Matching Conditions and Wavefunction Continuity

Every one-dimensional problem you have solved so far ran on the same quiet machinery. The Infinite Square Well pinned $\psi$ to zero at hard walls. The Finite Square Well glued an interior cosine to decaying tails and got a transcendental equation. The Delta-Function Potential Well kinked the slope by a prescribed amount. The Potential Step and the Rectangular Barrier stitched travelling waves across sudden edges. Five problems, five apparently ad-hoc boundary rules — continuous here, jumping there, vanishing somewhere else. This article is where the rules stop being ad-hoc. All of them follow from one source: the Schrödinger equation itself, integrated across the joint. Where the potential is finite, both $\psi$ and $\psi'$ must be continuous — not by decree, but because a broken wavefunction would demand an infinite term the equation does not contain. Where the potential holds a delta spike, the same integral delivers a finite, prescribed slope jump. Where the potential goes infinite over a region, the wave is expelled and only $\psi=0$ survives at the wall. One theorem, three regimes, zero exceptions. The payoff is more than tidiness. Matching is the engine of one-dimensional quantum mechanics: counting equations against unknowns tells you — before solving anything — why bound spectra come out discrete while scattering happens at every energy. And packaging the matching algebra into a transfer matrix turns chains of wells and barriers into simple matrix multiplication, the working method behind every quantum-well laser, superlattice and multilayer mirror ever designed.

Key takeaways

The time-independent Schrödinger equation is second-order: wherever $V$ is finite, both $\psi$ and $\psi'$ are continuous — integrate the equation across the joint and the jump terms vanish with the window.
A delta potential $g\,\delta(x-x_0)$ is the loophole: the same integral leaves a finite kick, $\Delta\psi'=(2mg/\hbar^2)\,\psi(x_0)$ — sign included, attractive or repulsive.
An infinite wall replaces matching by expulsion: $\psi=0$ at the wall, and the slope condition is dropped because there is nothing on the far side to match to.
Equation-counting decides spectra: bound problems have one condition too many, forcing a transcendental equation with discrete roots; scattering problems balance exactly, so every $E>0$ works and only amplitudes are determined.
The transfer matrix industrializes matching: each interface and each stretch of free propagation is a $2\times2$ matrix, structures compose by multiplication, and $\det M=1$ across any structure that starts and ends at the same potential.
Key idea: boundary behaviour is not postulated — it is read off the Schrödinger equation by integrating across the boundary, and the singularity strength of $V$ decides how much smoothness survives.

The core insight

Integrating $\psi''=(2m/\hbar^2)(V-E)\psi$ across a joint shows that $\psi$ and $\psi'$ stay continuous wherever $V$ is finite, that $\psi'$ jumps by exactly $(2mg/\hbar^2)\psi(x_0)$ where $V$ carries a delta of strength $g$ , and that $\psi$ itself must vanish where $V$ becomes infinite — the complete grammar of piecewise quantum mechanics.

Where this lives today

Matching conditions are the daily bread of device engineering. Every quantum-well laser, resonant-tunnelling diode and quantum-cascade structure is designed by transfer-matrix codes multiplying exactly the factors of equation (3) — with one refinement: in a heterostructure the effective mass changes between layers, and the honest junction condition becomes continuity of $\psi$ and $\psi'/m^*$ (BenDaniel–Duke), so that probability current, not bare slope, crosses the interface. The same matrices, with $k$ replaced by refractive index, design anti-reflection coatings and dielectric mirrors. And for smooth potentials the philosophy survives as the shooting method — quantization by matching, executed by computer.

The physical picture

Read the Schrödinger equation as a curvature budget: $\psi''=(2m/\hbar^2)(V-E)\,\psi$ . The wavefunction is a flexible string whose bending is bought by the local value of $V-E$ . A kink — a sudden slope change — is infinite curvature at a point, and the budget can pay for it only if the potential supplies something infinite there. A finite step, however tall, buys a finite curvature change: the string bends sharply, never creases. A delta spike supplies exactly one unit of concentrated infinity: one crease, of prescribed angle. An infinite wall supplies unbounded expulsion: the string is pressed flat against zero.

