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Rectangular barrier: scattering solution, flux balance and transmission

Physical scope

The line model of the pair with the declared rectangular barrier of height U and width a, in the scattering reading of the model: the unique bounded piecewise-classical stationary solution with prescribed asymptotics, its conserved probability current, the exact transmission and reflection coefficients with the flux balance T+R=1, the identification of T and R with ratios of the transmitted and reflected currents to the reference current of an incoming reference amplitude of one inverse square root metre, and the thick-barrier error bound. The scattering solution is a declared generalized stationary solution and not an element of the model state space; the statement is conditional on P1, P2, P4, P5, the declared barrier data and the cited mathematics, and asserts no wave-packet dynamics, no detector, no count and no empirical result.

Statement

Adopt the postulates P1, P2, P4, P5 and the declared line model The declared one-dimensional line model: state space, Hamiltonian, bound states and generalized stationary solutions, with $m > 0$ , $ℏ > 0$ and energy, potential and position read in the units of Dimensions, SI unit conventions and quantity types of the framework. Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Fix $a > 0$ , $U > E > 0$ and the incident reference amplitude $A_{i n} : = 1 m^{- 1 / 2}$ , and let $V (x) : = {\begin{cases} U, & 0 < x < a, \\ 0 & otherwise, \end{cases}$ let $k : = \sqrt{2 m E} / ℏ$ , $κ : = \sqrt{2 m (U - E)} / ℏ$ , and consider the declared generalized stationary solutions of the model at energy $E$ (clause (e) of the model) that are bounded, classically $C^{1}$ on $R$ and $C^{2}$ on each of $(- \infty, 0), (0, a), (a, \infty)$ . Then:

(a) Scattering solution. There is exactly one such $ψ$ with the transmission asymptotics $ψ (x) = A_{i n} (e^{i k x} + r e^{- i k x})$ for $x < 0$ and $ψ (x) = A_{i n} t e^{i k x}$ for $x > a$ , where the amplitude ratios $r, t \in C$ are dimensionless and uniquely determined; with $D : = 2 \cosh (κ a) + i (\frac{κ}{k} - \frac{k}{κ}) \sinh (κ a)$ one has $t = 2 e^{- i k a} / D$ , and $T : = ∣ t ∣^{2} = [1 + \frac{U^{2} \sinh^{2} (κ a)}{4 E (U - E)}]^{- 1}, R : = ∣ r ∣^{2} = 1 - T .$

(b) Conserved current and flux balance. The declared probability current $j (x) : = \frac{ℏ}{m} Im (\overline{ψ (x)} ψ^{'} (x))$ of the scattering solution is constant on $R$ , equal to $\frac{ℏ k}{m} ∣ A_{i n} ∣^{2} (1 - ∣ r ∣^{2}) = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} ∣ t ∣^{2}$ . With the reference current $j_{i n} : = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2}$ of the incoming term $x \mapsto A_{i n} e^{i k x}$ , the transmitted right-going current equals $T j_{i n}$ , the reflected left-going current has magnitude $R j_{i n}$ , and $T + R = 1.$ The pair reads $T$ and $R$ as the transmitted and reflected fractions of the reference current of the declared scattering solution; this reading is part of the model and is not a measurement.

(c) Thick-barrier form and error. With $q : = e^{- 2 κ a}$ and $C : = U^{2} / (4 E (U - E))$ one has $T = \frac{(4 / C) q}{1 + (\frac{4}{C} - 2) q + q^{2}};$ consequently, whenever $δ : = (\frac{4}{C} - 2) q + q^{2}$ satisfies $∣ δ ∣ \leq \frac{1}{2}$ , $∣ T - \frac{16 E (U - E)}{U^{2}} q ∣ \leq \frac{32 E (U - E)}{U^{2}} q ∣ δ ∣ .$

(d) Scope. The scattering solution $ψ$ is a declared generalized stationary solution, not an element of $L^{2} (R; C)$ ; it is not a preparation of the model, and no Born probability is read from it. The numbers $T, R$ are exact model quantities of the declared solution, read as ratios of currents under the model's scattering reading; the statement contains no wave-packet construction, no time-dependent limit, no detector response and no measured count, and it asserts neither that a particle passes through the barrier nor that it is reflected.

