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Tunnelling through a rectangular barrier: an exactly solvable scaling and its thick-barrier comparison

Physical scope

A finite check in the declared line model of the pair: the rectangular barrier with height twice the incident energy and a width scaling chosen so that the barrier exponent is the natural logarithm of ten. The example computes the exact transmission and reflection coefficients as rational numbers, brackets them by rationals, compares the exact value with the thick-barrier approximation and verifies the theorem's error bound. It is a computation in the model, not a measurement; it reports no count, no apparatus and no empirical value, and it makes no claim about nature.

Example

Work in the declared line model The declared one-dimensional line model: state space, Hamiltonian, bound states and generalized stationary solutions with mass parameter $m > 0$ , action constant $ℏ > 0$ , and lengths, energies and currents read in the units of Dimensions, SI unit conventions and quantity types of the framework. Declare the incident energy $E > 0$ and the barrier height $U : = 2 E$ , so that $κ : = \sqrt{2 m (U - E)} / ℏ = \sqrt{2 m E} / ℏ = : k$ ; declare further the barrier width $a : = \ln (10) / κ > 0$ , so that $κ a = \ln 10$ and $e^{κ a} = 10$ . Consider the scattering solution of Rectangular barrier: scattering solution, flux balance and transmission for this barrier and set $q : = e^{- 2 κ a}$ . Then, as exact identities in the model:

(i) $C : = U^{2} / (4 E (U - E)) = 1$ , $q = 1 / 100$ , and $T = \frac{400}{10201}, R = 1 - T = \frac{9801}{10201};$

(ii) with rational brackets, $\frac{3921}{100000} < T < \frac{3922}{100000}, \frac{96078}{100000} < R < \frac{96079}{100000};$

(iii) the thick-barrier approximation value is $\frac{16 E (U - E)}{U^{2}} q = \frac{1}{25}$ , the exact difference is $T - \frac{1}{25} = - \frac{201}{255025},$ which lies between $- 0.000789$ and $- 0.000788$ , and the error bound of the theorem reads $∣ T - \frac{1}{25} ∣ \leq \frac{201}{125000} = 0.001608;$

(iv) since $q = 1 / 100 > 0$ , the transmission $T$ is positive although the declared barrier height satisfies $U > E$ ; and by clause (b) of the barrier theorem the number $T$ is the transmitted fraction of the reference current $j_{i n} = (ℏ k / m) ∣ A_{i n} ∣^{2}$ , with $A_{i n} = 1 m^{- 1 / 2}$ , while $R$ is the reflected fraction.

The example is an exact computation in the declared barrier model with declared data $U, E, a, m, ℏ$ related as above; it records no measurement and no apparatus value.

Verification

Proof technique: direct; reduce the barrier formula under $U = 2 E$ , evaluate the hyperbolic cosine at $κ a = \ln 10$ with the exponential addition and reciprocal laws, and verify the thick-barrier comparison by rational arithmetic.

Given: The line model and the scattering theorem of the barrier item; the declared data $E > 0$ , $U = 2 E$ , $a = \ln (10) / κ$ with $κ = \sqrt{2 m E} / ℏ$ , and $A_{i n} = 1 m^{- 1 / 2}$ .

[F1] Barrier formulas. Under the hypotheses of Rectangular barrier: scattering solution, flux balance and transmission one has $T = ∣ t ∣^{2} = [1 + U^{2} \sinh^{2} (κ a) / (4 E (U - E))]^{- 1}$ , $R = 1 - T$ , $j = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} (1 - ∣ r ∣^{2}) = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} ∣ t ∣^{2}$ , the reading of $T$ and $R$ as fractions of the reference current $j_{i n} = (ℏ k / m) ∣ A_{i n} ∣^{2}$ , and, with $q = e^{- 2 κ a}$ and $C = U^{2} / (4 E (U - E))$ , the thick-barrier form $T = (4 / C) q / (1 + (\frac{4}{C} - 2) q + q^{2})$ with error bound $∣ T - \frac{16 E (U - E)}{U^{2}} q ∣ \leq \frac{32 E (U - E)}{U^{2}} q ∣ δ ∣$ whenever $∣ δ ∣ \leq \frac{1}{2}$ , $δ = (\frac{4}{C} - 2) q + q^{2}$ (clauses (a)-(c) of the theorem).

[F2] Exponential and hyperbolic laws. $\exp (x + y) = \exp (x) \exp (y)$ for all real $x, y$ ; $\exp (- x) = 1 / \exp (x)$ for every real $x$ ; and $\log : (0, \infty) \to R$ is the inverse of $\exp$ , so that $\exp (\log 10) = 10$ ; the real functions $\sinh$ and $\cosh$ are defined by $\sinh z = (e^{z} - e^{- z}) / 2$ and $\cosh z = (e^{z} + e^{- z}) / 2$ for real $z$ , and satisfy $\cosh^{2} - \sinh^{2} = 1$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The natural logarithm as the inverse of the exponential function, The six hyperbolic functions and their natural domains).

[F3] Rational arithmetic. For positive integers the usual order and field laws of the rationals hold; in particular $x < y$ if and only if $a x < a y$ for $a > 0$ , and $\frac{p}{q} < \frac{r}{s}$ if and only if $p s < r q$ for positive denominators.

