# Quantum Key Distillation Using Binary Frames

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## Abstract

**:**

## 1. Introduction

## 2. Related Work

- Binary protocol [16] is a reconciliation method that find and correct errors after the transmission of quantum pulses caused by the noise in the channel and possibly from the eavesdropper. After Alice and Bob obtain an error estimation based on a portion of their sifted key, they determine whether the error failure threshold has been breached. If the error rate is in excess of the fail threshold, Alice and Bob begin the raw key step again. If the estimated error rate is acceptable, Alice and Bob begin the first of a number of passes and use a predetermined random permutation, applying it to the sifted key bits.
- Cascade [17] is a reconciliation method that has become the de-facto standard for all QKD practical implementations. After a number of passes, permutations and cascades, the protocol finishes with low probability that errors still remain [21]. However, large communication overhead have raised methods based on error correcting codes which are more practical.
- The Winnow algorithm [18] is a reconciliation method based on Hamming codes which introduces additional errors because the Hamming algorithm can only reveal one single error in each block.
- LDPC [19] is a linear error correcting code that uses iterative decoding using the sum product soft decision decoder to correct transmission errors.

- Cascade exhibits great efficiency at low error rates but is still robust up to $18\%$ error rate if required.
- Effective estimation of the error rate in the quantum channel.
- Interactivity could be high intensive in the number of passes to check parity.
- The number of required permutations of the shared bits could demand a persistent computational effort.

## 3. BB84 Protocol

## 4. Quantum Flows

#### 4.1. Measurement of Non-Orthogonal Quantum States

- Single detection: One of the two qubits is detected at Bob’s station. It could be processed as usually in BB84 protocol. However, in our context, this kind of detection will not be included as part of the distillation process.
- Double detection: The two non-orthogonal states are detected at Bob’s station.

- 3.
- No detection: No pulse is registered.

#### 4.2. Quantum Photonic Gains

## 5. Distillation Based in Non-Orthogonal Quantum States

#### 5.1. Frames

#### 5.2. Matching Results (Mr)

#### 5.3. Sifting Protocol

- Alice prepares and sends to Bob a pair of non-orthogonal qubits over the quantum channel. She chooses a pair randomly between $({|0\rangle}_{X},{|0\rangle}_{Z})$, $({|0\rangle}_{X},{|1\rangle}_{Z})$, $({|1\rangle}_{X},{|0\rangle}_{Z})$ and $({|1\rangle}_{X},{|1\rangle}_{Z})$.
- Bob chooses randomly the measurement basis (X or Z) to measure the incoming pair of non-orthogonal qubits.
- After several rounds, using a classical channel, Bob announces to Alice the non-orthogonal pairs that produce double matching detection event (remember that Bob obtains a distribution of single and double detection events, which can be matching or non-matching). As indicated before, the states inside a quantum pair are temporally separated each other, so users must agree previously on the time difference.

#### 5.4. Sifting Bits Based in the Xor Function

- Alice prepares and sends to Bob a pair of non-orthogonal qubits over the quantum channel. She chooses a pair randomly between $({|0\rangle}_{X},{|0\rangle}_{Z})$, $({|0\rangle}_{X},{|1\rangle}_{Z})$, $({|1\rangle}_{X},{|0\rangle}_{Z})$ and $({|1\rangle}_{X},{|1\rangle}_{Z})$.
- Bob chooses randomly the measurement basis (X or Z) to measure the incoming pair of non-orthogonal qubits.
- After several rounds, using a classical channel, Bob announces to Alice the double matching detection events.
- Alice computes the usable frames ${f}_{i}$ where $i=1\dots 6$ (see Figure 3) and sends to Bob the required information to construct such frames (Alice knows which pairs of qubits are paired into a frame).
- Bob constructs the frames grouping the pairs of qubits, then he computes the sifting bits of each frame and sends them back to Alice over a public channel.
- Using the sifting bits and looking up Table 5 Alice identifies Bob’s Matching Results. Given a frame, the sifting bits are correlated with a unique MR because they conform a complete binary set $\{00,01,10,11\}$, thus Alice is allowed to recognize Bob’s MR. Then, Table 4 is used to derive the secret bits.
- On the other side, Bob uses Table 4 to get the shared bits.

#### 5.5. Security of the Sifting Bits

## 6. Error Correction Method

#### 6.1. Picking up Undetected Errors

- $|{0}_{X}\rangle $ is detected as $|{1}_{X}\rangle $ or $|{0}_{Z}\rangle $ results in $|{1}_{Z}\rangle $.
- $|{1}_{X}\rangle $ is detected as $|{0}_{X}\rangle $ or $|{1}_{Z}\rangle $ results in $|{0}_{Z}\rangle $.

