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The three-dimensional continuity identity, the vanishing stationary divergence and the incident and outgoing current readings

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $m, ℏ > 0$ , let $Ω \subseteq R^{3}$ and $I \subseteq R$ be nonempty open sets, and let $u, v : Ω \times I \to R$ be functions of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms) that satisfy the two real equations $ℏ \partial_{t} u = - \frac{ℏ^{2}}{2 m} Δ v + V v, - ℏ \partial_{t} v = - \frac{ℏ^{2}}{2 m} Δ u + V u$ pointwise on $Ω \times I$ for a real-valued function $V$ on $Ω \times I$ , where $x = (x_{0}, x_{1}, x_{2})$ , $\partial_{i}$ differentiates in spatial coordinate $x_{i}$ for $i \in {0, 1, 2}$ with $t$ fixed, $\partial_{t}$ differentiates in $t$ with $x$ fixed, and $Δ = \partial_{0}^{2} + \partial_{1}^{2} + \partial_{2}^{2}$ (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ). The equations are the componentwise reading of $i ℏ \partial_{t} ψ = - \frac{ℏ^{2}}{2 m} Δ ψ + V ψ$ for $ψ : = u + i v$ . Put $n : = u^{2} + v^{2} : Ω \times I \to R, j : = \frac{ℏ}{m} (u \nabla v - v \nabla u) : Ω \times I \to R^{3},$ with $\nabla$ the spatial gradient $(\partial_{0}, \partial_{1}, \partial_{2})$ of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case at each fixed $t$ and $div$ the divergence of Divergence and curl of a $C^{1}$ vector field. Then:

(a) Continuity identity. $\partial_{t} n + div j = 0$ pointwise on $Ω \times I$ .

(b) Stationary solutions. If $u, v : Ω \to R$ are $C^{2}$ and independent of $t$ , if $V : Ω \to R$ is real, and if there is a real $E$ with $- \frac{ℏ^{2}}{2 m} Δ u + V u = E u, - \frac{ℏ^{2}}{2 m} Δ v + V v = E v$ pointwise on $Ω$ , then $div j = 0$ on $Ω$ for the time-independent field $j = \frac{ℏ}{m} (u \nabla v - v \nabla u)$ , and for every closed ball $\overline{B (a, r)} \subseteq Ω$ with outward unit normal $ν$ on the sphere $\partial B (a, r)$ , $\int_{\partial B (a, r)} j \cdot ν d S = 0.$

(c) The two explicit readings. Let $k > 0$ , $d \in S^{2}$ and define $u (x) : = \cos (k d \cdot x)$ , $v (x) : = \sin (k d \cdot x)$ on $Ω = R^{3}$ (then $u, v$ are $C^{\infty}$ and satisfy the stationary equations with $V = 0$ , $E = ℏ^{2} k^{2} / (2 m)$ ). Then $j = \frac{ℏ k}{m} d on R^{3} .$ Let $r_{0} > 0$ , write $Ω : = {x \in R^{3} : ∣ x ∣ > r_{0}}$ , let $Ω ∋ x = r ω$ with $r = ∣ x ∣$ , $ω \in S^{2}$ , and let $a, b : S^{2} \to R$ be functions with $a, b \in C^{1} (S^{2})$ . Here $S^{2} : = {ω \in R^{3} : ∣ ω ∣ = 1}$ , and, for a function $g : S^{2} \to R$ , $g \in C^{1} (S^{2})$ means that its radial extension $\tilde{g} : R^{3} ∖ {0} \to R$ , $\tilde{g} (x) : = g (x / ∣ x ∣)$ , is $C^{1}$ in the open-domain sense of $C^{k}$ Euclidean maps and diffeomorphisms. Define $u (x) : = \frac{a (ω) \cos (k r) - b (ω) \sin (k r)}{r}, v (x) : = \frac{a (ω) \sin (k r) + b (ω) \cos (k r)}{r}$ on $Ω$ , so that $u + i v = \frac{f (ω)}{r} e^{i k r}$ with $f : = a + i b$ . Then $u, v \in C^{1} (Ω)$ and, with $\hat{r} : = ω$ , $j \cdot \hat{r} = \frac{ℏ k}{m} \frac{∣ f (ω) ∣^{2}}{r^{2}}, equivalently r^{2} j \cdot \hat{r} = \frac{ℏ k}{m} ∣ f (ω) ∣^{2},$ and the current identity holds for every declared $f \in C^{1} (S^{2})$ , with no equation required of the field.

