular $G (t)^{- 1} = M_{e^{- i θ (\cdot, t)}}$ and $∣ G (t) h ∣ = ∣ h ∣$ pointwise for chosen representatives.

[F2]

Classical partial derivatives of smooth functions are weak derivatives (Classical derivatives agree with weak derivatives).

[F3]

The complex exponential is entire with derivative $\exp^{'} (z) = \exp z$ (The complex exponential is entire and its complex derivative is itself). For an increment $h \in C$ , this derivative means $\exp (z + h) = \exp z + \exp (z) h + o (∣ h ∣)$ , so, identifying $C$ with $R^{2}$ , its real total derivative is multiplication by $\exp z$ . The chain and product rules applied to the real $C^{2}$ function $θ (\cdot, t)$ therefore give $D_{j} (e^{i θ (x, t)}) = i D_{j} θ e^{i θ}$ and $D_{k} D_{j} (e^{i θ}) = (i D_{k} D_{j} θ - (D_{k} θ) (D_{j} θ)) e^{i θ}$ pointwise (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F4]

Weak product rule: if $a \in L^{\infty}$ has weak first derivatives $D_{k} a \in L^{\infty}$ and $u \in W^{1, 2}$ , then $a u \in W^{1, 2}$ and $D_{k} (a u) = a D_{k} u + (D_{k} a) u$ (Weak product rule for a bounded coefficient with bounded weak derivative).

[F5]

$P_{j} f = - i ℏ D_{j} f$ for $f \in D (P_{j})$ , and $W^{1, 2} \subseteq D (P_{j})$ ; $W^{2, 2} = {f : D_{j} f \in W^{1, 2} \forall j}$ (Self-adjointness, Fourier representation and translation identities of position and momentum, Position and momentum operators on Euclidean L2, Integer-order Sobolev spaces and their norms).

[F6]

Dominated convergence for a sequence of measurable functions dominated by an $L^{2}$ class (Dominated convergence).

[F7]

For fixed $t \in I$ , $x \mapsto \partial_{t} χ (x, t)$ is measurable: choose $h_{n} \to 0$ with $t + h_{n} \in I$ ; each difference quotient $x \mapsto (χ (x, t + h_{n}) - χ (x, t)) / h_{n}$ is continuous and converges pointwise to $\partial_{t} χ (x, t)$ . The stated uniform bound therefore makes this derivative slice an $L^{\infty}$ class.

[F8]

If $g$ is continuous on the closed interval with endpoints $a, b$ , differentiable in its interior, and $∣ g^{'} ∣ \leq M$ there, then $∣ g (b) - g (a) ∣ \leq M ∣ b - a ∣$ : when $g (b) \neq g (a)$ , choose $ζ : = \overline{g (b) - g (a)} / ∣ g (b) - g (a) ∣$ and apply the real mean value theorem to $Re (ζ g)$ ; the zero-difference case is immediate (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

1.1givenF1

[Unitarity] For each $t$ , $∣ e^{i θ (x, t)} ∣ = 1$ for every $x$ , so [F1] gives that $G (t)$ is unitary on $L^{2}$ . This is the first assertion of (a).

1.2givenF2F3F4

[First-order preservation] Let $u \in W^{1, 2}$ . By [F3] the multiplier $a : = e^{i θ (\cdot, t)}$ is bounded with classical derivative $D_{j} a = i (D_{j} θ) a$ , and this classical derivative is its weak derivative by [F2]; the coefficient $D_{j} θ$ is bounded by hypothesis. So [F4] applies: $G (t) u \in W^{1, 2}$ and $D_{j} (G (t) u) = G (t) D_{j} u + i (D_{j} θ) G (t) u$ , which is (b).

2.1givenstep 1.2F2F3F4

[Second-order preservation] Let $u \in W^{2, 2}$ , so $D_{j} u \in W^{1, 2}$ for every $j$ . By step 1.2, $D_{j} (G (t) u) = G (t) D_{j} u + i (D_{j} θ) G (t) u$ . Apply [F4] to $a = G (t)$ and the class $D_{j} u \in W^{1, 2}$ : $G (t) D_{j} u \in W^{1, 2}$ , with $D_{k} (G (t) D_{j} u) = G (t) D_{k} D_{j} u + i (D_{k} θ) G (t) D_{j} u$ . Apply [F4] to the bounded coefficient $i D_{j} θ$ (whose weak derivative $i D_{k} D_{j} θ$ is bounded by [F2], [F3]) and $G (t) u \in W^{1, 2}$ from step 1.2: $i (D_{j} θ) G (t) u \in W^{1, 2}$ . Adding, $D_{j} (G (t) u) \in W^{1, 2}$ with an explicit formula, so $G (t) W^{2, 2} \subseteq W^{2, 2}$ ; applying the same argument to $- χ$ (which satisfies the same hypotheses) shows $G (t)^{- 1} W^{2, 2} \subseteq W^{2, 2}$ , hence equality, and with step 1.2 this proves (a).

2.2givenstep 1.2F5algebra

[Gauge covariance] Let $u \in W^{1, 2}$ . By [F5], $P_{j} (G (t) u) = - i ℏ D_{j} (G (t) u)$ and $P_{j} u = - i ℏ D_{j} u$ ; substituting step 1.2, $(P_{j} - q (A_{j} + D_{j} χ)) G (t) u = - i ℏ (G (t) D_{j} u + i (D_{j} θ) G (t) u) - q (A_{j} + D_{j} χ) G (t) u = G (t) P_{j} u + ℏ (D_{j} θ) G (t) u - q A_{j} G (t) u - q (D_{j} χ) G (t) u$ . Since $D_{j} θ = \frac{q}{ℏ} D_{j} χ$ , the middle two terms cancel and the right side equals $G (t) (P_{j} u - q A_{j} u) = G (t) (P_{j} - q A_{j}) u$ , which is (c).

3.1givenF1F3F6F7F8algebra∎

[Time continuity and differentiability] Fix $h \in L^{2}$ and $t \in I$ . For every sequence $s_{n} \to 0$ with $t + s_{n} \in I$ , $G (t + s_{n}) h \to G (t) h$ pointwise and $∣ G (t + s_{n}) h - G (t) h ∣^{2} \leq 4 ∣ h ∣^{2} \in L^{1}$ ; [F6] gives convergence in $L^{2}$ , and the sequential criterion gives continuity in (d). For the difference quotients, pointwise differentiability and [F3] give the limit $\frac{i q}{ℏ} (\partial_{t} χ (\cdot, t)) G (t) h$ . Put $M : = \sup_{R^{3} \times I} ∣ \partial_{t} θ ∣ < \infty$ . For each fixed $x$ , the complex-valued curve $r \mapsto e^{i θ (x, r)}$ has derivative of modulus $∣ \partial_{r} θ (x, r) ∣ \leq M$ by [F1], [F3]; [F8] applied on the interval between $t$ and $t + s_{n}$ therefore bounds each difference quotient by $M ∣ h (x) ∣$ . Its pointwise limit has the same bound, so the squared difference from the limit is dominated by $4 M^{2} ∣ h ∣^{2} \in L^{1}$ ; [F6] yields convergence in $L^{2}$ to the asserted derivative. This is the second half of (d).

Gauge multiplication preserves H2 and transports the magnetic momentum

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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