Figure

The three junction grammars — smooth join at a finite step, prescribed kink at a delta, expulsion at an infinite wall

The second half of the picture is bookkeeping, and it decides each problem's fate before any algebra. Each interface contributes two equations ( $\psi$ and $\psi'$ continuous, or the delta variant). Now count unknowns. In a bound-state problem, normalizability kills the growing exponentials and the overall scale of $\psi$ is unphysical — the coefficient count comes up one short of the equation count, and the leftover condition must constrain the energy itself: that is the transcendental equation, its isolated roots the discrete spectrum. In a scattering problem the incident amplitude is fixed at 1 and both travelling directions are allowed, so the books balance at every energy: nothing constrains $E$ , and matching delivers $r$ and $t$ instead. Discrete versus continuous spectra is not deep mystery — it is linear algebra counting.

The theorem: integrate across the joint

Everything hangs on one manoeuvre, so we do it honestly. Write the equation in curvature form, isolating what the boundary sees:

\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=\frac{2m}{\hbar^2}\,\big(V(x)-E\big)\,\psi(x) \qquad (1)

where: the right side is the curvature price list — finite wherever $V$ is finite. The question "how smooth is $\psi$ at $x_0$ ?" becomes "how big is the integral of the right side over a shrinking window around $x_0$ ?", and that is a question with a clean answer for each class of potential.

Derivation AContinuity of ψ and ψ′, and the delta exception, from one integral

We prove the complete boundary grammar: (i) $\psi'$ is continuous wherever $V$ is bounded, (ii) $\psi$ is continuous even more robustly, (iii) a delta contribution $g\,\delta(x-x_0)$ in $V$ produces the jump $\Delta\psi'=(2mg/\hbar^2)\psi(x_0)$ , and (iv) an infinite wall forces $\psi=0$ with no slope condition. The strategy is to integrate equation (1) over $[x_0-\varepsilon,\,x_0+\varepsilon]$ and watch which terms survive $\varepsilon\to0$ . The one point of care: keep track of which function's jump each integral controls — integrating once controls $\psi'$ , and the boundedness of $\psi'$ then controls $\psi$ .

Step 1. Integrate (1) once across the window:

\psi'(x_0+\varepsilon)-\psi'(x_0-\varepsilon)=\frac{2m}{\hbar^2}\int_{x_0-\varepsilon}^{x_0+\varepsilon}\big(V(x)-E\big)\,\psi(x)\,\mathrm{d}x

Step 2. Let $V$ be bounded near $x_0$ , say $|V-E|\le M$ there. Then the integrand is bounded by $M\max|\psi|$ and the integral by $2\varepsilon M\max|\psi|$ , which dies linearly as $\varepsilon\to0$ :

\big|\Delta\psi'\big|\le\frac{2m}{\hbar^2}\,2\varepsilon\,M\,\max|\psi|\;\longrightarrow\;0

Why even a huge finite step changes nothing: $M$ can be a billion electron-volts — the window still shrinks faster than the integrand can pay. Only a potential that grows as the window shrinks can leave a residue.

Step 3. $\psi$ itself is continuous under even weaker assumptions: $\psi'$ exists and is bounded on both sides (it solves a second-order ODE with locally integrable coefficients), and a function with bounded derivative cannot jump. Each integration smooths by one degree: $\psi''$ may contain a delta, then $\psi'$ a step at worst, then $\psi$ a kink at worst — never a break.

Step 4. Now let $V(x)=V_{\text{smooth}}(x)+g\,\delta(x-x_0)$ . The smooth part vanishes from the Step-1 integral as before, but the delta pays its fixed toll regardless of $\varepsilon$ :

\Delta\psi'=\frac{2m}{\hbar^2}\int_{x_0-\varepsilon}^{x_0+\varepsilon} g\,\delta(x-x_0)\,\psi(x)\,\mathrm{d}x=\frac{2mg}{\hbar^2}\,\psi(x_0)

Why the sign convention matters: with $g=-\alpha<0$ (attractive) the slopes are pulled together — the cusp of the Delta-Function Potential Well and Bound State, $\Delta\psi'=-(2m\alpha/\hbar^2)\psi(0)$ ; with $g=+\alpha>0$ (repulsive) they are pushed apart — the scattering kink of Delta-Function Barrier and Scattering. One formula, both signs, and it is worth checking each time that your jump has the sign your potential demands.

