University Physics V · Finite Potential Wells and Quantum Tunneling · 6.4

The Scattering Matrix & R + T = 1

Every number a scattering experiment can report is a ratio of probability currents, so this is where you learn to build one. Construct j from the wavefunction, notice it cannot change along the line, and watch R + T = 1 and S†S = 1 fall out as the same sentence written twice.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Everything a one-dimensional scattering experiment can report is a ratio of probability currents, and that current is fixed by the Schrödinger equation before any potential is named. Differentiate |ψ|² in time, substitute iℏ∂ψ/∂t = Hψ, and the kinetic term reorganises itself into a divergence: ∂ρ/∂t + ∂j/∂x = 0 with j = (ℏ/m) Im(ψ*ψ′). In a stationary state the density is frozen, so dj/dx = 0 and j is one number for the entire line, the same number in front of the barrier, inside it, and behind it.

Evaluate that one number in the two asymptotic regions, where ψ is only a pair of plane waves, and the interference terms cancel exactly, leaving j = (ℏk/m)(incoming squared minus outgoing squared) on each side. Equating the two sides is R + T = 1. Package the four asymptotic amplitudes as a matrix, outgoing = S · incoming in flux-normalised units, and the same statement reads S†S = 1.

That is the whole content of the topic: unitarity is a conservation law you derived, not an axiom you assumed, and it is bought with two assumptions you can violate. Give V an imaginary part and S†S falls short of 1 by exactly the absorbed fraction; put the scatterer somewhere the potential never flattens out, and the asymptotic amplitudes S is built from do not exist at all.

Simple definition: The scattering matrix S is the map from the flux-normalised amplitudes of the waves running into a scatterer to those running out of it, and its unitarity is nothing more than the statement that probability current is conserved.
Example: For a 1.00 eV electron crossing a step down to V = −3.00 eV, S = [[−1/3, 2√2/3], [2√2/3, 1/3]]. Each column has unit norm, which is exactly R = 0.111 and T = 0.889 adding to 1.

Probability current and continuityj = (ℏ/m) Im(ψ*ψ′)∂|ψ|²/∂t + ∂j/∂x = 0

A stationary state has a frozen density, so dj/dx = 0: j is one number for the whole line, not a field.

j is probability per second (s⁻¹ in 1D). For A exp(ikx), j = (ℏk/m)|A|² = v|A|². Continuity needs V real.

Asymptotic amplitudes and the S matrix(B, C)ᵀ = S (A, D)ᵀS = [[r, t′], [t, r′]]

In or out is decided by the sign of the group velocity, not by which side of the scatterer you stand on.

ψ → A exp(ik_L x) + B exp(−ik_L x) at x → −∞, and C exp(ik_R x) + D exp(−ik_R x) at x → +∞. A and D come in.

R and T are ratios of currentsR = |jᵣ| / |jᵢ| = |r|²T = |jₜ| / |jᵢ| = (k_R/k_L)|t|²

The velocity factor is why |t| can exceed 1 at a step down while T never does.

k_L = √(2m(E−V_L))/ℏ and k_R = √(2m(E−V_R))/ℏ, both in m⁻¹; the extra factor is the speed ratio v_R/v_L.

Unitarity, and what it forcesS†S = 1 ⟺ R + T = 1|r| = |r′| and |t| = |t′|

Transmission is identical from either side of any barrier, however lopsided its profile. That is reciprocity, not symmetry.

S acts on flux-normalised amplitudes √(k_L)A and √(k_R)D. For real V, time reversal adds Sᵀ = S, so t = t′ exactly.

Transfer matrix to scattering matrixS = (1/M₂₂) [[−M₂₁, 1], [det M, M₁₂]]

M composes across slabs but is only pseudo-unitary; S is unitary but does not compose. Multiply in M, read physics in S.

M is defined by (C, D)ᵀ = M (A, B)ᵀ, with |det M| = k_L/k_R. Every entry is dimensionless.

Absorption, and the size of the breachV = V_R − iW ⟹ dj/dx = −(2W/ℏ)|ψ|²A = 1 − R − T

The optical model ledger: a nonzero 1 − R − T is inelastic physics, not an arithmetic slip.

