# BB84 and Ekert91 protocols

Benett and Brassard proposed a protocol for a secret key exchange between Alice and Bob in 1984 (BB84).

Today's (symmetrical) cryptography algorithms rest upon secure key transmission. In general this key has to be transmitted through the internet (or some public channel) and an eavesdropper (Eve) can easily intercept the communication, catch the key and the whole encoding is for nothing. And what's worse: Alice and Bob never know that they have another listener. One solution to key distribution is to use asymmetrical algorithms like RSA to encode the symmetric key. RSA uses a public key for encoding and a private (secret) key for decoding. This way, the secret key doesn't have to be sent through the internet. It's as difficult for Eve to compute the inverse RSA algorithm as it is to factor large integers or discrete logarithms (which is widely believed to be prohibitively difficult with current technology).

Symmetrical algorithms would be safer, if only the key transmission problem weren't there. And that's where quantum cryptography comes in.

Two parties using BB84 know that Eve is listening and will not use this key for transmitting the actual message.

### BB84 protocol

Alice wants to send the key to Bob. She has two bases with polarized Photons:[see also http://de.geocities.com/me_dal_ges/bb84.pdf]

1.bases: 1.1 horizontally (0°) Bit:0 1.2 vertically (90°) Bit:1 '''2.bases: ''' 2.1 + 45° Bit:0 2.2 -45° Bit:1

She chooses an arbitrary basis for her bits and sends them over the Quantum Channel to Bob. Bob measures in an arbitrary chosen basis, too. If he has no detection he deletes this register. Then he sends this information and information on what bases he chose over a public channel to Alice and keeps the outcome of each measurement secret. After Alice gets Bob's info, she can compare it to her own chosen bases and select the coincidences. She sends the information on the coinciding bases back to Bob and he just looks in his protocol, checks whether he had 0 or 1 for this register and the result forms the key. Finally Alice and Bob can do a security check: Each party adds their Bits (0+1+0+0+0+1...) and both must have an either odd or even result. Now, they can use the key and encrypt the message. The key should be as long as the plain-text and be used only once (one-time-pad). Eve can wire-tap the public channel, but that won't do her any good. She gets information on the bases and not on the outcome of the measurement. In case Eve attempts to measure part of the Quantum Channel she betrays herself by a high Quantum Bit Error Rate (QBER) and Alice and Bob are warned.

#### Quantum Bit Error Rate

The Quantum Bit Error Rate (QBER) is the ratio of an error rate to the key rate and contains information on the existence of an eavesdropper and how much he knows.

QBER = $p_f + \frac{p_d n q \Sigma f_r t_l }{2}$ µ

pf : probability for a wrong 'click' (1-2%)

            pd : probability for a wrong photon signal (Si: 10 − 7; GaAs 10 − 5)
            n : number of detections
            q :  phase = 1/2 (better for optical fibres); polarizaton = 1 (better in the air)
            Σ: detector efficiency
            fr: pulse repeat frequency
            tl: transmission rate (for large distances small)
            µ : attenuation for light pulses (single photons = 1)

Usually the QBER is around 11%. It means that Eve didn't gain more information on the key as Bob.

#### Implementation

Image:Implementation_BB84.jpg

Phase gate: $\begin{pmatrix} exp(i \phi) &amp; 0 \\ 0 &amp; 1 \end{pmatrix}$

PBS: $\begin{pmatrix} 1 &amp; i \\ i &amp; 1 \end{pmatrix}$

(V/H) basis: 1. Operations on horizontal polarized light $|H\rang = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $\frac{1}{2} \begin{pmatrix} 1 &amp; i\\i &amp; 1 \end{pmatrix}\begin{pmatrix} exp(i \phi_A) &amp; 0\\0 &amp; 1 \end{pmatrix}\begin{pmatrix} 1 &amp; 0\\0 &amp; exp(i \phi_B) \end{pmatrix}\begin{pmatrix} 1 &amp; i\\i &amp; 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} exp(i \phi_A)- exp(i \phi_B)\\ i(exp(i \phi_A)+ exp(i \phi_B))\end{pmatrix}$ 2. Operations on vertically polarized light $|V\rang = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ $\frac{1}{2} \begin{pmatrix} 1 &amp; i\\i &amp; 1 \end{pmatrix}\begin{pmatrix} exp(i \phi_A) &amp; 0\\0 &amp; 1 \end{pmatrix}\begin{pmatrix} 1 &amp; 0\\0 &amp; exp(i \phi_B) \end{pmatrix}\begin{pmatrix} 1 &amp; i\\i &amp; 1 \end{pmatrix}\begin{pmatrix} 0 \\ 1\end{pmatrix} = \frac{1}{2} \begin{pmatrix} i(exp(i \phi_A)+ exp(i \phi_B))\\ - exp(i \phi_A)- exp(i \phi_B) \end{pmatrix}$