#### 6.2. Error-Correction Security Model

- –
- To distill secret bits, Alice will use only 4 types of frames: ${f}_{2}$, ${f}_{3}$, ${f}_{4}$ and ${f}_{6}$ which are represented in Table 12. Alice will incorporate 3 auxiliary frames: ${f}_{7}$, ${f}_{9}$, ${f}_{10}$.
- –
- In case of errors, SS are correctable as demonstrated in Table 10 and Table 11. As implied from these tables, half of the SS must be removed. After Alice informs to Bob which cases must be eliminated (those that come from $\mathrm{SS}=(10,01),(01,01),(01,10),(10,10)$), they keep $\frac{1}{7}$ of the total frames, half of the framing gain. Thus, we say that the secret framing gain is $\frac{1}{7}$. In addition, frames ${f}_{7}$, ${f}_{9}$ and ${f}_{10}$ must be discarded because they are used to detect errors and they do not add up secret bits.
- –
- Since each SS comes from two different frames it can be correlated to one secret bit, this property is demonstrated in Table 13.

## 7. Privacy Pre-Amplification

**Throughput.**The throughput of the framed reconciliation can be computed as $\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{N}{2}$. The throughput of the protocol varies quadratically $O\left({N}^{2}\right)$ with the number of the double matching detection events N.

**Effective Throughput Speed.**As discussed in the previous section, the secret framing gain is $\frac{1}{7}$, so the number of secret bits will be $\frac{1}{7}\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{N}{2}$. A running example of the framed reconciliation is shown in Appendix A. If $N=1000$, the number of secret bits is around 10${}^{5}$. Since the errors can be removed in no more than tens of milliseconds, the throughput speed achieves 10${}^{6}$ bps. Such speed can be further enhanced applying a bigger N and using more computational resources as shown in Table 14 (see also Figure 7).

**Efficiency.**The minimum number of required bits to reconcile the shared frames is $2({n}^{2}-n)$ bits (because there are four publicly revealed bits per frame), but also the total number of revealed bits is $2({n}^{2}-n)$, so the efficiency of the protocol achieves unity.

**Round Trips.**Although this protocol is an interactive reconciliation protocol, it only requires four rounds to be completed. Just a single transmission (from Alice to Bob) is needed for correction bits (the indices of events that must removed and those of the erroneous detection events). No redundant information is required. Other protocols require tens of parity check passes [21]. No extra permutation or interleaving is required to achieve reconciliation.

**Qber.**As we will show in the security analysis section, the protocol remains secure although the eavesdropper could be equipped with unlimited quantum memory and multiple copies of Bob’s quantum states. It is known that the Photon Number Splitting attack (PNS) can be detected when the QBER of the channel is beyond $25\%$ due to Eve’s erroneous basis selection. By contrast, the security of the framed reconciliation method is invariant despite the number of copies that Eve obtains from the quantum channel therefore immune to the PNS attack. In this case, no estimation of the QBER from the quantum channel is needed. Remarkably, we do not see any limit in the QBER of the channel because a single auxiliary null quantum pair is enough to detect all the errors. Remember that each double detection event is combined with each other.

## 8. The Intercept and Resend (IR) Attack

- –
- Alice sends to Bob the pair $(|{0}_{X}\rangle ,|{0}_{Z}\rangle )$ over the quantum channel. Eve measures them and let us assume she obtains a double matching detection event say $(|{0}_{Z}\rangle ,|{0}_{Z}\rangle )$.
- –
- Eve prepares and sends to Bob the quantum pair $(|{0}_{Z}\rangle ,|{0}_{Z}\rangle )$.
- –
- Suppose Eve makes sure that both quantum pulses arrive to Bob’s optical station. There are five possible outcomes: $\left\{\right(|{0}_{Z}\rangle ,|{0}_{Z}\rangle ),(|{0}_{X}\rangle ,|{0}_{X}\rangle ),(|{1}_{X}\rangle ,|{1}_{X}\rangle ),(|{1}_{X}\rangle ,|{0}_{X}\rangle ),(|{0}_{X}\rangle ,|{1}_{X}\rangle \left)\right\}$. Since only one case matches Eve’s double detection event, the probability to get the same result is $\frac{1}{5}$ (the same probability is present when Eve obtains $(|{0}_{Z}\rangle ,|{1}_{Z}\rangle )$ and she resends those states to Bob).

## 9. The Photon Number Splitting Attack

- –
- $\frac{1}{2}$ because of the probability to get a double matching detection event.
- –
- $\frac{1}{2}$ due to basis matching. Eve must measure choosing between two different measurement basis (X or Z).

**Secret Throughput Speed.**Let us represent the shared information between Alice and Bob after they executed privacy pre-amplification as ${I}_{ab}=\frac{1}{7}\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{N}{2}$ where N is the number of double matching detection events. As discussed previously, Eve can obtain $25\%$ of the shared secret information, so Eve can distill ${I}_{ae}=\frac{1}{7}\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{\frac{N}{4}}{2}$, now we can derive the secret throughput speed $\mathsf{\Delta}I={I}_{ab}-{I}_{ae}$ as indicated by Equation (6).