Step 5. The infinite wall is the limit the Finite Square Well rehearsed: the exterior tail $e^{-\kappa|x|}$ has $\kappa\to\infty$ as $V_0\to\infty$ , so the exterior wave is squeezed to nothing and $\psi(\text{wall})=0$ ; with the far side dead, no second condition remains — the slope is whatever the interior needs. That is why the Infinite Square Well imposed one condition per wall, not two. Two consistency checks are worth noting: numerically, the exact step-junction solution satisfies both continuity conditions to machine precision while the shrinking-window bound on $\Delta\psi'$ falls linearly with $\varepsilon$ ; symbolically, decaying tails satisfy the delta jump only at $\kappa=m\alpha/\hbar^2$ for the attractive sign (the known bound state) and never for the repulsive sign.

Matching at work: quantization or amplitudes

With the grammar proven, the counting argument from the physical picture becomes a working method. Consider its two canonical outcomes side by side.

Bound states: one equation too many. The Finite Square Well is the archetype, its solve — done in full there — pure matching: interior $\cos(lx)$ or $\sin(lx)$ , exterior $e^{-\kappa|x|}$ , continuity of both at the edge. Dividing the two matching equations eliminates the amplitudes and leaves conditions on the energy alone,

z\tan z=\sqrt{z_0^2-z^2}\;\;(\text{even}),\qquad -z\cot z=\sqrt{z_0^2-z^2}\;\;(\text{odd}) \qquad (2)

where: $z=la$ packages the interior phase and $z_0$ the well strength, exactly as defined there. The structural point: the amplitudes dropped out. Two homogeneous conditions on one free ratio can only be satisfied if the energy conspires — matching converts smoothness into quantization. For $z_0=5.12$ (that article's worked well) there are exactly four roots, alternating even/odd, the lowest at $z_1=1.312$ ; the count obeys $N=1+\lfloor 2z_0/\pi\rfloor=4$ .

Scattering: exact balance. Send in a wave from the left instead. Both directions are now allowed in the outer regions, the incident amplitude is fixed at 1, and each interface's two equations exactly consume the remaining unknowns. Nothing is left to constrain $E$ — every positive energy scatters — and matching instead yields $r$ and $t$ , whose honest probability interpretation (current ratios, velocity factors, $R+T=1$ ) is the doctrine of Transmission and Reflection Coefficients. The Potential Step and Reflection is the one-interface instance; the Rectangular Barrier and Quantum Tunnelling the two-interface one.

The transfer matrix: matching, industrialized

Solving four matching equations for the barrier was tolerable. A quantum-cascade laser stacks dozens of interfaces; eliminating hundreds of coupled equations is not a method, it is a punishment. The fix: make matching local and composable. In any region of constant potential, $\psi=A\,e^{ikx}+B\,e^{-ikx}$ is fixed by the coefficient pair $(A,B)$ . Every junction is then a linear map from the pair on one side to the pair on the other, and every stretch of free flight just advances phases. Two primitive matrices generate everything:

M_{\text{int}}(k_1\!\to\!k_2)=\frac{1}{2}\begin{pmatrix}1+\dfrac{k_1}{k_2} & 1-\dfrac{k_1}{k_2}\\[6pt] 1-\dfrac{k_1}{k_2} & 1+\dfrac{k_1}{k_2}\end{pmatrix},\qquad M_{\text{prop}}(k,L)=\begin{pmatrix}e^{ikL} & 0\\ 0 & e^{-ikL}\end{pmatrix} \qquad (3)

where: $M_{\text{int}}$ is just the two continuity equations at one interface, solved once and for all; $M_{\text{prop}}$ carries the pair across a distance $L$ of constant potential. A structure is the ordered product of its parts — first element rightmost, because matrices act in sequence:

\begin{pmatrix}A_{\text{out}}\\ B_{\text{out}}\end{pmatrix}=M_N\cdots M_2\,M_1\begin{pmatrix}A_{\text{in}}\\ B_{\text{in}}\end{pmatrix},\qquad T=\frac{1}{|M_{22}|^{\,2}}\;\;(\text{same }k\text{ at both ends; for }k_{\text{in}}\!\neq\!k_{\text{out}}, T=\frac{k_{\text{in}}}{k_{\text{out}}}\frac{1}{|M_{22}|^{\,2}}) \qquad (4)

where: the $T$ readout follows from imposing scattering boundary conditions — no incoming wave from the right — on the composite. Two structural facts make the method trustworthy: $\det M_{\text{prop}}=1$ exactly, and $\det M_{\text{int}}=k_1/k_2$ — each interface's determinant carries precisely the velocity factor of Transmission and Reflection Coefficients, so factors telescope to $\det M_{\text{total}}=1$ whenever the ends sit at the same potential. Unit determinant is current conservation wearing matrix clothes; the deep reason — Wronskian constancy, proven in that article — guarantees it for any real potential.