The probability to get horizontal polarized light is then: $P_H = \|H\|^2 = \frac{1}{2} (1 + cos \Delta \phi)$ with the Phase difference Δϕ = ϕA − ϕB. And the probability to get vertical polarized light$P_V= \|V\|^2 = \frac{1}{2} (1 - cos \Delta \phi)$

If Δϕ = Π/2 or 3Π/2 then $P_V = P_H = \frac{1}{2}$. The signal is only half as intensive as is should be and Bob knows: that's the wrong basis. If Δϕ = 0 then PV = 0 and PH = 1. The signal at the H exit has the highest intensity and the lowest intensity the V exit. That means the basis is correct. If Δϕ = Π then PV = 1 and PH = 0. The signal at the V exit has the highest intensity and the lowest intensity the H exit. That also means the basis is correct. The same holds for the (P/M) basis with states orthogonal to the (H/V) basis states.

### Dense Coding

#### Bell State Analysis

A Bell State analysis is a projection to an entangled sate. The Bell States are an example of entangled states. They behave either as bosons or as fermions. Fermions are antisymmetrical under permutation of two particles. Entangled Bell Pairs can be produced by parametric down conversion, change of polarization (lambda/4) and phase addition (lambda/2)

$|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|H\rangle_A \otimes |H\rangle_B + |V\rangle_A \otimes |V\rangle_B) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\0\\0\\1 \end{pmatrix}$ (Boson)

$|\Phi^-\rangle = \frac{1}{\sqrt{2}} (|H\rangle_A \otimes |H\rangle_B - |v\rangle_A \otimes |V\rangle_B) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\0\\0\\-1 \end{pmatrix}$ (Boson)

$|\Psi^+\rangle = \frac{1}{\sqrt{2}} (|H\rangle_A \otimes |V\rangle_B + |V\rangle_A \otimes |H\rangle_B) = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\1\\1\\0 \end{pmatrix}$ (Boson)

$|\Psi^-\rangle = \frac{1}{\sqrt{2}} (|H\rangle_A \otimes |V\rangle_B - |V\rangle_A \otimes |H\rangle_B) = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\1\\-1\\0 \end{pmatrix}$(Fermion)

Implementation Image:Dense Coding.jpg

After a Hadamard Operation antiymmetrical states and symmetrical states can be distinguished. In Quantum Optics a Hadamard Gate can be performed by a beam splitter. BS: $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1\end{pmatrix}$

Hadamard on ∣Φ + : $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1\end{pmatrix}\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\0\\0\\1 \end{pmatrix} = \frac{2}{\sqrt{2}} \begin{pmatrix} 1 \\0\\0\\1 \end{pmatrix} = 2 |\Phi^+\rangle$

Hadamard on ∣Φ − : $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1\end{pmatrix}\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\0\\0\\-1 \end{pmatrix} = \frac{2}{\sqrt{2}} \begin{pmatrix} 0\\1\\1\\0 \end{pmatrix} = 2 |\Psi^+\rangle$

Hadamard on ∣Ψ + : $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1\end{pmatrix}\frac{1}{\sqrt{2}} \begin{pmatrix} 0\\1\\1\\0 \end{pmatrix} = \frac{2}{\sqrt{2}} \begin{pmatrix} 1\\0\\0\\-1 \end{pmatrix} = 2 |\Phi^-\rangle$

Hadamard on ∣Ψ − : $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1\end{pmatrix}\frac{1}{\sqrt{2}} \begin{pmatrix} 0\\1\\-1\\0 \end{pmatrix} = \frac{2}{\sqrt{2}} \begin{pmatrix} 0\\-1\\1\\0 \end{pmatrix} = -2 |\Psi^-\rangle$

If the state is antisymmetrical both detectors on Alice side will click. In case the state is symmetrical both photons will end up in one detector. ∣Ψ −  can be easily distinguished. Either D(H) and D(V') or D(H') and D(V) have coincidences with half intensity. ∣Ψ +  can be distinguished from ∣Φ +  and ∣Φ −  by measuring the polarization with a PBS (polarization beam splitter). It will lead to a detection with full intensity in one detector. With ∣Ψ +  both photons have a different polarizations. There will be coincidences with D(H) and D(V) or with D(H') and D(V')

∣Φ +  and ∣Φ −  cannot be distinguished in this basis. Both states lead to detections at D(H), D(V), D(H') or D(V') with 1/4 intensity.

Category: Quantum Cryptography