Complementarity & Delayed Choice

The loop of the last lesson gave a binary answer: know the branch and interference dies, do not know it and interference survives. Nature is more generous than that. Which-way knowledge and interference trade off continuously, along a curve you can derive in a page — and the trade can be undone after the fact, by a choice made too late to have influenced anything.

Learning Objectives

After this lesson you will be able to:

1. State complementarity in its modern which-way form, and explain why it is a statement about available information rather than about clumsy apparatus.
2. Model a which-way marker as an entangled spin–marker state, compute the resulting visibility $V = |\langle{m_+}|{m_-}\rangle|$, and identify what the reduced density matrix loses.
3. Define the distinguishability $D$, prove $D^2 + V^2 \le 1$ for a spin-$\tfrac12$ interferometer, and locate the two detector models of Lesson 1 on that circle.
4. Argue rigorously that no human observer is required, using the partial trace to show that awareness never enters the prediction.
5. Analyze quantum erasure: recover full-visibility interference by post-selecting on a complementary marker measurement, and explain why the raw, unsorted data show nothing.
6. Run Wheeler's delayed-choice argument against a classical conspiracy theory, derive the constraint $P({+}x;\theta) = \tfrac12(1 + \cot\theta)$, and show why the surviving loophole closes.

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Intuition

Bohr's complementarity is usually taught as a slogan about waves and particles: a quantum object has both aspects, and any experiment that reveals one hides the other. As a slogan it is true and almost useless, because it does not say how much of one you give up for how much of the other, and it invites the wrong mechanism — the idea that the price is paid in disturbance, in photons banging into electrons and knocking them off course.

The modern way to say it is sharper and, it turns out, more fundamental. Consider a device with two paths. If the world contains a record from which the path could be determined — the which-way information — the object behaves as though it took one path. If no such record exists anywhere, the two paths interfere. That is the whole principle, and notice what it does not mention: forces, momentum kicks, energy exchange, or observers.

The reason the disturbance story had to be abandoned is experimental. Modern which-way markers can be made arbitrarily gentle — an internal atomic state flipped, a photon's polarization rotated, a nuclear spin flipped — carrying no momentum worth speaking of and changing no trajectory. The interference dies anyway. Whatever destroys it is not the shove; it is the correlation.

And correlation, unlike a shove, is reversible. That is the hinge on which this lesson turns. If the which-way record lives in a quantum system rather than a smashed photographic plate, we can go back and measure that system in a basis that carries no branch information at all — and the interference returns. Even, as Wheeler realized, if we decide to do so after the atom has finished its trip [Fre, §2.2].

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Theory

A which-way marker, done properly

Model the marker as a second quantum system. Prepare an atom in $\lvert{+z}\rangle$, send it into an $x$-oriented analyzer loop, and arrange that whatever branch it takes leaves the marker in a corresponding state: $\lvert{m_+}\rangle$ for the $+x$ branch and $\lvert{m_-}\rangle$ for the $-x$ branch. Since $\lvert{+z}\rangle = \tfrac{1}{\sqrt2}\left(\lvert{+x}\rangle + \lvert{-x}\rangle\right)$, the joint state after the loop is

$$\lvert{\Psi}\rangle = \frac{1}{\sqrt2}\Bigl(\lvert{+x}\rangle\otimes\lvert{m_+}\rangle + \lvert{-x}\rangle\otimes\lvert{m_-}\rangle\Bigr) ,$$

which we abbreviate $\tfrac{1}{\sqrt2}(\lvert{+x, m_+}\rangle + \lvert{-x, m_-}\rangle)$ and will keep writing that way. The marker states are normalized but need not be orthogonal; write their overlap

$$s = \langle{m_+}|{m_-}\rangle , \qquad 0 \le |s| \le 1 .$$

Now measure the atom along $z$ and ignore the marker entirely. Projecting,

$$\bigl(P_+^z \otimes \mathbb{1}\bigr)\lvert{\Psi}\rangle = \tfrac12\,\lvert{+z}\rangle\otimes\bigl(\lvert{m_+}\rangle + \lvert{m_-}\rangle\bigr) ,$$

using $\langle{+z}|{\pm x}\rangle = 1/\sqrt2$, and taking the squared norm gives

$$P(+z) = \frac{1}{4}\bigl\lVert \lvert{m_+}\rangle + \lvert{m_-}\rangle \bigr\rVert^2 = \frac{1 + \mathrm{Re}\,s}{2} .$$

Read the two limits. Orthogonal markers, $s = 0$: perfect which-way information, $P(+z) = \tfrac12$, no interference. Identical markers, $s = 1$: no information at all, $P(+z) = 1$, full interference and the loop is the identity of Lesson 1. In between, everything in between — and the interference term is governed by nothing but the overlap of the marker states.

