5. The photon number splitting attack

In the PNS attack, the eavesdropper captures no less than one photon from each of the multiphoton states with the purpose of storing them in quantum memory, at the same time that she hinders the single photon states in the quantum channel. When Bob has uncovered over public channels the measurement basis he has used, the eavesdropper executes the same measurements on the quantum states she has stored [25].

When the PNS attack is applied to the ack-state protocol, the eavesdropper captures no less than one photon of the multi-photon states (parallel and non-orthogonal), and she stands by Bob's declarations about the measurement bases he has utilized with the aim of applying the same measurements on her stored states. In Bob's side, a distribution over the following sifting events is achieved 2ð Þ M , 2ð Þ nM , ð Þ S-1 , ð Þ S-2 and 2ð Þ L , where every one may originate in parallel or non-orthogonal states; however, just Alice knows those outcomes.

After Bob declares both the measurement bases (X or Z) and the sifting occurrences, Eve executes the measurements utilizing the same measurement bases and she gets the same bits from the multi-photonic single sifting instances: ð Þ S-1 and ð Þ S-2 , parallel and non-orthogonal. Moreover, the same outcomes from the 2ð Þ M measurements of the parallel and (a half of the) non-orthogonal multi-photonic states are acquired by the eavesdropper. However, she cannot acquire the secret bits from the 1-state ð Þ S-i and 2ð Þ M sifting occurrences, given that the eavesdropped cannot discriminate parallel and non-orthogonal states.

In order to get the secret bits, Eve obstructs the 1-photon states which incorporate single and double-detection events from parallel and non-orthogonal states. In doing that, an error gain in the photonic gain of the single and double-detection events is introduced by Eve. At that point, Eve executes a channel substitution expanding the transmittance of the channel. The fiber channel transmittance among Alice and Bob is written as TAB <sup>¼</sup> <sup>10</sup>�α<sup>l</sup> <sup>10</sup> where α is the loss coefficient measured in dB=km and the length l is measured in km. Moreover, the local transmittance at Bob's side, ηB, is defined as tBη<sup>D</sup> where tB is the internal transmittance of optical components and η<sup>D</sup> is the quantum efficiency of Bob's detectors. Then, the general transmission and detection efficiency at Bob's side ηBT is computed as ηBT ¼ tBηDTAB [18]. A mathematical description of the gain of detection events will be presented in the following section.

#### 5.1. The gain of detection events

quantum channel, Bob measures two incoming states of a biqubit utilizing the same measure-

1. Alice is equipped with a photon source with an expected photon number μ showing Poisson distribution. A parallel or an orthogonal biqubit is selected at random by Alice, and

2. The biqubit (two incoming pulses) is measured by Bob using the same measurement basis X (or Z) that he selects haphazardly (in a further section, we discuss the convenience of avoiding consecutiveness of states and how it can be prevented if Alice forwards a burst of

4. Alice and Bob perform sifting utilizing single compatible events and double compatible matching detection events (from parallel states) in order to share secret bits. Likewise, sifting is applied to the double-detection events that contain a single compatible detection event. With this aim, Bob indicates if the single detection is the first or the second inside the

Table 3 exhibits a case of the nack state protocol. Here, two biqubits are transmitted to Bob from Alice. The first biqubit is the orthogonal pair ðj0Xi; j1XiÞ, and the second biqubit is the parallel pair ðj1Zi; j1ZiÞ. In case the two states sent by Alice reach Bob's detection system with no failure, a double-detection event is generated. In the situation that just one of the two states of

The nack-state protocol has been conceived of to use the same optical hardware of the BB84 protocol; thus, it can be configured in most QKD systems as a software module application. However, two additional tasks must be implemented: the random computation of biqubits before preparing and sending the quantum states and the sifting stage of the protocol, which must include (1) sifting of single matching (compatible or non-compatible), where Bob announces the number of the single-detections inside the biqubit and (2) sifting of double detection, matching or non-matching, from parallel or orthogonal states. The error correction

In the PNS attack, the eavesdropper captures no less than one photon from each of the multiphoton states with the purpose of storing them in quantum memory, at the same time that she hinders the single photon states in the quantum channel. When Bob has uncovered over public channels the measurement basis he has used, the eavesdropper executes the same measure-

When the PNS attack is applied to the ack-state protocol, the eavesdropper captures no less than one photon of the multi-photon states (parallel and non-orthogonal), and she stands by Bob's declarations about the measurement bases he has utilized with the aim of applying the

the first states of each pair, followed by a burst of the second states of each pair).

ment basis (X or Z). The following steps depict the nack state protocol:

48 Advanced Technologies of Quantum Key Distribution

3. Bob declares publicly his measurement basis decisions.

the biqubit reaches Bob's station, he gets a single-detection event.

5. The photon number splitting attack

ments on the quantum states she has stored [25].

and privacy amplification stages of the QKD protocol do not require changes.

biqubit.

she arranges the biqubit to be sent to Bob through the quantum channel.

In Table 4 (upper part), the gain of the single-detection events is depicted with the Qð Þ <sup>þ</sup> symbol. According to Ma et al. [18], the gain of detection events is acquired from two origins: the photon source and the quantum channel. The photon source presents an expected photon number μ, and it adopts Poisson distribution. Contrastively, the quantum channel exhibits a distribution that is computed for every i photons' state (where i is the quantity of photons in each pulse) that is named yield. The gain Qi of i photons' state is the product of the probability of Alice sending an i photons' state (that adopts Poisson distribution) and the yield of i photons' state (and background states). It will generate a gain at Bob's side provoked by the detection of events corresponding to the relation Qi ¼ Yi μi <sup>i</sup>! e�<sup>μ</sup> where Yi is the yield of i photons' state.

The yield Yi is computed across the following steps:

1. The fiber channel transmittance among Alice and Bob is denoted as TAB <sup>¼</sup> <sup>10</sup>�α<sup>l</sup> <sup>10</sup> where α is the loss coefficient measured in dB/km, and the length l is measured in km. Moreover, the local transmittance at Bob's side, ηB, is written as tB � η<sup>D</sup> where tB is the internal transmittance of optical components and η<sup>D</sup> is the quantum efficiency of Bob's detectors. Then, the overall transmission and detection efficiency at Bob's side, ηBT, is computed as <sup>η</sup>BT <sup>¼</sup> tB � <sup>η</sup><sup>D</sup> � TAB and typically <sup>η</sup>BT ranges to 10�<sup>3</sup> [18];


double-matching detection events 2ð Þ M and the ð Þ S-i sifting instances which are consistent with the states she fixed, to verify corresponding photonic gains, parallel and non-orthogonal. As mentioned before, the one-photon states are blocked by eavesdropper and she performs a channel substitution to adjust the transmittance of the channel, TAB. Nevertheless, this activity

The QPEG after Eve blocks the one-photon states and can be written as ΔQ ¼ Q<sup>1</sup> where Q<sup>1</sup> is the gain of the one-photon states and it must be computed for the single- and the double-

detection events and single-detection events, respectively, where <sup>Q</sup><sup>1</sup>ð Þ <sup>þ</sup> <sup>¼</sup> ð Þ <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>η</sup> � <sup>Y</sup>0<sup>η</sup> <sup>μ</sup>e�<sup>μ</sup>, <sup>Q</sup>ð Þ � <sup>¼</sup> <sup>e</sup>�μη � <sup>Y</sup>0, <sup>η</sup> is the transmittance of the channel, and the detectors at Bob's side of the

The eavesdropper must adjust the transmittance, TAB, in order to remain hidden in the channel to achieve the two reference photonic gains, Qð Þ <sup>þ</sup>;<sup>þ</sup> and Qð Þ �;<sup>∓</sup> , for the double-detection events and single-detection events, respectively. Given <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> 6¼ <sup>Q</sup>ð Þ �;<sup>∓</sup> Eve can adjust TAB to <sup>Q</sup><sup>2</sup>

