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University Physics V · Finite Potential Wells and Quantum Tunneling · 6.6

Poles of the Transmission Amplitude

Scattering happens on the real axis, but the potential's whole discrete content lives off it. Continue t(k) into the complex plane and read the bound spectrum off the positive imaginary axis, the resonances off the fourth quadrant, and every lifetime off how far below the axis a pole sits.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

The scattering calculation you run on the real axis is one line through a much larger object. Build t(k) for a real short-range V by matching e(ikx) at both ends: k enters only through exponentials and through linear matching conditions, all entire in k, so t continues to complex k as a meromorphic function with no new physical input. Where t has a pole, the incoming amplitude it was divided by has vanished, and what survives is a solution that is purely outgoing at both ends — the potential ringing with nothing driving it.

On the positive imaginary axis, k = iκ, outgoing means decaying in both directions, so that solution is square-integrable and the pole is a bound state at E = −ħ²κ²/2m. Nowhere else is it a state. At k = kᵣ − iγ in the fourth quadrant the same solution grows like e(γx), its energy is complex, E = (ħ²/2m)(kᵣ² − γ² − 2i kᵣγ) ≡ ER − iΓ/2 with Γ = 2ħ²kᵣγ/m, and its time factor decays as e(−Γt/2ħ).

That is a resonance, and the price is the Hilbert space: ER − iΓ/2 is not an eigenvalue of the self-adjoint H, whose spectrum stays real, but a pole of a continued amplitude on the second Riemann sheet of the energy — visible on the real axis only as a Breit–Wigner peak of width Γ and a state that lives for ħ/Γ.

Simple definition
A pole of the transmission amplitude is a complex wavenumber at which the Schrödinger equation admits a solution carrying only outgoing waves: on the positive imaginary axis that solution is a bound state, and in the lower half plane it is a resonance.
Example
An electron pole at k = (0.80 − 0.02i) Å⁻¹, with ħ²/2m = 3.810 eV⋅Ų, gives ER = 3.810(0.80² − 0.02²) = 2.44 eV and Γ = 3.810(4 × 0.80 × 0.02) = 0.244 eV, so the state survives τ = ħ/Γ = 2.70 fs.
Outgoing-only (Siegert) boundary conditionψ ∼ e(ikx), x → +∞ψ ∼ e(−ikx), x → −∞

It has a solution only at isolated complex k. Those k are exactly the poles of t, and they carry the whole discrete content of V.

k is complex here, and nothing arrives from either side, so this is the potential's source-free response

Pole condition: a vanishing Jost Wronskiant(k) = 2ik / W(k)W = f₋ f₊′ − f₋′ f₊

Turns pole-hunting into root-finding: solve W(k) = 0 in the complex plane by Newton or Muller iteration.

f₊ → e(ikx) as x → +∞ and f₋ → e(−ikx) as x → −∞; for V = 0, W = 2ik and t = 1 as it must

Bound-state pole, positive imaginary axisk = iκ, κ > 0E = −ħ²κ²/2mψ ∼ e(−κ|x|)

The only region of the k plane where a pole is a genuine eigenvector of the Hermitian H; a short-range V has finitely many.

κ in m⁻¹, E in J or eV. Outgoing at both ends here means decaying at both ends, so the state is normalisable

Resonance pole and its complex energyk = kᵣ − iγE = (ħ²/2m)(kᵣ² − γ² − 2i kᵣγ) ≡ ER − iΓ/2

Peak position and full width come straight off the pole's two coordinates: a narrow pole, γ ≪ kᵣ, has Γ/ER = 4γ/kᵣ.

kᵣ, γ in m⁻¹ with γ > 0; ER = ħ²(kᵣ² − γ²)/2m and Γ = 2ħ²kᵣγ/m, both energies

Exponential decay and the lifetime|ψ(t)|² = |ψ(0)|² e(−Γt/ħ)τ = ħ/Γ

The pole's distance below the real axis is a clock. Move it ten times closer and the state lives ten times longer.

τ = 6.582×10⁻¹⁶ eV⋅s ÷ Γ[eV]. Γ = 0.244 eV gives τ = 2.70 fs; Γ = 0.400 meV gives τ = 1.65 ps

Breit–Wigner line shape near one poleT(E) ≈ (Γ/2)² / [(E − ER)² + (Γ/2)²]

The inverse route, and the one a laboratory takes: a measured peak gives ER from its centre and Γ from its full width at half maximum.