The same statement in the language of Term 1: trace the marker out. In the $x$ basis the atom's reduced state is

$$\rho = \operatorname{Tr}_m \lvert{\Psi}\rangle\langle{\Psi}\rvert = \frac12\begin{pmatrix} 1 & s^{*} \\ s & 1 \end{pmatrix} ,$$

a density matrix whose diagonal is untouched and whose off-diagonal elements — the coherences — are multiplied by the marker overlap. Marking does not change the branch populations at all; it degrades the coherence between them, and that is exactly what interference is made of. This suppression of off-diagonal elements by entanglement with something else is decoherence, the mechanism that explains why macroscopic objects never show interference: their markers (air molecules, thermal photons, the rest of the universe) go orthogonal in femtoseconds.

How much do you know? Distinguishability

The visibility of the interference is the fractional swing of $P(+z)$ away from the featureless $\tfrac12$:

$$V \equiv 2P(+z) - 1 = \mathrm{Re}\,s ,$$

which for a marker with no gratuitous phase is just $|s|$. The complementary quantity measures how much the marker tells you. Given a single copy of the marker, the best possible guess of which branch was taken succeeds with probability $\tfrac12\left(1 + \sqrt{1 - |s|^2}\right)$ — the Helstrom bound for discriminating two equally likely pure states. Define the distinguishability as the bias above chance, doubled so that it runs from 0 to 1:

$$D \equiv 2\Bigl[\tfrac12\bigl(1+\sqrt{1-|s|^2}\bigr)\Bigr] - 1 = \sqrt{1 - |s|^2} .$$

Then, immediately,

$$D^2 + V^2 = \left(1 - |s|^2\right) + |s|^2 = 1 .$$

For a pure-state marker the trade-off is a circle: knowledge and interference are two coordinates of one unit vector. A marker that is partly mixed, or a discrimination strategy that is not optimal, sits strictly inside, so in general

$$D^2 + V^2 \le 1 ,$$

the duality relation of Greenberger and YaSin and of Englert. It is complementarity with numbers attached, and it says something stronger than Bohr's slogan: you may have a little of each, and the exchange rate is fixed.

It is worth locating Lesson 1's two detector models on this circle.

Model	Distinguishability $D$	Visibility $V$	$D^2 + V^2$
Coherent marker, overlap $s$	$\sqrt{1-s^2}$	$s$	$1$ (optimal)
Detector armed a fraction $\eta$ of the time	$\eta$	$1-\eta$	$\eta^2 + (1-\eta)^2 \le 1$

The armed-or-idle detector is inefficient in a precise sense: at $\eta = \tfrac12$ it achieves $D^2 + V^2 = \tfrac12$, only half the information–coherence budget a coherent marker of the same visibility would use. That is not a defect of the analysis, it is a fact about the apparatus — a device that is sometimes fully on and sometimes fully off wastes the middle of the range that a weakly-coupled marker exploits.

Is a human observer needed?

Freericks poses this as a problem [Fre, §2.2.1]: cover the readouts of the pass-through detectors so nobody can see them. Does the interference come back?

It does not, and the derivation above says why in one line. The prediction for the atom is $\rho = \operatorname{Tr}_m\lvert{\Psi}\rangle\langle{\Psi}\rvert$, and the partial trace contains no variable for "was anyone looking." The coherence is suppressed by $|s|$, the overlap of the marker states, from the moment the correlation is established. Covering the dial changes nothing, because the dial was never in the formula.

Three corollaries are worth stating explicitly, because they are the three most common misreadings of quantum measurement:

• Reading the record is irrelevant. The suppression happens when the correlation forms, not when a person consults it. (R.1.3's exercise E4 made this claim; the partial trace is the proof.)
• Destroying the record does not help. Smashing the detector, or letting its state leak into the lab, only entangles the marker with more systems, driving $|s|$ further toward zero. Coherence is lost by spreading, and spreading is irreversible in practice.
• Erasing the record is a completely different operation. To restore interference you must measure the marker in a basis that carries no branch information, and then use that outcome. That is not "not looking"; it is looking at a carefully chosen complementary observable. It is the subject of the next section.