Q<sup>1</sup> � Qð Þ � but not both simultaneously. In other words, she is not able to fulfill the conditions ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0 and ΔQð Þ �;<sup>∓</sup> ¼ 0; in this manner, the attack becomes detectable. If the eavesdropper adjusts TAB to make it produce a photonic deviation in one or in both gains, she will

Consequently, Eve knows that she must be careful and makes no changes in TAB; otherwise,

QBER of single-detection events is 0:5<sup>2</sup> as in BB84. In contrast, when no attack is produced

Given that the probability of obtaining a (compatible) matching measurement from the non-

orthogonal states decreases one-half for each copy of quantum states in Eve's memory. In contrast, no contribution is made by the multi-photonic parallel states to increase the QBER

What should Alice and Bob expect from the nonappearance of the IRFS attack? For illustrative purposes, consider the situation where μ ¼ 0:2, ηBT ¼ 0:8, which is the general efficiency among Alice and Bob and zero dark counts ð Þ Y<sup>0</sup> ¼ 0 . In such a case, the great majority of the total biqubits sent by Alice to Bob ends up in Bob's station as lost biqubits ð Þ � 72:61% ; singledetection events are � 25:2%, and just � 0:0219% of the measurement cases are doubledetection events. Despite the double-detection gain being very low, it ought not be viewed as

she will be detected. Now, the QBER that Eve produces is <sup>0</sup>:5Q0þ0:52Q1þ0:53Q2þ…

<sup>1</sup>ð Þ <sup>þ</sup> <sup>¼</sup> <sup>Q</sup><sup>2</sup>

<sup>1</sup> and ΔQð Þ �; <sup>∓</sup> ¼ Q<sup>1</sup> � Qð Þ � for double-

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

<sup>Q</sup>0þQ1þQ2þ… because the

<sup>Q</sup>0þQ1þQ2þ… where ed is the detection error

, we derived the error rate of the non-orthogonal

<sup>Q</sup>0þQ1þQ2þ… . The QBER from the multi-photonic non-

<sup>1</sup> or

51

produces error gains in the single- and double-detection events that Alice can verify.

one-photon states and Y<sup>0</sup> is the background noise according to Ma et al. [18].

detection events. The error gain is <sup>Δ</sup>Qð Þ <sup>þ</sup>;<sup>þ</sup> <sup>¼</sup> <sup>Q</sup><sup>2</sup>

introduce a detectable QBER to the system.

orthogonal double-detection events is 0:52

double-detection events as <sup>0</sup>:<sup>5</sup> <sup>Q</sup>0þQ<sup>1</sup> ð Þþ0:52Q2þ0:53Q3þ…

according to Ma et al. [18].

6. The IRFS attack

the QBER of the system is given by <sup>0</sup>:5Q0þedð Þ <sup>Q</sup>1þQ2þQ3þ…

because Bob makes public the basis measurements used by him.

Here, ηBT and ηET are the overall efficiency of Bob and Eve, respectively. In the IRFS attack, Eve remains undetected given that she meets the condition <sup>η</sup>ET <sup>≥</sup> ln 2e�μη ð Þ BT �Y0�<sup>1</sup> �<sup>μ</sup> . At the lower part of the table, the gain of the double ð Þ <sup>þ</sup>; <sup>þ</sup> -detection events is shown, which is denoted as Qð Þ <sup>þ</sup>;<sup>þ</sup> , and the gain of single ð Þ �; ∓ detection events is represented as Qð Þ �; <sup>∓</sup> . In the IRFS attack, Eve can effectively forward half of her biqubits to Bob's detectors. The "�" symbol denotes multiplication inside the Qð Þ �; <sup>∓</sup> relation. The factor of 1/2 is a result of Bob using an active basis choice, compelling Eve to blind his detector when his basis differs from her own (half the time), and considering that each pair of pulses is detected in the same basis, Bob will always be blinded by Eve for both pulses or neither pulses, resulting in the same factor 1/2 for both single and double-detection events

Table 4. The background noise is defined as the gain of the single (non-empty) and empty pulses, Qð Þ <sup>þ</sup> and Qð Þ � , respectively, where μ is the expected photon number of the source and Y0.


The overall gain <sup>Q</sup>ð Þ <sup>þ</sup> is the summation of each Qi contribution, thus: <sup>Q</sup>ð Þ <sup>þ</sup> <sup>¼</sup> <sup>P</sup><sup>∞</sup> <sup>i</sup>¼<sup>1</sup> Qi <sup>¼</sup> <sup>P</sup><sup>∞</sup> i¼1 Yi μi <sup>i</sup>! <sup>e</sup>�<sup>μ</sup>, which leads to the relation <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μηBT . Finally, the quantum bit error rate (QBER) between Alice and Bob has been derived by Ma et al. [18] through the relation QBERAB <sup>¼</sup> <sup>0</sup>:5Y0þed <sup>1</sup>�e�μη ð Þ BT <sup>Y</sup>0þ1�e�μηBT , where ed is the error probability of the detector ed � <sup>10</sup>�<sup>2</sup> � �.

With the aim to obtain the gain of double-detection events Qð Þ �;� , Qð Þ �;<sup>∓</sup> , and Qð Þ <sup>þ</sup>;<sup>þ</sup> , we consider that each gain has independence of any other, that is, Qð Þ �;� ¼ Qð Þ � � Qð Þ � , Qð Þ <sup>þ</sup>;� ¼ Qð Þ <sup>þ</sup> � Qð Þ � , Qð Þ <sup>þ</sup>;� � Qð Þ �;<sup>þ</sup> , and Qð Þ <sup>þ</sup>;<sup>þ</sup> ¼ Qð Þ <sup>þ</sup> � Qð Þ <sup>þ</sup> . From the previous discussion, we know that the gain of the double-detection events decreases quadratically: <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> � <sup>Q</sup><sup>2</sup> ð Þ <sup>þ</sup> . In practical implementations of QKD, the single-matching events have the order of 10�<sup>5</sup> , while the double-matching events reach the order of 10�10.

#### 5.2. Detecting the photon number splitting attack

In replacing TAB, the photonic gain of the single-detection events or the double-detection events can be adjusted by Eve but not both at the same time. In contrast, Alice utilizes the double-matching detection events 2ð Þ M and the ð Þ S-i sifting instances which are consistent with the states she fixed, to verify corresponding photonic gains, parallel and non-orthogonal.

As mentioned before, the one-photon states are blocked by eavesdropper and she performs a channel substitution to adjust the transmittance of the channel, TAB. Nevertheless, this activity produces error gains in the single- and double-detection events that Alice can verify.

The QPEG after Eve blocks the one-photon states and can be written as ΔQ ¼ Q<sup>1</sup> where Q<sup>1</sup> is the gain of the one-photon states and it must be computed for the single- and the doubledetection events. The error gain is <sup>Δ</sup>Qð Þ <sup>þ</sup>;<sup>þ</sup> <sup>¼</sup> <sup>Q</sup><sup>2</sup> <sup>1</sup>ð Þ <sup>þ</sup> <sup>¼</sup> <sup>Q</sup><sup>2</sup> <sup>1</sup> and ΔQð Þ �; <sup>∓</sup> ¼ Q<sup>1</sup> � Qð Þ � for doubledetection events and single-detection events, respectively, where <sup>Q</sup><sup>1</sup>ð Þ <sup>þ</sup> <sup>¼</sup> ð Þ <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>η</sup> � <sup>Y</sup>0<sup>η</sup> <sup>μ</sup>e�<sup>μ</sup>, <sup>Q</sup>ð Þ � <sup>¼</sup> <sup>e</sup>�μη � <sup>Y</sup>0, <sup>η</sup> is the transmittance of the channel, and the detectors at Bob's side of the one-photon states and Y<sup>0</sup> is the background noise according to Ma et al. [18].

