Scattering states and phase shifts

Last section, parity marshalled the bound states into a tidy even-odd-alternating queue. But think back to the scattering experiments of section 2.11: a wave packet hits a barrier, part bounces back, part gets through, we computed the transmission $T$ and reflection $R$, and the story seemed complete.

Was it? Consider a loose end left hanging there — the Ramsauer–Townsend effect: slow electrons passing through xenon atoms find, at certain energies, $T=1$; the gas turns nearly transparent to them. $T=1$ means “everything got through”. Does that mean at this energy the potential did nothing at all to the electron?

Not so cheap. The wave got through, yes, but it came out with its waveform bodily displaced — the crests are not where they would have been had there been no potential. Transmission is completely blind to this: $T$ compares only the moduli of amplitudes and throws all the phase information away. To see this invisible hand of the potential, we need a new quantity. It is called the phase shift.

First, tell the two kinds of state apart

The energy eigenstates of a one-dimensional potential problem fall naturally into two classes, handled in entirely different ways:

• Bound states: energy below the potential’s value at infinity. The wavefunction decays exponentially at both ends, can be normalised, and the energies take discrete values — the wells of chapter 2 and last section’s parity classification are all about these.
• Scattering states: energy above the potential at infinity. The wavefunction oscillates as a plane wave far away, does not decay and cannot be normalised, and every $E$ is allowed — the spectrum is continuous.

The question changes accordingly. For bound states we asked “which energies are allowed”. A scattering state can have any energy, so we ask instead: a wave of given energy goes in — what has changed when it comes out?

The full content of the answer is surprisingly small.

The cleanest stage: the half-axis problem

First reduce the problem until only one number is left. Imagine the particle confined to the half-axis $x>0$ (an impenetrable wall at $x=0$), with a square well of width $a$ and depth $V_0$ against the wall:

$$V(x)=\begin{cases}-V_0,&0<x<a\\0,&x>a\end{cases}$$

This “wall plus short-range potential” stage looks special, but it is precisely the core geometry of three-dimensional scattering from chapter 5 onward — the radial equation of a 3D problem lives on a half-axis, with the origin playing the wall. Learn it now, reuse it wholesale later.

With no potential, the solution at energy $E=\hbar^2k^2/2m$ must vanish at the wall:

$$\psi_{\text{free}}(x)\propto\sin(kx)\tag{4.2.1}$$

With the potential, the potential acts only for $x<a$; beyond that stretch the equation is identical to the free case and the solution is still a sine wave — but there is no longer any reason for it to happen to “line up” at $x=0$. Its most general form is

$$\psi(x)\propto\sin(kx+\delta),\qquad x>a\tag{4.2.2}$$

This angle $\delta$ is the phase shift: the presence of the potential has translated the outside waveform bodily by a distance $\delta/k$. The potential did not change the wavelength (energy is conserved; the wave comes out with the same $k$), and in steady state it did not change the amplitude either. The only thing it can do is slide the waveform — everything a short-range potential does to a wave of given energy is condensed into this one number.

The full derivation: δ(k) for the square well

Derivation: phase shift of the half-axis square welladvanced~10 minexpand

Step 1: write the solutions in the two regions.

Inside the well ($0<x<a$) the kinetic energy is larger, with wavenumber

$$k'=\frac{\sqrt{2m(E+V_0)}}{\hbar}>k=\frac{\sqrt{2mE}}{\hbar}\tag{4.2.5}$$

The solution must vanish at the wall, so inside it can only be

$$\psi_{\text{in}}(x)=A\sin(k'x)\tag{4.2.6}$$

Outside, by definition, we write

$$\psi_{\text{out}}(x)=B\sin(kx+\delta)\tag{4.2.7}$$

Step 2: join them at $x=a$. $V$ is finite, so both $\psi$ and $\psi'$ are continuous:

$$A\sin(k'a)=B\sin(ka+\delta)\tag{4.2.8}$$$$Ak'\cos(k'a)=Bk\cos(ka+\delta)\tag{4.2.9}$$

Step 3: divide the two equations to eliminate the amplitudes. We do not care about $A,B$ (continuum normalisation is a separate story) — only about the shape. Dividing gives

$$k'\cot(k'a)=k\cot(ka+\delta)\tag{4.2.10}$$

The left side is all known quantities — this is “the potential passing its information outward through the logarithmic derivative”: the outside waveform is required to connect at $x=a$ with a prescribed slope-to-value ratio.

Step 4: solve for δ. Invert the relation into a tangent:

$$\tan(ka+\delta)=\frac{k}{k'}\tan(k'a)\tag{4.2.11}$$$$\boxed{\ \delta(k)=-ka+\arctan\!\left[\frac{k}{k'}\tan(k'a)\right]\ }\tag{4.2.12}$$

Check 1: no potential. With $V_0=0$ we have $k'=k$, the bracket becomes $\tan(ka)$, and $\delta=-ka+ka=0$. No potential, no phase shift. Correct.