The reason this matters beyond bookkeeping is that it removes consciousness from physics without removing the mystery. Nothing here explains how an atom manages to behave as though it took both branches. It only establishes that the thing which decides the outcome is a physical correlation with a physical system, computable, quantitative, and entirely indifferent to who is in the room.

Quantum erasure

Now make the marker concrete, following Freericks' atomic version [Fre, §2.2.2]. Let the marker be the atom's own internal state: $\lvert{g}\rangle$ for the ground state and $\lvert{e}\rangle$ for an excited state, orthogonal, so $s = 0$ and interference is fully destroyed. An exciter — a light pulse tuned to the transition — sits in one branch of the loop and promotes the atom to $\lvert{e}\rangle$ if it passes that way.

Put the exciter in the $+x$ branch. The joint state after the loop is

$$\lvert{\Psi}\rangle = \frac{1}{\sqrt2}\Bigl(\lvert{+x, e}\rangle + \lvert{-x, g}\rangle\Bigr) .$$

This is an entangled state: neither the spin nor the internal state has a value of its own, but they are perfectly correlated. Freericks' terminology is exact and worth adopting — tagging is entangling. Rewriting the branches in the $z$ basis, with $\lvert{\pm x}\rangle = (\lvert{+z}\rangle \pm \lvert{-z}\rangle)/\sqrt2$,

$$\lvert{\Psi}\rangle = \frac12\Bigl(\lvert{+z}\rangle\otimes\bigl(\lvert{e}\rangle + \lvert{g}\rangle\bigr) + \lvert{-z}\rangle\otimes\bigl(\lvert{e}\rangle - \lvert{g}\rangle\bigr)\Bigr) ,$$

so a $z$ measurement that ignores the internal state gives $P(\pm z) = \tfrac12$ apiece — interference gone, exactly as $s = 0$ requires.

But look at the structure. Define the two complementary marker states

$$\lvert{\pm}\rangle_m = \frac{1}{\sqrt2}\bigl(\lvert{g}\rangle \pm \lvert{e}\rangle\bigr) ,$$

each of which is an equal superposition of "excited" and "not", and therefore says nothing whatever about which branch was taken. In terms of these,

$$\lvert{\Psi}\rangle = \frac{1}{\sqrt2}\Bigl(\lvert{+z}\rangle\otimes\lvert{+}\rangle_m - \lvert{-z}\rangle\otimes\lvert{-}\rangle_m\Bigr) .$$

The same state, rewritten. Measure the marker in the $\{\lvert{g}\rangle, \lvert{e}\rangle\}$ basis and you learn the branch and the $z$ result is random. Measure it in the $\{\lvert{+}\rangle_m, \lvert{-}\rangle_m\}$ basis and you learn nothing about the branch — but the outcome tells you the $z$ result with certainty: a $+$ marker means $\lvert{+z}\rangle$, a $-$ marker means $\lvert{-z}\rangle$. This is quantum erasure. The which-way information was not destroyed so much as traded, and the currency is the choice of measurement basis on the marker.

In hardware, the basis change is a "half-excitation" pulse — Freericks' superpositioner, which maps $\lvert{g}\rangle \mapsto (\lvert{g}\rangle + \lvert{e}\rangle)/\sqrt2$ and $\lvert{e}\rangle \mapsto (\lvert{g}\rangle - \lvert{e}\rangle)/\sqrt2$; anyone who has met Term 2 will recognize a Hadamard. Following it with a ground-state filter (his de-exciter, which de-excites and blocks any atom that was excited) projects the marker onto $\lvert{+}\rangle_m$. Half the atoms are discarded, and the survivors are all in $\lvert{+z}\rangle$ — the input state, restored, out of an ensemble that was 50–50 the moment before.

The half you throw away matters. Erasure never manufactures interference out of nothing; it sorts an existing ensemble into two subensembles that interfere oppositely. Keeping the $\lvert{+}\rangle_m$ half gives all $+z$; keeping the $\lvert{-}\rangle_m$ half gives all $-z$; add them back together and you recover the flat 50–50 you started with. The raw, unsorted data show no fringes and never will. That is precisely why erasure cannot be used to signal, a point Lesson 3 turns into a theorem.

Delayed choice

Wheeler's question was about timing. The marker measurement and the spin measurement act on different factors of the joint state, so their operators commute, so the order does not matter — the joint statistics are identical either way. That is a mathematical fact, and it has a startling reading: we may run the $z$ measurement first, record the results, and only afterwards decide which basis to measure the marker in [Fre, §2.2.2].

