1. Worked problem §
What is the single-photon probability at μ=0.1?
Hint 1
Use n=1 in the Poisson formula.
Worked solution
0.1e⁻⁰·¹≈0.090484.
Quantum Cryptography · Chapter 05
Which physical assumptions make the mathematical protocol applicable?
Starting point: Chapters 1–4 of this course.
Anchor example: A vacuum-plus-weak-decoy estimate through a lossy optical link.
Method: Trace the protocol, calculate an example, and identify the security assumptions.
Watch for: An ideal qubit model does not certify an optical implementation.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Chapters 1 to 4 analysed an abstract qubit, a noiseless detector, and a source that emits exactly one photon on demand. No laboratory provides all three. This chapter asks what changes when the source is an attenuated laser, the link is a lossy fibre, the detectors click for reasons of their own, and the measurement station might be under Eve's control. The aim is not to make the protocol sound fragile, but to keep the boundary between a proved statement and a modelled one in clear view.
The first idealisation to go is the single-photon source. This section derives what an attenuated laser actually emits, shows why the difference is a security problem rather than a technicality, and reconstructs the single-photon contribution from measurements at two intensities.
A phase-randomized coherent pulse with mean photon number \(\mu\) has photon-number probabilities
\[P_n=e^{-\mu}\frac{\mu^n}{n!},\]so the vacuum and multiphoton fractions are
\[P_{\text{vac}}=e^{-\mu},\qquad P_{\ge2}=1-e^{-\mu}(1+\mu).\]Attenuating a laser therefore does not produce one photon per trial. Most pulses are empty, some carry one photon, and a small fraction carry two or more. Only the last group is dangerous. In a multiphoton pulse Eve can, in an idealised attack, keep one photon and forward the rest: after the basis announcement her stored copy reveals the bit without the intercept–resend disturbance derived in Section 2.2.1. Channel loss helps her, because she can forward only the pulses that suit her and let the others disappear into the loss budget.
Decoy-state methods answer this by varying the intensity and using the observed detection statistics to bound the single-photon contribution. The inference has a precondition: conditioned on the photon number, the signal and decoy preparations must be indistinguishable in all modes that Eve can access. A source whose pulses differ in spectrum or timing between intensities provides no such inference, however good the statistics.
What follows in this section is a source calculation plus one decoy bound. It describes a model; it is not a decoy-state security proof, which additionally fixes the attack model, the finite-count statistics, and the failure budget.
Calculate the multiphoton probability at \(\mu=0.1\), per emitted pulse and among pulses that are not vacuum.
\(P_{\ge2}=1-e^{-0.1}(1.1)\approx0.004679\), about 0.468% of emitted pulses. Dividing by the nonvacuum probability \(1-e^{-0.1}\approx0.095163\) gives \(0.0492\), so roughly 4.9% of the pulses that reach the channel at all carry two or more photons. The two numbers answer different questions: the first sizes the hazard against the clock rate, the second against the surviving detections—and an adversary who benefits from loss cares about the second kind of fraction.
A coherent state of complex amplitude \(\alpha\) has the expansion
\[|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n\ge0}\frac{\alpha^n}{\sqrt{n!}}|n\rangle,\]which is a superposition, not a mixture: the relative phases between different \(n\) are fixed and observable in principle. Write \(\alpha=\sqrt\mu\,e^{i\theta}\) and randomize \(\theta\) uniformly, then discard it. Every off-diagonal term carries a factor
\[\frac{1}{2\pi}\int_0^{2\pi}e^{i(n-m)\theta}\,d\theta=\delta_{nm},\]so all coherence between different photon numbers vanishes and what remains is diagonal:
\[\rho_\mu=\sum_{n=0}^{\infty}e^{-\mu}\frac{\mu^n}{n!}|n\rangle\langle n|.\]Phase randomization is therefore a physical operation the model relies on, not a figure of speech. Calling a pulse “weak” establishes nothing by itself: an unrandomised weak pulse has a definite phase and is not described by \(\rho_\mu\).