W is positive, in J or eV. A is the flux lost to channels the one-dimensional elastic model does not carry.

Differentiate the density and a current appears

Start from ρ = |ψ|² and iℏ ∂ψ/∂t = −(ℏ²/2m)ψ″ + Vψ. Then ∂ρ/∂t = ψ*∂ψ/∂t + ψ∂ψ*/∂t, and substituting the equation splits it in two. The kinetic piece reorganises into (iℏ/2m)(ψ*ψ″ − ψψ*″), which is exactly −∂/∂x of j = (ℏ/2mi)(ψ*ψ′ − ψψ*′) = (ℏ/m) Im(ψ*ψ′). The potential piece is (2/ℏ) Im(V)|ψ|², and it vanishes only because V is real: that is the one and only place hermiticity of H is used. So ∂ρ/∂t + ∂j/∂x = 0. Now impose stationarity. An eigenvector of H evolves as ψ(x) exp(−iEt/ℏ), whose density does not move, so ∂ρ/∂t = 0 and dj/dx = 0. In one dimension that is far stronger than it looks: it is not a statement about a flow field, it says j is a single constant for the entire real line. Compute it at x = −∞ and you have computed it at x = +∞, through whatever sits in between.

Four amplitudes, and which two are incoming

S exists only where the potential flattens out. Require V → V_L as x → −∞ and V → V_R as x → +∞, and set k_L = √(2m(E−V_L))/ℏ, k_R = √(2m(E−V_R))/ℏ. For E above both floors the asymptotic solution is ψ → A exp(ik_L x) + B exp(−ik_L x) on the left and C exp(ik_R x) + D exp(−ik_R x) on the right. Four complex numbers, but the equation is second order, so only two are free. The split into incoming and outgoing is made by the sign of the group velocity, not by position: A runs right and lives on the left, D runs left and lives on the right, so A and D are what you prepare and B and C are what comes back. Define S by (B, C)ᵀ = S(A, D)ᵀ, so r = B/A and t = C/A when D = 0, while r′ = C/D and t′ = B/D when A = 0. Nothing in that definition mentions the shape of V in between. S is a boundary-value object, and two very different potentials sharing one S are indistinguishable to any scattering experiment.

The interference terms carry no current

Evaluate j on the left. With ψ = A exp(ikx) + B exp(−ikx), the product ψ*ψ′ = ik|A|² − ik|B|² + ik(A B* exp(2ikx) − A* B exp(−2ikx)). The bracket has the form z − z*, which is purely imaginary, so ik times it is purely real and drops out of Im(ψ*ψ′) exactly. The standing-wave ripple in |ψ|², the most conspicuous thing in any plot of the left region, therefore carries no current at all. What survives is j_L = (ℏk_L/m)(|A|² − |B|²), and identically j_R = (ℏk_R/m)(|C|² − |D|²). Set D = 0 for incidence from the left and impose dj/dx = 0: k_L|A|² = k_L|B|² + k_R|C|². Divide through by k_L|A|² and you have 1 = |r|² + (k_R/k_L)|t|², which is R + T = 1 with R and T read off as flux ratios. The factor k_R/k_L is not cosmetic: at the step in the worked examples it turns |t|² = 1.78 into T = 0.889.

S†S = 1 is that same sentence in matrix form

Rescale so the conserved quantity is an ordinary norm: set a = √(k_L)A, b = √(k_L)B, c = √(k_R)C and d = √(k_R)D. Then j_L = j_R reads |a|² + |d|² = |b|² + |c|², so the map S taking the pair (a, d) to the pair (b, c) preserves length, and a length-preserving linear map on ℂ² is unitary. Hence S†S = 1. Its diagonal entries are R + T = 1 and R′ + T′ = 1; its off-diagonal entry is r*t′ + t*r′ = 0. That last relation is not decorative. Take moduli, then use the two diagonal relations to eliminate |t| and |t′|, and you get |r|² = |r′|², hence |t| = |t′|: the transmission probability through a barrier is the same from either side, however lopsided the barrier. Numerically this is also the best diagnostic you have. Assemble S, form S†S − 1, and report its largest entry; on a healthy calculation it sits near 10⁻¹⁵.

