Unbound States, Phase Shift and Time Delay · Quantum Mechanics (Phillips)

Above the well the joining condition can always be met, so nothing is quantized. What survives is a single number: how far the outgoing wave has been pushed.

§5.1a found the bound states of Fig. 5.1’s well by demanding a smooth join at $x = a$ . That demand was restrictive — satisfiable only at particular energies, which is where discreteness came from.

Now take $E > 0$ , so the particle is not trapped. The algebra is almost identical. The conclusion is completely different.

Setting up

A particle with positive energy $E$ approaches the well from the right and is reflected at the wall. Write

E = \frac{\hbar^2k_0^2}{2m} - V_0 = \frac{\hbar^2k^2}{2m}\tag{5.13}

so classically the momentum would be $\hbar k_0$ inside the well and $\hbar k$ outside. The particle is faster inside — it has fallen into an attractive region.

step 1 of 4

Compare each line with §5.1a's. Only the third region differs.

1REGION 1, x < 0. The infinite wall is unchanged, so ψ still vanishes.
$ψ (x) = 0$
2REGION 2, inside the well. The equation is identical to the bound case, Eq. (5.5), so the solution is identical too.
$ψ (x) = C sin k_{0} x$
Equation (5.14). Continuity at x = 0 killed the cosine, exactly as before.
3REGION 3, x > a. Here is the ONE difference. E is now POSITIVE, so the equation is oscillatory instead of exponential — and nothing needs to be discarded to keep it finite at infinity.
$\frac{d ^{2} ψ}{d x ^{2}} = - k^{2} ψ$
Equation (5.15). In §5.1a this read +α²ψ and forced us to throw away the growing exponential. Here both solutions are perfectly well behaved.
4So the general solution survives intact, with an arbitrary phase.
$ψ (x) = D sin (k x + δ)$
Equation (5.16). The constant δ is the PHASE SHIFT — and it is new, because in the bound case there was nothing left over to be free.

Matching at $x = a$ gives, exactly as before,

C\sin k_0a = D\sin(ka + \delta)\tag{5.17}

k_0C\cos k_0a = kD\cos(ka + \delta)\tag{5.18}

and dividing,

k_0\cot k_0a = k\cot(ka + \delta)\tag{5.19}

The crucial difference

Equation (5.19) looks just like Eq. (5.11). It behaves nothing like it.

Compare the two joining conditions carefully

	bound, Eq. (5.11)	unbound, Eq. (5.19)
condition	$k_0\cot k_0a = -\alpha$	$k_0\cot k_0a = k\cot(ka+\delta)$
free parameters	none — $\alpha$ and $k_0$ are both fixed by $E$	one: $\delta$
what it restricts	the energy, to discrete values	nothing
what it determines	which $E$ are allowed	the value of $\delta$ at each $E$

In the bound case both sides were pinned by the energy, so the equation was a constraint on $E$ and most energies failed it. Here the right-hand side contains $\delta$ , which is not determined by anything else — so for any positive $E$ whatsoever you can find a $\delta$ that makes the join smooth.

Hence: bound energies are discrete, unbound energies form a continuum. The difference is not deep physics; it is the presence of one spare constant.

So for any $E > 0$ there is a smooth eigenfunction

\psi(x) = \begin{cases} 0 & \text{if } -\infty < x < 0\\ C\sin(k_0x) & \text{if } 0 < x < a\\ D\sin(kx + \delta) & \text{if } a < x < \infty\end{cases}\tag{5.20}

undulating with wave number $k_0$ where a classical particle would have momentum $\hbar k_0$ , and with $k$ where it would have $\hbar k$ .

ψ with the well
same energy, no well (δ = 0)

energy E4.93depth w2.00π/a

k (outside) 3.1416

k₀ (inside) 7.0248

δ (rad) -2.7528

τ = 2ħ dδ/dE -0.3792

Both curves have the same wavelength outside — the energy fixes that — and the same amplitude, because reflection is total. The well has changed nothing except where the wave sits, by δ/k = -0.8762 a. The time delay is negative: the particle spends less time in the encounter than a free one would, because it speeds up crossing the well.

Natural units ħ = m = a = 1, so E is in units of ħ²/ma² and the shaded strip is the well. Raise E and the wavelength shortens while k₀/k → 1: the wave inside comes to match the wave outside, and the phase shift tends to (k₀ − k)a ≈ V₀/√(2E), which goes to zero. A fast particle crosses the well almost as though it were not there. Note that δ is only defined modulo π, so the readout jumps by π as you sweep — that is the book's own parenthetical remark in problem 3, not a glitch, and it is why the limit above is stated for the principal value.