Check 2: shallow well, low energy. For small $V_0$, $k'\approx k$; expanding to first order gives $\delta>0$ and proportional to $V_0$: an attractive potential gives a positive phase shift, and the stronger the potential the harder it pulls. Matches intuition.

Plug in real numbers (an electron, $\hbar c=197.3$ eV·nm, $mc^2=511$ keV): take $V_0=1.0$ eV, $a=0.2$ nm, $E=0.5$ eV.

$$k=\frac{\sqrt{2\times511000\times0.5}}{197.3}=3.62\ \text{nm}^{-1},\qquad k'=\frac{\sqrt{2\times511000\times1.5}}{197.3}=6.28\ \text{nm}^{-1}\tag{4.2.13}$$

$ka=0.725$, $k'a=1.255$, $\tan(k'a)=3.06$, so

$$\delta=-0.725+\arctan(0.577\times3.06)=-0.725+1.056=0.331\ \text{rad}\approx19°\tag{4.2.14}$$

A positive shift — the well pulls the waveform in by about $\delta/k\approx0.09$ nm, roughly half the well width.

Parity returns the favour: scattering on the full line

The half-axis problem has only reflection, no transmission. Back on the full line with a symmetric potential (the square barriers and wells of section 2.11), waves can come in from the left or from the right, and the problem looks twice as complicated. Here last section’s parity makes good on its promise: it splits the scattering into two channels that never disturb each other.

Derivation: transmission T = cos²(δₑ − δₒ)advanced~9 minexpand

Step 1: change basis. In a symmetric potential $[\hat H,\hat P]=0$; the continuum is degenerate, but the two states at each energy can always be combined into one even and one odd (exactly what the warning box in section 4.1 said). Outside the range of the potential, the even and odd solutions are

$$\psi_{\ee}(x)\propto\cos(k|x|+\delta_{\ee}),\qquad \psi_{\text{o}}(x)\propto\operatorname{sgn}(x)\sin(k|x|+\delta_{\text{o}})\tag{4.2.15}$$

Each channel is a half-axis problem of its own: the odd channel vanishes at the origin (the $\sin$ type — precisely the derivation above, with condition $\psi(0)=0$); the even channel is flat there (the $\cos$ type, condition $\psi'(0)=0$). Each carries its own phase shift, $\delta_{\ee}$ and $\delta_{\text{o}}$, and neither meddles with the other.

Step 2: decompose “incident from the left” into the two channels. A plane wave coming from the left is half even, half odd:

$$\ee^{\ii kx}=\cos kx+\ii\sin kx\tag{4.2.16}$$

Step 3: each channel does exactly one thing. Within a channel, scattering just means “multiply the outgoing part by $\ee^{2\ii\delta}$”. Split the $\cos$ and $\sin$ into incoming and outgoing travelling waves, attach each outgoing term its own $\ee^{2\ii\delta}$, then recombine the channels: collect the coefficient of $\ee^{\ii kx}$ in the region $x>a$ (transmission) and of $\ee^{-\ii kx}$ in $x<-a$ (reflection), giving

$$t=\frac{\ee^{2\ii\delta_{\ee}}+\ee^{2\ii\delta_{\text{o}}}}{2},\qquad r=\frac{\ee^{2\ii\delta_{\ee}}-\ee^{2\ii\delta_{\text{o}}}}{2}\tag{4.2.17}$$

Step 4: read the result. After factoring out the common phase, the moduli depend only on the difference of the shifts:

$$T=|t|^2=\cos^2(\delta_{\ee}-\delta_{\text{o}}),\qquad R=|r|^2=\sin^2(\delta_{\ee}-\delta_{\text{o}})\tag{4.2.18}$$

$T+R=1$ holds automatically — no need for the closing algebraic miracle of section 2.11; current conservation is now guaranteed by the structure.

Numerical check: a symmetric square well of width $2a=0.4$ nm and depth 1.0 eV, at $E=0.5$ eV. The even-channel formula follows by the same route (swap the $\tan$ for the form matching that channel’s condition): $\delta_{\ee}=0.660$, $\delta_{\text{o}}=0.331$, hence $T=\cos^2(0.329)=0.896$. The transmission formula of section 2.11, $T=[1+V_0^2\sin^2(2k'a)/4E(E+V_0)]^{-1}$, gives $0.896$ — identical to the last digit.

Two numbers, $(\delta_{\ee},\delta_{\text{o}})$, hold $T$, $R$, and all the phase information they leave out — the whole of the scattering. $T=1$ is no longer mysterious: it only requires $\delta_{\ee}-\delta_{\text{o}}$ to be an integer multiple of $\pi$, while each channel separately may have moved the wave a long way. The “transparent” xenon atoms of the Ramsauer–Townsend effect in fact stamp a substantial phase shift onto the electron wave — in cold-atom experiments such shifts have been measured directly.