Figure

The transfer-matrix assembly line — coefficient pairs carried across interfaces and free stretches by 2×2 multiplication

Go deeper · Derivation B · 2 minComposing a two-interface structure and checking it against the direct solve

Derivation BComposing a two-interface structure and checking it against the direct solve

We assemble the transfer matrix for the simplest composite — a region of different potential, width $L$ , between identical outer regions — and verify it three ways: the composite determinant must be exactly 1, the matrix transmission must equal a direct solve of all four matching equations, and for the barrier case it must reproduce the $\sin^2$ formula of Rectangular Barrier and Quantum Tunnelling. The strategy is pure assembly: write the three factors, multiply in the correct order, impose scattering boundary conditions. The care point is ordering — matrices act rightmost first — and remembering $\det M_{\text{int}}=k_1/k_2$ , not 1: only full round trips return to unit determinant.

Step 1. Assemble. Reading the structure left to right — enter the inner region, cross it, exit — the composite is

M=M_{\text{int}}(k_2\!\to\!k_1)\;M_{\text{prop}}(k_2,L)\;M_{\text{int}}(k_1\!\to\!k_2)

Why this order: the incident pair is acted on first by the entry interface, so that matrix stands rightmost; composition then proceeds leftward, one physical element per factor.

Step 2. Determinant bookkeeping. $\det M=(k_2/k_1)\cdot1\cdot(k_1/k_2)=1$ : the two interfaces' velocity factors cancel because the structure returns to its starting potential. Any chain that ends where it began — however many layers — telescopes the same way.

Step 3. Impose scattering conditions. With $(A_1,B_1)=(1,r)$ and $(A_3,B_3)=(t',0)$ , the second row of $(t',0)^{\mathsf T}=M(1,r)^{\mathsf T}$ gives $0=M_{21}+M_{22}\,r$ , so $r=-M_{21}/M_{22}$ , and the first row then yields $t'=M_{11}-M_{12}M_{21}/M_{22}=\det M/M_{22}=1/M_{22}$ . Hence $T=|t'|^2=1/|M_{22}|^2$ — equation (4)'s readout, derived rather than decreed.

Step 4. Numerical verification, three ways. Electron at $E=12$ eV crossing a $V_0=10$ eV, $L=0.5$ nm region (over-the-barrier, all wavenumbers real): matrix composition $T=0.817656362$ ; direct four-equation solve $T=0.817656362$ (agreement $10^{-16}$ ); closed form $\big[1+V_0^2\sin^2(k_2L)/4E(E-V_0)\big]^{-1}$ the same digits. $R+T=1$ to $10^{-12}$ . Why a numerical check belongs here: the matrix earns its keep on structures too big for hand algebra, so the honest test is agreement with an independent method on one small enough to do both. *As a consistency check: $\det M_{\text{int}}=k_1/k_2$ , $\det M_{\text{prop}}=1$ , and the composite determinant equals $1$ to $10^{-12}$ ; matrix, direct, and analytic transmission values agree to $10^{-9}$ or better at the quoted energies, including the staircase example below.*

Limiting cases

Case	Result	Meaning
$V$ bounded at $x_0$ (any finite step)	$\psi,\ \psi'$ both continuous	the workhorse junction: two equations per interface
$V\supset g\,\delta(x-x_0)$	$\psi$ continuous; $\Delta\psi'=\dfrac{2mg}{\hbar^2}\psi(x_0)$	one crease, sign follows $g$ : attractive pulls slopes together, repulsive pushes apart
$V\to\infty$ over a region (hard wall)	$\psi=0$ at the wall; no slope condition	expulsion replaces matching; the ISW rule as a limit of FSW tails $\kappa\to\infty$
bound problem (decay both ends)	conditions outnumber coefficients by one	leftover condition quantizes $E$ : discrete spectrum
scattering problem (travelling ends)	conditions and coefficients balance	every $E$ allowed; matching returns $r,t$ instead
structure with equal end potentials	$\det M_{\text{total}}=1$	current conservation in matrix form; interface factors $k_1/k_2$ telescope away

Worked examples

Example 1 — one junction, by the general recipe (confidence builder). An electron of energy $E=5.0$ eV crosses a single potential step of height $V_0=3.0$ eV (the Potential Step and Reflection setup — here re-solved as an instance of the general grammar, to see the machinery run). Find the amplitudes and check the bookkeeping.