Do that. The $z$ detectors fire 50–50 and are done with. Then, later — arbitrarily later, and far enough away that no signal could have reached them in time — choose the marker basis. Choose $\{\lvert{g}\rangle, \lvert{e}\rangle\}$ and the records sort into two subensembles by branch, each 50–50 in $z$: no fringes. Choose $\{\lvert{+}\rangle_m, \lvert{-}\rangle_m\}$ and the same $z$ records sort into two subensembles that are pure $+z$ and pure $-z$: perfect fringes.

Wheeler wondered aloud whether this meant the choice propagated backward in time to instruct the atom. It does not, and the formalism says why without any appeal to interpretation. Nothing about the $z$ records changed — the same list of $+$ and $-$ symbols is on the page either way, with the same statistics. What the late choice selected was how to partition that list. A correlation that was present in the joint state from the moment of tagging is being revealed, not created, and revealing it requires the marker results, which travel forward in time at ordinary speed like everything else.

Wheeler's argument against a conspiracy

The delayed choice does more than dissolve a puzzle: it kills a class of classical explanations outright, and this is the sharpest thing in §2.2.

The explanation to be killed is a conspiracy theory: the atom does travel one definite branch, but the probability with which it picks each branch is allowed to depend on the whole configuration of the apparatus, including parts it has not reached yet. Let us give the theory every advantage and see where it breaks [Fre, §2.2].

Setup: a $\lvert{+z}\rangle$ source, an $x$-oriented loop with both gates open, and a final analyzer at angle $\theta$ from vertical. Quantum mechanics says the loop does nothing, so $P(+) = \cos^2(\theta/2)$. The conspiracy says the atom takes the $+x$ branch with some probability $p(\theta)$ and the $-x$ branch with probability $1-p(\theta)$, after which the final analyzer acts normally. The angle between the $x$ axis and the analyzer is $90° - \theta$, so by compound probability

$$P(+) = p\,\cos^2\!\left(45° - \frac{\theta}{2}\right) + (1-p)\,\sin^2\!\left(45° - \frac{\theta}{2}\right) .$$

Simplify with the double-angle identity $\cos^2 u - \sin^2 u = \cos 2u$, taking $u = 45° - \theta/2$ so that $\cos 2u = \cos(90° - \theta) = \sin\theta$, and with $\sin^2 u = \tfrac12\left(1 - \cos 2u\right) = \tfrac12(1 - \sin\theta)$:

$$P(+) = p\sin\theta + \frac{1 - \sin\theta}{2} = \left(p - \frac12\right)\sin\theta + \frac12 .$$

Demand that this equal the measured $\cos^2(\theta/2) = \tfrac12(1 + \cos\theta)$ and solve:

$$\left(p - \frac12\right)\sin\theta = \frac{\cos\theta}{2} \quad\Longrightarrow\quad p(\theta) = \frac12\bigl(1 + \cot\theta\bigr) .$$

Now impose the one constraint a probability cannot escape, $0 \le p \le 1$, which forces $-1 \le \cot\theta \le 1$ and hence

$$45° \le \theta \le 135° .$$

Outside that window there is no assignment of branch probabilities that reproduces the data. At $\theta = 0$, for instance, the conspiracy is stuck at $P(+) = \tfrac12$ while the experiment gives 1; that is just the $\tfrac12$-versus-$1$ result of Lesson 1 in a new costume. So a first pass already rules the theory out over most of its range.

But a determined skeptic keeps the window. Inside $45°$–$135°$ the "magic values" $p(\theta)$ exist, and there they save the classical picture. Wheeler's move closes it. Set the analyzer to $\theta_{\text{i}} = 90°$ while the atom crosses the loop, so the conspiracy must commit to $p = \tfrac12$; then rotate the analyzer to $\theta_{\text{f}} = 0°$ after the atom has left the loop and before it arrives. Quantum mechanics is unmoved: the loop did nothing, so $P(+) = \cos^2 0 = 1$. The conspiracy is stuck with the branch probabilities it already committed to and predicts

$$P(+) = \left(\tfrac12 - \tfrac12\right)\sin 0° + \tfrac12 = \frac12 .$$

One versus a half — as far apart as two predictions can be, in a regime where the static version of the theory was safe. Delayed choice therefore rules out every theory in which the unwatched atom travels one definite branch, whether or not the branch probabilities are allowed to peek at the apparatus, because the apparatus can be changed after the peeking is over.