The assertion is an identity for the declared differentiable functions; it constructs no probability law, no state, no beam and no cross section. The fields of clause (c) are only $C^{1}$ : an $f \in C^{1} (S^{2})$ need not admit two angular derivatives, and no outgoing field is claimed here to solve the free equation or to be stationary.

Facts & Assumptions

Given: the open sets $Ω, I$ , the functions $u, v, V$ , the constants $m, ℏ > 0$ , and the data of clause (c).

[F1]

Differentiation rules. For totally differentiable Euclidean maps the sum and scalar-multiple rules hold (Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives) and the chain rule $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ holds (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ), with $D f (a) e_{j} = \partial_{j} f (a)$ recovering the partial derivatives (A total derivative computes every directional derivative, and its matrix is the Jacobian); the partial derivatives themselves are those of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , the entries of the Jacobian of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. For scalar functions on an interval the sum, product and quotient rules hold (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 $g (c) \neq 0$ ), and $(\sin)^{'} = \cos$ and $(\cos)^{'} = - \sin$ on $R$ (The derivatives of sine and cosine are cosine and minus sine). Also $\sin^{2} s + \cos^{2} s = 1$ for every real $s$ (Parity and the Pythagorean identity for sine and cosine).

[F2]

Divergence of a product with a scalar. For $C^{1}$ functions $φ : Ω \to R$ and $F : Ω \to R^{3}$ , the equality $div (φ F) = ⟨ \nabla φ, F ⟩ + φ div F$ holds pointwise, the divergence and gradient being those of Divergence and curl of a $C^{1}$ vector field and The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case (Divergence and curl are linear and satisfy the scalar product rules).

[F3]

Divergence theorem. Assume ${A C}_{ω}$ . For a bounded $C^{1}$ domain $Ω_{0}$ and $F \in C^{1} (\overline{Ω_{0}}; R^{3})$ , $\int_{Ω_{0}} div F d x = \int_{\partial Ω_{0}} F \cdot ν d S$ (Divergence on a bounded C1 Euclidean domain).

Proof

technique · direct; differentiate $n$, use the two real equations to cancel the terms, then evaluate the stationary divergence and the two explicit fields

Given: the data of the statement and [F1]-[F3].

1.1F1given

By [F1], $\partial_{t} n = \partial_{t} (u^{2} + v^{2}) = 2 (u \partial_{t} u + v \partial_{t} v)$ pointwise on $Ω \times I$ .

1.2F2F1algebra

Put $F : = u \nabla v - v \nabla u$ . By [F2] applied to each of the two products, using $div (\nabla v) = \sum_{i = 0}^{2} \partial_{i} \partial_{i} v = Δ v$ and likewise for $u$ , one has $div (u \nabla v) = \nabla u \cdot \nabla v + u Δ v$ and $div (v \nabla u) = \nabla v \cdot \nabla u + v Δ u$ ; the mixed terms are equal by symmetry of the inner product and cancel in the difference, so $div F = u Δ v - v Δ u$ and hence $div j = \frac{ℏ}{m} (u Δ v - v Δ u)$ .

1.3givenalgebra

Reading the complex equation $i ℏ \partial_{t} (u + i v) = - \frac{ℏ^{2}}{2 m} Δ (u + i v) + V (u + i v)$ by real and imaginary parts gives $ℏ \partial_{t} u = - \frac{ℏ^{2}}{2 m} Δ v + V v$ and $- ℏ \partial_{t} v = - \frac{ℏ^{2}}{2 m} Δ u + V u$ , i.e. $\partial_{t} u = - \frac{ℏ}{2 m} Δ v + \frac{V}{ℏ} v$ and $\partial_{t} v = \frac{ℏ}{2 m} Δ u - \frac{V}{ℏ} u$ .

1.4F1givenalgebra

For $u = \cos (k d \cdot x)$ , $v = \sin (k d \cdot x)$ the chain rule [F1] gives $\nabla u = - k d \sin (k d \cdot x)$ and $\nabla v = k d \cos (k d \cdot x)$ ; hence $u \nabla v - v \nabla u = k d (\cos^{2} + \sin^{2}) = k d$ and $j = \frac{ℏ k}{m} d$ . The two functions are $C^{\infty}$ and satisfy the stationary equations with $V = 0$ , because $\partial_{i} \partial_{i} u = - k^{2} d_{i}^{2} \cos (k d \cdot x)$ and $\partial_{i} \partial_{i} v = - k^{2} d_{i}^{2} \sin (k d \cdot x)$ by [F1], so summing over $i = 0, 1, 2$ and using $∣ d ∣^{2} = 1$ gives $Δ u = - k^{2} u$ and $Δ v = - k^{2} v$ with $E = ℏ^{2} k^{2} / (2 m)$ .