Step 1. Wavenumbers. $k_1=\sqrt{2mE}/\hbar=1.146\times10^{10}\ \mathrm{m^{-1}}$ ; beyond the step $k_2=\sqrt{2m(E-V_0)}/\hbar=k_1\sqrt{2/5}=7.245\times10^{9}\ \mathrm{m^{-1}}$ .

Step 2. Two conditions at one interface, two unknowns — exact balance: scattering, no quantization. Continuity: $1+r=t$ ; slope: $ik_1(1-r)=ik_2t$ . Solve: $r=\dfrac{k_1-k_2}{k_1+k_2}=\mathbf{0.225}$ , $t=\dfrac{2k_1}{k_1+k_2}=\mathbf{1.225}$ .

Step 3. Read the red flag: $t>1$ . An amplitude above 1 is legal — the transmitted wave is taller — but it screams that $|t|^2$ is no probability. The honest currents (Transmission and Reflection Coefficients doctrine) give $T=(k_2/k_1)\,t^2=\mathbf{0.949}$ , $R=r^2=\mathbf{0.051}$ , summing to 1.000 to twelve digits in the verification script. Matching hands you amplitudes; currents turn them into probabilities.

Example 2 — a two-interface staircase by matrix assembly (exam level). An electron at $E=5.0$ eV crosses a "speed bump": the potential rises to $V_1=2.0$ eV for a stretch of $L=0.30$ nm, then returns to zero. Find $T$ by transfer matrix and check current conservation.

Step 1. Wavenumbers: $k_A=1.146\times10^{10}\ \mathrm{m^{-1}}$ outside, $k_B=k_A\sqrt{3/5}=8.874\times10^{9}\ \mathrm{m^{-1}}$ inside.

Step 2. Assemble $M=M_{\text{int}}(k_B\!\to\!k_A)\,M_{\text{prop}}(k_B,L)\,M_{\text{int}}(k_A\!\to\!k_B)$ and multiply numerically. Determinant check first: $\det M=1.000000000000$ — the $k_A/k_B$ and $k_B/k_A$ factors have telescoped, so no current leaks in the algebra.

Step 3. Read off the transmission: $T=1/|M_{22}|^2=\mathbf{0.986}$ . The direct four-equation solve agrees to $10^{-12}$ . So a 2 eV bump that a classical 5 eV electron would cross without slowing reflects 1.4% of the flux — abrupt potential changes scatter waves even when energy is plentiful, the same lesson the barrier taught above its top.

Step 4. Scale the lesson: a 40-layer quantum-cascade stack is just 40 more multiplications, where direct elimination would mean an 82-equation system re-solved at every trial energy. This is why band-structure and multilayer-optics codes speak transfer matrices.

The matcher of Leipzig

Origins · 1 minFelix Bloch and how this was discovered

Felix Bloch

1905–1983 · Zurich → Leipzig → Stanford

Bloch turned matching conditions into the theory of solids. His 1928 thesis under Heisenberg asked what the Schrödinger equation does in a periodic potential — a wavefunction matched consistently across every cell wall of an infinite crystal — and found plane waves modulated by lattice-periodic functions, the matching consistency condition carving the energy axis into allowed bands and forbidden gaps. Kronig and Penney made the mechanism transparent three years later with exactly this chapter's tools: a periodic array of deltas, one jump condition per site. Bloch went on to co-invent NMR (Nobel Prize, 1952) and serve as CERN's first director-general — but every band diagram in every semiconductor datasheet still runs on his boundary conditions.

History

1926

— Schrödinger's equation makes boundary behaviour a derivable consequence rather than a postulate

1928

— Bloch's thesis: consistent matching across a periodic lattice yields band structure

1930

— Dirac's Principles systematizes the delta function and its integral calculus

1931

— Kronig and Penney solve the delta-lattice band model — one jump condition per site

1950

— Abelès develops the 2×2 transfer-matrix method for optical multilayers, the same algebra at optical wavelengths

1970

— Esaki and Tsu propose the semiconductor superlattice: engineered matching problems, grown atom by atom

