# The Analyzer Loop > Bolt a Stern-Gerlach analyzer to its own mirror image and you get a machine that provably does > nothing: atoms come out exactly as they went in. That is the whole point. Once you own a device > that can be switched between *doing nothing* and *doing something* by dropping a shutter into a > beam you never look at, you can ask nature questions that a plain analyzer cannot even phrase. ## Learning Objectives After this lesson you will be able to: 1. Construct the **analyzer loop** from a Stern-Gerlach front end and an eraser back end, and prove from completeness that the open loop acts as the identity operator. 2. Predict the output of a loop for each gate configuration, and explain why blocking one branch gives $\tfrac12$ where leaving both open gives $1$. 3. Analyze a **partially working pass-through detector**, derive the interpolation $P(+z) = 1 - \eta/2$, and connect it to the coherence of the marker states. 4. Work three-analyzer-loop cascades systematically, tracking both the exit state and the surviving fraction at every stage. 5. Explain **interaction-free measurement** — how adding a blocker can *increase* the number of atoms reaching the detector — and compute its efficiency. 6. State clearly which of these results are thought experiments and which have been realized in the laboratory, and why the distinction does not weaken the argument. --- ## Intuition Course R.1 ended with a postulate: measuring an atom projects its state, and the projection is not a passive reading of something already there. That leaves a natural question hanging. If a Stern-Gerlach analyzer *destroys* the atom's memory of the previous axis, is the destruction caused by the physical violence of the magnet — the shove, the gradient, the tug on the magnetic moment — or by something more abstract, namely that the two beams have been *separated* in a way that would let someone read which is which? The two hypotheses can be told apart, and the device that tells them apart is beautifully simple. Take a normal analyzer, which splits an incoming beam into two spatially separated branches. Now attach a second magnet assembly, geometrically the mirror image of the first, that bends the two branches back together until they overlap and continue down a single pipe. Every atom has felt the full violence of the field gradient — twice, in fact. But the *separation* has been undone, and with it any possibility of telling which branch an atom used. The prediction is stark: the composite device restores the input state exactly. An atom prepared $\lvert{+z}\rangle$ that is sent through an $x$-oriented loop comes out $\lvert{+z}\rangle$, not 50–50. Turning on a strong magnet, splitting a beam in two, and bending it back has accomplished literally nothing. That "nothing" is the most useful object in this chapter. Because now put a shutter into one of the two branches, at the point where they are farthest apart. The device that did nothing now does something — it becomes a measurement — and the only thing that changed is the *availability of which-branch information*. Everything in the rest of this course grows out of comparing those two configurations [Fre, §2.1]. > **An honest warning up front.** No one has built an atomic Stern-Gerlach analyzer loop of the kind > drawn below; recombining two beams with the phase coherence this demands is brutally hard, and the > classic experiments in this family are done with light or with neutrons instead. These are *thought > experiments*, and their outcomes are computed from rules already tested to many decimal places > rather than read off a plate. That said, the ingredients are real: Frisch and Segrè studied > coherent spin transport between Stern-Gerlach stages in the 1930s, atom-chip interferometers have > since realized recombining spin interferometers in a different geometry (Machluf et al., 2013), and > the same circuits run on quantum computers today and behave exactly as described. We will flag the > thought experiments as we go and never lean on one where a real result exists. --- ## Theory ### Building the loop A Stern-Gerlach analyzer along $\hat{n}$ takes one input beam and produces two spatially separated output beams, labelled $+$ and $-$. Reverse the sign of the gradient in a second, identical magnet and the two beams bend back toward each other; a third stage removes the residual transverse velocity so that the beams merge and travel on together. The composite is the **analyzer loop**: one input port, one output port, and a region in the middle where the atom's two possible branches are macroscopically far apart. The back half is often called a **quantum eraser**, because it undoes exactly the separation the front half created. ```mermaid flowchart LR IN["input beam
state |psi>"] --> SPLIT["SG front end