This is the first of two no-go results in this course, and it is deliberately the easier one. It kills theories where the setting is the hidden input. The next lesson kills a much larger class — where the atoms themselves carry complete instruction sets — and needs a genuinely cleverer experiment to do it.

Where else this shows up

The same structure appears wherever two paths can be tagged. In a double slit, the "marker" is anything correlated with the slit — a recoiling screen, a scattered photon, a nuclear spin — and the fringes fade exactly as $|s|$ does; the Pre-Term's P.3.2 Matter as Waves treats the interference side of that story, and this lesson supplies the reason it is fragile. In a quantum computer, the "marker" is the environment, $|s|$ is what decoherence times measure, and the whole discipline of error correction is the art of keeping $|s|$ near 1 for long enough to finish the computation.

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Hands-on (Python)

import numpy as np

# ---- Spin-plus-marker state:  (|+x,m+> + |-x,m->)/sqrt(2) ----------------
pz, mz = np.array([1, 0], complex), np.array([0, 1], complex)
px, mx = (pz + mz) / np.sqrt(2), (pz - mz) / np.sqrt(2)

def tagged_state(s):
    """Marker overlap <m+|m-> = s, realized with 2-d marker states."""
    m_plus = np.array([1, 0], complex)
    m_minus = np.array([s, np.sqrt(max(0.0, 1 - s**2))], complex)
    return (np.kron(px, m_plus) + np.kron(mx, m_minus)) / np.sqrt(2)

def p_plus_z(psi):
    P = np.kron(np.outer(pz, pz.conj()), np.eye(2))
    return np.vdot(psi, P @ psi).real

print("   s     P(+z)   V=2P-1   D=sqrt(1-s^2)   D^2+V^2")
for s in (0.0, 0.25, 0.5, 0.75, 1.0):
    P = round(p_plus_z(tagged_state(s)), 12)       # trim float dust for display
    V, D = 2 * P - 1, np.sqrt(1 - s**2)
    print(f" {s:.2f}    {P:.3f}    {V:.3f}      {D:.3f}         {D**2 + V**2:.3f}")
# V == s exactly, and D^2 + V^2 == 1 on the nose: the duality circle.

# ---- The armed-or-idle detector sits strictly inside the circle ----------
print("\n eta    D      V      D^2+V^2")
for eta in (0.0, 0.25, 0.5, 0.75, 1.0):
    D, V = eta, 1 - eta
    print(f" {eta:.2f}   {D:.3f}  {V:.3f}   {D**2 + V**2:.3f}")

# ---- Quantum erasure: sorting a 50-50 ensemble into two pure halves ------
g, e = np.array([1, 0], complex), np.array([0, 1], complex)
psi = (np.kron(px, e) + np.kron(mx, g)) / np.sqrt(2)      # exciter on +x
m_plus, m_minus = (g + e) / np.sqrt(2), (g - e) / np.sqrt(2)

def sort_by_marker(psi, marker):
    """Project the marker onto `marker`; return (weight, P(+z) in that half)."""
    P = np.kron(np.eye(2), np.outer(marker, marker.conj()))
    out = P @ psi
    w = np.vdot(out, out).real
    return w, p_plus_z(out / np.sqrt(w))

print(f"\nunsorted             P(+z) = {p_plus_z(psi):.3f}")
for name, m in (("marker = ground   ", g), ("marker = excited  ", e),
                ("marker = |+>_m    ", m_plus), ("marker = |->_m    ", m_minus)):
    w, p = sort_by_marker(psi, m)
    print(f"{name}   weight = {w:.3f}   P(+z | this half) = {p:.3f}")
# g/e sorting: each half is still 50-50 in z  -> which-way kept, no fringes.
# +/- sorting: the halves are 1.000 and 0.000 -> which-way erased, fringes back.

# ---- Wheeler: the conspiracy theory's branch probability -----------------
print("\n theta   p(theta) = (1 + cot theta)/2      allowed?")
for deg in (0, 30, 45, 90, 135, 150, 180):
    th = np.deg2rad(deg)
    p = np.inf if deg in (0, 180) else 0.5 * (1 + 1 / np.tan(th))
    ok = "yes" if 0 <= p <= 1 else "NO"
    print(f" {deg:4d}    {p:>8.3f}                        {ok}")
# Solvable only for 45 <= theta <= 135; and a delayed switch from 90 to 0
# leaves the conspiracy predicting 0.5 where the experiment gives 1.