The transfer matrix is not the scattering matrix

M is defined by (C, D)ᵀ = M(A, B)ᵀ, and its virtue is that it composes: a staircase potential is the ordered product of one matrix per interface and per slab. Its vice is that it is not unitary. Flux conservation reads M†σ₃M = (k_L/k_R)σ₃ instead, which fixes |det M| = k_L/k_R and leaves M with one growing and one decaying direction, so its condition number climbs like exp(2κL). The two matrices are related by rearrangement, not by similarity. Solving (C, D)ᵀ = M(A, B)ᵀ for the outgoing pair gives r = −M₂₁/M₂₂, t′ = 1/M₂₂, t = det M/M₂₂ and r′ = M₁₂/M₂₂, that is S = (1/M₂₂)[[−M₂₁, 1],[det M, M₁₂]]. So the working rule is: multiply in M, then convert to S to read physics and to test unitarity. When a thick-barrier product has quietly lost its decaying component to round-off, the symptom that shows up is R + T ≠ 1.

Where unitarity is only approximate

Keep the derivation but let V = V_R − iW with W positive. The potential term no longer cancels: ∂ρ/∂t + ∂j/∂x = −(2W/ℏ)|ψ|², so in a stationary state dj/dx = −(2W/ℏ)|ψ|², a strictly negative number, and the current steps down as it crosses the absorbing region. Then S†S ≤ 1 as a matrix inequality, the eigenvalues of S†S are the survival probabilities of the two channels, and A = 1 − R − T is the flux removed. This is the optical model of nuclear physics: a nucleus is given a complex potential precisely because elastic scattering is not the only thing a neutron can do, and A folds capture, inelastic excitation and fission into the one number the elastic model refuses to resolve. The figure below makes the trade concrete. A delta of strength α − iW gives A = 2w/((1+w)² + β²) with w = mW/(ℏ²k), which peaks at exactly 0.5: a purely absorbing point scatterer can never eat more than half the beam, because to absorb it must first fail to reflect.

Change one variable at a time

Make the relationship visible.

Interactive model

Real strength β = mα/(ℏ²k)1.0

Absorption w = mW/(ℏ²k)0.0

Hold the absorption at 0 and sweep β: no step ever appears at the origin and R + T = 1 exactly, however strong the barrier. Now raise w and watch the current drop as it crosses x = 0 — then check that the absorbed fraction never passes 0.50, wherever you put the sliders.

REFLECTED R0.500

TRANSMITTED T0.500

R + T1.000

ABSORBED 1 − R − T0.000

Live interpretationREFLECTED R: 0.500. TRANSMITTED T: 0.500. R + T: 1.000. ABSORBED 1 − R − T: 0.000

Catch the common trap

Explain before calculating.

An electron of energy E = 1.00 eV crosses a step down at x = 0 into a region where V = −8.00 eV. Matching ψ and ψ′ at the step gives r = −1/2 and t = 1/2. Which pair of reflection and transmission probabilities is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron of energy E = 1.00 eV meets a potential step that rises to V₀ = 0.75 eV at x = 0. Find R and T, and say exactly what is wrong with quoting T = |t|².

Wavenumbers, from k = √(2mE′)/ℏ with E′ the energy above the local floor. Left: E′ = 1.00 eV gives k_L = 5.12 nm⁻¹. Right: E′ = 1.00 − 0.75 = 0.25 eV gives k_R = 2.56 nm⁻¹. So k_R/k_L = √(0.25/1.00) = 0.500.
Continuity of ψ and ψ′ at x = 0 gives r = (k_L − k_R)/(k_L + k_R) = 0.500/1.500 = 1/3 and t = 2k_L/(k_L + k_R) = 2/1.5 = 4/3.
Reflection compares two currents at the same k, so it needs no correction: R = |r|² = 1/9 = 0.111. Transmission compares currents at different k. Note first that |t|² = 16/9 = 1.78, and no probability exceeds 1 — the amplitude grew because the wave slowed down and piled up.
Carry the velocity factor: T = (k_R/k_L)|t|² = 0.500 × 16/9 = 8/9 = 0.889. Ledger: R + T = 0.111 + 0.889 = 1.000, which is dj/dx = 0 and nothing else.