The eavesdropper must adjust the transmittance, TAB, in order to remain hidden in the channel to achieve the two reference photonic gains, Qð Þ <sup>þ</sup>;<sup>þ</sup> and Qð Þ �;<sup>∓</sup> , for the double-detection events and single-detection events, respectively. Given <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> 6¼ <sup>Q</sup>ð Þ �;<sup>∓</sup> Eve can adjust TAB to <sup>Q</sup><sup>2</sup> <sup>1</sup> or Q<sup>1</sup> � Qð Þ � but not both simultaneously. In other words, she is not able to fulfill the conditions ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0 and ΔQð Þ �;<sup>∓</sup> ¼ 0; in this manner, the attack becomes detectable. If the eavesdropper adjusts TAB to make it produce a photonic deviation in one or in both gains, she will introduce a detectable QBER to the system.

Consequently, Eve knows that she must be careful and makes no changes in TAB; otherwise, she will be detected. Now, the QBER that Eve produces is <sup>0</sup>:5Q0þ0:52Q1þ0:53Q2þ… <sup>Q</sup>0þQ1þQ2þ… because the QBER of single-detection events is 0:5<sup>2</sup> as in BB84. In contrast, when no attack is produced the QBER of the system is given by <sup>0</sup>:5Q0þedð Þ <sup>Q</sup>1þQ2þQ3þ… <sup>Q</sup>0þQ1þQ2þ… where ed is the detection error according to Ma et al. [18].

Given that the probability of obtaining a (compatible) matching measurement from the nonorthogonal double-detection events is 0:52 , we derived the error rate of the non-orthogonal double-detection events as <sup>0</sup>:<sup>5</sup> <sup>Q</sup>0þQ<sup>1</sup> ð Þþ0:52Q2þ0:53Q3þ… <sup>Q</sup>0þQ1þQ2þ… . The QBER from the multi-photonic nonorthogonal states decreases one-half for each copy of quantum states in Eve's memory. In contrast, no contribution is made by the multi-photonic parallel states to increase the QBER because Bob makes public the basis measurements used by him.

## 6. The IRFS attack

2. The transmittance η<sup>i</sup> of i photons' state at Bob's, that is, ηBTi ¼ 1 � 1 � ηBT

Photonic-Gain Alice Alice � Bob Eve � Bob

<sup>Q</sup>ð Þ �; <sup>∓</sup> <sup>2</sup>e�<sup>μ</sup>� <sup>2</sup> <sup>e</sup>�μη ð Þ� BT � <sup>Y</sup><sup>0</sup> <sup>e</sup>�μη ð Þ� ET � <sup>Y</sup><sup>0</sup>

<sup>1</sup> � <sup>e</sup>�<sup>μ</sup> ð Þ <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μη ð Þ BT <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μη ð Þ ET

�<sup>μ</sup> . At the lower part of the table, the gain of the double ð Þ <sup>þ</sup>; <sup>þ</sup> -detection

Here, ηBT and ηET are the overall efficiency of Bob and Eve, respectively. In the IRFS attack, Eve remains undetected given

events is shown, which is denoted as Qð Þ <sup>þ</sup>;<sup>þ</sup> , and the gain of single ð Þ �; ∓ detection events is represented as Qð Þ �; <sup>∓</sup> . In the IRFS attack, Eve can effectively forward half of her biqubits to Bob's detectors. The "�" symbol denotes multiplication inside the Qð Þ �; <sup>∓</sup> relation. The factor of 1/2 is a result of Bob using an active basis choice, compelling Eve to blind his detector when his basis differs from her own (half the time), and considering that each pair of pulses is detected in the same basis, Bob will always be blinded by Eve for both pulses or neither pulses, resulting in the same factor 1/2 for both

<sup>Q</sup>ð Þ � <sup>e</sup>�<sup>μ</sup> <sup>e</sup>�μηBT � <sup>Y</sup><sup>0</sup> — <sup>Q</sup>ð Þ <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�<sup>μ</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μηBT <sup>1</sup>

<sup>Q</sup>ð Þ �;� <sup>e</sup>�2<sup>μ</sup> <sup>e</sup>�μη ð Þ BT � <sup>Y</sup><sup>0</sup> <sup>2</sup> —

<sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> <sup>1</sup> � <sup>e</sup>�<sup>μ</sup> ð Þ<sup>2</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μη ð Þ BT <sup>2</sup> <sup>1</sup>

The overall gain <sup>Q</sup>ð Þ <sup>þ</sup> is the summation of each Qi contribution, thus: <sup>Q</sup>ð Þ <sup>þ</sup> <sup>¼</sup> <sup>P</sup><sup>∞</sup>

3. The yield Yi of the i photons' state is acquired from two sources, the background noise ð Þ Y<sup>0</sup> and the true signal. Presuming that the background counts are independent from the signal photon detection, Yi is given by Yi ¼ Y<sup>0</sup> þ ηBTi � Y0ηBTi. However, assuming Y<sup>0</sup> is

Table 4. The background noise is defined as the gain of the single (non-empty) and empty pulses, Qð Þ <sup>þ</sup> and Qð Þ � ,

<sup>i</sup>! <sup>e</sup>�<sup>μ</sup>, which leads to the relation <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μηBT . Finally, the quantum bit error rate (QBER) between Alice and Bob has been derived by Ma et al. [18] through the relation

<sup>Y</sup>0þ1�e�μηBT , where ed is the error probability of the detector ed � <sup>10</sup>�<sup>2</sup> � �.

With the aim to obtain the gain of double-detection events Qð Þ �;� , Qð Þ �;<sup>∓</sup> , and Qð Þ <sup>þ</sup>;<sup>þ</sup> , we consider that each gain has independence of any other, that is, Qð Þ �;� ¼ Qð Þ � � Qð Þ � , Qð Þ <sup>þ</sup>;� ¼ Qð Þ <sup>þ</sup> � Qð Þ � , Qð Þ <sup>þ</sup>;� � Qð Þ �;<sup>þ</sup> , and Qð Þ <sup>þ</sup>;<sup>þ</sup> ¼ Qð Þ <sup>þ</sup> � Qð Þ <sup>þ</sup> . From the previous discussion, we know that the gain of the double-detection events decreases quadratically: <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> � <sup>Q</sup><sup>2</sup>

In replacing TAB, the photonic gain of the single-detection events or the double-detection events can be adjusted by Eve but not both at the same time. In contrast, Alice utilizes the

practical implementations of QKD, the single-matching events have the order of 10�<sup>5</sup>

assuming independence among the i photons of the i photons' state;

) and <sup>η</sup>BT � <sup>10</sup>�<sup>3</sup>

respectively, where μ is the expected photon number of the source and Y0.

the double-matching events reach the order of 10�10.

