## 4 Quantum Integrated Photonic Devices
### Integrated Quantum Circuits for Quantum Cryptography
#### 4.2.1 Quantum States Generator and Quantum Projectors On-Chip for BB84  
Figure 13 shows a device to generate four quantum states, which are required in the Alice system of the BB84 protocol. It consists of four paths, that is, four ion-exchanged channel guides fabricated in a hybrid glass substrate. Moreover, an OG in the rare-earth-doped glass section (DBR lasers) generates four weak laser signals (weak coherent states) obtained with pumping beams \(\mathrm{P}_{j}\), \((j=1,2,3,4)\), whereas in the passive glass section four directional couplers \(X_{\pi/4}\) and one coupler \(X_{\pi/2}\) along with a phase shifter \(Z_{\pi/2}\) generate the four states required for BB84 protocol at the outputs 2 and 3, namely, the 1-qubits \((1/\sqrt{2})(|1_{2}\rangle+e^{i\delta}|1_{3}\rangle)\), with \(\delta=\pm\pi/2,\pi,0\). The pumping beams \(\mathrm{P}_{j}\) are activated in a random way, and moreover, as shown later, the device (Alice device) emits each of the four states through the outputs of paths 2 and 3 with a probability equal to 50%.  
For example, if a weak coherent state is excited in channel waveguide 1 with pumping beam \(\mathrm{P}_{1}\), then it is obtained \(|L\rangle\approx|1_{1}\rangle\). Next, the concatenated couplers \(X_{\pi/4}\) and \(X_{\pi/2}\) along with the shifter \(Z_{\pi/2}\) (see Figure 13) perform a linear transformation, therefore, by taking into account Equation (5) for \(\theta=\pi/4\) and \(\theta=\pi/2\) and Equation (7) for \(\Phi=\pi/2\), the following output state is obtained  
\[|L_{1}\rangle=\frac{1}{2}(|1_{1}\rangle-i|1_{4}\rangle)+\frac{i}{2}(|1_{2} \rangle+i|1_{3}\rangle) \tag{9}\]  
Note that in channel guides 2 and 3 we have one of the states of the BB84 protocol, that is, in a vector form we have the state \((1/\sqrt{2})(1,i)^{t}\), with \(t\) indicating transposed. The other three states are obtained by using the other pumping beams. All of them are produced with a 50% of probability what is reasonable for quantum key distributions.  
On the other hand, the passive part of this device (the blue section, that is, without the active section where the OG is placed) is a quantum projector for the above states. Indeed, by considering that the inputs are the channel guides 1 and 3, then the first two couplers \(X_{\pi/4}\) and the coupler \(X_{\pi/2}\) make a state division. The next upper coupler \(X_{\pi/4}\) projects the states \((1/\sqrt{2})(1,\pm 1)^{t}\), whereas the phase shifter and the bottom coupler \(X_{\pi/4}Z_{\pi/2}\) project the states \((1/\sqrt{2})(1,\pm i)^{t}\). This means that the device (Bob device) produces a random choice of bases. For example, let us consider the state \((1/\sqrt{2})(|1_{1}\rangle+|1_{3}\rangle)\equiv(1/\sqrt{2})(1,1)^{t}\), then, the final state is  
\[|L_{1}\rangle=\frac{1}{\sqrt{2}}|1_{2}\rangle+\frac{1}{2}e^{-i\pi/4}|1_{3} \rangle+\frac{1}{2}e^{i\pi/4}|1_{4}\rangle \tag{10}\]  
Note that if the state chooses the channel 1 and 2, then it is projected on the output 2 (\(|1_{2}\rangle\)). If the input state is \((1/\sqrt{2})(|1_{1}\rangle-|1_{3}\rangle)\equiv(1/\sqrt{2})(1,-1)^{t}\), then we obtain  
\[|L_{1}\rangle=\frac{i}{\sqrt{2}}|1_{1}\rangle-\frac{1}{2}e^{-i\pi/4}|1_{3} \rangle+\frac{1}{2}e^{i\pi/4}|1_{4}\rangle \tag{11}\]  
namely, the state is projected on the output 1 (\(|1_{1}\rangle\)). The same procedure can be followed for the states \((1/\sqrt{2})(|1_{1}\rangle\pm i|1_{3}\rangle)\equiv(1/\sqrt{2})(1,\pm i)^{t}\) which are projected on the outputs 3 and 4.