One isolated pole, γ ≪ kᵣ, symmetric barriers. T = 1 at E = ER and T = ½ at E = ER ± Γ/2

01

Why t(k) continues off the real axis at all

Solve the scattering problem the usual way and k appears only inside e(±ikx) and inside linear matching conditions, all of which are entire functions of k. For a V of finite support every coefficient produced by the matching is therefore entire, and t(k) is a ratio of entire functions: one meromorphic function on the whole complex k plane, fixed by the same potential, with no extra assumption smuggled in. A potential that merely decays, like e(−μ|x|), buys the continuation only above the line Im k = −μ/2; a Coulomb tail breaks it altogether and needs a different frame. The real positive axis, where k = √(2mE)/ħ, is one line drawn across that object. Every bound level, every resonance and every virtual state the potential supports is encoded in where t stops being finite away from that line. The move of this topic is to stop treating negative and positive energies as two calculations and treat them as two regions of one function.

02

A pole is a solution with no incoming wave

Write ψ = e(ikx) + r e(−ikx) on the left and t e(ikx) on the right. Every amplitude has already been divided by the incoming one, so t = ∞ says a finite outgoing wave survives a vanishing input. Formalise it with the Jost solutions f₊(k, x) → e(ikx) as x → +∞ and f₋(k, x) → e(−ikx) as x → −∞, whose Wronskian gives t(k) = 2ik/W(k). A pole is a zero of W, and a vanishing Wronskian means f₊ and f₋ are the same solution up to a constant: one function, purely outgoing at both ends. That is the Siegert boundary condition, and it is the only place in this unit where a boundary condition is imposed at complex k. Read it as a resonator condition — the potential ringing on its own — and bound states and resonances stop being two problems. They are the same question asked in two regions of one plane.

03

The positive imaginary axis, and why nothing else is a state

Put k = iκ. Then e(ikx) = e(−κx) dies as x → +∞ and e(−ikx) = e(+κx) dies as x → −∞, so for κ > 0 the outgoing-only solution is square-integrable: a bound state at E = ħ²k²/2m = −ħ²κ²/2m, below the asymptotic floor. A short-range V supports finitely many, and Levinson's theorem counts them from the phase shift, δ(0) − δ(∞) = nb π. Everywhere else the same solution leaves L². On the negative imaginary axis, k = −iκ, it grows at both ends: a virtual or antibound state, no eigenvector at all, yet a strong influence on low-energy scattering — the neutron-proton singlet pole sits there and produces the anomalous scattering length aₛ = −23.7 fm. Reality of V forces t(−k*) = t(k)*, so a pole either sits on the imaginary axis or comes with its mirror at −k*; and for a fast-decaying V no pole is permitted in the open upper half plane off that axis, since it would be a state growing in time.

04

The fourth quadrant: reading ER and Γ off a pole

A pole at k = kᵣ − iγ with kᵣ, γ > 0 squares to E = (ħ²/2m)(kᵣ² − γ² − 2i kᵣγ). Call that ER − iΓ/2 and you have ER = ħ²(kᵣ² − γ²)/2m and Γ = 2ħ²kᵣγ/m: peak position and full width, from two coordinates. Because E = ħ²k²/2m is two-to-one, the k plane double-covers the E plane — the upper half is the physical sheet, and the lower half, where every resonance lives, is the second sheet, reached by circling the branch point at E = 0. That is the whole reason a Hermitian H can carry a complex number here without contradiction. Take an electron pole at k = (0.80 − 0.02i) Å⁻¹ with ħ²/2m = 3.810 eV⋅Ų: ER = 3.810 × 0.6396 = 2.44 eV, Γ = 3.810 × 0.0640 = 0.244 eV, and the γ² term moves ER by only 1.5 meV. Near an isolated pole T(E) collapses onto the Breit–Wigner form, which is why a fitted peak is a pole measurement.

05

The Gamow state decays in time by growing in space

Attach the time factor. e(−iEt/ħ) with E = ER − iΓ/2 gives |ψ(t)|² ∝ e(−Γt/ħ), so τ = ħ/Γ: Γ = 0.244 eV is 2.70 fs, Γ = 0.400 meV is 1.65 ps, and the 117 MeV width of the Δ(1232) is 5.6×10⁻²⁴ s. The same complex k makes the spatial part diverge, since |e(ikᵣx + γx)| = e(γx) blows up as x → +∞. So the Siegert or Gamow state is not a vector in L², and ER − iΓ/2 is not an eigenvalue of the self-adjoint H. The growth is bookkeeping, not paradox: the amplitude now sitting at distance x left the well a time x/v ago, when the state was more populated by exactly that factor. A physical packet truncates the tail at the causal radius x ≈ vt. Exponential decay is itself only asymptotic — quadratic at very short times, and power-law at very long ones, where the branch cut at E = 0 outlives the pole.