5.2. Detecting the photon number splitting attack

small (around 10�<sup>5</sup>

single and double-detection events

that she meets the condition <sup>η</sup>ET <sup>≥</sup> ln 2e�μη ð Þ BT �Y0�<sup>1</sup>

50 Advanced Technologies of Quantum Key Distribution

QBERAB <sup>¼</sup> <sup>0</sup>:5Y0þed <sup>1</sup>�e�μη ð Þ BT

Yi μi � �<sup>i</sup>

<sup>2</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μη ð Þ ET

<sup>2</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>�μη ð Þ ET <sup>2</sup>

, the above equation can be reduced to Yi � Y<sup>0</sup> þ ηBTi.

for i ¼ 0, 1, …,

<sup>i</sup>¼<sup>1</sup> Qi <sup>¼</sup> <sup>P</sup><sup>∞</sup>

i¼1

ð Þ <sup>þ</sup> . In

, while

What should Alice and Bob expect from the nonappearance of the IRFS attack? For illustrative purposes, consider the situation where μ ¼ 0:2, ηBT ¼ 0:8, which is the general efficiency among Alice and Bob and zero dark counts ð Þ Y<sup>0</sup> ¼ 0 . In such a case, the great majority of the total biqubits sent by Alice to Bob ends up in Bob's station as lost biqubits ð Þ � 72:61% ; singledetection events are � 25:2%, and just � 0:0219% of the measurement cases are doubledetection events. Despite the double-detection gain being very low, it ought not be viewed as insignificant given that the amount of pulses sent by Alice is high (1011 1013 [29]), and the transmission interim can be legitimately upgraded. However, for practical purposes, we will presume that the secret bits in the nack state protocol are delivered by single-detection events, and the key rate is at most the BB84 key rate. Nevertheless, we assert that double-detection events can be utilized to identify the IRFS attack, so in this section, we defend the security of the protocol, in spite of Eve's endeavors to enhance her attack.

6.2. Detecting the IRFS attack with quantum channel substitution

single- and the double-detection events.

ð Þ �; ∓ AB and ð Þ �; ∓ EB with the same intention.

ΔQð Þ �;<sup>∓</sup> EB . The following equation system is obtained:

�μη ð Þ BT � <sup>Y</sup><sup>0</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>

Y<sup>0</sup> þ 1 � e

absence. For double-detection events, we represent R as <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> EB

detection events. In addition, we will define the vacuum ratio r as <sup>e</sup>

�μηET �Y<sup>0</sup>

2 e

Solving the system for <sup>η</sup>ET, we get lnY<sup>0</sup>

6.3. The photon and the vacuum ratios

cases depicted in Figure 6.

gains QEB

<sup>R</sup>ð Þ �; <sup>∓</sup> <sup>¼</sup> <sup>r</sup>ffiffi

2 <sup>p</sup> , but <sup>r</sup> <sup>¼</sup> <sup>e</sup>

typical parameters, for example, <sup>Y</sup><sup>0</sup> <sup>¼</sup> <sup>10</sup>�<sup>5</sup>

It is expected that the eavesdropper would endeavor to adjust both gains, from single- and double-detection events, applying a quantum channel substitution and tuning it to a specific transmittance. We define the quantum photon error gain (QPEG or simply ΔQ) as the deviation from the reference gain that is caused by Eve's apparatus at Bob's receiver station when she performs the IRFS attack. In ordinary conditions, it is ideally expected that ΔQ � 0, for the

QPEG of double ð Þ þ; þ -detection events is written as ΔQð Þ <sup>þ</sup>;<sup>þ</sup> , while we denote the QPEG of single ð Þ �; ∓ -detection events as ΔQð Þ �;<sup>∓</sup> . ΔQð Þ <sup>þ</sup>;<sup>þ</sup> is computed as the difference Qð Þ <sup>þ</sup>;<sup>þ</sup> AB� Qð Þ <sup>þ</sup>;<sup>þ</sup> EB where the symbol ð Þ þ; þ AB defines the reference gain of the double-detection events and ð Þ þ; þ EB denotes the gain of the double-detection events at Bob's side but in the presence of Eve. Similarly, ΔQð Þ �;<sup>∓</sup> is computed as Qð Þ �;<sup>∓</sup> AB � Qð Þ �; <sup>∓</sup> EB , where we apply the sub-index of

Using the relations of Table 4, the possibility of the eavesdropper to fulfill simultaneously the conditions ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0 and ΔQð Þ �;<sup>∓</sup> ¼ 0 can be established. Allow Eve to adjust freely ηBT and ηET. Thus, the eavesdropper's goal is to make ΔQð Þ <sup>þ</sup>;<sup>þ</sup> AB ¼ ΔQð Þ <sup>þ</sup>;<sup>þ</sup> EB and ΔQð Þ �; <sup>∓</sup> AB ¼

�μη ð Þ ET � <sup>Y</sup><sup>0</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>

<sup>2</sup> <sup>Y</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> � <sup>e</sup>

�μη ð Þ ET (1)

Quantum Flows for Secret Key Distribution http://dx.doi.org/10.5772/intechopen.75964 53

�μη ð Þ ET <sup>2</sup> (2)

�<sup>μ</sup> , which, in the practice, cannot be satisfied,

, μ ¼ 0:1 produces ηET ¼ 1:15. Consider also the

, while <sup>Q</sup>ð Þ �;<sup>∓</sup> EB Qð Þ �; ∓ AB

�μηET �Y<sup>0</sup> <sup>e</sup>�μηBT �Y<sup>0</sup>

2

for single-

2 <sup>p</sup> .

.If the eavesdropper

<sup>p</sup> . To discard Eve's pres-

Qð Þ þ;þ AB

�μη <sup>ð</sup> BT Þ ¼ <sup>e</sup>

and ln 1ð Þ <sup>þ</sup>Y<sup>0</sup>

given that the second relation yields ηET as negative and the first relation cannot be fulfilled for

We will introduce a convenient method to detect the presence of the eavesdropper without requiring one to compute deviations from the reference gain, that is, ΔQð Þ¼ þ; þ 0 or ΔQð Þ¼ �; ∓ 0. For this purpose, let us define the photon ratio R as the relation between the

adjusts the channel to achieve Qð Þ þ; þ AB ¼ Qð Þ þ; þ EB, then Eq. (2) is satisfied. We get that

ence, it is not necessary to verify that <sup>Δ</sup>Qð Þ¼ �; <sup>∓</sup> 0, but it must be confirmed that <sup>R</sup>ð Þ �;<sup>∓</sup> <sup>&</sup>gt; <sup>1</sup>ffiffi

<sup>e</sup>�μηBT �Y<sup>0</sup> and <sup>η</sup>ET <sup>≥</sup> <sup>η</sup>BT; thus, <sup>r</sup> <sup>≤</sup> 1 and <sup>R</sup>ð Þ �; <sup>∓</sup> <sup>≤</sup> <sup>1</sup>ffiffi

QAB where the subscript EB denotes the presence of the eavesdropper and AB indicates its

�μη ð Þ BT <sup>2</sup> <sup>¼</sup> <sup>1</sup>

�μ

### 6.1. Detecting the IRFS attack with blinding pulses and quantum channel substitution

Within the sight of the IRFS attack with blinding pulses, Eve is amid the quantum channel utilizing an optical detection system comparable to Bob's station. Eve is challenged to reproduce gains of single- and double-detection events at Bob's side to pass unnoticed in the quantum channel. However, the gain of the single-detection events decreases directly with the channel efficiency, but the double-detection gain drops quadratically. In the next section we demonstrate that, for practical parameters of the quantum channel, the two gains cannot be adjusted by the eavesdropper at the same time. Eve cannot control the two gains because of the fact that:


Eve still has the possibility to adjust the efficiency of the quantum channel to the gain of the double-detection events. Therefore, with the purpose of removing the excess of the singledetection gain, Eve could eliminate pulses in proportion to some probability (e.g., 0.5). However, in accordance with the second statement given previously in this section, the eavesdropper would lose double-detection pulses (a quarter in this example). Eve could be more selective discarding only single-detection events on which the detection occurred in the second pulse. By using this scheme, the double-detection gain is unaltered for Eve. However, given that the number of single detections inside the biqubit, first or second (see Table 3), is announced by Bob publicly, the presence of Eve becomes evident.

