3. The ack-state protocol

previously measured. In 50% of the measurements, Eve successfully chooses the correct measurement basis, while Bob chooses the same basis as her half of the time. Given that, she generates a quantum bit error rate (QBER) of 50% � 50% ¼ 25% (see Figure 2 and a

In the intercept resend with faked states (IRFS) attack, the eavesdropper does not want to reconstruct the original states. Instead, it produces pulses of light controlled by her that are detectable by Bob as she stays unnoticed in the quantum channel. Due to imperfections in their optical system, Alice and Bob assume that the quantum states they are detecting are the original ones while they are actually detecting light pulses generated by the eavesdropper. Those light pulses are known as faked states [10]. There are several weaknesses in Bob's detector than can be exploited to perform this attack such as time shift [11–13] or quantum blinding [10–12]. When using quantum blinding (quantum blinding attack), the QKD system is controlled by an eavesdropper who uses bright photon pulses during the linear mode operation of the APDs. Using this attack, Eve can eavesdrop on the full secret key but it will not increase the QBER of the protocol. To do this, Eve sends bright pulses to Bob and those are detected by the APD. It will then operate like a classical photo diode

instead of operating in Geiger mode and allowing Eve to obtain the key [14, 15].

the classical post-processing, getting the same secret bit as Alice and Bob.

over a commercial QKD system was for the first time implemented [15].

Resulting from this, as shown in Figure 3a, when Bob selects the same measurement basis Eve has chosen, a detection event occurs in the corresponding APD detector. On the other hand, if Bob measures using the opposite basis, as in Figure 3b, the two detectors get a part of the optical power and no event is detected. In this way, the eavesdropper blinds Bob's APD detectors and makes them work as classical photo diodes. In the final stage of the protocol, Eve uses the announcements made by Bob on the public channel to execute

A watchdog detector that can detect bright faked states can be used as a very simple countermeasure and it can be applied in the electronic detection system [16]. In the University of Singapore an intercept resend attack with faked states and quantum blinding

Figure 2. An intercept resend (IR) attack toward the BB84 protocol causes a quantum bit error rate (QBER) of 25% that can be detected. The figure shows Alice sending a ∣0Zi state to Bob. In the middle of the quantum channel is Eve applying an X basis measurement and she gets ∣1Xi. Consequently, she makes a copy of that state and sends it to Bob who gets ∣1Zi as he used the Z basis measurement. The process introduces an error in the secret bit given that Alice expects Bob to get

study by Bennett et al. [7]).

40 Advanced Technologies of Quantum Key Distribution

∣0Zi.

3. Intercept resend with faked states (IRFS) attack.

Consider a BB84-based protocol encoding a classical bit that uses one of the four non-orthogonal quantum states ∣þXi, ∣�Xi, ∣þZi, and ∣�Zi (see Figure 1). When using the SARG04 protocol [25], Alice produces one of the four BB84 quantum states she will send to Bob, it means, she produces a state associated with two conjugate basis (X and Z). Classical bits on SARG04 protocol are encoded as follows: 0 is coded with∣þZi and ∣�Zi and 1 is coded with ∣þXi and ∣�Xi (see Figure 4) where black dots in the bidimensional Bloch sphere represent the qubits (the non-orthogonal states are right angled and the orthogonal states are represented as diametrically opposed and the parallel states have the same position in the sphere). The basis measurement X and Z appear as horizontal and vertical lines, respectively. In contraposition, the BB84 protocol encodes the bit 0 as ∣þZi and ∣�Xi and the bit 1 with ∣�Zi and ∣þXi.

result ∣þXi; and as this result can be obtained for both states in the set Sð Þ <sup>þ</sup>;<sup>þ</sup> ; he needs to dispose of the bit 1 from ∣þXi. In case Bob measures using the Z basis measurement and obtains ∣þZi, once more, he is not able to distinguish the state sent by Alice. In the opposite way, if he measures in the Z basis and gets ∣�Zi, he is sure Alice sent ∣þXi and adds a 0 to his key. On her side, Eve needs to perform a measurement using the conjugate basis X and Z to obtain the same secret bit as Bob, demanding multi-photonic pulses with at least three pho-

Quantum Flows for Secret Key Distribution http://dx.doi.org/10.5772/intechopen.75964 43