The trade is easiest to read as a table.
| \(\mu\) | Vacuum \(P_0\) | One photon \(P_1\) | Two or more \(P_{\ge2}\) | \(P_{\ge2}\) given nonvacuum |
|---|---|---|---|---|
| 0.1 | 0.9048 | 0.0905 | 0.0047 | 0.049 |
| 0.5 | 0.6065 | 0.3033 | 0.0902 | 0.229 |
| 1.0 | 0.3679 | 0.3679 | 0.2642 | 0.418 |
Conditioning changes what a small percentage means. At \(\mu=0.1\) the multiphoton fraction is 0.468% of pulses but 4.92% of nonvacuum pulses; at \(\mu=0.5\) it is 9.02% of pulses but 22.9% of nonvacuum pulses. A protocol that reasons only about the unconditional fraction will underestimate the danger in a lossy link.
Photon-number splitting shows why this is a security question rather than an engineering detail. Eve keeps one photon of a multiphoton signal, forwards the rest, waits for the basis announcement, and then measures her copy in the right basis. She gains full information with no additional disturbance to the single-photon events that the error test is calibrated on. A security proof is allowed to grant Eve this capability even when a laboratory implementation would find it hard; the proof must hold for the adversary, not for the most convenient one.
Let \(Y_n\) be the probability of a detection given an \(n\)-photon emission, and \(Q_\mu\) the measured detection probability per pulse at intensity \(\mu\). If the conditional \(n\)-photon preparations are identical across intensities—the assumption stated in Section 5.1.1—then the gains decompose by photon number:
\[Q_\mu e^{\mu}=\sum_{n\ge0}\frac{\mu^n}{n!}Y_n.\]Each intensity gives one such linear equation in the unknowns \(Y_n\); measuring two intensities gives two equations, not enough to solve for all of them. The decoy trick extracts a bound on \(Y_1\) instead. With \(0<\nu<\mu\), subtract \((\nu^2/\mu^2)\,Q_\mu e^{\mu}\) from \(Q_\nu e^{\nu}\). The \(n=2\) coefficient vanishes, and for \(n\ge3\) the coefficient is
\[\nu^n-\nu^2\mu^{n-2}=\nu^2\bigl(\nu^{n-2}-\mu^{n-2}\bigr)\le0,\]because \(\nu<\mu\). Yields are nonnegative, so those higher terms can only lower the difference. The difference is therefore bounded above by its vacuum and single-photon terms alone, and rearranging gives the asymptotic vacuum-plus-weak-decoy bound
\[Y_1\ge Y_1^L=\frac{\mu}{\mu\nu-\nu^2}\,D,\]where \(D\) collects the measured gains and the background estimate,
\[D=Q_\nu e^\nu-\frac{\nu^2}{\mu^2}Q_\mu e^\mu-\frac{\mu^2-\nu^2}{\mu^2}Y_0.\]Three practical rules follow from the algebra. Clip the result at zero: a negative lower bound means the data certify no single-photon yield, not a negative one. Treat a value above one as evidence of inconsistent inputs or of statistical fluctuations left out of the exact calculation, never as a yield. And remember that the bound is only as good as the indistinguishability assumption; exact arithmetic cannot repair a source that behaves differently at different intensities.
For a pure-loss channel with \(\eta=0.1\), \(\mu=0.5\), \(\nu=0.1\), and no background, compare the true \(Y_1\), the decoy bound, and the naive estimate \(Q_\nu e^\nu/\nu\).
Here \(Q_x=1-e^{-\eta x}\), so \(Q_\nu\approx0.009950\) and \(Q_\mu\approx0.048771\). The bound gives \(Y_1^L\approx0.09725\), safely below the true \(Y_1=\eta=0.1\) and corresponding to a single-photon gain \(Q_1^L=\mu e^{-\mu}Y_1^L\approx0.02949\). The naive estimate returns \(Q_\nu e^\nu/\nu\approx0.110\), which exceeds the physical value: it credits every detection at the weak intensity to single photons while multiphoton emissions contribute as well. The subtraction is precisely what removes that credit, and its conservatism—2.75% low here—is the price of not knowing the individual yields.
The decoy bound converts two measured gains into one certified quantity. It does not, by itself, bound Eve's information: that requires relating the single-photon phase error to the observed error rate, which is the entanglement argument of Chapter 3 applied to the single-photon component. Nor does it account for finite counts, which enter through confidence bounds whose signs must be tracked carefully in the subtraction—an upper bound on \(Q_\mu\) and a lower bound on \(Q_\nu\) both make \(Y_1^L\) smaller. Ma and colleagues develop the protocol analysis in detail, and the educational laboratory in this course recomputes the bound and checks it against the known pure-loss model.