06

Three numerical routes to a pole

Route one: build t(k) from a transfer or scattering matrix and Newton-iterate on 1/t(k) = 0 in complex k, seeded from a Breit–Wigner fit of the real-axis peak. Cheap, but the transfer-matrix product conditions like e(2κL), so a thick barrier destroys the root before the iteration converges; use the stable scattering recursion instead. Route two: complex scaling. Rotate x → x e(iθ), and the continuum swings down by 2θ about E = 0 while any resonance inside that wedge becomes a genuine square-integrable eigenvector of the rotated non-Hermitian Hamiltonian, so scipy.linalg.eig on a rotated finite-difference matrix returns ER − iΓ/2 directly. Route three: fit T(E) on a real grid. Fast and honest for one isolated narrow pole, and wrong the moment two resonances overlap or a background phase tilts the line into a Fano profile — then only the continuation gives an unambiguous ER and Γ.

02

Change one variable at a time

Make the relationship visible.

Interactive model
1.0 1/a
2.0 1/a
0.15 1/a

Hold kᵣ at 2.0 and drag γ from 0.60 down to 0.05: the pole climbs toward the real axis while the peak narrows from Γ = 4.80 to Γ = 0.40, twelve times sharper. Then move κ and watch the right-hand curve not move at all — a bound state sits below threshold and makes no peak.

Interactive physics modelLeft, the complex k plane at equal scales: the dot on the positive imaginary axis at k = i1.00/a is a bound state at E = −1.00, and the dot in the fourth quadrant at k = (2.00 − 0.15i)/a is a resonance whose dashed drop measures γ. Right, the Breit–Wigner transmission that pole makes on the real energy axis, peaking at E_R = 3.98, the arrow spanning Γ = 1.20.Im kRe kbound k = i1.00resonance|t|² = 1Γ = 1.20real E

BOUND STATE E = −κ²-1.00 ħ²/2ma²

RESONANCE ER3.98 ħ²/2ma²

FULL WIDTH Γ = 4kᵣγ1.20 ħ²/2ma²

QUALITY ER/Γ3.31

Live interpretationBOUND STATE E = −κ²: −1.00 ħ²/2ma². RESONANCE ER: 3.98 ħ²/2ma². FULL WIDTH Γ = 4kᵣγ: 1.20 ħ²/2ma². QUALITY ER/Γ: 3.31

03

Catch the common trap

Explain before calculating.

A real, short-range one-dimensional potential gives a transmission amplitude with a pole at k = (1.20 − 0.03i) Å⁻¹, for an electron with ħ²/2m = 3.810 eV⋅Ų. Which statement about this pole is correct?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron scattering off a short-range potential has a transmission amplitude whose pole nearest the real axis sits at k = (0.80 − 0.02i) Å⁻¹. Take ħ²/2mₑ = 3.810 eV⋅Ų and ħ = 6.582×10⁻¹⁶ eV⋅s. Find the resonance energy, the full width, the lifetime, and the quality factor ER/Γ.
  1. Square the pole: k² = (kᵣ − iγ)² = kᵣ² − γ² − 2i kᵣγ = 0.6400 − 0.0004 − 2i(0.80)(0.02) = 0.6396 − 0.0320i Å⁻².
  2. Multiply by ħ²/2m to get the complex energy: E = 3.810 × (0.6396 − 0.0320i) = 2.437 − 0.1219i eV.
  3. Match to E = ER − iΓ/2: ER = 2.44 eV and Γ/2 = 0.1219 eV, so Γ = 0.244 eV. The γ² term shifts ER down by only 3.810 × 0.0004 = 1.5 meV, 0.06% — which is what γ ≪ kᵣ buys you.
  4. Lifetime: τ = ħ/Γ = 6.582×10⁻¹⁶ eV⋅s ÷ 0.244 eV = 2.70×10⁻¹⁵ s = 2.70 fs.
  5. Quality factor: Q = ER/Γ = 2.437/0.244 = 9.99, and equivalently kᵣ/4γ = 0.80/0.08 = 10.