Similar to the BB84, in the ack-state protocol, Alice encodes a classical bit as: 0 is encoded with ∣þZi and ∣�Xi and 1 is encoded with ∣�Zi and ∣þXi. And also, in the same manner as the SARG04 protocol, the ack-state uses the four sets of non-orthogonal states Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi , Sð Þ <sup>þ</sup>;� ¼ jþf g <sup>X</sup>i; j�Zi , Sð Þ �;<sup>þ</sup> ¼ j�f g <sup>X</sup>i; jþZi , and Sð Þ �;� ¼ j�f g <sup>X</sup>i; j�Zi . But in the ack-state protocol the set Alice used, Sð Þ <sup>þ</sup>;<sup>þ</sup> , Sð Þ <sup>þ</sup>;� , Sð Þ �;<sup>þ</sup> or Sð Þ �;� , is never revealed. As an illustration, suppose Alice chooses the set Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi rather than transmitting one of the two states, say ∣þXi, and publishing the sifting instance, Sð Þ <sup>þ</sup>;<sup>þ</sup> , she transmits the two states ∣þXi and ∣þZi. At that point, Bob measures the states using the same basis, X or Z, one by one, as the two states reach successively. If Bob measures with the X basis, he surely will obtain ∣þXi (after he measures the first state) but he can obtain ∣þXi or ∣�Xi on the second measurement, with a probability of 0.5 for each event. If Bob obtains f g jþXi; j�Xi after the second measurement, the result is unclear to him and he has to discard it. On the other hand, if he gets f g jþXi; jþXi the result is unambiguous and he should add a bit 1 to his key. With the purpose of allowing Alice to recover the same bit, Bob makes the announcement of the basis measurement X and the matching condition in accordance with the following criterion: 2ð Þ M if the two detection events make clicks on the same detector; it includes the cases f g jþXi; jþXi , f g j�Xi; j�Xi , fjþZi; jþZig, f g j�Zi; j�Zi and (2nM) if the detection event makes clicks on the opposite detectors, for example, f g jþXi; j�Xi , f g j�Xi; jþXi , f g jþZi; j�Zi , f g j�Zi; jþZi . Alice obtains the secret bit given that the f g jþXi; jþZi states she sent, the X basis, and the 2ð Þ M measurement result permit her to conclude that Bob definitely got f g þX; þ<sup>X</sup> (consider the cases depicted in

Contrarily, in the case Bob measured the two states ∣þXi and ∣þZi with the Z basis, he would acquire one of the two possible results: 2ð Þ ¼ jþ M f g <sup>Z</sup>i; jþZi or 2ð Þ ¼ j� nM f g <sup>Z</sup>i; jþZi . In the first case, he publishes the Z basis and the 2ð Þ M result; then Alice and Bob add a 0 to the key. In the second case, Bob makes the announcement of the Z basis and the 2ð Þ nM result but in this case, they discard the result. When using the ack-state protocol the 2ð Þ M results can be used to

Alice sends Bob obtains a 2ð Þ M Secret bit

Table 1. Using the X basis, Bob measures the two states sent by Alice and he obtains a (2M) result.

f g jþXi; jþZi f g jþXi; jþXi 1 f g jþXi; j�Zi f g jþXi; jþXi 1 f g j�Xi; jþZi f g j�Xi; j�Xi 0 f g j�Xi; j�Zi f g j�Xi; j�Xi 0

tons.

Table 1).

In the sifting phase of the SARG04 protocol, the basis used by Alice is not revealed as this would reveal the bit. As a substitute, she declares to which sifting set the state belongs in accordance with the following four sifting sets: Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi , Sð Þ <sup>þ</sup>;� ¼ jþf g <sup>X</sup>i; j�Zi , Sð Þ �;<sup>þ</sup> ¼ j�f g <sup>X</sup>i; jþZi , and Sð Þ �;� ¼ j�f g <sup>X</sup>i; j�Zi . For instance, consider that Alice sends ∣þXi and she announces the set Sð Þ <sup>þ</sup>;<sup>þ</sup> . Bob makes his measurements on the X basis and he gets the

Figure 4. The non-orthogonal states used in the SARG04 protocol encodes the bit 0 with the states ∣þZi and ∣�Zi and the bit 1 is encoded with ∣þXi and ∣�Xi.