#### 9.1. Quantum Measurement Bases Choice Attack

#### Secret Throughput Speed

## 10. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## Appendix A

**Table A1.**Bob announces to Alice eight double matching detection events N = 8 (enumerated from ${i}_{1}$ to ${i}_{8}$).

Bob’s Detection | Bob’s Public Announcement | Alice’s Original Pair |
---|---|---|

$\left(\right)$ | ${i}_{1}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{2}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{3}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{4}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{5}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{6}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{7}$ | $\left(\right)$ |

$\left(\right)$ | ${i}_{8}$ | $\left(\right)$ |

**Table A2.**Alice constructs the set of useful frames, then she sends to Bob the frame arrangement information: $\{1.{f}_{3}=({i}_{1},{i}_{4}),2.{f}_{3}=({i}_{1},{i}_{6}),3.{f}_{3}=({i}_{3},{i}_{4}),4.{f}_{3}=({i}_{3},{i}_{6}),5.{f}_{6}=({i}_{4},{i}_{5}),6.{f}_{4}=({i}_{4},{i}_{7}),7.{f}_{6}=({i}_{4},{i}_{8}),8.{f}_{2}=({i}_{5},{i}_{6}),9.{f}_{4}=({i}_{6},{i}_{7}),10.{f}_{6}=({i}_{6},{i}_{8})\}$.

1. $\begin{array}{c}{i}_{1}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{4}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{3}$ | 5. $\begin{array}{c}{i}_{4}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{5}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ ${f}_{6}$ | 9. $\begin{array}{c}{i}_{6}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{7}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{4}$ |

2. $\begin{array}{c}{i}_{1}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{6}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{3}$ | 6. $\begin{array}{c}{i}_{4}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{7}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{4}$ | 10. $\begin{array}{c}{i}_{6}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{8}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ ${f}_{6}$ |

3. $\begin{array}{c}{i}_{3}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{4}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{3}$ | 7. $\begin{array}{c}{i}_{4}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{8}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ ${f}_{6}$ | |

4. $\begin{array}{c}{i}_{3}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{6}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{3}$ | 8. $\begin{array}{c}{i}_{5}\phantom{\rule{2.84526pt}{0ex}}\\ {i}_{6}\end{array}$ $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{0}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ ${f}_{2}$ |

**Table A3.**Bob publishes the Sifting String that contains the sifting bits and the measured bits. Alice deduces MR and associates the corresponding secret bit (sb) according to Table 7.

$\begin{array}{c}1.\end{array}$$\begin{array}{c}{i}_{1}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{4}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=00\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,01\end{array}$$\begin{array}{c}{f}_{3}\end{array}$ | $\begin{array}{c}5.\end{array}$$\begin{array}{c}{i}_{4}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{5}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=10\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$$\begin{array}{c}{f}_{6}\end{array}$ | $\begin{array}{c}9.\end{array}$$\begin{array}{c}{i}_{6}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{7}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=11\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$$\begin{array}{c}{f}_{4}\end{array}$ |

$\begin{array}{c}2.\end{array}$$\begin{array}{c}{i}_{1}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{6}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=10\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,01\end{array}$$\begin{array}{c}{f}_{3}\end{array}$ | $\begin{array}{c}6.\end{array}$$\begin{array}{c}{i}_{4}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{7}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=00\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$$\begin{array}{c}{f}_{4}\end{array}$ | $\begin{array}{c}10.\end{array}$$\begin{array}{c}{i}_{6}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{8}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=11\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=11,11\end{array}$$\begin{array}{c}{f}_{6}\end{array}$ |

$\begin{array}{c}3.\end{array}$$\begin{array}{c}{i}_{3}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{4}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=11\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=11,11\end{array}$$\begin{array}{c}{f}_{3}\end{array}$ | $\begin{array}{c}7.\end{array}$$\begin{array}{c}{i}_{4}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{8}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=00\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,11\end{array}$$\begin{array}{c}{f}_{6}\end{array}$ | |

$\begin{array}{c}4.\end{array}$$\begin{array}{c}{i}_{3}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{6}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=01\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,11\end{array}$$\begin{array}{c}{f}_{3}\end{array}$ | $\begin{array}{c}8.\end{array}$$\begin{array}{c}{i}_{5}\phantom{\rule{5.69054pt}{0ex}}\\ {i}_{6}\phantom{\rule{5.69054pt}{0ex}}\\ \mathrm{sifting}\phantom{\rule{4.pt}{0ex}}\mathrm{bits}\end{array}$$\begin{array}{c}\mathrm{MR}=01\\ \left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,01\end{array}$$\begin{array}{c}{f}_{2}\end{array}$ |

**Table A4.**Alice and Bob derive the secret bits according to Table 7. In is this example the number of secret bits is 5 which is consistent with the relation $\frac{1}{7}\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{8}{2}$.