1.1givenF2algebra

[Reduction of the data] $U = 2 E$ gives $U - E = E$ , so $C = U^{2} / (4 E (U - E)) = (2 E)^{2} / (4 E \cdot E) = 1$ and $κ = \sqrt{2 m E} / ℏ = k$ ; the declared width gives $κ a = \ln 10$ , hence $q = e^{- 2 κ a} = \exp (- 2 \ln 10) = (\exp (\ln 10))^{- 2} = 10^{- 2} = 1 / 100$ by [F2].

2.1step 1.1F1F2F3algebra

[Exact coefficients] With $C = 1$ the formula of [F1] reads $T = [1 + \sinh^{2} (κ a)]^{- 1} = \cosh^{- 2} (κ a)$ by $\cosh^{2} - \sinh^{2} = 1$ of [F2]. By [F2], $\cosh (κ a) = \frac{e^{κ a} + e^{- κ a}}{2} = \frac{10 + 1 / 10}{2} = \frac{101}{20}$ , so $T = (\frac{101}{20})^{- 2} = \frac{400}{10201}$ and $R = 1 - T = \frac{10201 - 400}{10201} = \frac{9801}{10201}$ .

3.1step 2.1F1F3algebra

[Thick-barrier comparison] With $C = 1$ and $q = \frac{1}{100}$ , [F1] gives the thickness coefficient $\frac{16 E (U - E)}{U^{2}} = \frac{16 E \cdot E}{4 E^{2}} = 4$ , so the approximation value is $4 q = \frac{4}{100} = \frac{1}{25}$ and the exact difference is $\frac{400}{10201} - \frac{1}{25} = \frac{400 \cdot 25 - 10201}{10201 \cdot 25} = \frac{10000 - 10201}{255025} = - \frac{201}{255025}$ . The parameter $δ = (4 - 2) q + q^{2} = \frac{2}{100} + \frac{1}{10000} = \frac{201}{10000}$ satisfies $∣ δ ∣ = \frac{201}{10000} \leq \frac{1}{2}$ , and the error bound is $\frac{32 E (U - E)}{U^{2}} q ∣ δ ∣ = \frac{32 E^{2}}{4 E^{2}} \cdot \frac{1}{100} \cdot \frac{201}{10000} = \frac{201}{125000}$ . The exact difference obeys $\frac{201}{255025} < \frac{201}{125000}$ by [F3] because $125000$ and $255025$ are positive and $\frac{201}{255025} < \frac{201}{125000}$ is equivalent to $201 \cdot 125000 < 201 \cdot 255025$ ; this verifies the stated bound. The rational brackets of (iii) follow from the exact fractions: $\frac{788}{1000000} < \frac{201}{255025} < \frac{789}{1000000}$ because $788 \cdot 255025 < 201 \cdot 1000000 < 789 \cdot 255025$ , and $\frac{201}{125000} = \frac{1608}{1000000}$ .

3.2step 2.1F3algebra

[Rational brackets] By [F3], $\frac{3921}{100000} < \frac{400}{10201}$ is equivalent to $3921 \cdot 10201 < 400 \cdot 100000$ , i.e. to $39 998 121 < 40 000 000$ , which holds; and $\frac{400}{10201} < \frac{3922}{100000}$ is equivalent to $400 \cdot 100000 < 3922 \cdot 10201$ , i.e. to $40 000 000 < 40 008 322$ , which holds. Hence $\frac{3921}{100000} < T < \frac{3922}{100000}$ ; subtracting from $1$ reverses the inequalities and gives $\frac{96078}{100000} < R < \frac{96079}{100000}$ , since $1 - \frac{3922}{100000} = \frac{96078}{100000}$ and $1 - \frac{3921}{100000} = \frac{96079}{100000}$ .

4.1step 1.1step 2.1F1algebra∎

[Reading] By the barrier theorem clause (b), the conserved current of the scattering solution is $j = \frac{ℏ k}{m} ∣ A_{i n} ∣^{2} ∣ t ∣^{2} = T j_{i n}$ , where $j_{i n} = (ℏ k / m) ∣ A_{i n} ∣^{2}$ ; $T$ is the transmitted fraction and $R$ is the reflected fraction of this reference current. Since $q > 0$ , $T = \frac{400}{10201} > 0$ even though $E < U$ , which is the tunnelling statement of the model. This is a model computation and asserts no measured count or apparatus behaviour.

Remarks

Why this scaling is exactly solvable. At $U = 2 E$ the factor $U^{2} / (4 E (U - E))$ equals one and $κ = k$ , so the general barrier formula collapses to ${sech}^{2} (κ a)$ , and the width $a = \ln (10) / κ$ makes the hyperbolic cosine rational.
What the check shows and what it does not. The computation verifies the exact formulas, the rational brackets and the published error bound of the barrier theorem at one declared parameter point. It does not establish the accuracy of the physical model, does not compare with data and produces no experimental conclusion.

Tunnelling through a rectangular barrier: an exactly solvable scaling and its thick-barrier comparison

Example

Verification

Remarks

Depends on

Dependency tree · 2 levels

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