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Exercises

E1 (easy). A which-way marker leaves the two branches in states with real overlap $s = 0.6$. Compute $P(+z)$, the visibility, and the distinguishability, and verify the duality relation. How often would the best possible guess of the branch be right?

Solution

$P(+z) = (1 + s)/2 = 0.80$. Visibility $V = 2P - 1 = s = 0.60$. Distinguishability $D = \sqrt{1 - s^2} = \sqrt{1 - 0.36} = 0.80$. Check: $D^2 + V^2 = 0.64 + 0.36 = 1$. ✓

The optimal guess succeeds with probability $\tfrac12(1 + D) = 0.90$ — better than the coin flip you would manage with no marker, but not certain, and the shortfall is exactly what buys back 60% of the interference.

E2 (easy). In the erasure setup with the exciter on the $+x$ branch, an experimenter measures the internal state in the $\{\lvert{g}\rangle, \lvert{e}\rangle\}$ basis and finds "excited." What is the atom's spin state, and what does a subsequent $z$ measurement give? Repeat for a marker measurement in the $\{\lvert{+}\rangle_m, \lvert{-}\rangle_m\}$ basis returning $+$.

Solution

Excited. From $\lvert{\Psi}\rangle = \tfrac{1}{\sqrt2}(\lvert{+x, e}\rangle + \lvert{-x, g}\rangle)$, projecting the marker onto $\lvert{e}\rangle$ leaves the spin in $\lvert{+x}\rangle$ — the branch the exciter sat in. A subsequent $z$ measurement is then 50–50, since $|\langle{\pm z}|{+x}\rangle|^2 = \tfrac12$. The which-way information is intact and the interference is gone.

Marker $+$. From the rewritten form $\lvert{\Psi}\rangle = \tfrac{1}{\sqrt2}(\lvert{+z}\rangle\lvert{+}\rangle_m - \lvert{-z}\rangle\lvert{-}\rangle_m)$, projecting onto $\lvert{+}\rangle_m$ leaves the spin in $\lvert{+z}\rangle$ exactly. The $z$ measurement then gives $+$ with probability 1 — full interference restored, at the cost of discarding the half of the runs that returned $-$.

E3 (medium). Show that the reduced density matrix of the spin, after tracing out a marker with overlap $s$, is $\rho = \tfrac12\begin{pmatrix} 1 & s^{*}\\ s & 1\end{pmatrix}$ in the $x$ basis. Compute its purity $\operatorname{Tr}\rho^2$ and relate it to $D$ and $V$. What does purity 1 mean physically?

Solution

Write $\lvert{\Psi}\rangle = \tfrac{1}{\sqrt2}(\lvert{+x, m_+}\rangle + \lvert{-x, m_-}\rangle)$ and trace the marker:

$$\rho = \tfrac12\Bigl(\lvert{+x}\rangle\langle{+x}\rvert + \lvert{-x}\rangle\langle{-x}\rvert + \langle{m_-}|{m_+}\rangle\,\lvert{+x}\rangle\langle{-x}\rvert + \langle{m_+}|{m_-}\rangle\,\lvert{-x}\rangle\langle{+x}\rvert\Bigr),$$

using $\operatorname{Tr}_m\lvert{m_+}\rangle\langle{m_-}\rvert = \langle{m_-}|{m_+}\rangle$. In the $x$ basis that is the stated matrix, with $s = \langle{m_+}|{m_-}\rangle$. Its eigenvalues are $(1 \pm |s|)/2$, so

$$\operatorname{Tr}\rho^2 = \frac{(1+|s|)^2 + (1-|s|)^2}{4} = \frac{1 + |s|^2}{2} = \frac{1 + V^2}{2} = 1 - \frac{D^2}{2} .$$

Purity 1 requires $|s| = 1$, i.e. identical marker states: the marker did not become correlated with the atom at all, the joint state is a product, and the atom is still in a pure state with full interference. Purity $\tfrac12$ (the minimum for a qubit) means orthogonal markers, a maximally mixed atom, and $D = 1$. Purity is the duality relation wearing Term 1's clothes.

E4 (medium). An experimenter runs the delayed-choice erasure, measuring $z$ first and the marker much later, and reports "the late choice changed the earlier data, so information travelled backward in time." Rebut this precisely. Then explain what would change if the de-exciter worked only half the time.