Facts & Assumptions

Given: The postulates P1, P2, P4, P5 and the choice hypotheses; the line model with $m > 0$ , $ℏ > 0$ ; barrier data $a > 0$ , $U > E > 0$ ; $A_{i n} = 1 m^{- 1 / 2}$ ; $k = \sqrt{2 m E} / ℏ$ , $κ = \sqrt{2 m (U - E)} / ℏ$ ; the piecewise constant $V$ of the statement; the declared generalized stationary solutions at energy $E$ .

[F1]

Flux identity. If separate real functions $u_{s}, v_{s} : Ω \to R$ with continuous second derivatives solve the stationary component equations with the same real $V$ and $E$ on an open interval $Ω$ , then $j_{s} = \frac{ℏ}{m} (u_{s} v_{s}^{'} - u_{s}^{'} v_{s})$ is constant there; in particular this holds for the real and imaginary parts of a classically $C^{2}$ stationary solution on an interval on which $V$ is constant (The one-dimensional continuity identity and the constant flux of stationary solutions clause (b)).

[F2]

The mathematical barrier theorem. For $a, U, E, m, ℏ > 0$ with $U > E$ , the bounded $C^{1}$ functions that are $C^{2}$ on each of $(- \infty, 0), (0, a), (a, \infty)$ and satisfy $- \frac{ℏ^{2}}{2 m} ψ^{''} + V ψ = E ψ$ on each interval: exactly one has the transmission asymptotics with uniquely determined $r, t$ ; $t = 2 e^{- i k a} / D$ with the displayed $D$ ; $T = ∣ t ∣^{2} = [1 + U^{2} \sinh^{2} (κ a) / (4 E (U - E))]^{- 1}$ ; $R = ∣ r ∣^{2} = 1 - T$ ; and with $q = e^{- 2 κ a}$ , $C = U^{2} / (4 E (U - E))$ the thick-barrier displays of the statement hold (Rectangular barrier: transmission coefficients and the thick-barrier error clauses (a)-(c)).

[F3]

The model's generalized-solution reading. A declared generalized stationary solution has real energy, is bounded and classically $C^{1}$ with continuous derivative, is $C^{2}$ on each interval where $V$ is continuous, and satisfies the stationary equation there. It also satisfies the global weak stationary equation of that definition; for a piecewise constant potential, the matching $C^{1}$ piecewise $C^{2}$ solutions satisfy it by integration by parts on the pieces, with internal boundary terms cancelling. A generalized solution can belong to $L^{2}$ ; exclusion from the preparation space must therefore be checked for the particular scattering solution. Its declared amplitude has dimension $ℓ^{- 1 / 2}$ , so its continuous current is read in $s^{- 1}$ , and the scattering reading uses the declared current ratios (The declared one-dimensional line model: state space, Hamiltonian, bound states and generalized stationary solutions clause (e), Dimensions, SI unit conventions and quantity types of the framework clauses (d)1 and (d)5).

Proof

technique · direct; instantiate the mathematical barrier theorem for the declared data, prove the current is constant by the flux identity across the two jumps, and evaluate the current on the two exterior intervals to obtain the flux balance and the fraction reading

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

1.1givenF2F3algebra

[The scattering solution] Dividing the declared solution by the fixed nonzero $A_{i n}$ gives a bounded $C^{1}$ profile, $C^{2}$ on each of the three intervals, satisfying the same stationary equation as [F2] with unit incident amplitude. Conversely multiplying any profile in [F2] by $A_{i n}$ gives the required regularity and classical equation, and the interface matching gives its global weak equation by [F3], hence a declared generalized solution. Hence [F2] clause (a) gives existence and uniqueness of the amplitude ratios $r, t$ and $t = 2 e^{- i k a} / D$ , while clause (b) gives the displays for $T$ and $R$ ; this is (a).