Resonance: the sharp turn in δ

Plot $\delta(E)$ as a curve, and over most stretches it varies gently. But near certain energies it surges by about $\pi$ within a very narrow window — this is a resonance. The standard local shape (the Breit–Wigner form) is

$$\tan\delta=\frac{\Gamma/2}{E_R-E}\tag{4.2.19}$$

As $E$ sweeps through the resonance energy $E_R$, $\delta$ passes through $\pi/2$ (mod $\pi$), with a width set by $\Gamma$.

Now watch all of this with your own eyes. In the simulation below, set $V_0$ negative (a well) and switch to the transmission curve: the spikes where $T$ returns to 1 are the moments $\delta_{\ee}-\delta_{\text{o}}$ crosses $\pi$. Then watch the transmitted wave packet near a resonance energy emerge later than off resonance — that is the Wigner delay:

◈Wave-packet evolution and scattering

A Gaussian packet hits a rectangular barrier. The time-dependent Schrödinger equation is solved live by the split-operator method, with ħ = m = 1.

E / V₀

0.80

Reflection R

0.000

Transmission T

0.000

Packet width Δx

3.00

Press play to send the packet into the barrier. R and T lock automatically once the packet has completely left the barrier region. The barrier width on the grid is a = 1.500.

Incident momentum k₀2.00

E = k₀²/2 = 2.00

Barrier height V₀2.5

positive = a barrier

Barrier width a1.5

T decays exponentially with a: T ~ e^(−2κa)

Initial packet width σ3.0

larger σ means better-defined momentum, closer to a plane wave

View

Live evolutionTransmission curve T(E)

Show Re ψ

Speed8×

PlayBack to t = 0

∎Key formulas

• Definition of the phase shift

  $$\psi(x)\propto\sin(kx+\delta),\quad x>a$$

  Everything a short-range potential does to a wave of given energy; δ>0 pulls in (attraction), δ<0 pushes out (repulsion)

• Square-well phase shift

  $$\delta(k)=-ka+\arctan\left[\frac{k}{k'}\tan(k'a)\right],\quad k'=\frac{\sqrt{2m(E+V_0)}}{\hbar}$$

  Mind the arctan branch; continue continuously from δ(∞)=0

• Parity-channel transmission

  $$T=\cos^2(\delta_{\ee}-\delta_{o}),\qquad R=\sin^2(\delta_{\ee}-\delta_{o})$$

  T+R=1 is guaranteed by the structure; T=1 ⟺ the channel shifts differ by an integer multiple of π

• Resonance and time delay

  $$\tan\delta=\frac{\Gamma/2}{E_R-E},\qquad \tau=2\hbar\frac{\dd\delta}{\dd E}$$

  A surge of π in δ marks a quasi-bound state; narrower Γ, longer lifetime

?Self-check4 questions

1. 1.

   Why does the phase shift "hold everything a short-range potential does to a scattering wave of given energy"?

   • Because the potential cannot change the distant wavelength or the steady-state amplitude, so the only thing left to move is the overall position of the waveform
   • Because the phase shift determines the particle's energy
   • Because the wavefunction can always be chosen real
   • Because the phase shift is just the transmission written another way

   Submit
2. 2.

   In a symmetric potential, T = cos²(δₑ − δₒ). The correct reading of T = 1 (perfect transmission) is:

   • The two parity channels' phase shifts differ by an integer multiple of π; the potential may still be acting on the wave substantially
   • Both phase shifts must be zero, so the potential might as well not exist
   • This only happens when the potential is zero
   • The reflected wave adds in phase with the incident wave

   Submit
3. 3.

   The phase-shift curve δ(E) surges by about π over a very narrow interval near some energy. This tells us: (select all that apply)

   Select all that apply

   • A quasi-bound state (resonance) exists near that energy
   • The dwell time τ = 2ħ dδ/dE in the potential region becomes markedly longer there
   • The narrower the surge interval, the longer the quasi-bound state lives
   • The transmission must be zero at that energy

   Submit
4. 4.

   An electron is incident on a square well against the wall of the half-axis: depth V₀ = 1.0 eV, width a = 0.2 nm, incident energy E = 0.5 eV. Compute the phase shift from δ = −ka + arctan[(k/k′)tan(k′a)], in degrees. (ħc = 197.3 eV·nm, mc² = 511000 eV)

   °200% relative tolerance

   Submit

What comes next

Phase shifts and transmission — we can now compute both, as long as the potential is one square well or barrier. Real potentials are rarely so obliging: a resonant-tunneling diode is a three-layer barrier-well-barrier stack, and semiconductor superlattices run to hundreds of layers. With the method of section 2.11, every added layer means four more matching equations — a hundred layers means solving several hundred simultaneously. Doing that by hand is suicide.

The next section brings in an assembly-line tool: package each slab of potential as a $2\times2$ matrix, and the whole multilayer structure becomes a product of matrices. The razor-sharp resonance in the double barrier — that $T$ rocketing from 0.04 to 1 — will be its first trophy.