A key rate is a ratio, and most confusion about QKD performance comes from mixing the ratios. This section separates transmission from error, adds background clicks, and then multiplies the pieces together in the right order.
For a fibre with attenuation \(\alpha\) decibels per kilometre over length \(L\), the transmission factor is
\[\eta=10^{-\alpha L/10}.\]Detector efficiency, connector loss, basis sifting, and protocol acceptance reduce usable counts further. None of these factors is an error rate. A signal that never arrives carries no bit and cannot be wrong: the QBER conditions on retained detections, and is therefore blind to loss by construction.
For an idealised single-photon source pulsed at rate \(F\) through transmission \(\eta\) into detectors of efficiency \(d\), with unbiased bases, the sifted count rate is approximately
\[R_{\text{sift}}\approx\frac{F\eta d}{2},\]if further losses and dead time are neglected. Multiplying this by a secret fraction produces a planning model, not a performance claim. A reported number should carry its assumed parameters explicitly—attenuation, length, pulse rate, detector efficiency, background, and dead time—and any parameter that was chosen rather than measured should be labelled as hypothetical.
| Length | Loss | \(\eta\) | Sifted rate |
|---|---|---|---|
| 10 km | 2 dB | 0.63 | 63,000 /s |
| 50 km | 10 dB | 0.10 | 10,000 /s |
| 100 km | 20 dB | 0.010 | 1,000 /s |
| 200 km | 40 dB | 0.00010 | 10 /s |
Every 50 km costs a factor of ten in rate; every 100 km costs a factor of a hundred. That exponential is why loss, rather than the error rate, sets the practical distance record of a QKD link—until background clicks take over, which is the subject of the next subsection.
Use \(\alpha=0.2\) dB/km, \(L=50\) km, \(F=10^{6}\)/s, and \(d=0.2\). Find the simplified sifted rate, and the rate at 100 km.
The loss is \(0.2\times50=10\) dB, so \(\eta=10^{-1}=0.1\). The sifted rate is \(10^{6}\times0.1\times0.2/2=10{,}000\) bits/s before testing and extraction. At 100 km the loss doubles to 20 dB, \(\eta=0.01\), and the rate falls to 1,000 bits/s. Both figures assume unit quantum efficiency beyond \(d\), no dark counts, and an ideal single-photon source; a weak-pulse source with realistic intensities does considerably worse.
Let \(d\) be detector efficiency and \(\eta\) the channel transmission. For a phase-randomized coherent pulse, the probability that at least one signal photon is detected is
\[p_s=1-e^{-\mu\eta d}.\]If the background-click probability per time window is \(Y_0\), an independent-click model gives the total gain
\[Q=1-(1-Y_0)e^{-\mu\eta d}\approx Y_0+\mu\eta d\]in the low-click regime. If signal clicks carry misalignment error \(e_d\) while background bits are random, the observed error rate is the mixture of the two contributions:
\[E\approx\frac{\tfrac12Y_0+e_d\,p_s}{Y_0+p_s}.\]The expression neglects simultaneous clicks and the rule used to assign them, so it is a model rather than a measurement formula. It still shows a physical limit. As \(p_s\to0\), the numerator and denominator are both dominated by \(Y_0\), so \(E\to1/2\) even though nothing has changed in the detector's alignment. Loss does not flip bits; it dilutes the signal until the background is all that is left, and a rate model that reduces the count while holding the observed error fixed will therefore overstate the achievable distance.
For \(Y_0=10^{-5}\), \(p_s=10^{-3}\), and \(e_d=0.01\), compute \(E\). Repeat for \(p_s=10^{-5}\) and \(p_s\to0\).
At \(p_s=10^{-3}\): \((5\times10^{-6}+10^{-5})/(1.01\times10^{-3})\approx0.0149\), still close to the misalignment value. At \(p_s=10^{-5}\) the same formula gives \((5\times10^{-6}+10^{-7})/(1.1\times10^{-5})\approx0.255\). As \(p_s\to0\) it tends to \(1/2\). A detector whose alignment never changed has become useless because the signal now competes with its own dark counts, and a 25% error rate is far above the 11% threshold of the ideal model (Section 3.3.2).