Both strategies could be combined by Eve to increase the efficiency of the channel to produce an overabundance of the double-detection gain, but it would also increase the single-detection gain. The issue for Eve is that once a strategy to remove pulses is chosen, it affects equally the singleand the double-detection gains. Such gains obey diverse rates: while the first decreases linearly, the second fluctuates quadratically with the transmittance of the channel. Moreover, at the receiver station, the single- and double-detection events are registered as haphazard interleaved events.

In the following sections, a convenient method to compute the photon gain deviation caused by the IRFS attack at a practical level is discussed.

#### 6.2. Detecting the IRFS attack with quantum channel substitution

insignificant given that the amount of pulses sent by Alice is high (1011 1013 [29]), and the transmission interim can be legitimately upgraded. However, for practical purposes, we will presume that the secret bits in the nack state protocol are delivered by single-detection events, and the key rate is at most the BB84 key rate. Nevertheless, we assert that double-detection events can be utilized to identify the IRFS attack, so in this section, we defend the security of

6.1. Detecting the IRFS attack with blinding pulses and quantum channel substitution

Within the sight of the IRFS attack with blinding pulses, Eve is amid the quantum channel utilizing an optical detection system comparable to Bob's station. Eve is challenged to reproduce gains of single- and double-detection events at Bob's side to pass unnoticed in the quantum channel. However, the gain of the single-detection events decreases directly with the channel efficiency, but the double-detection gain drops quadratically. In the next section we demonstrate that, for practical parameters of the quantum channel, the two gains cannot be adjusted by the eavesdropper at the same time. Eve cannot control the two gains because of the

1. the transmittance of the channel can be adjusted to a unique value by the eavesdropper

2. Eve's station receives Alice's optical pulses sequentially. In this manner, once a pulse is detected in the eavesdropper station, she is not able to know whether the next pulse will be likewise detected or lost. That is, Eve has no form to know when a single or a double-

Eve still has the possibility to adjust the efficiency of the quantum channel to the gain of the double-detection events. Therefore, with the purpose of removing the excess of the singledetection gain, Eve could eliminate pulses in proportion to some probability (e.g., 0.5). However, in accordance with the second statement given previously in this section, the eavesdropper would lose double-detection pulses (a quarter in this example). Eve could be more selective discarding only single-detection events on which the detection occurred in the second pulse. By using this scheme, the double-detection gain is unaltered for Eve. However, given that the number of single detections inside the biqubit, first or second (see Table 3), is announced by

Both strategies could be combined by Eve to increase the efficiency of the channel to produce an overabundance of the double-detection gain, but it would also increase the single-detection gain. The issue for Eve is that once a strategy to remove pulses is chosen, it affects equally the singleand the double-detection gains. Such gains obey diverse rates: while the first decreases linearly, the second fluctuates quadratically with the transmittance of the channel. Moreover, at the receiver station, the single- and double-detection events are registered as haphazard interleaved events.

In the following sections, a convenient method to compute the photon gain deviation caused

the protocol, in spite of Eve's endeavors to enhance her attack.

52 Advanced Technologies of Quantum Key Distribution

either to adjust the single or the double-detection gain and

fact that:

detection event will occur.

Bob publicly, the presence of Eve becomes evident.

by the IRFS attack at a practical level is discussed.

It is expected that the eavesdropper would endeavor to adjust both gains, from single- and double-detection events, applying a quantum channel substitution and tuning it to a specific transmittance. We define the quantum photon error gain (QPEG or simply ΔQ) as the deviation from the reference gain that is caused by Eve's apparatus at Bob's receiver station when she performs the IRFS attack. In ordinary conditions, it is ideally expected that ΔQ � 0, for the single- and the double-detection events.

QPEG of double ð Þ þ; þ -detection events is written as ΔQð Þ <sup>þ</sup>;<sup>þ</sup> , while we denote the QPEG of single ð Þ �; ∓ -detection events as ΔQð Þ �;<sup>∓</sup> . ΔQð Þ <sup>þ</sup>;<sup>þ</sup> is computed as the difference Qð Þ <sup>þ</sup>;<sup>þ</sup> AB� Qð Þ <sup>þ</sup>;<sup>þ</sup> EB where the symbol ð Þ þ; þ AB defines the reference gain of the double-detection events and ð Þ þ; þ EB denotes the gain of the double-detection events at Bob's side but in the presence of Eve. Similarly, ΔQð Þ �;<sup>∓</sup> is computed as Qð Þ �;<sup>∓</sup> AB � Qð Þ �; <sup>∓</sup> EB , where we apply the sub-index of ð Þ �; ∓ AB and ð Þ �; ∓ EB with the same intention.

Using the relations of Table 4, the possibility of the eavesdropper to fulfill simultaneously the conditions ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0 and ΔQð Þ �;<sup>∓</sup> ¼ 0 can be established. Allow Eve to adjust freely ηBT and ηET. Thus, the eavesdropper's goal is to make ΔQð Þ <sup>þ</sup>;<sup>þ</sup> AB ¼ ΔQð Þ <sup>þ</sup>;<sup>þ</sup> EB and ΔQð Þ �; <sup>∓</sup> AB ¼ ΔQð Þ �;<sup>∓</sup> EB . The following equation system is obtained:

$$\mathcal{Q}(e^{-\mu\eta\_{\text{ET}}} - Y\_0)(Y\_0 + 1 - e^{-\mu\eta\_{\text{ET}}}) = (e^{-\mu\eta\_{\text{ET}}} - Y\_0)(Y\_0 + 1 - e^{-\mu\eta\_{\text{ET}}}) \tag{1}$$

$$\left( \left( Y\_0 + 1 - e^{-\mu \eta\_{\text{BT}}} \right)^2 = \frac{1}{2} \left( Y\_0 + 1 - e^{-\mu \eta\_{\text{ET}}} \right)^2 \tag{2}$$

Solving the system for <sup>η</sup>ET, we get lnY<sup>0</sup> �μ and ln 1ð Þ <sup>þ</sup>Y<sup>0</sup> �<sup>μ</sup> , which, in the practice, cannot be satisfied, given that the second relation yields ηET as negative and the first relation cannot be fulfilled for typical parameters, for example, <sup>Y</sup><sup>0</sup> <sup>¼</sup> <sup>10</sup>�<sup>5</sup> , μ ¼ 0:1 produces ηET ¼ 1:15. Consider also the cases depicted in Figure 6.

#### 6.3. The photon and the vacuum ratios

We will introduce a convenient method to detect the presence of the eavesdropper without requiring one to compute deviations from the reference gain, that is, ΔQð Þ¼ þ; þ 0 or ΔQð Þ¼ �; ∓ 0. For this purpose, let us define the photon ratio R as the relation between the gains QEB QAB where the subscript EB denotes the presence of the eavesdropper and AB indicates its absence. For double-detection events, we represent R as <sup>Q</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> EB Qð Þ þ;þ AB , while <sup>Q</sup>ð Þ �;<sup>∓</sup> EB Qð Þ �; ∓ AB for singledetection events. In addition, we will define the vacuum ratio r as <sup>e</sup> �μηET �Y<sup>0</sup> <sup>e</sup>�μηBT �Y<sup>0</sup> .If the eavesdropper adjusts the channel to achieve Qð Þ þ; þ AB ¼ Qð Þ þ; þ EB, then Eq. (2) is satisfied. We get that <sup>R</sup>ð Þ �; <sup>∓</sup> <sup>¼</sup> <sup>r</sup>ffiffi 2 <sup>p</sup> , but <sup>r</sup> <sup>¼</sup> <sup>e</sup> �μηET �Y<sup>0</sup> <sup>e</sup>�μηBT �Y<sup>0</sup> and <sup>η</sup>ET <sup>≥</sup> <sup>η</sup>BT; thus, <sup>r</sup> <sup>≤</sup> 1 and <sup>R</sup>ð Þ �; <sup>∓</sup> <sup>≤</sup> <sup>1</sup>ffiffi 2 <sup>p</sup> . To discard Eve's presence, it is not necessary to verify that <sup>Δ</sup>Qð Þ¼ �; <sup>∓</sup> 0, but it must be confirmed that <sup>R</sup>ð Þ �;<sup>∓</sup> <sup>&</sup>gt; <sup>1</sup>ffiffi 2 <sup>p</sup> .