What a phase shift is

The constant $\delta$ left over in Eq. (5.20) is the phase shift . Before asking what it means, it is worth seeing exactly what it does to the wave.

Rewrite Eq. (5.20) in complex exponentials using $\sin\theta = (e^{i\theta} - e^{-i\theta})/2i$ , and define $A_0 = -C/2i$ and $A = -De^{-2i\delta}/2i$ :

\psi(x) = \begin{cases} A_0e^{-ik_0x} - A_0e^{+ik_0x} & \text{if } 0 < x < a\\ Ae^{-ikx} - Ae^{2i\delta}e^{+ikx} & \text{if } a < x < \infty\end{cases}\tag{5.21}

A stationary state describes no dynamics at all

Phillips is careful here, and it is easy to skate past. Equation (5.21) uses words like incoming, outgoing, travelling back and forth — but it is a state of definite energy, so by §4.6 it has no observable time dependence whatsoever. Nothing moves. Nothing arrives. Nothing leaves.

The “incoming” and “outgoing” labels describe the mathematical decomposition of a standing wave into two travelling pieces, not a sequence of events. To describe an actual reflection — a particle that approaches, interacts, and departs — you need a non-stationary state of uncertain energy: a wave packet.

The book’s own reference to that machinery cites “Section 4.7”, which does not exist — chapter 4 ends at §4.6, and the correct reference is §4.5. One of several stray cross-references in this book; the site flags each where it occurs.

From phase shift to time delay

Build a packet from a narrow band of energies around $E$ , with $c(E')$ the energy probability amplitude of §4.5. Incoming and outgoing packets differ by the factor $e^{2i\delta(E')}$ , and expanding $\delta$ to first order about $E$ gives

\Psi_i(x,t) \propto F(t + x/v)\tag{5.22}

\Psi_f(x,t) \propto F(t - x/v - 2\hbar\,\mathrm d\delta/\mathrm dE)\tag{5.23}

Same shape $F$ , same speed $v$ — but the outgoing packet’s argument carries an extra constant. That constant is a time delay :

\tau \approx 2\hbar\frac{\mathrm d\delta}{\mathrm dE}\tag{5.24}

Why it is the derivative, and why it is negative here

Why a derivative. A single phase does nothing observable — §4.3 made that point at length. What a packet responds to is how the phase varies across the band of energies it contains. A phase that is constant in $E$ shifts nothing; a phase that varies linearly in $E$ displaces the packet in time. So the delay must involve $\mathrm d\delta/\mathrm dE$ , and could not involve $\delta$ alone.

Why negative. For this potential the particle speeds up on entering the well at $x = a$ , since the well is attractive. It therefore covers the interaction region faster than a free particle would, and comes back early. A negative delay is an advance.

That is a genuine and slightly disquieting result — but note what it is not. Nothing travels faster than the packet’s own group velocity outside the well, and no information arrives before it could have. The advance is bounded by the time the particle spends in the well.

Ψ(x,t)
envelope S(x), Eq. (2.5)
a crest — moves at ω/k
envelope peak — moves at dω/dk

band Δkk/8components41

ω(k): ω = ħk²/2m
length 2π/Δk: 8.4
ω/k: 3.00
dω/dk: 6.00(2.00×)

This widget shows a free packet, with no well in it: build one from a band of energies and it localizes, moves at the group velocity, and slowly spreads. That is the object Eqs. (5.22) and (5.23) are about. The well’s effect is to shift its arrival time by $\tau$ — a displacement of the whole envelope, not a change in its shape, at least to the first order this treatment keeps.

Two things carry over directly. The packet must contain a spread of energies, so it is exactly the non-stationary state §4.6 described — and by $\delta t\,\Delta E \approx \hbar$ , a well-defined arrival time requires an uncertain energy. And it spreads as it travels, which is the book’s own caveat that “in practice wave packets change in shape as they move”: Eq. (5.23) keeps only the first-order term.

Engineer's bridge — phase shift is a scattering parameter, time delay is group delay

Both halves of this section are standard signal-processing quantities wearing physics names.

The phase shift is $\angle S_{11}$ . A one-port network that reflects everything has $|S_{11}| = 1$ and all its information in the phase — exactly Eq. (5.21), where $|e^{2i\delta}| = 1$ and $\delta$ carries everything. A short-circuited transmission line stub is the same problem: the reflection coefficient’s phase tells you the stub’s electrical length.

The time delay is group delay. $\tau_g = -\mathrm d\phi/\mathrm d\omega$ is the standard definition, and Eq. (5.24) is that with $\phi = -2\delta$ and $\omega = E/\hbar$ . The reasoning is identical: a flat phase does nothing, a linear phase displaces the envelope, and curvature distorts it. The book even notes that “in practice wave packets change in shape as they move” — which is dispersion, the second-order term this first-order treatment drops.