Item | SS | Alice’s Frame | Bob’s MR | sb |
---|---|---|---|---|

1. | SS${}_{33}=10,01$ | ${f}_{3}$ | 00 | remove |

2. | SS${}_{32}=01,01$ | ${f}_{3}$ | 10 | remove |

3. | SS${}_{34}=11,11$ | ${f}_{3}$ | 11 | 0 |

4. | SS${}_{31}=00,11$ | ${f}_{3}$ | 01 | 1 |

5. | SS${}_{63}=10,10$ | ${f}_{6}$ | 10 | remove |

6. | SS${}_{43}=10,10$ | ${f}_{4}$ | 00 | 0 |

7. | SS${}_{61}=00,11$ | ${f}_{6}$ | 00 | 0 |

8. | SS${}_{22}=01,01$ | ${f}_{2}$ | 01 | remove |

9. | SS${}_{42}=01,10$ | ${f}_{4}$ | 11 | remove |

10. | SS${}_{64}=11,11$ | ${f}_{6}$ | 11 | 0 |

Item | Events | Error-Free SS | Frame | Erroneous SS | Operation to Be Implemented |
---|---|---|---|---|---|

1. | (${i}_{1},{i}_{4}$) | SS${}_{33}=10,01$ | ${f}_{3}$ | 00,11 | correct applying SS${}_{33}$ |

2. | (${i}_{1},{i}_{6}$) | SS${}_{32}=01,01$ | ${f}_{3}$ | 11,11 | correct applying SS${}_{32}$ |

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**Figure 1.**BB84 pulses represented in the Bloch sphere. The quantum states prepared by Alice (

**left**) could be $|{0}_{X}\rangle $, $|{1}_{X}\rangle $, $|{0}_{Z}\rangle $, $|{1}_{Z}\rangle $ and the measurement bases Bob could apply are X and Z (

**right**).

**Figure 2.**We represent pairs of quantum states: (

**a**) orthogonal pairs ($|{0}_{Z}\rangle $, $|{1}_{Z}\rangle $) and ($|{0}_{X}\rangle $, $|{1}_{X}\rangle $), (

**b**) non-orthogonal pairs ($|{0}_{X}\rangle $, $|{0}_{Z}\rangle $), ($|{1}_{X}\rangle $, $|{0}_{Z}\rangle $), ($|{0}_{X}\rangle $, $|{1}_{Z}\rangle $) and ($|{1}_{X}\rangle $, $|{1}_{Z}\rangle $) and (

**c**) parallel pairs ($|{0}_{Z}\rangle $, $|{0}_{Z}\rangle $), ($|{1}_{X}\rangle $, $|{1}_{X}\rangle $),($|{0}_{X}\rangle $, $|{0}_{X}\rangle $) and ($|{1}_{Z}\rangle $, $|{1}_{Z}\rangle $).

**Figure 3.**Alice sends the non-orthogonal pair $(|{0}_{X}\rangle ,|{1}_{Z}\rangle )$ to Bob. After a double matching detection event is produced at Bob’s optical system it could register $|{0}_{X}\rangle $ or $|{1}_{Z}\rangle $.

**Figure 4.**Quantum states inside a non-orthogonal pair are separated temporally to avoid losses due to consecutive detection events. The order between two non-orthogonal states is not relevant for the present discussion.

**Figure 5.**We see (at left) the states prepared by Alice, two pairs of non-orthogonal states: $(|{0}_{X}\rangle ,|{1}_{Z}\rangle )$ and $(|{1}_{X}\rangle ,|{0}_{Z}\rangle )$. After a double matching detection event is produced at Bob’s side (in this example two double detection events) the possible matching results are exhibited at the right.

**Figure 6.**The exchange of messages assuming an error free protocol. $NO$ represents the pairs of non-orthogonal qubits, the sub-indices k denote the double matching detection events at Bob’s station, f represents the the required information to construct the frames and s denotes the sifting bits computed by Bob.

**Figure 7.**From the experimental simulation in Table 14 we show the frame contribution when QBER = 30%, 605,060 frames have been created and 237,762 correspond to auxiliary frames. For each frame ${f}_{2}$, ${f}_{3}$, ${f}_{4}$ and ${f}_{6}$ we show the number of frames created (left) and the frames after error correction (right).

**Figure 8.**Alice sends a pair of non-orthogonal states to Bob who obtains a double matching detection at his optical detectors. Eve has a copy of such states, however she has a 0.5 chance to choose the correct measurement basis (X or Z). Furthermore, the probability to get a double matching detection event is 0.5. Therefore, Eve’s probability to get Bob’s result is just 0.25.