The ideal single-photon expression \(1-h_2(e_b)-h_2(e_p)\) is a fraction per sifted single-photon key bit. Multiplying it by every detected coherent pulse would credit vacuum events (which carry no bit) and multiphoton events (whose information Eve may share) with secrecy they do not have. In a conventional asymptotic decoy analysis, a common lower-rate form per emitted signal pulse is
\[R\ge q\left[Q_1^L\bigl(1-h_2(e_1^U)\bigr)-Q_\mu f_{\rm EC}h_2(E_\mu)\right],\]where \(q\) is the basis and key-selection factor, \(Q_1^L\) a certified single-photon gain from Section 5.1.3, and \(e_1^U\) an upper bound on the single-photon phase error in the stated symmetric model. The structure is worth reading aloud: the positive term credits only the certified single-photon part, while reconciliation is paid on all retained signal detections, because that is what was actually discussed in public. Substituting a single-photon rate into the second term would understate the leakage.
This is a stated asymptotic rate model, not a finite-size prescription. It has no smoothing terms, no confidence intervals on the gains, and no acceptance probability.
With the pure-loss gains above, take \(e_1^U=E_\mu=0.02\), \(f_{\rm EC}=1.1\), and \(q=1/2\). Compute the rate and interpret the two terms.
\(Q_\mu\approx0.048771\), \(Q_1^L\approx0.029493\), and \(h_2(0.02)\approx0.14144\). The privacy term is \(0.029493\times0.85856\approx0.025322\) and the reconciliation term is \(0.048771\times1.1\times0.14144\approx0.007588\). Their difference, scaled by \(q=0.5\), is about \(0.00887\) secret bits per signal pulse. Roughly a third of the privacy credit is spent on reconciliation, which is why reconciliation efficiency matters even when the error rate is small. If decoy pulses occupy some clock slots, multiply by the signal-intensity selection probability as well; using the full clock rate would overcount.
Every result so far assumed that a detector implements the measurement the proof describes. Real detectors can be blinded, saturated, or made to click on command, so the question of which devices a proof must trust is a security question in its own right.
A detector is a physical apparatus, and its behaviour outside the assumed input range need not match the textbook measurement. Tailored bright illumination, for instance, has been used to force commercial detectors into click patterns of an attacker's choosing, which is why “the detector is fine because it passes its calibration” is not a security argument.
Two labels are often used to describe protocols that reduce this trust. Measurement-device-independent QKD places a joint measurement at an untrusted relay while keeping characterised sources and the rest of the laboratory model. Device-independent QKD uses Bell-test statistics to certify security under a different model, which requires suitable isolation, trusted randomness for setting choices, and control of detection loopholes.
Neither label removes every assumption, and the labels are not nested in a way that makes one strictly safer. A Bell violation by itself is not a finite-key certificate, and a conventional entanglement-based protocol is not automatically device independent just because it uses entanglement.
| Model | Removed from trust | Still assumed | Evidence used |
|---|---|---|---|
| Prepare-and-measure BB84 | Nothing in the receiver or source, beyond the stated proof model | Sources and detectors behave as specified for the states and ranges used | Basis-resolved detection and error statistics |
| MDI QKD | The relay's measurement, which may be arbitrarily malicious | Characterised sources, basis announcements, authentication, and decoy inference | Joint gains, errors, and single-photon bounds |
| Device-independent QKD | Both measuring devices, and the internal workings of the sources | Laboratory isolation, free random choices, loophole-free detection, and a key-extraction theorem | Loophole-controlled Bell statistics plus key correlations |
For any proposed experiment, list four things before interpreting a click as key material: which components Eve controls, what information can leave each laboratory, how the random choices are generated, and which statistics the proof consumes. Those four answers determine which row of the table the implementation occupies—and whether its security claim applies at all.
A relay reports a successful Bell-state measurement. Does that alone certify a secret key?
No. Even in an MDI protocol, Alice and Bob must authenticate their messages, estimate errors and the single-photon contribution as applicable, and run the prescribed reconciliation and extraction under valid source assumptions. The relay's announcement is an input to those steps, not a substitute for them—and a malicious relay is free to announce anything it likes, which is exactly why the proof must be written for that case.