axis n"] SPLIT -->|"+ branch"| G1{"gate +
open / closed"} SPLIT -->|"- branch"| G2{"gate -
open / closed"} G1 --> REC["eraser back end
recombine"] G2 --> REC REC --> OUT["output beam
single pipe"] ``` Two more pieces of equipment complete the toolkit. A **gate** is a shutter that can be dropped into either branch, absorbing every atom that takes it. A **pass-through detector** sits in a branch and registers an atom going by without stopping it; its inert twin, a plain pipe, does nothing at all. The gates are what turn the loop from a null device into an experiment. Finally, a **polarized source**: a random oven followed by a loop with exactly one gate closed. Every atom that survives has been forced down a single known branch, so it emerges in a known state. This is the source we use from here on, drawn simply as an arrow, and it is nothing but an analyzer whose unwanted output has been thrown away. ### The open loop is the identity Why does the open loop do nothing? Not by decree — by arithmetic, and the arithmetic is one line. Let the loop be oriented along $\hat{n}$, with branch states $\lvert{+\hat n}\rangle$ and $\lvert{-\hat n}\rangle$. An atom entering in state $\lvert{\psi}\rangle$ has amplitude $\langle{+\hat n}|{\psi}\rangle$ to take the upper branch and $\langle{-\hat n}|{\psi}\rangle$ to take the lower one. If the recombination is coherent — if it introduces no relative phase and leaves no record anywhere of which branch was used — the output amplitude is the *sum* over the two routes, and the loop acts as the operator $$ U_{\text{loop}} = \lvert{+\hat n}\rangle\langle{+\hat n}\rvert + \lvert{-\hat n}\rangle\langle{-\hat n}\rvert = P_+^{\hat n} + P_-^{\hat n} = \mathbb{1} . $$ That is the **completeness relation** for an orthonormal basis, $\sum_i \lvert{i}\rangle\langle{i}\rvert = \mathbb{1}$, and it is the entire content of "the loop does nothing." Note what it does *not* depend on: the orientation $\hat n$, the incoming state, or the strength of the field. Any complete set of branches, recombined coherently, sums to the identity. Lesson R.1.3 already met this calculation as the interference of two indistinguishable routes; the loop is that interferometer built out of magnets. Close one gate and the sum loses a term. With the $-$ branch blocked the device implements $P_+^{\hat n}$ — not a unitary but a projector, which both attenuates the beam and re-prepares the survivors: $$ \text{both open} \Rightarrow \mathbb{1}, \qquad \text{only } + \text{ open} \Rightarrow P_+^{\hat n}, \qquad \text{only } - \text{ open} \Rightarrow P_-^{\hat n}, \qquad \text{both closed} \Rightarrow 0 . $$ ### The $\tfrac12$ versus $1$ experiment Now the experiment that makes the point. Prepare $\lvert{+z}\rangle$, send it through an $x$-oriented loop, and finish with a vertical analyzer. *Block the $-x$ branch.* Half the atoms are absorbed. The survivors are in $\lvert{+x}\rangle$, and the final analyzer splits them evenly: $P(+z) = |\langle{+z}|{+x}\rangle|^2 = \tfrac12$. Blocking the $+x$ branch gives the same $\tfrac12$, for the mirror-image reason. *Open both branches.* Now $U_{\text{loop}} = \mathbb{1}$, the atom leaves the loop in $\lvert{+z}\rangle$, and *every* atom exits the $+z$ port: $P(+z) = 1$. The tempting classical story — the atom really did take one branch or the other, we simply do not know which, so we should average the two blocked results — predicts $\tfrac12\cdot\tfrac12 + \tfrac12\cdot\tfrac12 = \tfrac12$. It is wrong, and it is worth being precise about *where* it goes wrong. Its arithmetic is unimpeachable; its premise is not. Adding the two blocked outcomes with classical weights is exactly the step that assumes a fact of the matter about the branch. Since the conclusion is false, the premise is false: an unwatched atom in an open loop does not take one branch or the other. Physicists say the atom is in a **superposition** of the two branches, or, more cautiously, that the branch is *indeterminate* — language chosen to describe the amplitude bookkeeping without pretending to a mechanical picture underneath it. Freericks makes the same point with a comparison worth keeping: for a decade a very popular sitcom advertised itself as a show about nothing, and the analyzer loop is an apparatus about nothing. What we learn from it comes entirely from what it *would* have done had a gate been closed [Fre, §2.1]. ### Watching the branches If the atom "goes both ways," we should be able to see it doing so. Replace the gates with pass-through detectors and look. Two facts come back, and they are both worth staring at. First, the detectors never both fire. An atom is registered on one branch or the other, always whole, never half on each; nothing splits. Second — and this is the price — the final analyzer now reads 50–50 instead of 100%. Watching the branches restored the classical prediction exactly. Replace the detectors with plain pipes and the 100% comes back. Now the sharpest version. Put a detector on the $+x$ branch only and a pipe on the $-x$ branch. The result is 50–50 again, *even for the atoms the detector never registered*, because a silent detector is as informative as a loud one: if it did not fire, the atom went the other way. What matters is not disturbance, not photons absorbed, not momentum kicks — it is whether a record exists anywhere in the world from which the branch could be inferred. The next lesson makes that statement quantitative and gives it a name. ### The partially working detector The interpolation between the two regimes is instructive, and it is the substance of the first of Freericks' problems [Fre, §2.1.1]. Keep the pipe on the $-x$ branch and put a detector on the $+x$ branch whose **efficiency** $\eta$ is adjustable: a fraction $\eta$ of the time it is armed and will register an atom on that branch, and a fraction $1-\eta$ of the time it is idle and records nothing at all. Treat the two cases with compound probability, exactly as in R.1.3. When the detector is armed, the branch is knowable and the classical analysis holds, giving $P(+z) = \tfrac12$. When it is idle, nothing anywhere records the branch, the loop is the identity, and $P(+z) = 1$. Averaging, $$ P(+z) = \eta\cdot\frac12 + (1-\eta)\cdot 1 = 1 - \frac{\eta}{2} . $$ At $\eta = 0$ we recover 100%, at $\eta = 1$ we recover 50%, and in between the answer slides smoothly. It is convenient to quote the result as a **visibility**, the fractional swing of the interference away from the flat 50–50 background, $$ V \equiv 2P(+z) - 1 = 1 - \eta , $$ so the visibility falls off linearly as the detector is armed more often. There is no threshold, no sudden collapse: partial which-way information buys partial loss of interference, and the trade is continuous. Lesson 2 shows that this linear law is a special case of a general inequality relating how much you know to how much interference survives. > **Careful.** This "armed or idle" detector is a *classical mixture of two experiments*, and it is > not the only way a detector can be inefficient. A marker that couples weakly to the atom on every > shot — leaving two non-orthogonal marker states rather than sometimes-orthogonal ones — degrades > the interference by a different mechanism and is the case treated in Lesson 2. The two models agree > on the endpoints and on the shape of the answer, and they differ in what a single shot means. Do > not conflate them. ### Cascaded loops, and the surviving fraction Chaining loops is where the bookkeeping starts to pay off. The rules are mechanical: 1. A loop with both gates open contributes $\mathbb{1}$: skip it entirely. 2. A loop with one gate closed contributes the projector onto the open branch. The surviving fraction is $|\langle{\text{open branch}}|{\text{incoming}}\rangle|^2$, and every survivor emerges in the open-branch state. 3. A loop with both gates closed contributes $0$. Take the standard crossed arrangement: a $\lvert{+z}\rangle$ source, then an $x$-oriented loop **A**, then a $z$-oriented loop **B** [Fre, §2.1]. | Gates closed | Analysis | Survivors | Exit state | | --- | --- | --- | --- | | none | both loops are the identity | 100% | $\lvert{+z}\rangle$ | | A's $-x$ | project onto $\lvert{+x}\rangle$, then B does nothing | 50% | $\lvert{+x}\rangle$ | | B's $-z$ | A does nothing, project onto $\lvert{+z}\rangle$ | 100% | $\lvert{+z}\rangle$ | | B's $+z$ | A does nothing, project $\lvert{+z}\rangle$ onto $\lvert{-z}\rangle$ | **0%** | — | | A's $-x$ and B's $+z$ | project onto $\lvert{+x}\rangle$ (50%), then onto $\lvert{-z}\rangle$ (50%) | **25%** | $\lvert{-z}\rangle$ | Read the last two rows again. With one shutter in the beam, nothing gets through. Add a *second* shutter and a quarter of the atoms come out. **Adding an obstruction increased the throughput from zero to 25%.