Where it breaks. In a filter, group delay is a delay: signals come out later. Here it is routinely negative, and that would be alarming in a causal network. It is not alarming here because $\tau$ is measured against free flight over the same path, not against zero — the particle covers the well faster than it would have covered empty space. Negative group delay does occur in electronics too (in gain-assisted media, or across a narrow band near an absorption line) and is equally not a causality violation for the same reason: the envelope is reshaped, not advanced past its own front.

Put a number on it — how early does the packet come back?

Take an electron, a well $5\ \mathrm{eV}$ deep and $1\ \mathrm{nm}$ wide, and send it in with $E = 1\ \mathrm{eV}$ .

Step 1 — the slope of the phase shift. Differentiating Eq. (5.19) numerically about that energy gives $\mathrm d\delta/\mathrm dE = -2.14\ \mathrm{eV^{-1}}$ . Negative, because a deeper well advances the wave.

Step 2 — the delay, from Eq. (5.24):

\tau = 2\hbar\frac{\mathrm d\delta}{\mathrm dE} = 2(1.055\times10^{-34})(-2.14/1.602\times10^{-19}) = -2.81\times10^{-15}\ \mathrm{s}

about $-2800$ attoseconds.

Step 3 — compare it with something. A classical electron of that energy crosses the well and back in $2a/v = 3.37\times10^{-15}\ \mathrm{s}$ . So the delay is $-0.83$ of a crossing time — the packet leaves nearly a full round trip early, because inside the attractive well it moves faster than it did outside.

Attoseconds are not hypothetical: attosecond pulse metrology measures exactly this kind of shift, and photoemission time delays of tens of attoseconds have been resolved experimentally. The phase shift is a real, measurable clock reading.

What §5.1 established in general

Phillips closes by naming three features that outlive this particular potential:

Wave functions undulate in classically allowed regions and fall off exponentially in classically forbidden ones.
Sufficiently attractive potentials give bound states with discrete energies.
Unbound particles have a continuous range of energies, and scattering imprints a phase shift that corresponds to a time delay.

Can an unbound eigenfunction be normalized?

It cannot — $\psi = D\sin(kx+\delta)$ oscillates forever and $\int|\psi|^2\mathrm dx$ diverges, exactly as §3.2 found for a plane wave.

The book’s footnote offers a workaround: put the whole system in a very large box, which “to some extent represents the laboratory containing the system”. It then admits that in practice the normalization of unbound states is seldom considered — they are characterized instead by a particle flux or a probability current density.

That is honest but unsatisfying, and it is deferred rather than solved. Chapter 7 handles it properly with Dirac delta normalization. Note also that the site’s own numerics inherit exactly this problem: qm.ts diagonalizes $\hat H$ in a finite box, so it turns the true continuum into a dense ladder of discrete levels whose spacing is set by the box size rather than by physics. For bound states that is harmless; for these, the box is the answer you get.

Reproduce it in NumPy

That any positive energy is allowed, what δ actually is for the book's own example, the negative time delay, and the high-energy limit — where the phase shift tends to (k₀−k)a and vanishes.

import numpy as np

# natural units hbar = m = a = 1; the well of Fig. 5.1 with w = 2 pi/a
V0 = 2*np.pi**2

def phase(E):
    """delta from Eq. (5.19): k0 cot k0 a = k cot(ka + delta)"""
    k  = np.sqrt(2*E)
    k0 = np.sqrt(2*(E+V0))
    return k, k0, np.arctan2(1.0, (k0/k)/np.tan(k0)) - k

print("unbound states: ANY positive energy is allowed")
print("=" * 66)
print("  the same joining condition that quantized the bound states")
print("  imposes nothing here -- it just fixes delta.")
print(f"  {'E':>9} {'k':>8} {'k0':>8} {'delta':>10} {'delta mod pi':>13} {'shift d/k':>11}")
for E in (0.5*np.pi**2, np.pi**2, 2*np.pi**2, 8*np.pi**2):
    k, k0, d = phase(E)
    print(f"  {E:9.4f} {k:8.4f} {k0:8.4f} {d:10.4f} {np.mod(d, np.pi):13.4f} {d/k:11.4f}")

print("\nproblem 3's number, and what energy actually gives it")
for lbl, E in [("hbar^2 pi^2 / 2ma^2", 0.5*np.pi**2), ("hbar^2 pi^2 / ma^2", np.pi**2)]:
    k, k0, d = phase(E)
    print(f"  E = {lbl:22}: delta = {d:+.4f}, +2pi -> {d+2*np.pi:.4f}")
print("  the book states delta = 3.53, which the FIRST of these reproduces")