AnswerR = 0.111 and T = 0.889. |t|² = 1.78 is an amplitude ratio, not a probability; T = (k_R/k_L)|t|².

MediumA 1.00 eV electron scatters from a delta barrier V(x) = α δ(x) with α = 0.30 eV⋅nm. Find R and T, confirm the ledger closes for every α, and check the phase relation that unitarity forces on a symmetric scatterer.

Dimensionless strength β = mα/(ℏ²k). With ℏ²/2m = 0.0381 eV⋅nm², m/ℏ² = 13.12 eV⁻¹nm⁻², and k = 5.12 nm⁻¹ at 1.00 eV, so β = 13.12 × 0.30 / 5.12 = 0.768.
Continuity of ψ plus the jump ψ′(0⁺) − ψ′(0⁻) = (2mα/ℏ²)ψ(0) gives t = 1/(1 + iβ) and r = t − 1 = −iβ/(1 + iβ).
Equal floors on both sides, so k_R = k_L and there is no velocity factor: T = |t|² = 1/(1 + β²) = 1/1.591 = 0.629 and R = |r|² = β²/(1 + β²) = 0.591/1.591 = 0.371.
Ledger: R + T = (1 + β²)/(1 + β²) = 1, identically in β. Unitarity is not a numerical coincidence at this α; it holds at every α.
The potential is even, so parity gives r = r′ and t = t′, making S = [[r, t],[t, r]]. Unitarity needs the columns orthogonal: r*t + t*r = 2 Re(r*t) = 0, so r and t must be 90° apart. Check: arg t = −arctan β = −37.5°, arg r = −90° − arctan β = −127.5°, a difference of exactly −90°.

Answerβ = 0.768, R = 0.371, T = 0.629, and R + T = 1 for every α. arg r − arg t = −90°, as unitarity plus even parity demands.

HardA 1.00 eV electron crosses a step down at x = 0 into a region where V = −3.00 eV. Write the raw amplitude matrix [[r, t′],[t, r′]], show it is not unitary, flux-normalise it, and verify that S is unitary and symmetric. Then compute T′ for incidence from the right.

Wavenumbers: E′ = 1.00 eV on the left gives k_L = 5.12 nm⁻¹; E′ = 4.00 eV on the right gives k_R = 10.25 nm⁻¹, so k_R/k_L = 2 exactly.
From the left: r = (k_L − k_R)/(k_L + k_R) = −1/3 and t = 2k_L/(k_L + k_R) = 2/3. From the right: r′ = (k_R − k_L)/(k_L + k_R) = +1/3 and t′ = 2k_R/(k_L + k_R) = 4/3.
Raw amplitude matrix [[−1/3, 4/3],[2/3, 1/3]]. Its column norms squared are 1/9 + 4/9 = 5/9 and 16/9 + 1/9 = 17/9. Neither is 1, because amplitudes are not the conserved quantity.
Flux-normalise, Sᵢⱼ = √(kᵢ/kⱼ) times the amplitude for j → i. So S₂₁ = √2 × 2/3 = 2√2/3 = 0.9428 and S₁₂ = (1/√2) × 4/3 = 2√2/3 = 0.9428, while the diagonal entries are unchanged: S = [[−1/3, 0.9428],[0.9428, 1/3]].
Test it. Each column gives 1/9 + 8/9 = 1. The cross term gives (−1/3)(0.9428) + (0.9428)(1/3) = 0. So S†S = 1, det S = −1/9 − 8/9 = −1, and S is symmetric, which is reciprocity and follows because V is real.
Probabilities: R = |r|² = 1/9 = 0.111 and T = (k_R/k_L)|t|² = 2 × 4/9 = 8/9 = 0.889. From the right, R′ = |r′|² = 1/9 and T′ = (k_L/k_R)|t′|² = (1/2) × 16/9 = 8/9. So T′ = T even though the step is completely asymmetric.

AnswerS = [[−1/3, 2√2/3],[2√2/3, 1/3]], unitary and symmetric with det S = −1. R = 0.111 and T = 0.889 = T′. The raw amplitude matrix is neither unitary nor a probability statement.