2.1step 1.1F1F3algebra

[Constant current on each interval] On each of the three open intervals $V$ is constant, hence real and continuous; the scattering solution is classically $C^{2}$ there and satisfies the stationary equation with real energy $E$ , so [F1] clause (b) gives $j^{'} = 0$ on each interval and $j$ constant on each. Since $ψ$ is classically $C^{1}$ on $R$ with continuous derivative, $j = \frac{ℏ}{m} Im (\overline{ψ} ψ^{'})$ is continuous on $R$ , so the three constants agree and $j$ is constant on $R$ .

3.1step 2.1F2algebra

[Evaluation on the exteriors] For $x < 0$ , $ψ = A_{i n} (e^{i k x} + r e^{- i k x})$ , so $Im (\overline{ψ} ψ^{'}) = ∣ A_{i n} ∣^{2} k (1 - ∣ r ∣^{2})$ : after expansion the two oscillating cross terms have real sum. Hence $j = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} (1 - ∣ r ∣^{2})$ on $(- \infty, 0)$ . For $x > a$ , $ψ = A_{i n} t e^{i k x}$ and $Im (\overline{ψ} ψ^{'}) = ∣ A_{i n} ∣^{2} k ∣ t ∣^{2}$ , so $j = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} ∣ t ∣^{2}$ on $(a, \infty)$ ; by step 2.1 the two values are equal, which reproduces $1 - ∣ r ∣^{2} = ∣ t ∣^{2}$ , i.e. $R + T = 1$ by (a).

4.1step 3.1F2F3algebra

[Fraction reading] The incoming term $A_{i n} e^{i k x}$ has current $\frac{ℏ k}{m} ∣ A_{i n} ∣^{2} = j_{i n}$ , while the reflected term $A_{i n} r e^{- i k x}$ has current $- \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} ∣ r ∣^{2} = - R j_{i n}$ . Their sum is the left exterior value of $j$ from step 3.1; the cross-term contribution vanishes. The transmitted right-going current is $T j_{i n}$ , so $T$ and $R$ are the transmitted and reflected fractions of the reference current. Clause (c) is [F2] clause (c).

5.1step 1.1step 4.1F2F3givenalgebra∎

[Scope] By step 1.1 the solution is a declared generalized stationary solution. Its transmission coefficient in [F2] satisfies $T > 0$ , since its displayed denominator is finite and positive. For $x > a$ , $∣ ψ (x) ∣^{2} = ∣ A_{i n} ∣^{2} T > 0$ , so $\int_{a}^{b} ∣ ψ (x) ∣^{2} d x = ∣ A_{i n} ∣^{2} T (b - a) \to \infty$ as $b \to \infty$ . Thus this particular solution is not in $L^{2} (R)$ and is not a preparation, as qualified by the model's clause (e). The numbers $T, R$ remain the declared current ratios of step 4.1, with no wave-packet, detector or empirical conclusion.

Remarks

Cross term of the current. The identity $Im (\overline{e^{i k x} + r e^{- i k x}} (i k e^{i k x} - i k r e^{- i k x})) = k (1 - ∣ r ∣^{2})$ shows that the incoming and reflected contributions add without interference in the current, although the density does carry the cross term; this is why the flux balance is an exact statement for the declared solution.
What is not asserted. No time-dependent wave packet, no arrival distribution, no resonant or above-barrier case ( $E \geq U$ is excluded by $U > E$ ), and no material barrier is treated. The thick-barrier clause is an error bound of the cited mathematical theorem, not a measurement.

Rectangular barrier: scattering solution, flux balance and transmission

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources