Quantum Repeaters

A quantum repeater is the architecture that makes long-distance entanglement possible at all. The obstacle it answers is loss: in optical fiber a single photon survives a length $L$ with probability $\eta(L)=e^{-L/L_{\mathrm{att}}}$ , where $L_{\mathrm{att}}\approx 22$ km at telecom wavelength, so a photon sent over $1000$ km arrives with probability $\sim 2\times10^{-20}$ — at a $10$ GHz source, roughly one photon every two centuries. A classical network drops in amplifiers, but the no-cloning theorem forbids deterministic copying of an unknown quantum state, and any phase-insensitive amplifier injects at least one noise photon per mode, shredding the entanglement it is meant to relay. The repeater refuses to amplify. Instead it divides and conquers, an idea due to Briegel, Dür, Cirac, and Zoller in 1998. Split the line into $N$ short elementary links; establish a heralded entangled pair across each, where the loss exponent involves only the short segment length; then fuse neighbouring pairs by entanglement swapping — a Bell-state measurement on the two inner qubits that leaves the outer qubits entangled even though they never met — nested over $\log_2 N$ levels to reach the whole span. Two more ingredients keep it honest: entanglement purification trades several noisy pairs for one cleaner pair whenever the fidelity clears $\tfrac12$ , and quantum memories store each heralded half while its neighbour is generated, so the segments' small success probabilities add in parallel rather than multiplying. The payoff is an exponential cost in $L$ converted into a polynomial one — and the benchmark that certifies success is beating the repeaterless PLOB bound $-\log_2(1-\eta)$ , the ultimate secret-key rate of a bare lossy channel.

Key takeaways

Direct transmission loses photons exponentially, $\eta=e^{-L/L_{\mathrm{att}}}$ with $L_{\mathrm{att}}\approx 22$ km, so quantum signalling past $\sim 1000$ km is hopeless — and no-cloning forbids the amplifiers that rescue classical light.
A repeater cuts the line into short elementary links, heralds entanglement across each, and stitches links together by entanglement swapping — a Bell-state measurement that teleports entanglement onto the distant end qubits.
Entanglement purification upgrades noisy pairs, sacrificing several to distil one of higher fidelity whenever $F>\tfrac12$ ; interleaving it with swapping holds fidelity above threshold.
Quantum memories store heralded halves while neighbouring segments retry, decoupling the segments in time — this is what turns the exponential loss cost into a polynomial overhead in distance.
The certificate of a genuine repeater is beating the repeaterless PLOB bound $-\log_2(1-\eta)\approx 1.44\,\eta$ ; three generations trade two-way heralding for one-way error correction as memories give way to loss-tolerant codes.
Key idea: loss and noise both grow exponentially with distance for direct transmission, and a repeater converts this into polynomial overhead by nesting two primitives — entanglement swapping to extend range without sending qubits far, and entanglement purification to restore fidelity — with quantum memories synchronizing the heralded, probabilistic steps.

The core insight

Loss and noise both scale exponentially with distance for direct transmission; a repeater converts this into polynomial overhead by nesting two primitives — entanglement swapping (to extend range without sending qubits far) and entanglement purification (to restore fidelity) — with quantum memories synchronizing the heralded, probabilistic steps.

Where this lives today

The repeater has crossed into the laboratory, with milestone experiments organized around the PLOB line. In 2020 the Lukin group at Harvard reported memory-enhanced quantum communication with a single silicon-vacancy centre in a diamond nanocavity that beat the repeaterless bound for its channel — the first physical memory node to outperform an equivalent bare fiber. On the network side, QuTech in Delft brought a three-node processor network online (Pompili and coworkers, 2021), swapping entanglement between remote nitrogen-vacancy nodes to entangle qubits that never interacted. A parallel line sidesteps memories: twin-field QKD (Lucamarini and colleagues, 2018), demonstrated over more than $600$ km, achieves a rate scaling like $\sqrt{\eta}$ rather than $\eta$ with no stored qubit. The frontier now is scaling — rare-earth crystals and trapped ions pushing memory coherence toward the seconds needed for continental spacing, and all-photonic, error-corrected designs trading heralding latency for the cost of large photonic graph states, the substrate a future quantum internet would run on.