**Figure 9.**Alice sends a pair of non-orthogonal states to Bob who obtains a double matching detection event at his optical detectors. Eve has a copy of such states, however he has a 0.5 chance to choose the optimal measurement basis, in this case the $X-Z$ basis. Despite Eve choose between bases $X+Z$ or $X-Z$, the chance to guess Bob’s result is $\frac{9}{16}=0.5625$ so she obtains an inconclusive result with $\frac{6}{16}=0.375$. From here, the probability for Eve to obtain Bob’s measurement result is 0.28125.

**Table 1.**Comparison of reconciliation methods as presented in [22].

Reconciliation | Method | Advantages | Disadvantages |
---|---|---|---|

Interactive | Binary [16] | Easy and simple | Large communication overhead |

Cascade [17] | Easy and simple | ||

Strong ability of error correction | |||

Code based | Winnow [18] | Communication time depending on the rate | Additional errors (Hamming) |

Great Efficiency | |||

LDPC [19] | Correction of errors as Cascade | ||

Improvement of the safety of the protocol |

**Table 2.**Measurement results after a double detection event (matching and non-matching). In the matching case, Bob measures the two quantum states with coincident results. Only double matching events encode a bit. The shared bit comes from the compatible measurement.

Bob’s Basis Measurement | ||||
---|---|---|---|---|

Alice’s Non-Orthogonal Pairs | Matching Event | Non-Matching Event | ||

X | Z | X | Z | |

($|{0}_{X}\rangle $, $|{0}_{Z}\rangle $) | ($|{0}_{X}\rangle $, $|{0}_{X}\rangle $) | ($|{0}_{Z}\rangle $, $|{0}_{Z}\rangle $) | ($|{0}_{X}\rangle $, $|{1}_{X}\rangle $) | ($|{1}_{Z}\rangle $, $|{0}_{Z}\rangle $) |

($|{0}_{X}\rangle $, $|{1}_{Z}\rangle $) | ($|{0}_{X}\rangle $, $|{0}_{X}\rangle $) | ($|{1}_{Z}\rangle $, $|{1}_{Z}\rangle $) | ($|{0}_{X}\rangle $, $|{1}_{X}\rangle $) | ($|{0}_{Z}\rangle $, $|{1}_{Z}\rangle $) |

($|{1}_{X}\rangle $, $|{0}_{Z}\rangle $) | ($|{1}_{X}\rangle $, $|{1}_{X}\rangle $) | ($|{0}_{Z}\rangle $, $|{0}_{Z}\rangle $) | ($|{1}_{X}\rangle $, $|{0}_{X}\rangle $) | ($|{1}_{Z}\rangle $, $|{0}_{Z}\rangle $) |

($|{1}_{X}\rangle $, $|{1}_{Z}\rangle $) | ($|{1}_{X}\rangle $, $|{1}_{X}\rangle $) | ($|{1}_{Z}\rangle $, $|{1}_{Z}\rangle $) | ($|{1}_{X}\rangle $, $|{0}_{X}\rangle $) | ($|{0}_{Z}\rangle $, $|{1}_{Z}\rangle $) |

**Table 3.**There are 6 useful frames: ${f}_{i}$, where $i=1,\dots ,6$ and 8 useless frames ${f}_{j}$, where $j=7,\dots ,14$.

Useful Frames | Useless Frames | ||
---|---|---|---|

${f}_{1}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ | ${f}_{2}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{0}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{7}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{0}_{Z}\rangle \\ |{0}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ | ${f}_{11}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ |

${f}_{3}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{4}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{8}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{0}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{12}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ |

${f}_{5}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{0}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{6}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ | ${f}_{9}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ | ${f}_{13}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{0}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ |

${f}_{10}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{0}_{Z}\rangle \\ |{0}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ | ${f}_{14}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{0}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ |

**Table 4.**There exist four possible Matching Results (MR) for $2\times 2$ frames. The bit produced by a double matching event is represented inside the ket notation with the symbol •. Additionally, each MR has been identified with a binary code left to each frame. After the sifting process such MR code will become part of the secret key.

MR=00 $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{\u2022}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\\ |{\u2022}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\end{array}\right)$ | MR=10 $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{\u2022}_{X}\rangle & -\\ -& |{\u2022}_{Z}\rangle \end{array}\right)$ |

MR=01 $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{\u2022}_{Z}\rangle \\ -\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{\u2022}_{Z}\rangle \end{array}\right)$ | MR=11 $\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{\u2022}_{Z}\rangle \\ |{\u2022}_{X}\rangle & -\end{array}\right)$ |

**Table 5.**To the left we see the 6 usable frames that Alice can prepare to be sent over the quantum channel. Provided Bob obtains the two (required) Matching Results he computes the sifting bits applying the xor function to each column (they are written at the bottom of each frame). The sifting bits which are publicly announced, conform the set $\{00,01,10,11\}$ that does not contain redundancy, so that Alice can identify without ambiguity Bob’s Matching Results.