Solution

Rebuttal. The $z$ data are unchanged by the late choice: both marker bases give 50–50 at each $z$ detector, and the recorded list of $\pm$ symbols has the same distribution either way. The spin and marker observables act on different tensor factors, so they commute and the marginal distribution of the spin outcome is independent of the marker measurement — the no-signalling statement of Lesson 3. What the late choice selects is the partition of the already-recorded list into subensembles. Comparing the sorted halves requires the marker outcomes, which are ordinary classical data travelling forward at ordinary speed. Nothing was changed; something was revealed.

Half-efficient de-exciter. After the superpositioner the state is $\tfrac{1}{\sqrt2}(\lvert{+z}, g\rangle - \lvert{-z}, e\rangle)$, so a perfect ground-state filter keeps the $\lvert{+z}\rangle$ half and blocks the other. A filter of efficiency $\varepsilon$ blocks an excited atom only a fraction $\varepsilon$ of the time, so the surviving ensemble carries weight $\tfrac12$ of pure $\lvert{+z}\rangle$ and weight $\tfrac12(1-\varepsilon)$ of pure $\lvert{-z}\rangle$:

$$P(+z) = \frac{\tfrac12}{\tfrac12 + \tfrac12(1-\varepsilon)} = \frac{1}{2 - \varepsilon} .$$

At $\varepsilon = \tfrac12$ this is $\tfrac23$, a visibility of $\tfrac13$ rather than 1. Partial erasure buys partial fringes — but note the schedule is not the linear one of Lesson 1's partial detector: a leaky filter contaminates the sorted subensemble rather than randomizing the coherence, so the two mechanisms degrade differently even though both interpolate between the same endpoints.

E5 (hard). Take the conspiracy theory seriously and audit it completely. (a) Derive $p(\theta) = \tfrac12(1 + \cot\theta)$ and the admissible window from $0 \le p \le 1$. (b) Verify explicitly at $\theta = 45°$ and $\theta = 120°$ that the magic values reproduce the quantum answer. (c) Design the cheapest delayed-choice experiment that refutes the theory even inside the window, and compute both predictions. (d) A last-ditch defender proposes that the branch probabilities track the analyzer angle continuously, updating right up to the moment of measurement. What does that concede, and why is it no longer a local classical theory?

Solution

(a) With the atom on the $+x$ branch, the analyzer at $\theta$ is at relative angle $90° - \theta$, so it fires $+$ with probability $\cos^2(45° - \theta/2)$; on the $-x$ branch it fires $+$ with the complementary $\sin^2(45° - \theta/2)$. Averaging, $P(+) = p\cos^2 u + (1-p)\sin^2 u$ with $u = 45° - \theta/2$. Using $\cos^2 u - \sin^2 u = \sin\theta$ and $\sin^2 u = (1 - \sin\theta)/2$,

$$P(+) = \left(p - \tfrac12\right)\sin\theta + \tfrac12 \stackrel{!}{=} \frac{1 + \cos\theta}{2} \;\Longrightarrow\; p = \frac12\left(1 + \cot\theta\right) .$$

Requiring $0\le p\le1$ gives $|\cot\theta| \le 1$, i.e. $45° \le \theta \le 135°$ on $(0°,180°)$.

(b) At $\theta = 45°$: $\cot 45° = 1$, so $p = 1$ — the atom must always take the $+x$ branch. Then $P(+) = \tfrac12\sin 45° + \tfrac12 = 0.8536 = \cos^2 22.5°$. ✓ At $\theta = 120°$: $\cot 120° = -1/\sqrt3$, so $p = 0.2113$. Then $P(+) = (0.2113 - 0.5)\sin 120° + 0.5 = -0.2887\times0.8660 + 0.5 = 0.25 = \cos^2 60°$. ✓

Note how strained the fits are — at $45°$ the theory is pinned to a deterministic branch choice, and at $135°$ to the opposite one, with no margin left. A theory forced to the boundary of its parameter space is usually about to die.

(c) Set the analyzer at $\theta_{\text{i}} = 90°$ (safely inside the window, forcing $p = \tfrac12$ as the atom crosses the loop), then swing it to $\theta_{\text{f}} = 0°$ after the atom has exited. Quantum: the loop is the identity, the atom is still $\lvert{+z}\rangle$, so $P(+) = 1$. Conspiracy: the branch was chosen with $p = \tfrac12$, and $P(+) = (\tfrac12 - \tfrac12)\sin 0° + \tfrac12 = \tfrac12$. A gap of $0.5$ in probability — E5 of R.1.2 says a couple of dozen atoms settle it. Any $\theta_{\text{f}}$ outside the window works; $0°$ is merely the most extreme.