In MDI QKD, Alice and Bob each prepare signals and send them to a relay, which performs a joint measurement and announces its outcome. Consider an ideal projection onto the antisymmetric Bell state
\[|\psi^-\rangle=\frac{|01\rangle-|10\rangle}{\sqrt2}.\]For opposite Z inputs, \(|\langle\psi^-|01\rangle|^2=|\langle\psi^-|10\rangle|^2=1/2\); for equal Z inputs, both amplitudes vanish. A successful announcement therefore selects anticorrelated bits, and one party flips to align the raw strings. The same projection is antisymmetric in X, so the property holds in both bases—which is why the relay's announcement is useful to Alice and Bob whatever the relay intended.
What makes the arrangement measurement-device-independent is that the security argument never needs to know what the relay actually did. A malicious relay can announce arbitrary results, so the proof uses the observed basis-dependent statistics together with source characterisation to bound what accepted announcements can imply. The successful projection is a correlation-generating operation, not a trusted declaration of secrecy. In an optical implementation, decoy methods bound the events in which both sources emitted exactly one photon, which is where the single-photon inference of Section 5.1.3 reappears. Lo, Curty, and Qi establish the measurement-device-independent approach.
Give the projection probabilities onto \(|\psi^-\rangle\) for the four Z-basis input pairs, and for the two X-basis antipodal pairs.
For \(|01\rangle\) and \(|10\rangle\) the probabilities are \(1/2\) each; for \(|00\rangle\) and \(|11\rangle\) they are zero. In X, \(|\psi^-\rangle=(|+-\rangle-|-+\rangle)/\sqrt2\), so \(|+-\rangle\) and \(|-+\rangle\) each project with probability \(1/2\), while \(|++\rangle\) and \(|−−\rangle\) project to zero. In both bases the relay's success signals opposite, not equal, inputs—so the relay cannot force agreement, and any successful announcement leaves Alice and Bob with a correlation they can check against their own data.
For outputs \(a,b\in\{-1,+1\}\), define the correlators \(E_{xy}=\langle ab\rangle\) for settings \(x,y\) and the CHSH combination
\[S=E_{00}+E_{01}+E_{10}-E_{11}.\]A deterministic local assignment \(a_x,b_y\in\{\pm1\}\) gives
\[S=a_0(b_0+b_1)+a_1(b_0-b_1).\]Exactly one parenthesis vanishes and the other equals \(\pm2\), so \(|S|=2\). Probabilistic local strategies are convex mixtures of such assignments and cannot exceed the bound, provided the settings are chosen independently of any hidden variables and each response depends only on its own setting and shared randomness. That proviso is the substance of the word “independent”: without it the bound does not apply.
For \(|\Phi^+\rangle\), take \(A_0=Z\), \(A_1=X\), \(B_0=(Z+X)/\sqrt2\), and \(B_1=(Z-X)/\sqrt2\). The ZZ and XX correlations equal one and the cross correlations vanish, so each term contributes \(1/\sqrt2\):
| Correlator | Settings | Value | Contribution to \(S\) |
|---|---|---|---|
| \(E_{00}\) | \(A_0=Z\), \(B_0=(Z+X)/\sqrt2\) | \(1/\sqrt2\) | \(+1/\sqrt2\) |
| \(E_{01}\) | \(A_0=Z\), \(B_1=(Z-X)/\sqrt2\) | \(1/\sqrt2\) | \(+1/\sqrt2\) |
| \(E_{10}\) | \(A_1=X\), \(B_0=(Z+X)/\sqrt2\) | \(1/\sqrt2\) | \(+1/\sqrt2\) |
| \(E_{11}\) | \(A_1=X\), \(B_1=(Z-X)/\sqrt2\) | \(-1/\sqrt2\) | \(+1/\sqrt2\) |
The total is \(S=4/\sqrt2=2\sqrt2\approx2.828\), the largest value quantum theory allows for two-outcome measurements on a qubit—the Tsirelson bound. Mixing the pair with maximally mixed noise,
\[\rho_v=v\,|\Phi^+\rangle\langle\Phi^+|+(1-v)\frac{I}{4},\]scales every correlator by \(v\), so \(S=2\sqrt2\,v\) and the local bound is violated only when \(v>1/\sqrt2\approx0.707\). A visibility below that value is compatible with a local model, and no device-independent claim can be based on it.