The physical picture

Picture a relay race run in the dark. You cannot photocopy the baton — that is no-cloning — and the runners keep dropping it, which is loss. If one runner carried it the whole way, the chance of arriving is the product of surviving every stretch, collapsing exponentially with distance. The repeater's trick is to let neighbouring runners hand off over short, well-lit stretches, and to give each a memory so one who has succeeded can wait, baton in hand, while the next stretch is retried. Because the stretches now succeed independently and in parallel, the total effort grows only polynomially in the number of stations rather than exponentially in distance. The hand-off itself is entanglement swapping, teleportation turned on its target: two adjacent Bell pairs share a middle station, and a joint Bell-state measurement on the inner qubits projects the far-apart outer qubits into an entangled state, teleporting entanglement across the gap without any qubit traversing it. The first figure shows the chain and the number that makes the scheme worth building: the elementary-link survival probability dwarfs the end-to-end one.

Figure

A repeater chain over 1000 km split into short elementary links: heralded Bell pairs form across each segment then fuse by Bell-state measurements at the interior nodes, so the survival probability that matters is the short-link value η₀ ≈ 0.24 rather than the end-to-end η ≈ 2×10⁻²⁰

Errors accumulate: every elementary pair starts imperfect and every swap compounds the noise. Purification is the counter-move — measure two noisy pairs against each other and keep only the heralded coincidences, which certify a cleaner survivor. Interleaving it with swapping at each level pins fidelity above threshold all the way up. The second figure is the scoreboard: secret-key rate versus distance, the PLOB bound falling like a stone while a repeater curve stays aloft and overtakes it.

Figure

Secret-key rate versus distance on a log scale: the repeaterless PLOB capacity −log₂(1−η) ≈ 1.44 e^(−L/22) falls exponentially, reaching ≈ 2×10⁻²⁰ ebits per use at 1000 km, while a repeater decays only polynomially and overtakes the bound beyond about 100 km — beating PLOB is the operational certificate of a genuine repeater

Why memories are indispensable is worth stating plainly. Without storage, every segment would have to succeed in the same instant, and that probability is again the product of the per-segment ones — the exponential creeps back in. Memory decouples the segments in time: each retries until it heralds, then holds its qubit until its neighbour is ready to swap, which is why memory coherence time and segment length are locked together. The third-generation viewpoint drops the waiting by recasting loss as a located erasure and correcting it one-way with a photonic code, so light keeps moving forward through correcting stations at gate speed.

The core equations

The whole subject begins with one exponential. Attenuation is quoted in decibels per kilometre; converting that into a survival probability fixes the length scale $L_{\mathrm{att}}$ every later estimate leans on.

\eta(L)=e^{-L/L_{\mathrm{att}}},\qquad L_{\mathrm{att}}=\frac{10}{\alpha\ln 10}\approx 22\ \mathrm{km}\ (1550\ \mathrm{nm}) \qquad (1)

where: $\eta$ — single-photon survival probability; $L$ — fiber length; $L_{\mathrm{att}}$ — attenuation length, the distance for $1/e$ survival; $\alpha\approx 0.2$ dB/km — telecom attenuation coefficient. Reading it: survival is exponentially small in $L$ , so doubling the distance squares the loss — the entire adversary a repeater is built to defeat. Entanglement swapping extends range without sending any qubit the full way. Two Bell pairs share a middle node; a Bell-state measurement (BSM) on the inner qubits fuses them, projecting the outer pair into a Bell state up to a Pauli correction fixed by the two classical outcome bits.

|\Phi^+\rangle_{AB}\otimes|\Phi^+\rangle_{CD}\;\xrightarrow{\ \mathrm{BSM\ on}\ B,C\ }\;|\Phi^+\rangle_{AD}\ (\text{up to Pauli}) \qquad (2)

where: $|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ — a Bell state; $A,D$ — outer, distant qubits, which end up entangled; $B,C$ — inner qubits, jointly measured. Reading it: the arrow doubles the entangled span at each application, so $n$ swaps reach across $2^n$ links. Swaps degrade fidelity, so it must be replenished. One round of recurrence purification (BBPSSW/DEJMPS) takes two identical Werner states of fidelity $F$ and, on a heralded coincidence, outputs one pair of higher fidelity.