result ∣þXi; and as this result can be obtained for both states in the set Sð Þ <sup>þ</sup>;<sup>þ</sup> ; he needs to dispose of the bit 1 from ∣þXi. In case Bob measures using the Z basis measurement and obtains ∣þZi, once more, he is not able to distinguish the state sent by Alice. In the opposite way, if he measures in the Z basis and gets ∣�Zi, he is sure Alice sent ∣þXi and adds a 0 to his key. On her side, Eve needs to perform a measurement using the conjugate basis X and Z to obtain the same secret bit as Bob, demanding multi-photonic pulses with at least three photons.

protocol are encoded as follows: 0 is coded with∣þZi and ∣�Zi and 1 is coded with ∣þXi and ∣�Xi (see Figure 4) where black dots in the bidimensional Bloch sphere represent the qubits (the non-orthogonal states are right angled and the orthogonal states are represented as diametrically opposed and the parallel states have the same position in the sphere). The basis measurement X and Z appear as horizontal and vertical lines, respectively. In contraposition, the

In the sifting phase of the SARG04 protocol, the basis used by Alice is not revealed as this would reveal the bit. As a substitute, she declares to which sifting set the state belongs in accordance with the following four sifting sets: Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi , Sð Þ <sup>þ</sup>;� ¼ jþf g <sup>X</sup>i; j�Zi , Sð Þ �;<sup>þ</sup> ¼ j�f g <sup>X</sup>i; jþZi , and Sð Þ �;� ¼ j�f g <sup>X</sup>i; j�Zi . For instance, consider that Alice sends ∣þXi and she announces the set Sð Þ <sup>þ</sup>;<sup>þ</sup> . Bob makes his measurements on the X basis and he gets the

Figure 4. The non-orthogonal states used in the SARG04 protocol encodes the bit 0 with the states ∣þZi and ∣�Zi and the bit

1 is encoded with ∣þXi and ∣�Xi.

BB84 protocol encodes the bit 0 as ∣þZi and ∣�Xi and the bit 1 with ∣�Zi and ∣þXi.

42 Advanced Technologies of Quantum Key Distribution

Similar to the BB84, in the ack-state protocol, Alice encodes a classical bit as: 0 is encoded with ∣þZi and ∣�Xi and 1 is encoded with ∣�Zi and ∣þXi. And also, in the same manner as the SARG04 protocol, the ack-state uses the four sets of non-orthogonal states Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi , Sð Þ <sup>þ</sup>;� ¼ jþf g <sup>X</sup>i; j�Zi , Sð Þ �;<sup>þ</sup> ¼ j�f g <sup>X</sup>i; jþZi , and Sð Þ �;� ¼ j�f g <sup>X</sup>i; j�Zi . But in the ack-state protocol the set Alice used, Sð Þ <sup>þ</sup>;<sup>þ</sup> , Sð Þ <sup>þ</sup>;� , Sð Þ �;<sup>þ</sup> or Sð Þ �;� , is never revealed. As an illustration, suppose Alice chooses the set Sð Þ <sup>þ</sup>;<sup>þ</sup> ¼ jþf g <sup>X</sup>i; jþZi rather than transmitting one of the two states, say ∣þXi, and publishing the sifting instance, Sð Þ <sup>þ</sup>;<sup>þ</sup> , she transmits the two states ∣þXi and ∣þZi. At that point, Bob measures the states using the same basis, X or Z, one by one, as the two states reach successively. If Bob measures with the X basis, he surely will obtain ∣þXi (after he measures the first state) but he can obtain ∣þXi or ∣�Xi on the second measurement, with a probability of 0.5 for each event. If Bob obtains f g jþXi; j�Xi after the second measurement, the result is unclear to him and he has to discard it. On the other hand, if he gets f g jþXi; jþXi the result is unambiguous and he should add a bit 1 to his key. With the purpose of allowing Alice to recover the same bit, Bob makes the announcement of the basis measurement X and the matching condition in accordance with the following criterion: 2ð Þ M if the two detection events make clicks on the same detector; it includes the cases f g jþXi; jþXi , f g j�Xi; j�Xi , fjþZi; jþZig, f g j�Zi; j�Zi and (2nM) if the detection event makes clicks on the opposite detectors, for example, f g jþXi; j�Xi , f g j�Xi; jþXi , f g jþZi; j�Zi , f g j�Zi; jþZi . Alice obtains the secret bit given that the f g jþXi; jþZi states she sent, the X basis, and the 2ð Þ M measurement result permit her to conclude that Bob definitely got f g þX; þ<sup>X</sup> (consider the cases depicted in Table 1).

Contrarily, in the case Bob measured the two states ∣þXi and ∣þZi with the Z basis, he would acquire one of the two possible results: 2ð Þ ¼ jþ M f g <sup>Z</sup>i; jþZi or 2ð Þ ¼ j� nM f g <sup>Z</sup>i; jþZi . In the first case, he publishes the Z basis and the 2ð Þ M result; then Alice and Bob add a 0 to the key. In the second case, Bob makes the announcement of the Z basis and the 2ð Þ nM result but in this case, they discard the result. When using the ack-state protocol the 2ð Þ M results can be used to


Table 1. Using the X basis, Bob measures the two states sent by Alice and he obtains a (2M) result.

distill secret bits but 2ð Þ nM is unclear causing those measurement outcomes to be useless and so they have to be discarded.

The ack-state protocol was introduced in [4]. In such a reference, the non-orthogonal states are called protocol states while parallel states are named decoy states. The ack-state protocol encodes one classical bit using two quantum states. Such encoding is done by means of non-orthogonal or parallel states. In quantum physics, if X ¼ j f i 0<sup>X</sup> ; j1Xig and Z ¼ j f i 0<sup>Z</sup> ; j1Zig are orthonormal bases, then the magnitude of each basis vector is the unity and any vector in such a space can be written as a linear combination of such basis. For instance, ∣0Xi can be rewritten as 1ffiffi 2 <sup>p</sup> <sup>∣</sup>0Zi þ <sup>1</sup>ffiffi 2 <sup>p</sup> ∣1Zi. Two qubits ∣0Xi and ∣0Zi are non-orthogonal if the inner product between them is different from zero, symbolically 0X∣0<sup>Z</sup> 6¼ 0. In consequence, 0X∣0<sup>Z</sup> <sup>¼</sup> <sup>1</sup>ffiffi 2 <sup>p</sup> ð Þþ <sup>1</sup> <sup>1</sup>ffiffi 2 <sup>p</sup> ð Þ0 and <sup>0</sup>X∣0<sup>Z</sup> <sup>¼</sup> <sup>1</sup>ffiffi 2 <sup>p</sup> . The inner product of orthogonal qubits is zero, for example, 0X∣1<sup>X</sup> ¼ 0 and identical (or parallel) qubits produce the unity under the inner product; thus, 0X∣0<sup>X</sup> ¼ 1.

Using this protocol, Alice chooses at random between sending a pair of parallel or nonorthogonal states. At the opposite side, Bob makes the measurement of the two successive pulses he receives with the same basis measurement, X or Z (see Figure 5). In this context, the pair of quantum states sent by Alice is called biqubit. Parallel biqubits define the parallel quantum flow and non-orthogonal biqubits define the non-orthogonal quantum flow. Summarizing the ack-state protocol with non-orthogonal and parallel states, we have the following:


Table 2 shows the results after Bob measures two consecutive states. Thus, one of the following detection events can be obtained:

the qubit ∣0Zi. As the X basis is used by Bob to measure both qubits, the qubit ∣0Xi is measured as ∣0Xi but the qubit ∣0Zi is measured as ∣0Xi or ∣1Xi with an equal probability of 50%. When Bob's measurement generates ∣0Xi, we say that this measurement is the ack of the first ∣0Xi state. Vice versa, if Bob gets ∣1Xi, we say that ∣1Xi is the negative

Figure 5. In this representation, two concentric circles define the order in which the states are prepared and sent. Therefore, the state that is first sent is contained in the inner circle state, and the outer circle state is prepared and transmitted. Alice at random interleaves orthogonal (non-orthogonal) and parallel states, given that she can verify the matching cases after Bob measurements. In the ack-state protocol, Bob uses the basis X Zð Þ to measure the two Alice's

f i j0<sup>Z</sup> ; j1Zig or f i j1<sup>Z</sup> ; j1Zig with the same probability. Alice decides to send at random two consecutive non-orthogonal states from the set: f iÞ ð i j0<sup>X</sup> ; j0<sup>Z</sup> ;ð i j0<sup>X</sup> ; j1ZiÞ;ð i j0<sup>Z</sup> ; j1XiÞ;ð i j1<sup>Z</sup> ; j1XiÞg. Bob will measure those states using the same measurement basis (X or Z). The parallel biqubits involve the following states: f iÞ ðj0Xi; j0<sup>X</sup> ;ðj1Xi; j1XiÞ;ðj0Zi; j0ZiÞ;ðj1Zi; j1ZiÞg. In the nack-state protocol Alice chooses randomly two consecutive parallel states as the case depicted in (c) ðj1Zi; j1ZiÞ. They produce a compatible measurement if Bob chooses, X for ∣iXi or Z for ∣iZi where i ¼ 0, 1. We represent

<sup>Z</sup>ig. He effectively gets the bit i jð Þ provided he measures f i jiX ; jiXig or jj

<sup>2</sup> probability. For instance, if Bob uses the Z basis to measure the incoming states f i j0<sup>X</sup> ; j1Zig he can obtain

Z ; <sup>j</sup><sup>j</sup>

Quantum Flows for Secret Key Distribution http://dx.doi.org/10.5772/intechopen.75964 45

<sup>Z</sup>ig which

ii. The single-detection event occurs when one state is lost and Bob obtains only one detection event. The symbol ð Þ �; ∓ is used to designate the single-detection event. More specifically, Bob uses the symbol ð Þ S-i to represent the single-detection event, where i can be 1 or

acknowledgment (the nack) of ∣0Xi.

non-orthogonal states f i jiX ; jj

occurs with <sup>1</sup>

In a channel with losses, we have two more possible results.

in (b) the case of quantum orthogonal states. Two cases are possible here: f iÞ ðj0Xi; j1<sup>X</sup> ;ðj0Zi; j1ZiÞg.

i. The states generate a double-detection event: The symbol ð Þ þ; þ is used to designate the photonic gain in a double-detection event. When both events are registered in a same detector, we call it a double-matching 2ð Þ M detection event. If the results of the measurements of the states are opposite, then we face a double non-matching 2ð Þ M detection event. Whereas 2ð Þ M non-orthogonal outcomes are useful to distill secret bits, the 2ð Þ M results cannot be used and are disposed. When we have a 2ð Þ M detection event, we may say that the second measurement is the acknowledgment (the ack) of the first measurement. In Figure 5 (top-right) the qubit ∣0Xi is the first one sent by Alice and then she sends

distill secret bits but 2ð Þ nM is unclear causing those measurement outcomes to be useless and

The ack-state protocol was introduced in [4]. In such a reference, the non-orthogonal states are called protocol states while parallel states are named decoy states. The ack-state protocol encodes one classical bit using two quantum states. Such encoding is done by means of non-orthogonal or parallel states. In quantum physics, if X ¼ j f i 0<sup>X</sup> ; j1Xig and Z ¼ j f i 0<sup>Z</sup> ; j1Zig are orthonormal bases, then the magnitude of each basis vector is the unity and any vector in such a space can be written as a linear combination of such basis. For instance, ∣0Xi can be rewritten as

is different from zero, symbolically 0X∣0<sup>Z</sup> 6¼ 0. In consequence, 0X∣0<sup>Z</sup> <sup>¼</sup> <sup>1</sup>ffiffi

(or parallel) qubits produce the unity under the inner product; thus, 0X∣0<sup>X</sup> ¼ 1.

2. At random, Bob chooses the basis X or Z to measure the received biqubit.

4. After analyzing those results, Alice tells Bob which cases to discard.

detected event (S-1 or S-2), or a lost biqubit 2ð Þ L (see the discussion below).

<sup>p</sup> ∣1Zi. Two qubits ∣0Xi and ∣0Zi are non-orthogonal if the inner product between them

<sup>p</sup> . The inner product of orthogonal qubits is zero, for example, 0X∣1<sup>X</sup> ¼ 0 and identical

Using this protocol, Alice chooses at random between sending a pair of parallel or nonorthogonal states. At the opposite side, Bob makes the measurement of the two successive pulses he receives with the same basis measurement, X or Z (see Figure 5). In this context, the pair of quantum states sent by Alice is called biqubit. Parallel biqubits define the parallel quantum flow and non-orthogonal biqubits define the non-orthogonal quantum flow. Summariz-

1. Alice randomly selects between a non-orthogonal biqubit and a parallel bi-qubit. In case she selects a non-orthogonal biqubit, she has to select at random one of the following states: fð Þ j i 0<sup>X</sup> ; j i 0<sup>Z</sup> ;ð Þ j i 0<sup>X</sup> ; j i 1<sup>Z</sup> ;ðj1Xi; j i 0<sup>Z</sup> Þ;ð Þ j i 1<sup>X</sup> ; j i 1<sup>Z</sup> g, where the order between states X or Z is as well picked at random. In case she selects a parallel biqubit, she should randomly choose a biqubit from the set: 0 fð Þ j i <sup>X</sup> ; j i 0<sup>X</sup> ;ð Þ j i 1<sup>X</sup> ; j i 1<sup>X</sup> ;ðj0Zi; j i 0<sup>Z</sup> Þ;ð Þ j i 1<sup>Z</sup> ; j i 1<sup>Z</sup> g. and then she gets it

3. Bob's basis of measurement is announced by him over the public channel and he also declares if the result obtained is either a double-detected event (2M or 2nM), a single-

Table 2 shows the results after Bob measures two consecutive states. Thus, one of the follow-

i. The states generate a double-detection event: The symbol ð Þ þ; þ is used to designate the photonic gain in a double-detection event. When both events are registered in a same detector, we call it a double-matching 2ð Þ M detection event. If the results of the measurements of the states are opposite, then we face a double non-matching 2ð Þ M detection event. Whereas 2ð Þ M non-orthogonal outcomes are useful to distill secret bits, the 2ð Þ M results cannot be used and are disposed. When we have a 2ð Þ M detection event, we may say that the second measurement is the acknowledgment (the ack) of the first measurement. In Figure 5 (top-right) the qubit ∣0Xi is the first one sent by Alice and then she sends

ing the ack-state protocol with non-orthogonal and parallel states, we have the following:

2 <sup>p</sup> ð Þþ <sup>1</sup> <sup>1</sup>ffiffi

2 <sup>p</sup> ð Þ0 and

so they have to be discarded.

44 Advanced Technologies of Quantum Key Distribution

2

2

ready and transmits it to Bob.

ing detection events can be obtained:

1ffiffi 2 <sup>p</sup> <sup>∣</sup>0Zi þ <sup>1</sup>ffiffi

<sup>0</sup>X∣0<sup>Z</sup> <sup>¼</sup> <sup>1</sup>ffiffi

Figure 5. In this representation, two concentric circles define the order in which the states are prepared and sent. Therefore, the state that is first sent is contained in the inner circle state, and the outer circle state is prepared and transmitted. Alice at random interleaves orthogonal (non-orthogonal) and parallel states, given that she can verify the matching cases after Bob measurements. In the ack-state protocol, Bob uses the basis X Zð Þ to measure the two Alice's non-orthogonal states f i jiX ; jj <sup>Z</sup>ig. He effectively gets the bit i jð Þ provided he measures f i jiX ; jiXig or jj Z ; <sup>j</sup><sup>j</sup> <sup>Z</sup>ig which occurs with <sup>1</sup> <sup>2</sup> probability. For instance, if Bob uses the Z basis to measure the incoming states f i j0<sup>X</sup> ; j1Zig he can obtain f i j0<sup>Z</sup> ; j1Zig or f i j1<sup>Z</sup> ; j1Zig with the same probability. Alice decides to send at random two consecutive non-orthogonal states from the set: f iÞ ð i j0<sup>X</sup> ; j0<sup>Z</sup> ;ð i j0<sup>X</sup> ; j1ZiÞ;ð i j0<sup>Z</sup> ; j1XiÞ;ð i j1<sup>Z</sup> ; j1XiÞg. Bob will measure those states using the same measurement basis (X or Z). The parallel biqubits involve the following states: f iÞ ðj0Xi; j0<sup>X</sup> ;ðj1Xi; j1XiÞ;ðj0Zi; j0ZiÞ;ðj1Zi; j1ZiÞg. In the nack-state protocol Alice chooses randomly two consecutive parallel states as the case depicted in (c) ðj1Zi; j1ZiÞ. They produce a compatible measurement if Bob chooses, X for ∣iXi or Z for ∣iZi where i ¼ 0, 1. We represent in (b) the case of quantum orthogonal states. Two cases are possible here: f iÞ ðj0Xi; j1<sup>X</sup> ;ðj0Zi; j1ZiÞg.

the qubit ∣0Zi. As the X basis is used by Bob to measure both qubits, the qubit ∣0Xi is measured as ∣0Xi but the qubit ∣0Zi is measured as ∣0Xi or ∣1Xi with an equal probability of 50%. When Bob's measurement generates ∣0Xi, we say that this measurement is the ack of the first ∣0Xi state. Vice versa, if Bob gets ∣1Xi, we say that ∣1Xi is the negative acknowledgment (the nack) of ∣0Xi.