A loophole-controlled violation constrains local explanations and can support a device-independent security proof, but converting \(S\) into a key length needs a separate theorem plus treatment of finite statistics, device memory, isolation, free setting choices, and detection events. Acín and colleagues derive a collective-attack security bound; that scope should not be relabelled as a general finite-key result.
A source produces \(|\Phi^+\rangle\) with visibility \(v=0.9\) against white noise. Compute \(S\), the Z-basis error rate, and say what still must be proved before a key can be extracted.
\(S=2\sqrt2(0.9)\approx2.546\), comfortably above the local bound of 2. The white-noise part gives a Z disagreement probability of \(1/2\), so the observed error is \((1-0.9)/2=0.05\). Both numbers look promising, and neither is a key length: a device-independent theorem must turn the violation into a min-entropy bound, finite statistics must cover the estimated \(S\), the experiment must close the detection and locality loopholes, and setting choices must be independent. Two favourable expectations are not a proof.
These sources cover the source model, the decoy analysis, the two device-trust frameworks, and the detector attacks that motivate them.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
What is the single-photon probability at μ=0.1?
Use n=1 in the Poisson formula.
0.1e⁻⁰·¹≈0.090484.
Why can a small multiphoton emission fraction still matter in a lossy channel?
Compare emissions with detections.
Eve can selectively transmit those pulses, so their fraction among detected events can be much larger.
How does transmission change over another 50 km at the same attenuation?
Add decibels, multiply transmission.
It gains a factor 0.1; total transmission over 100 km is 0.01.
Do 90% lost pulses imply 90% QBER?
Check the QBER denominator.
No. QBER counts incorrect retained outcomes divided by retained outcomes, rather than lost emissions.
Does MDI QKD certify an arbitrary defective source?
Identify the part of the apparatus removed from trust.
No. It removes trust in the measurement station under its proof model, not all source assumptions.
Why is ignoring undetected Bell trials dangerous?
Consider selection correlated with measurement settings.
Selective detection can bias the retained correlations and open a detection loophole unless handled by the proof.
Derive the lower bound on Y1 from the two gain series at μ and ν. Explain the sign of every multiphoton coefficient.
Subtract (ν/μ)² times the μ series from the ν series.
The vacuum coefficient is 1−ν²/μ²; the single-photon coefficient is ν−ν²/μ; the n=2 coefficient is zero. For n≥3, νⁿ−ν²μ^(n−2)≤0 since ν<μ. Nonnegative Yn means these higher terms only lower the difference. Therefore the difference is no greater than the vacuum plus single-photon terms; subtract the vacuum contribution and divide by the positive single-photon coefficient to obtain Y1L.
At μ=0.1, calculate the multiphoton probability both per emitted pulse and conditioned on a nonvacuum pulse. Explain which denominator matters for a selective-forwarding attack.
Divide 1−exp(−μ)(1+μ) by 1−exp(−μ) for the conditional probability.
The per-emission value is about 0.00467884. Dividing by 0.09516258 gives about 0.0491668. Neither is automatically the fraction among Bob’s detections: adversarial selection and loss can enrich those detections in multiphoton events. The security analysis needs a bound on the contribution to detected key events.
For a noisy Bell pair with visibility v=0.8, compute the CHSH value and the error rate when both parties use Z. Does a Bell violation itself supply a finite secret-key length?
The white-noise component has zero correlations and gives Z disagreement with probability one half.
S=2√2×0.8≈2.26274, exceeding two. The Z error is (1−v)/2=0.1. A key length still requires a specific device-independent theorem, finite-statistical treatment, isolation, setting independence, and a detection/acceptance model. The two expectations alone do not constitute a finite-key proof.
Using Y0=10⁻⁵ and ed=0.01, compare the low-click QBER estimates at signal-click probabilities 10⁻³ and 10⁻⁵. Explain the limiting behavior as signal vanishes.
Weight random background bits by 1/2 and signal bits by ed.
At 10⁻³, E=(5×10⁻⁶+10⁻⁵)/(0.00101)≈0.01485. At 10⁻⁵, E=(5×10⁻⁶+10⁻⁷)/(2×10⁻⁵)=0.255. As the signal tends to zero, the numerator tends to Y0/2 and denominator to Y0, so E tends to 1/2. This approximation neglects simultaneous-click handling.