F'=\frac{F^2+\tfrac19(1-F)^2}{F^2+\tfrac23 F(1-F)+\tfrac59(1-F)^2} \qquad (3)

where: $F$ — input fidelity of each copy with respect to $|\Phi^+\rangle$ ; $F'$ — output fidelity of the survivor, conditioned on success. Reading it: the map has fixed points at $F=\tfrac12$ and $F=1$ and gives $F'>F$ for every $F>\tfrac12$ , so any pair better than a coin-flip can be driven toward a perfect Bell state. The benchmark a repeater must clear is the repeaterless capacity of the lossy channel — a tight bound, proved by Pirandola, Laurenza, Ottaviani, and Banchi, on the secret-key rate of any point-to-point protocol, however clever.

K(\eta)\le -\log_2(1-\eta)\;\xrightarrow{\ \eta\to 0\ }\;\frac{\eta}{\ln 2}\approx 1.44\,\eta \qquad (4)

where: $K$ — two-way secret-key (and entanglement-distribution) capacity in ebits per use; $\eta=e^{-L/L_{\mathrm{att}}}$ — end-to-end transmissivity. Reading it: at long distance the bound decays linearly in $\eta$ , hence exponentially in $L$ ; a device whose end-to-end rate beats this value is, by definition, a working repeater. The nesting turns the exponential into a polynomial. Over $N=2^n$ links, the dominant time cost multiplies the elementary generation time by a per-level waiting prefactor and the swap and purification success probabilities.

N=2^n,\qquad T_{\mathrm{tot}}\sim \Big(\tfrac{3}{2}\Big)^{\!n}\,\frac{L_0}{c}\,\frac{1}{\eta_0\,P_s\,P_p^{\,n}},\qquad \Big(\tfrac32\Big)^{\!n}=N^{0.585} \qquad (5)

where: $L_0=L/N$ — link length; $\eta_0=e^{-L_0/L_{\mathrm{att}}}$ — link transmission; $c$ — light speed in fiber; $P_s$ — swap success probability; $P_p$ — per-level purification success probability. Reading it: the exponent now rides only the short $L_0$ , and the $(3/2)^n=N^{0.585}$ prefactor is sub-linear in $N$ , so the total cost is polynomial in $L$ . Finally the memory requirement, which ties the hardware to the geometry: a heralded half must survive at least long enough for its partner segment to report success across an elementary link.

\tau_{\mathrm{mem}}\;\gtrsim\;\frac{L_0}{c}\;=\;\frac{L}{N\,c} \qquad (6)

where: $\tau_{\mathrm{mem}}$ — required memory coherence time; $L_0/c$ — one-way signalling time across a link. Reading it: more, shorter links relax the memory demand but multiply the swaps and the fidelity loss they cause — the central engineering trade of first-generation design.

Derivation AEntanglement swapping as teleportation of one Bell-pair half

The goal is to prove that a Bell-state measurement on the inner qubits of two adjacent Bell pairs leaves the outer qubits maximally entangled, and that two classical bits of outcome convert the result into a chosen Bell state — so entanglement is teleported across a gap no qubit crosses, causality intact. We start from two independent Bell pairs, re-expand the inner pair in the Bell basis, project, and read off the Pauli correction.

Step 1. Write both pairs in the computational basis, $|\Phi^+\rangle_{AB}|\Phi^+\rangle_{CD}=\tfrac12\sum_{i,j=0}^{1}|i\rangle_A|i\rangle_B|j\rangle_C|j\rangle_D$ . Why: everything follows from re-grouping this fixed four-qubit amplitude, laid out with outer labels $A,D$ and inner labels $B,C$ explicit.

Step 2. Introduce the Bell basis of the inner qubits, $|\beta_\mu\rangle_{BC}=(\mathbb{1}\otimes U_\mu)|\Phi^+\rangle_{BC}$ with $U_\mu\in\{\mathbb{1},X,Z,XZ\}$ . Why: the BSM is a projective measurement onto these four states, so this basis reads off directly what the measurement leaves on $A,D$ .

Step 3. Re-express the sum: $|\Phi^+\rangle_{AB}|\Phi^+\rangle_{CD}=\tfrac12\big(|\Phi^+\rangle_{BC}|\Phi^+\rangle_{AD}+|\Phi^-\rangle_{BC}|\Phi^-\rangle_{AD}+|\Psi^+\rangle_{BC}|\Psi^+\rangle_{AD}+|\Psi^-\rangle_{BC}|\Psi^-\rangle_{AD}\big)$ , pairing each inner Bell outcome with a definite outer Bell state. Why: this is the whole mechanism — measuring $B,C$ forces $A,D$ into the matching entangled state.

Step 4. Project onto outcome $\mu$ : $\langle\beta_\mu|_{BC}\big(|\Phi^+\rangle_{AB}|\Phi^+\rangle_{CD}\big)=\tfrac12(\mathbb{1}_A\otimes U_\mu^\dagger)|\Phi^+\rangle_{AD}$ , so each of the four outcomes occurs with $p=\tfrac14$ and leaves $A,D$ one Pauli from $|\Phi^+\rangle$ ; Alice or Bob applies $U_\mu$ to $D$ after receiving the two bits. Why: before the bits arrive $A$ is maximally mixed and carries no information, so no signalling is possible — the correction makes the entanglement usable while respecting relativity. *The four outcomes are equiprobable at $\tfrac14$ and sum to unity, as the standard teleportation algebra requires.*

Go deeper · Derivation B · 1 minThe repeaterless bound and why a repeater must beat it

Derivation BThe repeaterless bound and why a repeater must beat it

The goal is to state the exact upper limit on the secret-key rate of a pure-loss fiber with no intermediate quantum processing, and to show it decays exponentially at long distance — which pins down the operational meaning of repeater as any device whose end-to-end rate exceeds this value. We model the fiber as a pure-loss bosonic channel, invoke the relative-entropy-of-entanglement bound on its Choi state, and expand at high loss.

Step 1. Model the fiber as a pure-loss channel of transmissivity $\eta$ ; the PLOB bound holds against the most general two-way LOCC-assisted protocol, so it covers every repeaterless scheme, adaptive or not. Why: a benchmark is meaningful only if nothing point-to-point can beat it, so the bound must quantify over all such strategies.

Step 2. The two-way assisted secret-key capacity is bounded by the relative entropy of entanglement of the channel's Choi state, which for pure loss gives $K(\eta)\le E_R(\rho_\chi)=-\log_2(1-\eta)$ — a value PLOB show is achievable, hence the exact capacity. Why: an upper bound that is also attainable turns an inequality into the true ceiling every repeaterless device runs into.

Step 3. Expand at high loss: $-\log_2(1-\eta)=\dfrac{\eta}{\ln 2}+O(\eta^2)\approx 1.44\,e^{-L/L_{\mathrm{att}}}$ , exponentially small in $L$ . Why: over $1000$ km the ceiling is $\sim 10^{-20}$ ebits per use, far below any usable rate — the quantitative statement of hopelessness.

Step 4. Inserting even one entangling station partitions the channel and lifts the achievable rate above this end-to-end value, so a rate above $-\log_2(1-\eta)$ at the full $\eta$ is the certificate of a repeater. Why: it converts a vague aspiration into a measurable pass/fail test experiments report against. *The small- $\eta$ expansion uses $-\log_2(1-\eta)=\eta/\ln 2+O(\eta^2)$ .*

Go deeper · Derivation C · 1 minPolynomial overhead of the nested repeater

Derivation CPolynomial overhead of the nested repeater

The goal is to show quantitatively why nesting swaps and purification replaces the exponential cost of direct transmission with a polynomial one, by counting the expected time to distribute one end-to-end pair across $N=2^n$ links. We estimate the elementary generation time, add the waiting penalty when two lower-level pairs must both be ready to swap, accumulate it over levels, and compare with the direct exponential.

Step 1. For one link, heralded generation over the short length $L_0$ needs $\sim 1/\eta_0$ attempts, giving $t_0\sim (L_0/c)/\eta_0$ with $\eta_0=e^{-L_0/L_{\mathrm{att}}}$ . Why: the exponential survives, but its exponent now contains only the short $L_0$ , not the full $L$ — the seed of the whole advantage.

Step 2. To swap two segments, both lower-level pairs must be ready at once; waiting for the slower of two Poisson successes multiplies the time per level by the harmonic-type prefactor $\sim 3/2$ . Why: this is the real cost of synchronization, exactly what memories pay for by letting a finished segment wait instead of restarting.

Step 3. Accumulate over the $n$ levels: $(3/2)^n=N^{\log_2(3/2)}=N^{0.585}$ , sub-linear in $N$ , while the swap and purification successes add polynomial factors $1/P_s$ and $1/P_p^{\,n}$ , bounded because purification holds the per-level fidelity above threshold. Why: a sub-linear prefactor times an elementary exponential in $L_0$ is a polynomial in $L$ when $N$ scales with $L$ .

Step 4. Compare totals: $T_{\mathrm{tot}}\sim L\cdot\mathrm{poly}(\log N)$ for the repeater versus $e^{L/L_{\mathrm{att}}}$ for direct transmission. Why: the exponential in the full $L$ is demoted to one in the much smaller $L_0$ plus polynomial bookkeeping — the defining win. *Note that $(3/2)^n=N^{0.585}$ since $\log_2(3/2)=0.585$ .*

Limiting cases

Repeater performance is a contest of competing exponentials, and the useful regimes fall out of asking which one dominates.

Case	Result	Meaning
Direct transmission (no repeater)	$R_{\mathrm{direct}}\sim\eta=e^{-L/L_{\mathrm{att}}}$	rate falls exponentially in total $L$ ; capacity capped by PLOB $-\log_2(1-\eta)$
Ideal first-generation repeater	$T_{\mathrm{tot}}\sim L_0\,e^{L_0/L_{\mathrm{att}}}\,\mathrm{poly}(L/L_0)$	cost polynomial in $L$ because the exponential rides only the short $L_0$ ; wins for large $L$
Third-generation (one-way) repeater	$R_{\mathrm{1way}}\sim c_{\mathrm{gate}}/L_0$ , loss corrected as erasure	rate set by the local gate clock, not round-trip latency; needs per-segment loss below $50\%$
High-loss elementary link	$\eta_0=e^{-L_0/L_{\mathrm{att}}}\to 0$	generation time $\sim 1/\eta_0$ blows up; favours shorter, more numerous links at the cost of more swaps
Very many links	$N\to\infty$ , $\eta_0\to 1$	elementary loss vanishes but swap, memory, and gate-error overheads dominate; optimum $N$ exists

Worked examples

1 · The tyranny of loss and the divide-and-conquer cure (fully worked)

A $1000$ km telecom fiber has $L_{\mathrm{att}}\approx 22$ km. Compare direct single-photon transmission with a first-generation repeater that splits the line into $N=32$ elementary links, and say concretely why the split helps.

Step 1. Direct survival: $\eta=e^{-1000/22}=e^{-45.5}$ ; since $45.5\times\log_{10}e=19.7$ , $\eta\approx \mathbf{1.8\times10^{-20}}$ . At a $10$ GHz source the arrival rate is $10^{10}\times 1.8\times10^{-20}=1.8\times10^{-10}$ per second — about one photon every 170 years. This is not slow; it is impossible.

Step 2. Split into $N=32=2^5$ links, so $L_0=31.25$ km and $n=5$ nesting levels. Elementary survival: $\eta_0=e^{-31.25/22}=e^{-1.42}\approx \mathbf{0.24}$ , so each short link heralds a pair in $\sim 1/\eta_0\approx \mathbf{4}$ attempts — a $19$ -order-of-magnitude gain in the probability that governs the clock.

Step 3. Stitching cost: connecting the links needs $N-1=31$ Bell-state measurements in $n=5$ levels, with synchronization penalty $(3/2)^5=\mathbf{7.59}=32^{0.585}$ , sub-linear in $N$ — so the exponential $e^{-45.5}$ of the whole line becomes $e^{-1.42}$ of one short link plus polynomial overhead.

Step 4. The catch: memories. With $\tau_{\mathrm{mem}}\gtrsim L_0/c\approx (31.25\ \mathrm{km})/(2\times10^{8}\ \mathrm{m/s})\approx \mathbf{0.16\ ms}$ , a heralded half can wait instead of forcing all $32$ links to succeed at once; without that wait the probabilities multiply back to $\sim(0.24)^{32}\approx 10^{-20}$ and the advantage evaporates. Elementary survival 0.24 versus end-to-end 2×10⁻²⁰ — the split turns an impossible link into an engineering problem, provided the memory holds ≳ 0.16 ms.

2 · Two rounds of entanglement purification

Elementary generation delivers Werner pairs of fidelity $F=0.80$ . Apply the recurrence map of equation (3) twice and report the fidelities, the per-round success probabilities, and the yield.

Step 1. First round. Numerator $=F^2+\tfrac19(1-F)^2=0.64+\tfrac19(0.04)=0.6444$ ; denominator $=0.64+\tfrac23(0.8)(0.2)+\tfrac59(0.04)=0.7689$ . So $F'=\mathbf{0.838}$ with success probability $P_p=\mathbf{0.769}$ — cleaner because $0.80>\tfrac12$ .

Step 2. Second round, $F=0.838$ . Numerator $=0.7025+\tfrac19(0.1618)^2=0.7054$ ; denominator $=0.7025+\tfrac23(0.838)(0.162)+\tfrac59(0.1618)^2=0.8075$ . So $F''=\mathbf{0.874}$ with success probability $\mathbf{0.807}$ ; the gain shrank from $+0.038$ to $+0.036$ near $F=1$ .

Step 3. Yield. Each round consumes two pairs to make one, so a twice-purified pair needs four raw pairs, and the tree heralds with probability $0.769^2\times 0.807=\mathbf{0.48}$ — about eight raw pairs of fidelity 0.80 buy one heralded pair of fidelity 0.87. That cost, paid at every level, is the overhead $P_p^{\,n}$ of equation (5). 0.80 → 0.838 → 0.874, at ≈ 48% yield per four raw pairs.

Physics of the portrait

The quantum repeater is not one person's discovery but a chain of ideas — teleportation, distillation, and a scaling insight — assembled into an architecture, and it is worth seeing the number that certifies it in the same frame as the people who built each link.

Origins · 1 minquantum repeater and how this was discovered

quantum repeater

the architecture that turns the loss-length exponential into a polynomial · long-distance quantum communication

The quantum repeater is younger than most of the physics it uses and is best read as a synthesis. Its two primitives were in hand by the mid-1990s: quantum teleportation (Bennett, Brassard, Crépeau, Jozsa, Peres, and Wootters, 1993), which made it conceivable to move quantum information without sending a carrier through the channel; and entanglement distillation, formalized in 1996 by Bennett and coworkers and sharpened into an efficient recurrence protocol by Deutsch and colleagues. The decisive assembly came in 1998, when Hans Briegel, Wolfgang Dür, Ignacio Cirac, and Peter Zoller showed that nesting swapping and purification, buffered by quantum memories, converts the exponential cost of distance into a polynomial one — the BDCZ repeater. In 2001 Duan, Lukin, Cirac, and Zoller mapped it onto realistic hardware with the DLCZ atomic-ensemble scheme, and Sangouard, Simon, de Riedmatten, and Gisin reviewed the resulting decade of experiment in 2011. The number that made the enterprise falsifiable arrived only in 2017, when Pirandola, Laurenza, Ottaviani, and Banchi proved the exact repeaterless capacity $-\log_2(1-\eta)$ — the line every genuine repeater must cross.

History

1993

— Bennett and coworkers introduce quantum teleportation, the seed of entanglement swapping

1996

— Bennett, Bernstein, Popescu, Schumacher and the Deutsch group formalize entanglement distillation

1998

— Briegel, Dür, Cirac, and Zoller propose the nested quantum repeater (BDCZ)

2001

— Duan, Lukin, Cirac, and Zoller give the DLCZ atomic-ensemble scheme

2017

— Pirandola, Laurenza, Ottaviani, and Banchi prove the PLOB repeaterless bound

Going further

References. Briegel, Dür, Cirac, Zoller, "Quantum Repeaters: The Role of Imperfect Local Operations in Quantum Communication", Physical Review Letters 81, 5932 (1998) · Duan, Lukin, Cirac, Zoller, "Long-distance quantum communication with atomic ensembles and linear optics", Nature 414, 413 (2001) · Sangouard, Simon, de Riedmatten, Gisin, "Quantum repeaters based on atomic ensembles and linear optics", Reviews of Modern Physics 83, 33 (2011) · Pirandola, Laurenza, Ottaviani, Banchi, "Fundamental limits of repeaterless quantum communications", Nature Communications 8, 15043 (2017) · Muralidharan et al., "Optimal architectures for long distance quantum communication", Scientific Reports 6, 20463 (2016) · Nielsen and Chuang, Quantum Computation and Quantum Information, chapter 12 · Wilde, Quantum Information Theory, on LOCC and distillation. Key numerical checkpoints: (1) $L_{\mathrm{att}}=10/(\alpha\ln 10)\approx 21.7$ km and direct survival $e^{-1000/22}\approx 1.8\times10^{-20}$ ; (2) the Bell-basis re-expansion of the swap yields four equiprobable outcomes $p=\tfrac14$ with their Pauli corrections; (3) the purification map has fixed points at $F=\tfrac12$ and $F=1$ , threshold $F'>F\Leftrightarrow F>\tfrac12$ , and trajectory $0.80\to0.838\to0.874$ at successes $0.769$ and $0.807$ ; (4) the expansion $-\log_2(1-\eta)\approx 1.44\,\eta$ ; (5) the prefactor $(3/2)^5=7.59=32^{0.585}$ .