** Nothing about that sentence is possible classically. A blocker can only remove atoms from a beam; it cannot conjure them. The resolution is that the second blocker did not add atoms, it destroyed an interference — the perfect cancellation that was keeping the $-z$ port dark. In operator language, $P_-^z P_+^x \ne 0$ while $P_-^z\,\mathbb{1} \lvert{+z}\rangle = 0$; the shutter in loop A replaced the identity with a projector, and the projector has a component the identity did not. ### Three loops in a row Freericks' §2.1.2 puts three loops in a row — $x$-oriented **A**, $z$-oriented **B**, $x$-oriented **C** — fed by a $\lvert{+z}\rangle$ source, and asks for the survivors and the exit state in six gate configurations. The mechanics are the three rules above applied left to right. Working them all is the fastest way to become fluent, so here they are in full. ```mermaid flowchart LR SRC["polarized source
|+z>"] --> A["loop A
x-oriented"] A --> B["loop B
z-oriented"] B --> C["loop C
x-oriented"] C --> DET["final detector"] ``` | Gates closed | Survivors | Exit state | | --- | --- | --- | | A's $+x$ | 50% | $\lvert{-x}\rangle$ | | C's $-x$ | 50% | $\lvert{+x}\rangle$ | | B's $+z$ | 0% | — | | B's $-z$ and C's $+x$ | 50% | $\lvert{-x}\rangle$ | | A's $-x$ and C's $+x$ | **0%** | — | | A's $-x$, B's $+z$ and C's $+x$ | **12.5%** | $\lvert{-x}\rangle$ | The third row is the same extinction as before: B's open branch is $\lvert{-z}\rangle$, the incoming state is $\lvert{+z}\rangle$, and $\langle{-z}|{+z}\rangle = 0$. The fifth row is the mirror statement one axis over: A leaves the atom in $\lvert{+x}\rangle$ and C's only open branch is $\lvert{-x}\rangle$, again orthogonal, again dark. The last row is the punchline. Start from the dark configuration of row five and add a *third* shutter, inside loop B. Now the chain is $\lvert{+z}\rangle \to \lvert{+x}\rangle$ (50%) $\to \lvert{-z}\rangle$ (another 50%) $\to \lvert{-x}\rangle$ (another 50%), so one atom in eight arrives — at a detector that was receiving *none* before the extra blocker was installed. ### Quantum seeing in the dark Turn that last observation into an instrument and it becomes something genuinely useful [Fre, §2.1.3]. Set up a $\lvert{+z}\rangle$ source, an $x$-oriented loop with both gates open, and a final vertical analyzer. With the loop clear, every atom exits the $+z$ port and the $-z$ port is *perfectly* dark — not "rarely visited," but forbidden, because the amplitudes for the two branches cancel exactly. Now suppose someone may or may not have placed an absorber in the $-x$ branch, and we want to find out which, using atoms, without any atom touching the absorber. Send one atom and enumerate: - with probability $\tfrac12$ the atom takes the $-x$ branch and is absorbed — we learn the absorber is there, but the hard way; - with probability $\tfrac12$ it takes the $+x$ branch and survives, emerging in $\lvert{+x}\rangle$. The final analyzer then sends it to the $+z$ port with probability $\tfrac12$ and to the $-z$ port with probability $\tfrac12$. So the three outcomes have probabilities $\tfrac12$ (absorbed), $\tfrac14$ ($+z$ click) and $\tfrac14$ ($-z$ click). A $+z$ click is inconclusive — it happens whether or not the absorber is present. But a $-z$ click is *impossible* with the branch clear. Every $-z$ click therefore proves an absorber is sitting in the $-x$ branch, and proves it with an atom that demonstrably went the other way and never came near it. This is an **interaction-free measurement**, proposed by Elitzur and Vaidman in 1993 and often called "quantum seeing in the dark." Its efficiency in this simplest form is $$ P(\text{interaction-free detection}) = \frac14 , $$ or $\tfrac12$ conditioned on the atom surviving at all. The usual figure of merit compares interaction-free detections against the *conclusive* outcomes only, $P(\text{detect})/[P(\text{detect}) + P(\text{absorbed})] = \tfrac13$ here, and it can be improved: an unbalanced loop that sends only a small amplitude down the risky branch raises it to $\tfrac12$ in the limit, and a Zeno-style scheme that repeats a weak version of the test $N$ times drives it toward 1 as $N\to\infty$. Kwiat and collaborators demonstrated the effect with single photons in 1995, and that experiment — unlike the atomic loop drawn here — is real hardware. What makes it possible is precisely the fact established at the top of this lesson. A dark port is dark because of a cancellation between two amplitudes, and an object in *either* branch spoils the cancellation whether or not it is ever struck. The atom does not need to interact with the absorber; the *amplitude* does. --- ## Hands-on (Python) Every result above is a product of $2\times2$ projectors, so we can check them all in a few lines — including the ones the table asserts without derivation. ```python import numpy as np ket_pz, ket_mz = np.array([1, 0], complex), np.array([0, 1], complex) def branch(theta, sign=+1): """Unit vector for the +/- branch of a loop at angle theta (radians).""" t = theta if sign > 0 else theta + np.pi return np.array([np.cos(t / 2), np.sin(t / 2)], complex) def proj(theta, sign=+1): v = branch(theta, sign) return np.outer(v, v.conj()) def loop(theta, open_plus=True, open_minus=True): """Operator implemented by an analyzer loop with the given gates open.""" M = np.zeros((2, 2), complex) if open_plus: M += proj(theta, +1) if open_minus: M += proj(theta, -1) return M Z, X = 0.0, np.pi / 2 # --- 1. The open loop is the identity, at any orientation ----------------- for name, th in (("z", Z), ("x", X), ("37 deg", np.deg2rad(37))): print(f"open loop ({name:6s}) == identity :", np.allclose(loop(th), np.eye(2))) # --- 2. The 1/2 vs 1 experiment ------------------------------------------ def outcomes(op, psi=ket_pz): """(fraction surviving the loop, P(+z) among the survivors).""" out = op @ psi survive = np.vdot(out, out).real joint = abs(np.vdot(ket_pz, out)) ** 2 # |<+z| M |psi>|^2 return survive, joint / survive print() for label, op in (("-x blocked", loop(X, True, False)), ("+x blocked", loop(X, False, True)), ("both open ", loop(X))): survive, cond = outcomes(op) print(f"x-loop, {label} : survive = {survive:.3f} P(+z | survived) = {cond:.3f}") # --- 3. Partially working detector: P(+z) = 1 - eta/2 -------------------- print("\n eta P(+z) visibility") for eta in (0.0, 0.25, 0.5, 0.75, 1.0): p = eta * 0.5 + (1 - eta) * 1.0 print(f" {eta:.2f} {p:.3f} {2 * p - 1:.3f}") # --- 4. Three-loop cascades (Freericks 2.1.2) ----------------------------- def cascade(gates): """gates = [(theta, open_plus, open_minus), ...] applied left to right.""" psi, frac = ket_pz.copy(), 1.0 for th, op, om in gates: psi = loop(th, op, om) @ psi n2 = np.vdot(psi, psi).real frac *= n2 if n2 > 1e-12: psi /= np.sqrt(n2) return frac, psi cases = { "A +x closed ": [(X, False, True), (Z, True, True), (X, True, True)], "C -x closed ": [(X, True, True), (Z, True, True), (X, True, False)], "B +z closed ": [(X, True, True), (Z, False, True), (X, True, True)], "B -z, C +x closed ": [(X, True, True), (Z, True, False), (X, False, True)], "A -x, C +x closed ": [(X, True, False), (Z, True, True), (X, False, True)], "A -x, B +z, C +x closed ": [(X, True, False), (Z, False, True), (X, False, True)], } names = {"+z": branch(Z, +1), "-z": branch(Z, -1), "+x": branch(X, +1), "-x": branch(X, -1)} print("\nthree-loop cascades (A: x, B: z, C: x), source |+z>") for label, g in cases.items(): frac, psi = cascade(g) tag = "-" if frac < 1e-12 else max(names, key=lambda k: abs(np.vdot(names[k], psi))) print(f" {label} survivors = {frac:6.1%} exit = {tag}") # Note the last two lines: blocking A's -x AND C's +x extinguishes the beam, # but ADDING a third blocker inside loop B lets 12.5% through. # --- 5. Interaction-free measurement ------------------------------------- p_absorbed = abs(np.vdot(branch(X, -1), ket_pz)) ** 2 psi_survive = branch(X, +1) p_minus_z = abs(np.vdot(ket_mz, psi_survive)) ** 2 * (1 - p_absorbed) print(f"\nabsorbed = {p_absorbed:.3f}") print(f"interaction-free detection = {p_minus_z:.3f}") print(f" ... conditioned on survival = {p_minus_z / (1 - p_absorbed):.3f}") ``` Running it prints `True` for every orientation in part 1; in part 2, a blocked branch gives `survive = 0.500, P(+z | survived) = 0.500` while both branches open give `1.000` and `1.000`; part 3 prints the linear visibility ramp; part 4 prints the six cascade rows, ending `0.0%` and then `12.5%`; and part 5 prints `0.500 / 0.250 / 0.500`. --- ## Exercises **E1 (easy).** An atom in state $\lvert{+z}\rangle$ enters an analyzer loop oriented at $37°$ from vertical, with both gates open, and is then measured along $z$. What is $P(+z)$? Now repeat with the loop oriented at $90°$ and at $180°$. What general statement do the three answers support?