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.5–2.6 (delta function and finite square well) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement H_I · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 2 · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6 Key results collected here for reference: the delta-potential slope jump $\Delta\psi'=-(2m\alpha/\hbar^2)\psi(0)$ and bound-state energy $E=-m\alpha^2/2\hbar^2$ ; the finite-well transcendental conditions $k\tan(ka)=\kappa$ and $-k\cot(ka)=\kappa$ ; the infinite-well energies $E_n=n^2\pi^2\hbar^2/2mL^2$ ; and probability-current continuity requiring $|r|^2+|t|^2=1$ . Numerical checks confirm: the shrinking-window integral, both delta-jump signs, the $z_0=5.12$ root count and lowest root $1.312$ , all transfer-matrix determinants, and the matrix-vs-direct-vs-analytic transmission agreements quoted in the derivation and worked examples.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Free Particle and Plane Waves · Bound States versus Scattering States · Delta-Function Potential Well and Bound State · Finite Square Well

Leads to: Transmission and Reflection Coefficients · Resonant Transmission and Tunnelling Resonances · Infinite Square Well · Wave Packets and Their Construction

Why exactly can't ψ itself ever jump, even at an infinite wall?

Two ways. Mathematically: $\psi'$ stays bounded near every junction on this menu, and a function with bounded derivative cannot jump — a break in $\psi$ would need a delta derivative in the potential. Physically: the probability current $j\propto\mathrm{Im}(\psi^*\psi')$ of Probability Current in One Dimension would be undefined at the break, creating or destroying probability at the joint. Even at the infinite wall $\psi$ doesn't jump — it reaches zero continuously and stays zero beyond.

Is ψ″ also continuous?

No — and that is fine. From $\psi''=(2m/\hbar^2)(V-E)\psi$ , curvature inherits every discontinuity of $V$ : at a finite step $\psi''$ jumps by $(2m/\hbar^2)\Delta V\,\psi$ while $\psi$ and $\psi'$ stay smooth. Any finite-well eigenfunction shows it — bending-toward-axis flips to bending-away exactly at the well edge, a break in curvature but not in slope. Smoothness stops at first order because the equation is second-order.

Where does the infinite wall's ψ = 0 rule actually come from? It looks like a decree.

It is a limit the Finite Square Well already contains. Outside a finite well the wave leaks as $e^{-\kappa|x|}$ ; crank $V_0\to\infty$ and the leak depth shrinks to zero, so continuity forces $\psi(\text{wall})\to0$ , while the slope condition dies with the dead exterior — no constraint on the interior slope survives. The check is quantitative: FSW levels converge to ISW levels as $z_0$ grows, exactly as that article's deep-well limit showed.

What if the potential has a finite step AND a delta spike at the same point?

Superpose the grammars — the integral treats each term of $V$ separately. The step contributes nothing to the jump integral (bounded); the delta pays its full toll: $\psi$ continuous, $\Delta\psi'=(2mg/\hbar^2)\psi(x_0)$ , the step's only role being different wavenumbers on the two sides. The hybrid is real — a heterojunction with an interface charge sheet is exactly a step-plus-delta.

Do these conditions survive in real semiconductor devices?

With one honest amendment. In a heterostructure the effective mass $m^*$ differs between materials, and blind $\psi'$ continuity would let probability current pile up at the interface. The repaired condition — continuity of $\psi$ and $\psi'/m^*$ (BenDaniel–Duke) — makes the current $j\propto\mathrm{Im}(\psi^*\psi'/m^*)$ cross smoothly. The lesson: the fundamental matched quantity is the current; slope continuity is its constant-mass special case.

Why bother with transfer matrices when I can just solve the linear system directly?

For one barrier, don't bother. The matrix wins on scaling and reuse: $N$ layers means $2N$ globally coupled equations, but only ever $2\times2$ matrices locally — and the same factors are reused at every trial energy when scanning $T(E)$ , which device design does thousands of times per structure. It also wins conceptually: $\det M=1$ makes current conservation automatic, and resonances announce themselves as near-zeros of $|M_{22}|$ — the language of Resonant Transmission and Tunnelling Resonances.

My potential is smooth — there are no interfaces anywhere. Is this article useless to me?

The theorem still speaks: for smooth $V$ there is nothing to match — which licenses the shooting method: integrate numerically from both ends and tune $E$ until value and slope agree at a meeting point. That agreement is a matching condition at an interface you invented for convenience. The semiclassical cousin — joining oscillatory to exponential behaviour across a smooth turning point — is the connection-formula problem of The WKB Approximation, previewed by Bohr–Sommerfeld and Semiclassical Quantization in 1D.