Prerequisites. Entanglement and Bell States · Quantum Teleportation · No-Cloning Theorem · LOCC and Local Operations · Magic State Distillation. Leads to. Magic State Distillation · Topological Quantum Computation · Measurement-Based Quantum Computation.

Ask the tutor: Ask me to walk through the swap algebra line by line, run the purification recurrence to a target fidelity, or estimate for your own distance and hardware whether a given repeater design clears the PLOB bound.

Why can't I just amplify the signal like a classical optical repeater does?

Because the state is unknown and quantum. A classical repeater copies the incoming light and re-emits a fresh, strong copy, but no-cloning proves that no device can copy an arbitrary unknown quantum state faithfully. The best a physical amplifier can do is phase-insensitive amplification, which the quantum limit forces to inject at least one noise photon per mode — enough to wash out the phase coherence entanglement depends on. A repeater never copies or amplifies; it generates fresh entanglement over short hops and splices them together with measurements, sidestepping the theorem entirely.

Does entanglement swapping let Alice and Bob communicate instantly?

No. After the Bell-state measurement the two end qubits are in one of four Bell states, correlated with a random outcome the middle node obtained. Until Alice and Bob learn that outcome — two classical bits that travel no faster than light — each end qubit is maximally mixed and carries no information. Only after applying the communicated Pauli correction do they hold a known, usable pair. So swapping distributes entanglement, not messages, and the classical side-channel enforces causality. Entanglement is a resource for teleportation and key distribution, never a signalling channel by itself.

What is the difference between entanglement swapping and entanglement purification?

They solve orthogonal problems and the repeater needs both. Swapping is about reach: it fuses two shorter links into one longer one via a Bell-state measurement, extending range without sending any qubit the full distance. Purification is about quality: it consumes several noisy pairs to distil fewer of higher fidelity by LOCC, and works whenever the input fidelity exceeds one half. Left alone, swapping compounds errors as links grow, while purification cannot extend range. Interleaving them at every level — swap to reach farther, purify to clean up — keeps fidelity above threshold across the span.

Why does the cost become polynomial instead of exponential in distance?

The exponential never disappears; it gets confined to a short length. Direct transmission pays $e^{-L/L_{\mathrm{att}}}$ in the full $L$ , astronomically small. A repeater with $N$ links pays $e^{-L_0/L_{\mathrm{att}}}$ per link with $L_0=L/N$ — an exponential in a much smaller number — then stitches links with overheads polynomial in $N$ : the sub-linear $(3/2)^n=N^{0.585}$ prefactor plus polynomial swap and purification factors. Because those overheads are polynomial while the elementary exponential is tamed by short $L_0$ , the total scales like $L$ times polynomial corrections — the game is moving the exponential from $L$ to $L_0$ .

How good does a quantum memory have to be?

Its coherence time must exceed the classical signalling delay across a link, $\tau_{\mathrm{mem}}\gtrsim L_0/c$ , because a heralded qubit must wait, phase intact, while news of its neighbour's success arrives. For a $30$ km link that is $0.15$ ms — already demanding — and continental spacings push it toward tens of milliseconds. This is why memory research drives repeater hardware: atomic ensembles, trapped ions, rare-earth crystals, and diamond colour centres are judged by coherence time and time-bandwidth product, and better memories permit longer links and fewer swaps.

Has anyone actually built a quantum repeater that beats the PLOB bound?

Repeater segments have, in controlled settings. In 2020 a memory-enhanced node built from a silicon-vacancy centre in a diamond cavity beat the repeaterless bound for its channel, and processor networks in Delft have performed the swap that entangles previously unconnected end qubits. Twin-field QKD over hundreds of kilometres achieves the repeater-like $\sqrt{\eta}$ scaling with no stored qubit at all. What does not yet exist is a full multi-link repeater delivering high-rate entanglement over continental distance with many nested levels in concert — awaiting memories with long enough coherence and efficient interfaces. The physics is settled; the engineering of scale is the open problem.