In a channel with losses, we have two more possible results.

ii. The single-detection event occurs when one state is lost and Bob obtains only one detection event. The symbol ð Þ �; ∓ is used to designate the single-detection event. More specifically, Bob uses the symbol ð Þ S-i to represent the single-detection event, where i can be 1 or


ðj1Zi; j1ZiÞ and two orthogonal biqubits ðj0Xi; j1XiÞ,ðj0Zi; j1ZiÞ. The parallel and orthogonal biqubits are interleaved at random by Alice. The performance of the protocol is not altered by order of the quantum states within the biqubit (see Figure 5). On the opposite side of the

X ∣0Xi, ∣1Xi X, 2nM Compatible double non-matching, useful

Z ∣0Zi, ∣0Zi Z, 2M Non-compatible double matching, useless Z ∣1Zi, ∣1Zi Z, 2M Non-compatible double matching, useless

Z ∣1Zi, ∣0Zi Z, 2M Non-compatible double non-matching, useless Z ∣0Zi, � Z, S<sup>1</sup> Non-compatible single matching, useless Z ∣1Zi, � Z, S<sup>1</sup> Non-compatible single matching, useless Z �, ∣0Zi Z, S<sup>2</sup> Non-compatible single matching, useless Z �, ∣1Zi Z, S<sup>2</sup> Non-compatible single matching, useless

X ∣0Xi, � X, S<sup>1</sup> Compatible single matching, useful X �, ∣1Xi X, S<sup>2</sup> Compatible single matching, useful

∣0Xi, ∣1Xi Z ∣0Zi, ∣1Zi Z, 2M Non-compatible double non-matching, useless

Z ∣1Zi, ∣1Zi Z, 2M Compatible double matching, useful Z ∣1Zi, � Z, S<sup>1</sup> Compatible single matching, useful Z �, ∣1Zi Z, S<sup>2</sup> Compatible single matching, useful

X ∣0Xi, ∣0Xi Z, 2M Non-compatible double matching, useless X ∣1Xi, ∣1Xi Z, 2M Non-compatible double matching, useless

X ∣1Xi, ∣0Xi Z, 2nM Non-compatible double non-matching, useless X ∣0Xi, � X, S<sup>1</sup> Non-compatible single matching, useless X ∣1Xi, � X, S<sup>1</sup> Non-compatible single matching, useless X �, ∣0Xi X, S<sup>2</sup> Non-compatible single matching, useless X �, ∣1Xi X, S<sup>2</sup> Non-compatible single matching, useless

∣1Zi, ∣1Zi X ∣0Xi, ∣1Xi Z, 2nM Non-compatible double non-matching, useless

We expect Alice to send the biqubits ∣0Xi, ∣1Xi and ∣1Zi, ∣1Zi; at that point, every conceivable measurement result at Bob's detector is written. We exhibit the detection event and Bob's corresponding advertisement over the public channel according to Bob's basis selection. Notice that the number of the single detections inside the biqubit, first or second, is

Table 3. The nack-state protocol running without blunders in the quantum channel is shown with each of the possible

X �, � X, Lost Biqubit lost

Z �, � Z, Lost Biqubit lost

Z �, � Z, Lost Biqubit lost

X �, � X, Lost Biqubit lost

openly declared by Bob.

measurement results at Bob's detectors.

As two compatible single-detection events

Quantum Flows for Secret Key Distribution http://dx.doi.org/10.5772/intechopen.75964 47

Alice's Bob's Detection Public Description

Biqubit Basis Event Disclosure

Table 2. Alice sends to Bob the non-orthogonal states ðj0Xi; j0ZiÞ and it shows all the possible measurement results at Bob's side.

2, depending on the state number that makes clicks after the basis measurement X or Z is applied to the two consecutive incoming states. This way, the number i will be published by Bob.

iii. The two pulses are lost. This case is denoted as ð Þ �; � or alternatively as 2L.

When applying the ack-state protocol, two consecutive non-orthogonal states are used by Alice and Bob to distill one secret bit. The basis measurement X or Z is declared publicly by Bob along with the sifting instances; he obtained 2ð Þ M , 2ð Þ M , ð Þ S-1 , ð Þ S-2 , and 2ð Þ L . Furthermore, the bits acquired from the single-detection events ð Þ S-1 and ð Þ S-2 are used by Alice to confirm the single photonic gain of the quantum channel.