Solution

$P(+z) = 1$ in all three cases. With both gates open the loop implements $P_+^{\hat n} + P_-^{\hat n} = \mathbb{1}$ for *any* axis $\hat n$, because the two branch states form an orthonormal basis of $\mathbb{C}^2$ and completeness does not care which basis. The atom therefore leaves in $\lvert{+z}\rangle$ and the final analyzer confirms it with certainty. The general statement: an open loop is the identity regardless of its orientation, its field strength, or the state fed into it. This is why the device is worth building — it gives us a way to split a beam in two without measuring anything, so that the *only* difference between the open and blocked configurations is the availability of which-branch information.

**E2 (easy).** A $\lvert{+z}\rangle$ beam passes through an $x$-oriented loop with the $-x$ gate closed, then through a $z$-oriented loop with the $-z$ gate closed. What fraction of the original beam emerges, and in what state? Repeat with the second loop's $+z$ gate closed instead.

Solution

*First case.* Loop A projects onto $\lvert{+x}\rangle$, surviving fraction $|\langle{+x}|{+z}\rangle|^2 = \tfrac12$. Loop B then projects onto $\lvert{+z}\rangle$, surviving fraction $|\langle{+z}|{+x}\rangle|^2 = \tfrac12$. Total $\tfrac12\times\tfrac12 = 25\%$, emerging in $\lvert{+z}\rangle$. *Second case.* Loop A gives the same $\tfrac12$ and $\lvert{+x}\rangle$. Loop B now projects onto $\lvert{-z}\rangle$, with $|\langle{-z}|{+x}\rangle|^2 = \tfrac12$. Total again $25\%$, this time in $\lvert{-z}\rangle$. The symmetry is the point: once loop A has erased the $z$ information, the two $z$ branches are on exactly equal footing, and which one you keep no longer matters for the count.

**E3 (medium).** In the partially working detector setup, suppose you measure $P(+z) = 0.82$ over a long run. What is the detector's efficiency $\eta$? How many atoms must you send to determine $\eta$ to $\pm 0.02$ at one standard error?

Solution

Invert $P(+z) = 1 - \eta/2$: $\eta = 2(1 - P) = 2(1 - 0.82) = 0.36$. For the shot budget, use the binomial standard error from R.1.3. With $p = 0.82$, $\sigma_{\hat p} = \sqrt{p(1-p)/N} = \sqrt{0.1476/N}$. Since $\eta = 2(1-\hat p)$, errors double: $\sigma_\eta = 2\sigma_{\hat p}$. Requiring $\sigma_\eta \le 0.02$ gives $$ 2\sqrt{\frac{0.1476}{N}} \le 0.02 \quad\Longrightarrow\quad N \ge \frac{4 \times 0.1476}{0.0004} = 1476 . $$ About 1500 atoms — modest, because the quantity being estimated swings over a full factor of two between the extreme configurations. Note the factor-of-two error amplification: quoting the answer as an efficiency rather than a probability costs you a factor of 4 in shots for the same precision.

**E4 (medium).** Prove the extinction in the crossed-loop experiment as an operator identity: show that a $\lvert{+z}\rangle$ beam through an open $x$-loop followed by a $z$-loop with the $+z$ gate closed gives exactly zero, and that closing A's $-x$ gate as well gives exactly $\tfrac14$. Explain in one sentence why "adding a blocker admits more atoms" is not a violation of anything.

Solution

*Both A gates open.* The composite operator acting on the input is $P_-^{z}\,\mathbb{1}\,\lvert{+z}\rangle = P_-^{z}\lvert{+z}\rangle = \lvert{-z}\rangle\langle{-z}|{+z}\rangle = 0$, since the $z$ basis is orthonormal. Not "small" — exactly zero, for every atom. *A's $-x$ gate closed.* Now the composite is $P_-^{z}P_+^{x}\lvert{+z}\rangle$. Step by step, $P_+^x\lvert{+z}\rangle = \lvert{+x}\rangle\langle{+x}|{+z}\rangle = \tfrac{1}{\sqrt2}\lvert{+x}\rangle$, then $P_-^z\left(\tfrac{1}{\sqrt2}\lvert{+x}\rangle\right) = \tfrac{1}{\sqrt2}\lvert{-z}\rangle\langle{-z}|{+x}\rangle = \tfrac12 \lvert{-z}\rangle$. The surviving probability is $\lVert\tfrac12\lvert{-z}\rangle\rVert^2 = \tfrac14$. ✓ *Why it is not paradoxical.* The blocker did not create atoms; it destroyed the destructive interference between the two $x$ branches that was cancelling all amplitude at the $-z$ port, and once one of the two cancelling terms is removed the other is free to contribute. Classically a shutter can only subtract because classical intensities are non-negative and simply add; quantum amplitudes are complex and can cancel, so removing one can raise a total.