Alice | Bob | |||
---|---|---|---|---|

${f}_{1}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ |

${f}_{2}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{0}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ |

${f}_{3}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ |

${f}_{4}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ |

${f}_{5}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{0}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ |

${f}_{6}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$ |

**Table 6.**The sifting bits obtained by Bob (written at the bottom of each frame) must be produced from at least two different Matching Results. At the right of each frame we have indicated the corresponding original Alice’s frame.

$\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{1}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{0}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{5}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{2},{f}_{6}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{3},{f}_{4}\end{array}$ |

$\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{1},{f}_{6}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{3}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{2},{f}_{5}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{4}\end{array}$ |

$\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{4},{f}_{5}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{1},{f}_{3}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{0}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{2}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\end{array}$$\begin{array}{c}{f}_{6}\end{array}$ |

$\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{1}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{1},{f}_{3},{f}_{6}\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{1}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\end{array}$$\begin{array}{c}{f}_{2},{f}_{4},{f}_{5}\end{array}$ |

**Table 7.**The Sifting String (SS) which is publicly announced is constructed with the sifting bits and the measured bits. To achieve a secret bit (sb) each SS must be correlated at least to two Matching Results (MR).

SS | MR | Frame | sb | MR | Frame | sb | |
---|---|---|---|---|---|---|---|

Sifting | Measured | (See Table 4) | (See Table 3) | (See Table 4) | (See Table 3) | ||

00 | 00 | 10 | ${f}_{1}$ | 0 | 11 | ${f}_{5}$ | 1 |

00 | 11 | 00 | ${f}_{2},{f}_{6}$ | 0 | 01 | ${f}_{3},{f}_{4}$ | 1 |

01 | 10 | 01 | ${f}_{1},{f}_{6}$ | 0 | 11 | ${f}_{4}$ | 1 |

01 | 01 | 10 | ${f}_{3}$ | 0 | 01 | ${f}_{2},{f}_{5}$ | 1 |

10 | 01 | 00 | ${f}_{1},{f}_{3}$ | 0 | 11 | ${f}_{2}$ | 1 |

10 | 10 | 00 | ${f}_{4},{f}_{5}$ | 0 | 10 | ${f}_{6}$ | 1 |

11 | 11 | 11 | ${f}_{1},{f}_{3},{f}_{6}$ | 0 | 10 | ${f}_{2},{f}_{4},{f}_{5}$ | 1 |

**Table 8.**We list the set of valid Sifting Strings (SS) for each frame ${f}_{i}$. Provided Alice has sent an specific frame to Bob, he returns the SS which must be one of the listed here, otherwise an error is detected. We analyze if an error is detectable when occurs in the 1st (or 2nd) measured bit.

Frame | Valid Sifting Strings (SS) | MR | 1st Bit | Detection | 2nd Bit | Detection | 1st and 2nd Bits | Detection |
---|---|---|---|---|---|---|---|---|