that in the BB84 protocol, the probability to get the correct bit is pc ¼ ð Þ 1 þ V =2, and the probability to obtain an erroneous bit is pe ¼ ð Þ 1 � V =2, where V is the visibility of the optical system. To calculate the QBER of the one-photon states, the relation QBER ¼ pe=ð Þ pe þ pc is

Now, suppose that the two parallel states are sent by Alice ðj1Zi; j1ZiÞ to Bob who measures them using the Z basis. Those states are depicted in Figure 7a. The probability to get the two

Figure 7a. Since the measurement cases ðj0Zi; j1ZiÞ and ðj1Zi; j0ZiÞ, Cases III and IV of Figure 7a, are always disposed because they are non-matching cases, the final probabilities

Those relations forward us to the QBER of the parallel and orthogonal states QBER <sup>¼</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup>

Figure 8 gives an illustration of the QBER of one-photon states of such protocols. Considering the QBER of the nack state is lower than BB84, it is interesting to acknowledge that the double-

Figure 7. The QBER of parallel and orthogonal states: Cases III and IV of (a) and (b) can be discarded by Alice, so they do

<sup>c</sup> , and the probability to get the opposite values ðj0Zi; <sup>j</sup>0ZiÞ is <sup>p</sup><sup>2</sup>

. The same reasoning can be applied to the orthogonal

<sup>e</sup> , case II of

55

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

þð Þ <sup>1</sup>þ<sup>V</sup> 2.

ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup>

applied [31].

states ðj1Zi; <sup>j</sup>1ZiÞ is <sup>p</sup><sup>2</sup>

c p2 cþp<sup>2</sup> e

biqubits case as depicted in Figure 7b.

and peparallel <sup>¼</sup> <sup>p</sup><sup>2</sup>

e p2 cþp<sup>2</sup> e

are pcparallel <sup>¼</sup> <sup>p</sup><sup>2</sup>

not produce errors.

Figure 6. The deviation from the reference gain is shown on the y-axis. The upper and bottom left graphs represent double detections, while the right graphs correspond to single detections. Considering that ηBT ¼ 0:001 and Eve uses ηET ¼ 0:0014, she accomplishes in (a), ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0, however, in (b), ΔQð Þ �;<sup>∓</sup> 6¼ 0. Conversely, if Eve adjusts ηET ¼ 0:002, she gets in (d)) ΔQð Þ �; <sup>∓</sup> ¼ 0, but in (c), she provokes simultaneously that ΔQð Þ <sup>þ</sup>;<sup>þ</sup> 6¼ 0.

Contrarily, if Eve modifies the channel to achieve Qð Þ �; ∓ AB ¼ Qð Þ �; ∓ EB, we get that <sup>R</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> <sup>¼</sup> <sup>2</sup> <sup>r</sup>2. Since r ≤ 1, we obtain that the IRFS attack causes Rð Þ <sup>þ</sup>;<sup>þ</sup> ≥ 2. To make sure that the system is protected against the IRFS attack, it is not necessary to check ΔQð Þ¼ þ; þ 0 but it is enough verifying its equivalent Rð Þ <sup>þ</sup>;<sup>þ</sup> < 2.

#### 6.4. The QBER of one-photon states

As quoted previously, in the nack state protocol, the great majority of the pulses sent by Alice to Bob behave as BB84 signal pulses. Each time a compatible basis measurement is applied by Bob, the result, either from single detection or double detection, is useful as in BB84. Thus, for practical purposes, the nack state protocol has an efficiency comparable to the BB84. However, a partial reduction of the bit rate can be expected, as Alice reduces the optical pulse rate to avoid the eavesdropper to record double-detection events. In this way, Eve is detected if she stays waiting for double-detection events before she can forward them.

Given that it decreases quadratically, the rate of the double-detection event is small. Nevertheless, at the same time, it is extraordinary that the QBER of the double-matching detection events from parallel and orthogonal states also decreases quadratically. To see this, let us recall that in the BB84 protocol, the probability to get the correct bit is pc ¼ ð Þ 1 þ V =2, and the probability to obtain an erroneous bit is pe ¼ ð Þ 1 � V =2, where V is the visibility of the optical system. To calculate the QBER of the one-photon states, the relation QBER ¼ pe=ð Þ pe þ pc is applied [31].

Now, suppose that the two parallel states are sent by Alice ðj1Zi; j1ZiÞ to Bob who measures them using the Z basis. Those states are depicted in Figure 7a. The probability to get the two states ðj1Zi; <sup>j</sup>1ZiÞ is <sup>p</sup><sup>2</sup> <sup>c</sup> , and the probability to get the opposite values ðj0Zi; <sup>j</sup>0ZiÞ is <sup>p</sup><sup>2</sup> <sup>e</sup> , case II of Figure 7a. Since the measurement cases ðj0Zi; j1ZiÞ and ðj1Zi; j0ZiÞ, Cases III and IV of Figure 7a, are always disposed because they are non-matching cases, the final probabilities are pcparallel <sup>¼</sup> <sup>p</sup><sup>2</sup> c p2 cþp<sup>2</sup> e and peparallel <sup>¼</sup> <sup>p</sup><sup>2</sup> e p2 cþp<sup>2</sup> e . The same reasoning can be applied to the orthogonal biqubits case as depicted in Figure 7b.

Those relations forward us to the QBER of the parallel and orthogonal states QBER <sup>¼</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup> þð Þ <sup>1</sup>þ<sup>V</sup> 2. Figure 8 gives an illustration of the QBER of one-photon states of such protocols. Considering the QBER of the nack state is lower than BB84, it is interesting to acknowledge that the double-

Contrarily, if Eve modifies the channel to achieve Qð Þ �; ∓ AB ¼ Qð Þ �; ∓ EB, we get that

Figure 6. The deviation from the reference gain is shown on the y-axis. The upper and bottom left graphs represent double detections, while the right graphs correspond to single detections. Considering that ηBT ¼ 0:001 and Eve uses ηET ¼ 0:0014, she accomplishes in (a), ΔQð Þ <sup>þ</sup>;<sup>þ</sup> ¼ 0, however, in (b), ΔQð Þ �;<sup>∓</sup> 6¼ 0. Conversely, if Eve adjusts ηET ¼ 0:002,

system is protected against the IRFS attack, it is not necessary to check ΔQð Þ¼ þ; þ 0 but it is

As quoted previously, in the nack state protocol, the great majority of the pulses sent by Alice to Bob behave as BB84 signal pulses. Each time a compatible basis measurement is applied by Bob, the result, either from single detection or double detection, is useful as in BB84. Thus, for practical purposes, the nack state protocol has an efficiency comparable to the BB84. However, a partial reduction of the bit rate can be expected, as Alice reduces the optical pulse rate to avoid the eavesdropper to record double-detection events. In this way, Eve is detected if she stays

Given that it decreases quadratically, the rate of the double-detection event is small. Nevertheless, at the same time, it is extraordinary that the QBER of the double-matching detection events from parallel and orthogonal states also decreases quadratically. To see this, let us recall

<sup>r</sup>2. Since r ≤ 1, we obtain that the IRFS attack causes Rð Þ <sup>þ</sup>;<sup>þ</sup> ≥ 2. To make sure that the

<sup>R</sup>ð Þ <sup>þ</sup>;<sup>þ</sup> <sup>¼</sup> <sup>2</sup>

enough verifying its equivalent Rð Þ <sup>þ</sup>;<sup>þ</sup> < 2.