**E5 (hard).** Design an interaction-free measurement with better than $\tfrac14$ efficiency. Consider a loop whose front end sends amplitude $\cos\alpha$ down the safe branch and $\sin\alpha$ down the branch that may contain the absorber, recombining coherently as usual, followed by an analyzer that distinguishes the input state from its orthogonal partner. **(a)** Compute, as functions of $\alpha$, the probabilities of absorption, of an inconclusive result, and of an interaction-free detection. **(b)** Find the limiting efficiency as $\alpha \to 0$, both unconditionally and conditioned on survival, and explain the trade-off. **(c)** Sketch how repeating a weak test $N$ times can push the efficiency toward 1.

Solution

**(a)** Write the input as $\lvert{\psi}\rangle$ and the two branch states as $\lvert{s}\rangle$ (safe) and $\lvert{r}\rangle$ (risky), with $\lvert{\psi}\rangle = \cos\alpha\lvert{s}\rangle + \sin\alpha\lvert{r}\rangle$. *Absorber present.* The risky component is removed, with probability $\sin^2\alpha$ of absorbing the atom. The surviving (unnormalized) state is $\cos\alpha\lvert{s}\rangle$. The final analyzer asks whether the atom is still $\lvert{\psi}\rangle$ or has become $\lvert{\psi^\perp}\rangle$. Using $\langle{\psi}|{s}\rangle = \cos\alpha$ and $\langle{\psi^\perp}|{s}\rangle = -\sin\alpha$: $$ P(\text{absorbed}) = \sin^2\alpha, \qquad P(\text{inconclusive}) = \cos^4\alpha, \qquad P(\text{detected}) = \cos^2\alpha\sin^2\alpha . $$ These sum to $\sin^2\alpha + \cos^2\alpha(\cos^2\alpha + \sin^2\alpha) = 1$. ✓ With $\alpha = 45°$ we recover the balanced case: $\tfrac12$, $\tfrac14$, $\tfrac14$. *Absorber absent.* The loop is the identity, the atom stays $\lvert{\psi}\rangle$, and the "$\lvert{\psi^\perp}\rangle$" port is dark — which is what makes a detection conclusive. **(b)** Unconditionally, $P(\text{detected}) = \cos^2\alpha\sin^2\alpha = \tfrac14\sin^2 2\alpha$ is *maximized* at $\alpha = 45°$, giving the familiar $\tfrac14$; as $\alpha\to0$ it vanishes like $\alpha^2$. Conditioned on the atom surviving, $$ \frac{P(\text{detected})}{P(\text{detected}) + P(\text{inconclusive})} = \frac{\cos^2\alpha\sin^2\alpha}{\cos^2\alpha\left(\sin^2\alpha + \cos^2\alpha\right)} = \sin^2\alpha , $$ which also shrinks as $\alpha\to0$. So by *both* of those measures a weak probe is worse: it makes each atom unlikely to be absorbed and equally unlikely to reveal anything, one for one. The standard figure of merit conditions instead on getting a *conclusive* answer — a detection or an absorption, discarding the inconclusive runs, which cost nothing but an atom: $$ \frac{P(\text{detected})}{P(\text{detected}) + P(\text{absorbed})} = \frac{\cos^2\alpha\sin^2\alpha}{\cos^2\alpha\sin^2\alpha + \sin^2\alpha} = \frac{\cos^2\alpha}{1 + \cos^2\alpha} \;\xrightarrow[\alpha\to0]{}\; \frac12 , $$ against $\tfrac13$ at $\alpha = 45°$. That is the sense in which an unbalanced loop helps: it does not find the absorber more often per atom, but of the times it settles the question, it settles it non-destructively half the time instead of a third. **(c)** The escape is to make the small effect *accumulate coherently* while the absorption stays second order. Split the rotation into $N$ steps of angle $\alpha = \pi/2N$, with a chance for the absorber to act after each. With the absorber *absent*, the $N$ small rotations compose into a full $90°$ rotation, so the atom ends in the orthogonal state with certainty. With the absorber *present*, each step is reset — the risky amplitude is either absorbed (probability $\sin^2\alpha \approx \alpha^2$) or projected back onto the safe branch — so after $N$ steps the atom survives with probability $(\cos^2\alpha)^N \to 1 - \pi^2/4N \to 1$ and is found *unrotated*, which is conclusive. Total failure probability falls like $1/N$: $N$ chances at $\alpha^2 = (\pi/2N)^2$ each. This is exactly the staircase of R.1.4 E5 run in reverse, using the quantum Zeno effect to make the presence of the absorber freeze the state; Kwiat and collaborators implemented it optically and reached efficiencies well above $\tfrac12$.