(d) Letting $p$ track the analyzer angle continuously concedes that the atom's branch choice responds to a setting fixed after it has passed the loop — a setting that can be placed at spacelike separation from the branch choice. The "hidden variable" is then not carried by the atom at all: it is an instantaneous dependence on a distant, later fact. That is no longer a local classical theory; it is nonlocality with extra steps, and it has given up the one property that made the classical picture attractive. This is the same escape hatch that reappears in Lesson 3 as the setting-dependence loophole, and it is why Bell's theorem is stated for local hidden variables: nonlocal ones were never in contention.

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Checkpoint

1. State complementarity in its which-way form. Why was the "disturbance" explanation abandoned?
2. For a marker with overlap $s$, what are $V$ and $D$, and what does $D^2 + V^2 \le 1$ say that Bohr's slogan does not?
3. Cover the detector readouts so no human can see them. Does interference return? Give the one-line argument.
4. In the erasure experiment, which marker basis destroys the fringes and which restores them? Why do the unsorted data show nothing either way?
5. What does the delayed-choice experiment rule out, and what is the admissible window for the conspiracy theory before delayed choice is invoked?

Answers

1. If a record exists anywhere from which the path could be determined, the object behaves as though it took one path; if no such record exists, the paths interfere. Disturbance was abandoned because markers can be made arbitrarily gentle — an internal state flip carrying negligible momentum — and the interference dies anyway. What matters is correlation, not force.
2. $V = |s|$ and $D = \sqrt{1-|s|^2}$. The inequality makes complementarity quantitative: you may have partial knowledge and partial interference, but only along a fixed exchange rate, with the pure-marker case saturating the circle. Bohr's version is the two endpoints; this is the whole curve.
3. No. The atom's predicted statistics are $\rho = \operatorname{Tr}_m\lvert{\Psi}\rangle\langle{\Psi}\rvert$, whose coherences are suppressed by $|s|$ from the moment the correlation forms; the partial trace contains no variable for who is looking.
4. The $\{\lvert{g}\rangle, \lvert{e}\rangle\}$ basis keeps the which-way information and leaves both subensembles 50–50; the $\{\lvert{+}\rangle_m, \lvert{-}\rangle_m\}$ basis erases it and sorts the runs into pure $+z$ and pure $-z$ halves. The unsorted data are the sum of the two halves, which is 50–50 in both cases — erasure sorts, it does not create.
5. It rules out every theory in which the unwatched atom travels one definite branch, even when the branch probabilities are allowed to depend on the apparatus setting, because the setting can be changed after the atom has passed. Before delayed choice, such a theory survives only for $45° \le \theta \le 135°$, where $p(\theta) = \tfrac12(1+\cot\theta)$ lies in $[0,1]$.

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Further Reading

• [Fre] J. K. Freericks, Quantum Mechanics Done Right, §2.2 — complementarity, the conspiracy-theory calculation, tagging with excited states, and the delayed-choice eraser this lesson follows.
• [NC] Nielsen & Chuang, §§2.4 and 8.3 — density operators, the partial trace, and decoherence as the loss of off-diagonal elements.
• [Mer] Mermin, Quantum Computer Science, §1.10 — measurement, entanglement with a pointer, and why the pointer's basis is a choice.
• [Gri] Griffiths & Schroeter, §12.1 — the EPR/complementarity discussion in wave-mechanics language.
• [Pre] Preskill, Ph219 Ch. 3 — the density operator, distinguishability of quantum states, and the Helstrom bound used for $D$.
• B.-G. Englert, "Fringe visibility and which-way information: an inequality", Phys. Rev. Lett. 77, 2154 (1996); D. Greenberger & A. YaSin, Phys. Lett. A 128, 391 (1988) — the duality relation derived here.
• The Analyzer Loop — the apparatus, and the armed-or-idle detector this lesson places on the duality circle.
• P.3.2 Matter as Waves — the double slit, whose fringes fade by exactly this mechanism.
• 1.5.1 The Density Operator and 1.5.2 Partial Trace & Reduced States — the formal machinery behind $\rho = \operatorname{Tr}_m\lvert{\Psi}\rangle\langle{\Psi}\rvert$.

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