54 Advanced Technologies of Quantum Key Distribution

waiting for double-detection events before she can forward them.

she gets in (d)) ΔQð Þ �; <sup>∓</sup> ¼ 0, but in (c), she provokes simultaneously that ΔQð Þ <sup>þ</sup>;<sup>þ</sup> 6¼ 0.

6.4. The QBER of one-photon states

Figure 7. The QBER of parallel and orthogonal states: Cases III and IV of (a) and (b) can be discarded by Alice, so they do not produce errors.

would be 500 microseconds, forcing a source rate of 2 kHz. Given the conservative fiber link loss of 0.2 dB/km the detection rate after 100 km (20 dB) would be less than 20/s, not counting detection efficiency. Shorter distances would be more favorable, but this implies the protocol is limited to short distances. There also is not any point in randomly adding delays as Eve would still be able to perfectly replicate the gains when the delay is insufficient and could choose to simply not intercept when the delay is too long, giving her partial information without any

Quantum Flows for Secret Key Distribution http://dx.doi.org/10.5772/intechopen.75964 57

Unfortunately for Eve, Alice can apply a reduction in the optical pulse rate forcing Eve to introduce a delay in the arrival time of the pulses at Bob's station. As a matter of fact, Alice could adjust such delay sending slow pulses as a random burst. Furthermore, slowing pulses

However, there is no reason why each pair must be sent in sequence. We call this protocol the non-structured nack-state. If Alice were to transmit a burst of the first states of each pair, followed by a burst of the second states of each pair, she would create a separation between the pairs equal to the length of the bursts and she would not reduce the pulse rate. Consider a 100 km fiber optic link; it would be able to send the first states of each pair for 500 microseconds, followed by the second state of each pair for the next 500 microseconds, with Bob rechoosing the same basis for both 500 microsecond bursts. Since the 500 microsecond delay is at least the full travel time in the quantum channel, Eve would always be compelled to fake the first state of each pair before receiving the second. If there is no issue with this approach, the authors can use it to justify Point 2 of Section 6.1, which in turn justifies Point 1 of the same

Another possibility for the eavesdropper is to fake double-detection events. After all, we may inquire why Eve cannot fake double-detection events as she stays covered up in the channel. First of all, let us recall that Alice knows which biqubits contain parallel or orthogonal states. Second, consider the cases portrayed in Table 5. Assume the ðj0Zi; j0ZiÞ biqubit has been sent to Bob by Alice. The first pulse reaches Eve's station, who measures it with the X (or Z) basis, but the second pulse arrives as a vacuum state either by the effect of the quantum channel, the detection system, or the photon source. Thus, Eve gets a single-detection event. In this moment, Eve determines to fake the second state, but she realizes that there are six potential outcomes to fake the ð i j0<sup>Z</sup> ; j0ZiÞ biqubit; such cases are listed in Table 5. Additionally, one of those cases is erroneous because no orthogonal measurement can be derived from parallel states. In this example, ðj1Zi; j0ZiÞ cannot be obtained from ðj0Zi; j0ZiÞ. Likewise, ðj0Zi; j0ZiÞ cannot be derived from ðj1Zi; j0ZiÞ. Consequently, if Eve tries to fake a double-detection event, she will

non-matching event but Bob announces a double-matching event or vice versa.

According to Collins et al. [30], Bob's visibility of Alice's quantum state is computed as

2 � Y0. Here, Vopt is the optical visibility with a perfect source and detectors; η is the probability

P total ð Þ where P signal ð Þ¼ TAB � <sup>η</sup> � Vopt and P total ð Þ¼ TAB � <sup>η</sup> <sup>þ</sup> <sup>ð</sup><sup>1</sup> � TAB � <sup>η</sup>Þ�

6. In this situation, a bit error is produced when Alice expects a double

can enhance the double-detection rate at Bob's side by reducing after-pulsing errors.

hint of her presence.

section.

6.6. Faking double-detection events

produce a bit error of <sup>1</sup>

VAB <sup>¼</sup> P signal ð Þ

Figure 8. The nack state protocol uses pairs of parallel and orthogonal states. The QBER of parallel and orthogonal states is derived using the probabilities of two consecutive BB84 measurements.

detection gain could be increased by future technologies. Even though there is not yet a formal derivation of the secret key rate for double-detection events, we can expect that the small QBER would lead to reaching longer QKD distances.

#### 6.5. The non-structured nack-state protocol

In the argument of Point 2 of Section 6.1, it is implicit that Eve uses only a single station, but this is not a practical restriction. Eve could use two stations, one near to Alice to detect and one near to Bob to generate fake pulses. In the event that quantum channel utilizes optical fibers (the most widely recognized useful channel for ground-based QKD), everything required by Eve is a radio connection between her two stations to "catch up" with the quantum link. Even assuming a low source rate of 1 MHz, the time delay between pulses is only 1 microsecond, which can be easily compensated using a 600 m link (traveling in free space takes 2 microseconds; traveling in fiber takes 3 microseconds). Any practical QKD system will operate over distances greater than 600 m, making it entirely achievable for Eve to detect both pulses of a pair before transmitting her fake state to Bob using a second station.

A 100 km link in optical fiber would limit the source rate to 6 kHz, and much less if the fiber is not straight, which is almost always the case. To truly be secure the period between two pulses would have to be the full travel time of the pulse over the quantum channel. For 100 km, it would be 500 microseconds, forcing a source rate of 2 kHz. Given the conservative fiber link loss of 0.2 dB/km the detection rate after 100 km (20 dB) would be less than 20/s, not counting detection efficiency. Shorter distances would be more favorable, but this implies the protocol is limited to short distances. There also is not any point in randomly adding delays as Eve would still be able to perfectly replicate the gains when the delay is insufficient and could choose to simply not intercept when the delay is too long, giving her partial information without any hint of her presence.

Unfortunately for Eve, Alice can apply a reduction in the optical pulse rate forcing Eve to introduce a delay in the arrival time of the pulses at Bob's station. As a matter of fact, Alice could adjust such delay sending slow pulses as a random burst. Furthermore, slowing pulses can enhance the double-detection rate at Bob's side by reducing after-pulsing errors.

However, there is no reason why each pair must be sent in sequence. We call this protocol the non-structured nack-state. If Alice were to transmit a burst of the first states of each pair, followed by a burst of the second states of each pair, she would create a separation between the pairs equal to the length of the bursts and she would not reduce the pulse rate. Consider a 100 km fiber optic link; it would be able to send the first states of each pair for 500 microseconds, followed by the second state of each pair for the next 500 microseconds, with Bob rechoosing the same basis for both 500 microsecond bursts. Since the 500 microsecond delay is at least the full travel time in the quantum channel, Eve would always be compelled to fake the first state of each pair before receiving the second. If there is no issue with this approach, the authors can use it to justify Point 2 of Section 6.1, which in turn justifies Point 1 of the same section.