--- ## Checkpoint 1. Why is the open analyzer loop equal to the identity operator, and on which property of the branch states does the argument rest? 2. State the outcome of the $\lvert{+z}\rangle \to$ ($x$-loop) $\to$ ($z$-analyzer) experiment for each of the four gate configurations, and identify the false premise in the classical averaging argument. 3. A detector sits on one branch only and fires half the time it should. What is $P(+z)$, and why does a *silent* detector still count as which-way information? 4. In the crossed-loop experiment, which configuration is dark and which configuration becomes bright when an *extra* blocker is added? Explain the mechanism in terms of amplitudes. 5. Describe the interaction-free measurement, state its efficiency in the balanced case, and say precisely which detector click certifies the absorber.

Answers

1. Because the coherent sum over the two routes is $\lvert{+\hat n}\rangle\langle{+\hat n}\rvert + \lvert{-\hat n}\rangle\langle{-\hat n}\rvert$, which is the completeness relation for an orthonormal basis and therefore $\mathbb{1}$. The argument needs only that the two branch states are orthonormal and span the space — it is independent of the loop's orientation and of the input state. 2. Both open: $P(+z) = 1$. Only $+x$ open: $\tfrac12$ (of the 50% that survive). Only $-x$ open: $\tfrac12$ likewise. Both closed: no output. The classical argument averages the two blocked results with weights $\tfrac12$ and gets $\tfrac12$; its false premise is that the atom took one branch or the other when nothing recorded which. 3. $P(+z) = 1 - \eta/2 = 1 - 0.25 = 0.75$, with visibility $V = 1 - \eta = \tfrac12$. A silent detector is informative because there are only two branches: no click on the watched branch means the atom went down the other one, so the record exists either way. 4. Closing only loop B's $+z$ gate is dark — the open branch is $\lvert{-z}\rangle$ and $\langle{-z}|{+z}\rangle = 0$. Also closing loop A's $-x$ gate makes 25% get through. The extra blocker removed one of the two amplitudes whose cancellation was keeping the $-z$ port dark, so the survivor is no longer cancelled. 5. Send $\lvert{+z}\rangle$ through an open $x$-loop into a $z$-analyzer. With the branches clear the $-z$ port is forbidden; with an absorber in the $-x$ branch it fires with probability $\tfrac14$ ($\tfrac12$ conditioned on survival). A $-z$ click certifies the absorber, and the atom that produced it went down the *other* branch and never touched it.

--- ## Further Reading - **[Fre]** J. K. Freericks, *Quantum Mechanics Done Right*, §2.1 — the analyzer loop, the crossed- and three-loop gate tables, and the "quantum seeing in the dark" problem this lesson follows. - **[Sak]** Sakurai & Napolitano, §1.1 — the same "recombine the beams" argument used to motivate the completeness relation, in the canonical treatment. - **[Mer]** Mermin, *Quantum Computer Science*, Ch. 1 — projectors, measurement, and the operator bookkeeping used for the cascades here. - **[NC]** Nielsen & Chuang, §§2.2.3–2.2.5 — projective measurement and the distinction between a coherent branch and a recorded one. - A. Elitzur & L. Vaidman, "Quantum mechanical interaction-free measurements", *Found. Phys.* **23**, 987 (1993); P. Kwiat et al., *Phys. Rev. Lett.* **74**, 4763 (1995) — the proposal and its optical realization, including the Zeno-enhanced version of E5. - [The Conundrum of Projections](../c01-freericks-intro-quantum-world/04-conundrum-of-projections.md) — the projection postulate and the Wigner sequence formula this lesson runs on. - [Probability in the Quantum World](../c01-freericks-intro-quantum-world/03-probability-quantum-world.md) — the amplitude-versus-probability rule, in its two-route form. - [1.3.1 Projective Measurement](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md) — the formal version of the projector algebra used throughout. --- ← Prev: [The Conundrum of Projections](../c01-freericks-intro-quantum-world/04-conundrum-of-projections.md) · Up: [Resources](../README.md) · Next: [Complementarity & Delayed Choice](02-complementarity-delayed-choice.md) →