AnswerER = 2.44 eV, Γ = 0.244 eV, τ = 2.70 fs, Q ≈ 10. The peak in T(E) is only ten times narrower than its own centre energy, so this is a broad resonance, not a near-bound state.

MediumFor V(x) = λδ(x) the matching conditions give t(k) = k/(k + imλ/ħ²). Locate its pole, then classify it for λ = −4.00 eV⋅Å and for λ = +4.00 eV⋅Å acting on an electron, with ħ²/2mₑ = 3.810 eV⋅Ų.
  1. t blows up where its denominator vanishes, at kₚ = −imλ/ħ². From ħ²/2m = 3.810 eV⋅Ų, m/ħ² = 1/(2 × 3.810) = 0.13123 (eV⋅Ų)⁻¹, so kₚ = −0.13123 λ i, in Å⁻¹ when λ is in eV⋅Å.
  2. Attractive well, λ = −4.00 eV⋅Å: kₚ = +0.13123 × 4.00 i = +0.5249i Å⁻¹. Im k > 0, so this is a bound state with κ = 0.5249 Å⁻¹.
  3. Its energy: E = −ħ²κ²/2m = −3.810 × 0.5249² = −3.810 × 0.27552 = −1.050 eV, and the state decays with length 1/κ = 1.905 Å. Cross-check against the closed form E = −mλ²/2ħ² = −0.13123 × 16.0/2 = −1.050 eV.
  4. Repulsive barrier, λ = +4.00 eV⋅Å: kₚ = −0.5249i Å⁻¹, on the negative imaginary axis. The outgoing solution now grows as e(+0.5249|x|), so it is not normalisable — a virtual (antibound) state, and the barrier has no bound state at all.
  5. Both poles sit on the imaginary axis, as t(−k*) = t(k)* demands for a real V: a pole anywhere off that axis must be accompanied by its mirror at −k*.

Answerkₚ = −imλ/ħ². λ = −4.00 eV⋅Å puts a bound state at k = +0.525i Å⁻¹, E = −1.05 eV; λ = +4.00 eV⋅Å puts a virtual state at k = −0.525i Å⁻¹ and supports no bound state.

HardA GaAs double-barrier diode with a 5.0 nm well (m* = 0.067 mₑ, so ħ²/2m* = 56.87 eV⋅Ų) transmits with T = 1 at E = 80.0 meV and T = 0.500 at 79.8 meV and 80.2 meV. Locate the pole in the complex k plane, give the lifetime, and estimate how many round trips the electron makes inside the well before escaping.
  1. Breit–Wigner puts T = ½ at E = ER ± Γ/2, so Γ = 80.2 − 79.8 = 0.400 meV and ER = 80.0 meV, giving Q = ER/Γ = 200.
  2. Real part of the pole: γ ≪ kᵣ here, so ER ≈ (ħ²/2m*)kᵣ², and kᵣ = √(0.0800 eV ÷ 56.87 eV⋅Ų) = √(1.4067×10⁻³ Å⁻²) = 3.751×10⁻² Å⁻¹.
  3. Imaginary part: Γ = (ħ²/2m*)(4kᵣγ), so γ = 4.00×10⁻⁴ eV ÷ (4 × 3.751×10⁻² × 56.87) = 4.00×10⁻⁴/8.532 = 4.69×10⁻⁵ Å⁻¹. The pole sits at k = (3.751×10⁻² − 4.69×10⁻⁵ i) Å⁻¹, a fraction γ/kᵣ = 1/4Q = 1.25×10⁻³ of its own real part below the axis.
  4. Lifetime: τ = ħ/Γ = 6.582×10⁻¹⁶ eV⋅s ÷ 4.00×10⁻⁴ eV = 1.65×10⁻¹² s = 1.65 ps.
  5. Speed in the well: v = ħkᵣ/m* = 2(ħ²/2m*)kᵣ/ħ = 2 × 56.87 × 3.751×10⁻² ÷ 6.582×10⁻¹⁶ = 6.48×10¹⁵ Å s⁻¹, so one round trip of 2 × 50 Å takes 1.54×10⁻¹⁴ s.
  6. Round trips: τ ÷ 1.54×10⁻¹⁴ s ≈ 107. The escape probability per round trip is therefore about 1%, which is the transmission each barrier must have — the narrow pole and the opaque barrier are the same fact.

Answerk = (3.751×10⁻² − 4.69×10⁻⁵ i) Å⁻¹, so E = (80.0 − 0.200i) meV; τ = 1.65 ps, about 107 round trips of the 5.0 nm well, near 1% escape per trip.