#### 6.6. Faking double-detection events

detection gain could be increased by future technologies. Even though there is not yet a formal derivation of the secret key rate for double-detection events, we can expect that the small QBER

Figure 8. The nack state protocol uses pairs of parallel and orthogonal states. The QBER of parallel and orthogonal states is

In the argument of Point 2 of Section 6.1, it is implicit that Eve uses only a single station, but this is not a practical restriction. Eve could use two stations, one near to Alice to detect and one near to Bob to generate fake pulses. In the event that quantum channel utilizes optical fibers (the most widely recognized useful channel for ground-based QKD), everything required by Eve is a radio connection between her two stations to "catch up" with the quantum link. Even assuming a low source rate of 1 MHz, the time delay between pulses is only 1 microsecond, which can be easily compensated using a 600 m link (traveling in free space takes 2 microseconds; traveling in fiber takes 3 microseconds). Any practical QKD system will operate over distances greater than 600 m, making it entirely achievable for Eve to detect both pulses of a

A 100 km link in optical fiber would limit the source rate to 6 kHz, and much less if the fiber is not straight, which is almost always the case. To truly be secure the period between two pulses would have to be the full travel time of the pulse over the quantum channel. For 100 km, it

pair before transmitting her fake state to Bob using a second station.

would lead to reaching longer QKD distances.

56 Advanced Technologies of Quantum Key Distribution

derived using the probabilities of two consecutive BB84 measurements.

6.5. The non-structured nack-state protocol

Another possibility for the eavesdropper is to fake double-detection events. After all, we may inquire why Eve cannot fake double-detection events as she stays covered up in the channel. First of all, let us recall that Alice knows which biqubits contain parallel or orthogonal states. Second, consider the cases portrayed in Table 5. Assume the ðj0Zi; j0ZiÞ biqubit has been sent to Bob by Alice. The first pulse reaches Eve's station, who measures it with the X (or Z) basis, but the second pulse arrives as a vacuum state either by the effect of the quantum channel, the detection system, or the photon source. Thus, Eve gets a single-detection event. In this moment, Eve determines to fake the second state, but she realizes that there are six potential outcomes to fake the ð i j0<sup>Z</sup> ; j0ZiÞ biqubit; such cases are listed in Table 5. Additionally, one of those cases is erroneous because no orthogonal measurement can be derived from parallel states. In this example, ðj1Zi; j0ZiÞ cannot be obtained from ðj0Zi; j0ZiÞ. Likewise, ðj0Zi; j0ZiÞ cannot be derived from ðj1Zi; j0ZiÞ. Consequently, if Eve tries to fake a double-detection event, she will produce a bit error of <sup>1</sup> 6. In this situation, a bit error is produced when Alice expects a double non-matching event but Bob announces a double-matching event or vice versa.

According to Collins et al. [30], Bob's visibility of Alice's quantum state is computed as VAB <sup>¼</sup> P signal ð Þ P total ð Þ where P signal ð Þ¼ TAB � <sup>η</sup> � Vopt and P total ð Þ¼ TAB � <sup>η</sup> <sup>þ</sup> <sup>ð</sup><sup>1</sup> � TAB � <sup>η</sup>Þ� 2 � Y0. Here, Vopt is the optical visibility with a perfect source and detectors; η is the probability


of detecting the photon when it arrives; TAB is the transmittance between Alice and Bob; and Y<sup>0</sup> is the background noise. On practical experimental parameters: <sup>α</sup> <sup>¼</sup> <sup>0</sup>:25 dB� km�1, <sup>η</sup> <sup>¼</sup> <sup>0</sup>:3,

, and Vopt ¼ 0:99. Figure 9 shows the visibility as a function of the distance.

probability to get, correctly or erroneously, the quantum bit sent by Alice, respectively. If we write such probabilities as a function of the optical visibility V, we have pc ¼ ð Þ 1 þ V =2 and

If QBER of double-detection events produced by the quantum channel is compared against the

<sup>6</sup> error rate caused by the eavesdropper, we can find that the maximum secure distance for detecting the IRFS attack when the eavesdropper fakes double-detection events is 176 km,

In the quantum flows approach, the transmitter interleaves pairs of quantum states, parallel and orthogonal (non-orthogonal), while the receiver applies active basis selection to perform state measurement. The QKD protocols based on quantum flows uses the same optical hardware of the BB84 protocol, and they can be implemented in most QKD systems as a software module

The ack-QKD protocol can be useful to detect the PNS attack. If the eavesdropper adjusts the transmittance TAB of the channel it produces a deviation in one or in both photonic gains; thus,

On the other side the intercept resend with faked (blinding) states (IRFS) attack is detected by the nack-state protocol using the gain of single- and double-detection events where the QBER of

Although double-detection events represent a small fraction of the total detection events, they are useful to detect the IRFS attack. In addition, the smaller QBER can be useful in future

We would like to mention that a major portion of this chapter has been borrowed from our previous publications: "Quantum Flows for Secret Key Distribution in the Presence of the

peþpc, where pc (pe) is the

59

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

<sup>6</sup> error rate caused by

, and we derived the QBER of the parallel and orthogonal

On the other hand, the QBER in BB84 can be computed as QBER <sup>¼</sup> pe

e p2 cþp<sup>2</sup> e

which is within the range of the BB84 key rate, as it appears in Figure 9.

double-detection events of the quantum channel is compared against the <sup>1</sup>

the eavesdropper, so the maximum secure distance results in 176 km.

<sup>Y</sup><sup>0</sup> <sup>¼</sup> <sup>10</sup>�<sup>4</sup>

pe ¼ ð Þ 1 � V =2.

1

Therefore, pc <sup>¼</sup> <sup>p</sup><sup>2</sup>

7. Conclusions

application.

Acknowledgements

states as QBER <sup>¼</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup>

c p2 cþp<sup>2</sup> e

ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup>

and pe <sup>¼</sup> <sup>p</sup><sup>2</sup>

þð Þ <sup>1</sup>þ<sup>V</sup> 2.

she will introduce a detectable QBER to the system.

implementations to distill secret bits at longer distances.

However, she can use six possible states, but one of them is erroneous, so she introduces an error probability of <sup>1</sup> 6. Here, the six choices for ðj0Zi; j0ZiÞ and ðj1Zi; j0ZiÞ biqubits are shown

Table 5. As soon as Eve detects the first state of a biqubit, she tries to fake the second state.

Figure 9. The error rate of double-detection events caused by the IRFS attack is <sup>1</sup> 6. When it is compared to the QBER of the quantum channel, the maximum secure distance to detect the IRFS attack is 176 km. In the presence of the IRFS attack, perfect visibility and zero dark counts are assumed in the link between Alice and Eve and from her to Bob.

of detecting the photon when it arrives; TAB is the transmittance between Alice and Bob; and Y<sup>0</sup> is the background noise. On practical experimental parameters: <sup>α</sup> <sup>¼</sup> <sup>0</sup>:25 dB� km�1, <sup>η</sup> <sup>¼</sup> <sup>0</sup>:3, <sup>Y</sup><sup>0</sup> <sup>¼</sup> <sup>10</sup>�<sup>4</sup> , and Vopt ¼ 0:99. Figure 9 shows the visibility as a function of the distance.

On the other hand, the QBER in BB84 can be computed as QBER <sup>¼</sup> pe peþpc, where pc (pe) is the probability to get, correctly or erroneously, the quantum bit sent by Alice, respectively. If we write such probabilities as a function of the optical visibility V, we have pc ¼ ð Þ 1 þ V =2 and pe ¼ ð Þ 1 � V =2.

Therefore, pc <sup>¼</sup> <sup>p</sup><sup>2</sup> c p2 cþp<sup>2</sup> e and pe <sup>¼</sup> <sup>p</sup><sup>2</sup> e p2 cþp<sup>2</sup> e , and we derived the QBER of the parallel and orthogonal states as QBER <sup>¼</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup> ð Þ <sup>1</sup>�<sup>V</sup> <sup>2</sup> þð Þ <sup>1</sup>þ<sup>V</sup> 2.

If QBER of double-detection events produced by the quantum channel is compared against the 1 <sup>6</sup> error rate caused by the eavesdropper, we can find that the maximum secure distance for detecting the IRFS attack when the eavesdropper fakes double-detection events is 176 km, which is within the range of the BB84 key rate, as it appears in Figure 9.