print("\ntime delay tau = 2 hbar d(delta)/dE   (Eq. 5.24)")
h = 1e-6
print(f"  {'E':>9} {'tau':>11} {'a/v':>9} {'tau / (a/v)':>13}")
for E in (0.5*np.pi**2, np.pi**2, 2*np.pi**2, 8*np.pi**2):
    dd = (phase(E+h)[2] - phase(E-h)[2])/(2*h)
    v = np.sqrt(2*E)
    print(f"  {E:9.4f} {2*dd:+11.5f} {1/v:9.4f} {2*dd*v:+13.4f}")
print("  negative throughout: the particle speeds up in the well, so the")
print("  encounter takes LESS time than free flight over the same distance.")

print("\nwhy a fast particle barely notices the well")
print("  delta is only defined MOD PI (the book says so in problem 3), so take")
print("  the principal value in (-pi/2, pi/2] before asking about limits:")
principal = lambda d: ((d + np.pi/2) % np.pi) - np.pi/2
print(f"  {'E':>9} {'k0/k':>8} {'(k0-k)a':>10} {'V0/sqrt(2E)':>13} {'delta principal':>16}")
for E in (1e2, 1e3, 1e4, 1e5):
    k, k0, d = phase(E)
    print(f"  {E:9.0f} {k0/k:8.4f} {k0-k:10.5f} {V0/np.sqrt(2*E):13.5f} {principal(d):16.5f}")
print("  k0/k -> 1, and the phase shift tends to (k0-k)a = V0/sqrt(2E) -> 0.")
print("  A fast particle crosses the well almost as if it were not there.")

printsunbound states: ANY positive energy is allowed ================================================================== the same joining condition that quantized the bound states imposes nothing here -- it just fixes delta. E k k0 delta delta mod pi shift d/k 4.9348 3.1416 7.0248 -2.7528 0.3888 -0.8762 9.8696 4.4429 7.6953 -3.1425 3.1407 -0.7073 19.7392 6.2832 8.8858 -3.5417 2.7415 -0.5637 78.9568 12.5664 14.0496 -11.0934 1.4730 -0.8828 problem 3's number, and what energy actually gives it E = hbar^2 pi^2 / 2ma^2 : delta = -2.7528, +2pi -> 3.5304 E = hbar^2 pi^2 / ma^2 : delta = -3.1425, +2pi -> 3.1407 the book states delta = 3.53, which the FIRST of these reproduces time delay tau = 2 hbar d(delta)/dE (Eq. 5.24) E tau a/v tau / (a/v) 4.9348 -0.37923 0.3183 -1.1914 9.8696 -0.00402 0.2251 -0.0179 19.7392 -0.14410 0.1592 -0.9054 78.9568 -0.00006 0.0796 -0.0007 negative throughout: the particle speeds up in the well, so the encounter takes LESS time than free flight over the same distance. why a fast particle barely notices the well delta is only defined MOD PI (the book says so in problem 3), so take the principal value in (-pi/2, pi/2] before asking about limits: E k0/k (k0-k)a V0/sqrt(2E) delta principal 100 1.0943 1.33295 1.39577 1.35238 1000 1.0098 0.43923 0.44138 0.43576 10000 1.0010 0.13951 0.13958 0.13933 100000 1.0001 0.04414 0.04414 0.04410 k0/k -> 1, and the phase shift tends to (k0-k)a = V0/sqrt(2E) -> 0. A fast particle crosses the well almost as if it were not there.

Where this leaves us

A wall behind the well made reflection total, so the encounter had only one possible outcome and the phase shift was the whole story.

§5.2 removes the wall and replaces the well with a barrier. Now there are two outcomes — reflection and transmission — the encounter becomes genuinely uncertain, and $\delta$ is replaced by something with a probability attached.

Check yourself

0 / 7 answered

1.Eq. (5.19) looks just like Eq. (5.11), yet one quantizes the energy and the other does not. What is the difference?
2.Why is the phase shift the only thing a scattering experiment can measure at fixed energy?
3.Eq. (5.21) speaks of incoming and outgoing waves. What is wrong with reading that as a description of motion?
4.Why does the time delay involve $d δ / d E$ rather than $δ$ itself?
5.The time delay here is negative. Is that a problem for causality?
6.At high energy the principal phase shift tends to $(k_{0} - k) a \approx V_{0} / 2 E$ . What does that say physically?
7.Unbound eigenfunctions cannot be normalized. How does the book handle it, and what does the site's own numerics inherit?