1.5F1givenalgebra

Write $c_{k} : = \cos (k r)$ , $s_{k} : = \sin (k r)$ and $f = a + i b$ . Then $u = r^{- 1} (a c_{k} - b s_{k})$ , $v = r^{- 1} (a s_{k} + b c_{k})$ , so by the chain and product rules for functions of the single variable $r$ [F1], $\partial_{r} u = - k r^{- 1} (a s_{k} + b c_{k}) - r^{- 2} (a c_{k} - b s_{k})$ and $\partial_{r} v = k r^{- 1} (a c_{k} - b s_{k}) - r^{- 2} (a s_{k} + b c_{k})$ . Multiplying out, $u \partial_{r} v - v \partial_{r} u = k r^{- 2} [(a c_{k} - b s_{k})^{2} + (a s_{k} + b c_{k})^{2}] = k r^{- 2} (a^{2} + b^{2}) = k r^{- 2} ∣ f (ω) ∣^{2}$ ; since $\hat{r}$ is a unit vector and the radial directional derivative of an angular function is zero, $j \cdot \hat{r} = \frac{ℏ}{m} (u \partial_{r} v - v \partial_{r} u)$ , which is the first display of (c), and multiplying by $r^{2} > 0$ gives the equivalent second display. Both $u$ and $v$ are $C^{1}$ on $Ω$ : they are sums of products of the $C^{\infty}$ radial factors $r^{- 1}$ , $\cos (k r)$ , $\sin (k r)$ with the $C^{1}$ angular factors $a, b$ , and the radial extensions of $a, b$ are $C^{1}$ by the explicit convention in (c), so the Euclidean product rules give the stated regularity.

2.1step 1.1step 1.2step 1.3algebra

Substituting 1.3 into 1.1 and using that $V$ is real, $2 (u \partial_{t} u + v \partial_{t} v) = 2 (- \frac{ℏ}{2 m} u Δ v + \frac{V}{ℏ} u v + \frac{ℏ}{2 m} v Δ u - \frac{V}{ℏ} v u) = - \frac{ℏ}{m} (u Δ v - v Δ u)$ . Adding the value of $div j$ from 1.2 gives $\partial_{t} n + div j = 0$ , which is (a).

2.2step 1.2givenalgebra

If $u, v, V$ do not depend on $t$ and satisfy the stationary equations, then $Δ u = \frac{2 m}{ℏ^{2}} (V - E) u$ and $Δ v = \frac{2 m}{ℏ^{2}} (V - E) v$ pointwise, so 1.2 gives $div j = \frac{ℏ}{m} \cdot \frac{2 m}{ℏ^{2}} (V - E) (u v - v u) = 0$ on $Ω$ .

3.1F3step 1.4step 1.5step 2.2algebra∎

Let $\overline{B (a, r)} \subseteq Ω$ . The ball $B (a, r)$ is a bounded $C^{1}$ domain and $j$ is $C^{1}$ on a neighbourhood of its closure, so [F3] gives $\int_{\partial B (a, r)} j \cdot ν d S = \int_{B (a, r)} div j d x = 0$ by 2.2; with 1.4 for the plane-wave reading, 1.5 for the outgoing current identity, 2.1 for (a) and this step for (b), the proof of (a), (b) and (c) is complete.

Remarks

Role. The lemma supplies the exact current reading used by the pair's stationary beam model: $j$ is the pair of clauses (a) and (b) that makes the current of a stationary solution divergence-free and flux-free through spheres, and clause (c) computes the two currents -- the incident plane wave with magnitude $ℏ k / m$ and the outgoing spherical wave with radial density $(ℏ k / m) ∣ f ∣^{2} / r^{2}$ -- that the model's cross-section reading compares. Clause (c) asserts the outgoing current identity without any equation for the field: the model's amplitude $f_{k}$ appears in the outgoing coefficient of the full stationary solution (incident wave plus outgoing wave plus remainder), whose existence and equation are established by the pair's stationary theorem, not here.
Choice. The divergence theorem used in 3.1 assumes countable choice; no additional choice principle is used.

The three-dimensional continuity identity, the vanishing stationary divergence and the incident and outgoing current readings

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources