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Qm One Dimensional Potentials — Examples

1 · Prerequisites

2 · Summary

The companion page to qm-one-dimensional-potentials collects three exact finite computations in the declared models. Tunnelling through a rectangular barrier: an exactly solvable scaling and its thick-barrier comparison treats the rectangular barrier at the exactly solvable scaling U=2E with κa=ln⁡10: it obtains T=400/10201 and R=9801/10201, brackets them by rationals, compares the exact value with the thick-barrier value 1/25 with difference −201/255025, and verifies the theorem's error bound 201/125000=0.001608 (Rectangular barrier: scattering solution, flux balance and transmission). Finite square well: an exactly solvable bound state and exact bound-state counts exhibits a declared well depth at which the unique bound state is exactly solvable, with k=κ=π/(4a) and binding energy E=−U/2, verifies the root condition with tan⁡(π/4)=1, and verifies the exact bound-state counts and branch parities at declared depths, including the always-present ground state (Finite square well: bound-state conditions, simplicity and count). The Dirichlet box: exact spectrum, nodes, constant shift and gaps records the exact free Dirichlet box eigenpairs and nodal points, the exact spectrum under a constant potential, the attainment of the min-max bound and the positive gaps (Interval models: discrete spectra, complete eigenbases, min-max bounds and nodal counts). What the one-dimensional potentials examples establish records the combined scope: the comparisons are exact internal checks of the cited theorems at declared parameter values, not simulations, not measurements and not statistical inference. The displayed decimals are rational bracket endpoints or exact terminating decimal expansions of rational values, justified in the verifications.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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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:=2E, so that κ:=2m(U−E)/ℏ=2mE/ℏ=:k; declare further the barrier width a:=ln⁡(10)/κ>0, so that κa=ln⁡10 and eκa=10. Consider the scattering solution of Rectangular barrier: scattering solution, flux balance and transmission for this barrier and set q:=e−2κa. Then, as exact identities in the model:

(i) C:=U2/(4E(U−E))=1, q=1/100, and T=40010201,R=1−T=980110201;

(ii) with rational brackets, 3921100000<T<3922100000,96078100000<R<96079100000;

(iii) the thick-barrier approximation value is 16E(U−E)U2q=125, the exact difference is T−125=−201255025, which lies between −0.000789 and −0.000788, and the error bound of the theorem reads ∣T−125∣≤201125000=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 jin=(ℏk/m)∣Ain∣2, with Ain=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=2E, evaluate the hyperbolic cosine at κa=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=2E, a=ln⁡(10)/κ with κ=2mE/ℏ, and Ain=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+U2sinh⁡2(κa)/(4E(U−E))]−1, R=1−T, j=ℏkm∣Ain∣2(1−∣r∣2)=ℏkm∣Ain∣2∣t∣2, the reading of T and R as fractions of the reference current jin=(ℏk/m)∣Ain∣2, and, with q=e−2κa and C=U2/(4E(U−E)), the thick-barrier form T=(4/C)q/(1+(4C−2)q+q2) with error bound ∣T−16E(U−E)U2q∣≤32E(U−E)U2q∣δ∣ whenever ∣δ∣≤12, δ=(4C−2)q+q2 (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,∞)→R is the inverse of exp⁡, so that exp⁡(log⁡10)=10; the real functions sinh⁡ and cosh⁡ are defined by sinh⁡z=(ez−e−z)/2 and cosh⁡z=(ez+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 ax<ay for a>0, and pq<rs if and only if ps<rq for positive denominators.

1.1givenF2algebra

[Reduction of the data] U=2E gives U−E=E, so C=U2/(4E(U−E))=(2E)2/(4E⋅E)=1 and κ=2mE/ℏ=k; the declared width gives κa=ln⁡10, hence q=e−2κa=exp⁡(−2ln⁡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(κa)]−1=cosh⁡−2(κa) by cosh⁡2−sinh⁡2=1 of [F2]. By [F2], cosh⁡(κa)=eκa+e−κa2=10+1/102=10120, so T=(10120)−2=40010201 and R=1−T=10201−40010201=980110201.

3.1step 2.1F1F3algebra

[Thick-barrier comparison] With C=1 and q=1100, [F1] gives the thickness coefficient 16E(U−E)U2=16E⋅E4E2=4, so the approximation value is 4q=4100=125 and the exact difference is 40010201−125=400⋅25−1020110201⋅25=10000−10201255025=−201255025. The parameter δ=(4−2)q+q2=2100+110000=20110000 satisfies ∣δ∣=20110000≤12, and the error bound is 32E(U−E)U2q∣δ∣=32E24E2⋅1100⋅20110000=201125000. The exact difference obeys 201255025<201125000 by [F3] because 125000 and 255025 are positive and 201255025<201125000 is equivalent to 201⋅125000<201⋅255025; this verifies the stated bound. The rational brackets of (iii) follow from the exact fractions: 7881000000<201255025<7891000000 because 788⋅255025<201⋅1000000<789⋅255025, and 201125000=16081000000.

3.2step 2.1F3algebra

[Rational brackets] By [F3], 3921100000<40010201 is equivalent to 3921⋅10201<400⋅100000, i.e. to 39 998 121<40 000 000, which holds; and 40010201<3922100000 is equivalent to 400⋅100000<3922⋅10201, i.e. to 40 000 000<40 008 322, which holds. Hence 3921100000<T<3922100000; subtracting from 1 reverses the inequalities and gives 96078100000<R<96079100000, since 1−3922100000=96078100000 and 1−3921100000=96079100000.

4.1step 1.1step 2.1F1algebra∎

[Reading] By the barrier theorem clause (b), the conserved current of the scattering solution is j=ℏkm∣Ain∣2∣t∣2=Tjin, where jin=(ℏk/m)∣Ain∣2; T is the transmitted fraction and R is the reflected fraction of this reference current. Since q>0, T=40010201>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=2E the factor U2/(4E(U−E)) equals one and κ=k, so the general barrier formula collapses to sech⁡2(κa), 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.
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Finite square well: an exactly solvable bound state and exact bound-state counts

Physical scope

Finite checks in the declared line model of the pair for the finite square well: an exactly solvable depth at which the unique bound state has energy minus one half of the depth, and exact bound-state counts and branch parities for declared depths. The example is a conditional computation in the model; it reports no measurement, no material well and no empirical value, and asserts no conclusion 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 and action constant ℏ>0, energies, lengths and the depth U read in the units of Dimensions, SI unit conventions and quantity types of the framework, and consider the finite square well V(x)=−U for ∣x∣≤a, V(x)=0 for ∣x∣>a of Finite square well: bound-state conditions, simplicity and count with half-width a>0 and depth U>0.

(i) Exact boundary case of the first branch. Declare U:=ℏ2π216ma2. Then K=2mU/ℏ=π2/(4a), the count parameter is 2aK/π=2/2, there is exactly one bound state, it is an even (cosine-type) state, and its binding energy is exactly E=−U2=−ℏ2π232ma2. For this declared depth the root of the even condition is exactly k=κ=π/(4a).

(ii) Exact counts at declared depths. With U:=25ℏ2π232ma2 one has K=5π/(4a) and 2aK/π=5/2, so there are exactly three bound states, with types even, odd and even in order of increasing energy. With U:=ℏ2π2/(8ma2) one has K=π/(2a) and exactly one bound state; with U:=ℏ2π2/(2ma2) one has K=π/a and exactly two bound states. For every declared depth U>0 there is at least one bound state.

(iii) Rational bracket of the first case. 7071000<2aKπ=22<7081000, so the count ⌈2aK/π⌉ equals one in case (i).

The example is a conditional computation in the declared well model and asserts no measured binding energy; the identities in (i) and (ii) are exact in the model at the displayed declared data.

Verification

Proof technique: direct; verify the exact root condition at the declared depth using the special value tan⁡(π/4)=1, read the count and branch criterion from the finite-well theorem with exact rational comparisons, and bracket the count parameter by rationals.

Given: The line model and the finite-well theorem Finite square well: bound-state conditions, simplicity and count; half-width a>0, mass m>0, action ℏ>0; the depths declared in (i) and (ii).

[F1] Finite-well conclusions. For every a,U>0 the assigned well operator is self-adjoint, its negative eigenvalues are exactly the numbers E∈(−U,0) whose parameters κ=2m(−E)/ℏ, k=2m(U+E)/ℏ satisfy ktan⁡(ka)=κ or −kcot⁡(ka)=κ, each such eigenvalue is simple with the stated parity of the eigenfunction, there are no eigenvalues outside (−U,0), the count is ⌈2aK/π⌉ with K=2mU/ℏ, the solutions lie one in each branch (nπ/2a,(n+1)π/2a) whose left endpoint satisfies nπ/2a<K and nowhere else, and the types alternate with the parity of n (Finite square well: bound-state conditions, simplicity and count clauses (a)-(c)).

[F2] Special value of the tangent. tan⁡(π/4)=1: by Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions, cos⁡(π/4)=cos⁡(π/2−π/4)=sin⁡(π/4); by Pi is the first positive zero of sine, sin⁡(π/4)>0; applying the half-angle identity of Half-angle identities with the sign determined by the quadrant at x=π/2, where sin⁡(π/2)=1 and cos⁡(π/2)=0 by Quarter-turn values and shifts by pi/2 and pi, gives sin⁡(π/4)=(1−0)/2=2/2=cos⁡(π/4); hence tan⁡(π/4)=sin⁡(π/4)/cos⁡(π/4)=1 by Tangent, cotangent, secant, and cosecant on their exact natural domains and the positivity of cos⁡(π/4).

[F3] Rational order. For positive integers, pq<rs iff ps<rq, and 0<12<1; if 0<x and x2<1 then x<1.

1.1givenF1F2algebra

[First case: data and root] With U=ℏ2π2/(16ma2) one has K=2mU/ℏ=π2/(4a), hence K2=π2/(8a2). Put k=κ=π/(4a); then k2+κ2=2⋅π2/(16a2)=π2/(8a2)=K2, and ktan⁡(ka)=κtan⁡(π/4)=κ by [F2], so the even condition of [F1] holds with E=−ℏ2κ2/(2m)=−ℏ2π2/(32ma2)=−U/2. By [F1] this E is a simple eigenvalue with an even eigenfunction.

1.2F1F3algebra

[Second case: three states] With U=25ℏ2π2/(32ma2) one has K=2mU/ℏ=5π/(4a), so 2aK/π=5/2 and ⌈5/2⌉=3. The admissible branches are those with left endpoint nπ/(2a)<5π/(4a), i.e. n2<54; by [F3] this holds exactly for n=0,1,2 (for n=2, 1<5/4; for n=3, 3/2>5/4), so by [F1] there are exactly three negative eigenvalues, one in each of the branches n=0,1,2, with types even, odd, even.

2.1step 1.1F1F3algebra

[First case: count] The count parameter is 2aK/π=2aπ2/(4a)/π=2/2. Since (2/2)2=1/2<1 and 2/2>0, [F3] gives 2/2<1; and 2/2>0. Hence ⌈2aK/π⌉=1 by [F1]; the branch with n=0 has left endpoint 0<K, while the branch with n=1 has left endpoint π/(2a) and π/(2a)>K=π2/(4a) because 12>24=122, equivalently 2>1, which holds since 22=2>1 and 2>0. So the unique branch is n=0, the even branch, consistent with step 1.1.

2.2step 1.2F1F3algebra

[Thresholds and the ground state] For K=π/(2a) the count parameter is 2aK/π=1 and the branch n=0 is admissible while n=1 has left endpoint π/(2a)=K, not strictly smaller, so exactly one bound state; for K=π/a the count parameter is 2 and the admissible branches are n=0,1 (left endpoints 0<π/a and π/(2a)<π/a), so exactly two. Since ⌈x⌉≥1 for every x>0 and the branch n=0 is always admissible, every well with a,U>0 has at least one bound state.

3.1step 2.1step 1.2F1F3algebra∎

[Bracket and reading] The bracket 7071000<22<7081000 follows from [F3] because (7071000)2=4998491000000<12<5012641000000=(7081000)2; hence 7071000<2aK/π<7081000 in case (i) and the count is one. No measured quantity is produced: the identities above are exact relations between the declared model data and the model's binding energies.

Remarks

  • Why the first case is exactly solvable. At the declared depth K=π2/(4a) the radius circle of the even root condition passes through the point k=κ=π/(4a) of the first branch, where tan⁡(ka)=1, so the transcendental root condition is satisfied by an exact rational multiple of 1/a and the signed binding energy is exactly E=−U/2 (its positive magnitude is U/2).
  • What the counts show and what they do not. The comparisons verify the branch criterion and the ceiling formula at declared depths. They are not a numerical solution of the transcendental equations at generic depths, do not compare with data and assert nothing about any physical well.
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The Dirichlet box: exact spectrum, nodes, constant shift and gaps

Physical scope

Finite checks in the declared Dirichlet interval model of the pair: the free box eigenpairs, their explicit nodal points, the exact spectrum under a constant potential shift, the attainment of the min-max bound and the eigenvalue gaps. The example is an exact conditional computation in the declared confined model; it asserts no material wall, no measured spectrum and no empirical result.

Example

Work in the declared Dirichlet interval model The declared interval models with Dirichlet and Robin boundary conditions, and the declared zero-range (delta) model clause (b) with 0<L<∞, mass m>0, action constant ℏ>0, and energies in joules under Dimensions, SI unit conventions and quantity types of the framework. Then, as exact identities in the model:

(i) Free box. For V=0 the eigenpairs of the assigned energy operator HD,0 are exactly (En, en),En=ℏ2π2n22mL2,en(x)=2L sin⁡ ⁣(nπxL),n≥1, the family (en)n≥1 is a complete orthonormal family of L2(0,L;C), each eigenvalue is simple, and en has exactly n−1 zeros in (0,L), namely the points xk=kL/n, k=1,…,n−1.

(ii) Constant shift. If V=c is the constant potential of value c∈R (joules), then HD,c has exactly the same eigenfunctions en and the eigenvalues En+c; consequently the min-max comparison bound of the interval theorem is attained with equality, ∣En+c−En∣=∣c∣=∥V∥∞ for every n.

(iii) Gaps. En+1−En=ℏ2π2(2n+1)2mL2>0, so the spectrum is strictly increasing.

The example is an exact conditional computation in the declared box model and asserts no measured transition energy.

Verification

Proof technique: direct; read the free-box clause of the interval theorem, evaluate the sine zeros explicitly, and shift the operator by a constant multiple of the identity to verify the exact spectrum under a constant potential.

Given: The declared Dirichlet interval model with 0<L<∞ and m,ℏ>0; the interval spectrum theorem Interval models: discrete spectra, complete eigenbases, min-max bounds and nodal counts; a real constant potential value c.

[F1] Dirichlet box conclusions. For every V∈C([0,L];R) the assigned operator HD,V=(ℏ2/2m)(AD+MV~) with V~=(2m/ℏ2)V is an assigned sharp observable, self-adjoint with compact resolvent and a complete orthonormal eigenfamily, with strictly increasing simple eigenvalues En and the nodal count: the eigenvector belonging to En has exactly n−1 zeros in (0,L); the min-max comparison ∣En−ℏ2π2n2/(2mL2)∣≤∥V∥∞ holds; and for V=0 a complete orthonormal family of eigenpairs is (ℏ2π2n2/(2mL2), en) with the displayed en, a complete orthonormal family (Interval models: discrete spectra, complete eigenbases, min-max bounds and nodal counts clause (a)).

[F2] Sine zeros. For n≥1 and x∈(0,L), sin⁡(nπx/L)=0 if and only if nπx/L=kπ for an integer k, i.e. x=kL/n; among these only k=1,…,n−1 lie in (0,L) (The zero sets of sine and cosine and the least positive common period 2 pi).

1.1givenF1algebra

[Free eigenpairs and completeness] For V=0 the hypotheses of [F1] hold with V=0, and its free-box clause gives the complete orthonormal family of eigenpairs (En,en) with En=ℏ2π2n2/(2mL2) and en(x)=2/Lsin⁡(nπx/L), together with completeness of (en) and simplicity of the eigenvalues.

2.1step 1.1F1F2algebra

[Nodal points] By [F2] the zeros of en in (0,L) are exactly xk=kL/n, k=1,…,n−1, so their number is n−1, in agreement with the nodal count of [F1].

2.2step 1.1F1algebra

[Constant shift] Let c∈R and V=c; then V~=2mc/ℏ2 is constant and HD,c=(ℏ2/2m)(AD+MV~)=(ℏ2/2m)AD+cI on D(AD), because multiplication by the constant V~ is the scalar V~ times the identity and (ℏ2/2m)(2mc/ℏ2)=c. Applying this identity to an eigenvector en of AD with eigenvalue λn=(nπ/L)2 gives HD,cen=(ℏ2/2m)λnen+cen=(En+c)en; by [F1] the eigenfamily (en) is complete and the eigenvalues simple, so the spectrum of HD,c is exactly {En+c:n≥1}.

3.1step 2.2F1algebra

[Attainment of the min-max bound] For this constant potential the bound of [F1] reads ∣En+c−En∣=∣c∣, and ∥V∥∞=∣c∣; hence the inequality is an equality for every n.

4.1step 3.1F1algebra∎

[Gaps and reading] Since (n+1)2−n2=2n+1, the successive eigenvalue difference is En+1−En=ℏ2π2(2n+1)/(2mL2)>0, consistent with strictness and simplicity in [F1]; every displayed number is an exact function of the declared data m,ℏ,L and the declared shift c, and no measured spectrum is asserted.

Remarks

  • Attainment is not generic. The equality in (ii) is a property of the constant potential, for which the perturbation commutes with the free box operator. For a nonconstant V the min-max bound is an inequality and its attainment is not claimed.
  • What is not asserted. The example treats the free and constantly shifted Dirichlet box only. It does not treat Robin conditions, singular perturbations or a material wall, and it reports no data.
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What the one-dimensional potentials examples establish

Physical scope

Scope of the examples page of the one-dimensional potentials pair: exact finite checks of the barrier transmission, the finite-well root conditions and counts, and the Dirichlet box spectrum and nodes. The remark asserts no new physical result, records no measurement and makes no claim about nature.

Scope of the examples page

The three computations. Tunnelling through a rectangular barrier: an exactly solvable scaling and its thick-barrier comparison treats the rectangular barrier of Rectangular barrier: scattering solution, flux balance and transmission at the exactly solvable scaling U=2E with κa=ln⁡10: it obtains T=400/10201 and R=9801/10201 as exact rationals, brackets them by 3921/100000<T<3922/100000, compares the exact value with the thick-barrier value 1/25 (the difference is exactly −201/255025) and verifies the theorem's error bound 201/125000=0.001608; the computation is a finite check of the model formulas, including the positivity of transmission through a classically forbidden barrier. Finite square well: an exactly solvable bound state and exact bound-state counts exhibits a declared well depth at which the unique bound state is exactly solvable (even state, k=κ=π/(4a), binding energy E=−U/2=−ℏ2π2/(32ma2)), verifies the even root condition with the special value tan⁡(π/4)=1, and verifies the count and branch criterion at declared depths: exactly three bound states with types even, odd, even when 2aK/π=5/2, exactly one when 2aK/π=1, exactly two when 2aK/π=2, and at least one for every a,U>0. The Dirichlet box: exact spectrum, nodes, constant shift and gaps records the exact free Dirichlet box eigenpairs En=ℏ2π2n2/(2mL2) with sine eigenfunctions, their explicit n−1 nodal points kL/n, the exact spectrum En+c of a constant potential with the min-max bound attained, and the positive gaps ℏ2π2(2n+1)/(2mL2).

What is and is not compared. The comparisons are internal: the exact barrier formula and its thick-barrier approximation, the exact well root condition and the exact branch count, and the exact box spectrum with the min-max bound. They verify the cited theorems at declared parameter points by exact rational and algebraic checks; they are not numerical approximations of generic transcendental roots, they contain no simulation presented as data, and they report no measured value, no uncertainty and no significance statement. The displayed decimals are rational bracket endpoints or exact terminating decimal expansions of rational values, including 201/125000=0.001608; each is justified in the verifications.

Evidence status. All three items are conditional computations in the declared line, well and box models of the pair's A page, under the same P1, P2, P4, P5 assumptions and the same cited mathematics; the empirical qualifications of the A page carry over unchanged. Nothing on this page is an observation, an apparatus model or a statistical inference, and agreement of the displayed identities with any measurement would be evidence about the combined model and apparatus assumptions rather than a proof of the model.

Sources