${f}_{1}$ | SS${}_{11}=00,00$ | 10 | 10,10 | yes | 01,01 | yes | 11,11 | no |

SS${}_{12}=01,10$ | 01 | 00,00 | no | 00,11 | yes | 01,01 | yes | |

SS${}_{13}=10,01$ | 00 | 00,11 | yes | 00,00 | no | 10,10 | yes | |

SS${}_{14}=11,11$ | 11 | 10,01 | no | 01,10 | no | 00,00 | no | |

${f}_{2}$ | SS${}_{21}=00,11$ | 00 | 10,01 | no | 10,10 | yes | 00,00 | yes |

SS${}_{22}=01,01$ | 01 | 00,11 | no | 00,00 | yes | 01,10 | yes | |

SS${}_{23}=10,01$ | 11 | 11,11 | no | 00,00 | yes | 01,10 | yes | |

SS${}_{24}=11,11$ | 10 | 01,01 | no | 10,10 | yes | 00,00 | yes | |

${f}_{3}$ | SS${}_{31}=00,11$ | 01 | 01,01 | no | 01,10 | yes | 00,00 | yes |

SS${}_{32}=01,01$ | 10 | 11,11 | no | 00,00 | yes | 10,10 | yes | |

SS${}_{33}=10,01$ | 00 | 00,11 | no | 00,00 | yes | 10,10 | yes | |

SS${}_{34}=11,11$ | 11 | 10,01 | no | 01,10 | yes | 00,00 | yes | |

${f}_{4}$ | SS${}_{41}=00,11$ | 01 | 01,01 | yes | 01,10 | no | 00,00 | yes |

SS${}_{42}=01,10$ | 11 | 00,00 | yes | 11,11 | no | 10,01 | yes | |

SS${}_{43}=10,10$ | 00 | 00,00 | yes | 00,11 | no | 10,01 | yes | |

SS${}_{44}=11,11$ | 10 | 01,01 | yes | 10,10 | no | 00,00 | yes | |

${f}_{5}$ | SS${}_{51}=00,00$ | 11 | 01,10 | yes | 10,01 | yes | 11,11 | no |

SS${}_{52}=01,01$ | 01 | 00,11 | yes | 00,00 | no | 01,10 | yes | |

SS${}_{53}=10,10$ | 00 | 00,00 | no | 00,11 | yes | 10,01 | yes | |

SS${}_{54}=11,11$ | 10 | 01,01 | no | 10,10 | no | 00,00 | no | |

${f}_{6}$ | SS${}_{61}=00,11$ | 00 | 10,01 | yes | 10,10 | no | 00,00 | yes |

SS${}_{62}=01,10$ | 01 | 00,00 | yes | 00,11 | no | 01,01 | yes | |

SS${}_{63}=10,10$ | 10 | 00,00 | yes | 11,11 | no | 01,01 | yes | |

SS${}_{64}=11,11$ | 11 | 10,01 | yes | 01,10 | no | 00,00 | yes |

**Table 9.**Bob measures the quantum pair ($|{0}_{X}\rangle ,|{1}_{Z}\rangle $), but $|{0}_{X}\rangle $ produces $|{1}_{X}\rangle $. This error is identified half of the times using ${f}_{9}$. Similarly, Bob receives ($|{0}_{X}\rangle ,|{0}_{Z}\rangle $) but $|{0}_{Z}\rangle $ results in $|{1}_{Z}\rangle $. This error is successfully managed half of the times using ${f}_{10}$. The second and fourth rows show the erroneous cases. The errors are represented with bars above the bits: $|{\overline{0}}_{X}\rangle $ in ${f}_{9}$ and $|{\overline{0}}_{Z}\rangle $ in ${f}_{10}$.

Alice | Bob | |||
---|---|---|---|---|

${f}_{9}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,00\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,00\end{array}$ |

${f}_{9}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{\overline{0}}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ |

${f}_{10}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,00\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{0}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=00,00\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ |

${f}_{10}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{\overline{0}}_{Z}\rangle \\ {|0\rangle}_{X}\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}{|0\rangle}_{Z}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\\ |{0}_{X}\rangle \phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{1}_{Z}\rangle \\ \phantom{\rule{5.69054pt}{0ex}}-\phantom{\rule{5.69054pt}{0ex}}& \phantom{\rule{5.69054pt}{0ex}}|{0}_{Z}\rangle \phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}-& |{1}_{Z}\rangle \\ |{0}_{X}\rangle & -\phantom{\rule{5.69054pt}{0ex}}\end{array}\right)\\ 0\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}1\\ \mathrm{SS}=01,10\end{array}$ | $\begin{array}{c}\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & -\\ -& |{0}_{Z}\rangle \end{array}\right)\phantom{\rule{5.69054pt}{0ex}}\\ 1\phantom{\rule{5.69054pt}{0ex}}\phantom{\rule{5.69054pt}{0ex}}0\\ \mathrm{SS}=10,10\end{array}$ |

**Table 10.**Error correction map for undetected errors. From Table 8 we list all erroneous cases that keep undetected. The bit underlined in the ket notation flips into the the bit underlined in the Sifting String. Some errors are identified using the auxiliary frames ${f}_{9}$ and ${f}_{10}$. If detection of error is not possible the frame must be removed. For this reason frame ${f}_{1}$ is ambiguous and must be removed.

Frame | Quantum Pair | Sifting String | Detection Frame | Sifting String | Error-Bit | Correction Code |
---|---|---|---|---|---|---|

${f}_{1}$ | $\left(\right)$ | $00,\underline{0}0$ | - | - | 1st | remove |

$10,\underline{0}1$ | ||||||

$\left(\right)$ | $00,0\underline{0}$ | - | - | 2nd | remove | |

$01,1\underline{0}$ | ||||||

${f}_{2}$ | $\left(\right)$ | $10,\underline{0}1$ | - | - | 1st | remove |

$01,\underline{0}1$ | ||||||

$\left(\right)$ | $00,\underline{1}1$ | ${f}_{10}$ | 01,10 | 1st | SS${}_{22}$ | |

$11,\underline{1}1$ | SS${}_{23}$ | |||||

${f}_{3}$ | $\left(\right)$ | $01,\underline{0}1$ | - | - | 1st | remove |

$10,\underline{0}1$ | ||||||

$\left(\right)$ | $11,\underline{1}1$ | ${f}_{9}$ | 10,10 | 1st | SS${}_{32}$ | |

$00,\underline{1}1$ | SS${}_{33}$ |

**Table 11.**Error correction map for undetected errors (cont). Frame ${f}_{5}$ is ambiguous and will be discarded.

Frame | Quantum Pair | Sifting String | Detection Frame | Sifting String | Error-Bit | Correction |
---|---|---|---|---|---|---|

${f}_{4}$ | $\left(\right)$ | $01,1\underline{0}$ | - | - | 2nd | remove |

$10,1\underline{0}$ | ||||||

$\left(\right)$ | $11,1\underline{1}$ | ${f}_{9}$ | 10,10 | 2nd | SS${}_{42}$ | |

$00,1\underline{1}$ | SS${}_{43}$ | |||||

${f}_{5}$ | $\left(\right)$ | $00,\underline{0}0$ | - | - | 1st | remove |

$01,\underline{0}1$ | ||||||

$\left(\right)$ | $00,0\underline{0}$ | - | - | 2nd | remove | |

$10,1\underline{0}$ | ||||||

${f}_{6}$ | $\left(\right)$ | $00,1\underline{1}$ | ${f}_{10}$ | 01,10 | 2nd | SS${}_{62}$ |

$11,1\underline{1}$ | SS${}_{63}$ | |||||

$\left(\right)$ | $10,1\underline{0}$ | - | - | 2nd | remove | |

$01,1\underline{0}$ |

**Table 12.**We list usable (4) frames, it must be included (4) frames ${f}_{7}$, ${f}_{8}$, ${f}_{9}$ and ${f}_{10}$ to verify errors by means of the null quantum pairs.

${f}_{2}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{0}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{3}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{0}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{4}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{0}_{X}\rangle & |{1}_{Z}\rangle \end{array}\right)$ | ${f}_{6}=$$\left(\begin{array}{cc}\phantom{\rule{5.69054pt}{0ex}}|{1}_{X}\rangle & |{1}_{Z}\rangle \\ |{1}_{X}\rangle & |{0}_{Z}\rangle \end{array}\right)$ |

SS | MR | Frame | sb | MR | Frame | sb |
---|---|---|---|---|---|---|

${\mathrm{SS}}_{21}={\mathrm{SS}}_{31}={\mathrm{SS}}_{41}={\mathrm{SS}}_{61}=00,11$ | 00 | ${f}_{2},{f}_{6}$ | 0 | 01 | ${f}_{3},{f}_{4}$ | 1 |

${\mathrm{SS}}_{24}={\mathrm{SS}}_{34}={\mathrm{SS}}_{44}={\mathrm{SS}}_{64}=11,11$ | 11 | ${f}_{3},{f}_{6}$ | 0 | 10 | ${f}_{2},{f}_{4}$ | 1 |

${\mathrm{SS}}_{42}={\mathrm{SS}}_{62}=01,10$ | 01 | ${f}_{6}$ | 0 | 11 | ${f}_{4}$ | 1 |

${\mathrm{SS}}_{22}={\mathrm{SS}}_{32}=01,01$ | 10 | ${f}_{3}$ | 0 | 01 | ${f}_{2}$ | 1 |

${\mathrm{SS}}_{23}={\mathrm{SS}}_{33}=10,01$ | 00 | ${f}_{3}$ | 0 | 11 | ${f}_{2}$ | 1 |

${\mathrm{SS}}_{43}={\mathrm{SS}}_{63}=10,10$ | 00 | ${f}_{4}$ | 0 | 10 | ${f}_{6}$ | 1 |

**Table 14.**Simulation of the protocol when have been registered 1000 double matching detection events. Tests were performed in an Intel Core i7-8750H 2.2 GHz, 12GB RAM.

QBER | Time (ms) | Secret Bits | Throughput (Kbps) |
---|---|---|---|

$5\%$ | 54.0146 | 59,873.7 | 1108.90435 |

$10\%$ | 57.6022 | 58,911.2 | 1025.13229 |

$15\%$ | 54.0614 | 52,630.1 | 972.532054 |

$20\%$ | 55.6709 | 48,830.9 | 877.799205 |

$25\%$ | 60.9381 | 46,706.4 | 773.520113 |

$30\%$ | 78.4297 | 40,960.0 | 522.251137 |

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Lizama-Perez, L.A.; López, J.M.
Quantum Key Distillation Using Binary Frames. *Symmetry* **2020**, *12*, 1053.
https://doi.org/10.3390/sym12061053

**AMA Style**

Lizama-Perez LA, López JM.
Quantum Key Distillation Using Binary Frames. *Symmetry*. 2020; 12(6):1053.
https://doi.org/10.3390/sym12061053

**Chicago/Turabian Style**

Lizama-Perez, Luis A., and J. Mauricio López.
2020. "Quantum Key Distillation Using Binary Frames" *Symmetry* 12, no. 6: 1053.
https://doi.org/10.3390/sym12061053