Let us again consider two photons with two different external states labeled one and two. For each photon, the polarization is described in a two-dimensional space, with basis for instance, x and y. Any quantum state of these two photons belongs to a four-dimensional space of which obvious basis vectors are: x_1 x_2, x_1 y_2, y_1 x_2, and y_1 y_2. These four two-photon states form a complete orthonormal basis. As in any vector space, we can form another basis by linear combination of the first basis vectors. A particularly important basis consists of the four so-called Bell states Psi-plus, Psi minus, Phi plus, and Phi minus. You can check that they are mutually orthogonal, so they constitute a complete basis. In contrast to the initial basis vectors which are factorized, the Bell states are entangled. You can check that it is impossible to factorize them as a product of a state of photon nu_1 by a state of photon nu_2, as we did last week for the EPR state. In fact, one of these Bell states is the EPR state of last week. Which one? Bell states play an important role in quantum information where entanglement is a key resource. From a practical point of view, it is possible to produce all these four states. In contrast, there is no practical method to implement a full measurement of the Bell observable of which the four Bell states are eigenstates. Fortunately, it is easy to implement a partial measurement, very useful in quantum information. Let us see these points in more details. There exist remarkable tools to manipulate the polarization of light beams, they are birefringent plates, which allow one to realize any transformation of the polarization. They can for instance, change the direction of a linear polarization, or change a linear polarization into circular polarization. It means that one can change at will, the polarization state of a single photon. If we take the most general polarization of a photon expressed in a linear polarization basis by two complex components alpha and beta, the plate will realize a transformation described by a unitary matrix M. In the language of quantum information, such a transformation is a single qubit gate. Some gates are particularly important for our discussion. With a half-wave plate aligned at 45 degrees from the x and y-axes, one can realize a gate that changes x into y, and vice-versa. It is represented by a matrix 0 1, 1 0, and called a Pauli X-gate. A quarter wave plate with axes along x and y can change a linear polarization at 45 degrees from x and y into a circular polarization. It is represented by a matrix 1 0, 0 i and called a phase gate. Using these gates on one or both photons of an entangled pair, allows one to transform one Bell state into another Bell state. This is a key to the production of any of the Bell states. Using non-linear interaction between light and matter, as we will see in a future lesson, one can produce a pair of entangled photons in a Bell state, for instance, the Phi plus state, which is the EPR state considered in the previous lesson. If now you apply a Pauli X-gate to photon nu_2, you obtain the state Psi plus. I let you find how by applying twice a phase gate to Phi plus you can obtain Phi minus. Finally, it is not difficult to obtain the last Bell state, Psi minus. Can you find how? Surprisingly, although there is a mathematical observable associated with the four Bell states since they form a complete orthogonal basis of the state space of the two polarization entangled photons, one does not know how to implement an apparatus that would make a complete measurement of that observable, that is to say, has four outputs that correspond to each of the Bell states of the two photons. One can, however, implement measuring setups that perform partial Bell measurements, that is to say, have outputs associated with at least one of the four Bell states. We consider a pair of photons nu_alpha, nu_beta, entering the two input ports alpha and beta of a beam splitter. Their polarization state belongs to the four-dimensional space just mentioned, and you are going to see that the device shown here, allows one to make a measurement associated with the projection onto the Bell state Psi minus. It means that the apparatus will sometimes return a signal if the input state has a non-null component on Psi minus. The corresponding probability is the squared modulus of that component. Moreover, when that signal is emitted, the initial state is projected onto the eigenspace associated with Psi minus. The corresponding projector is P_Psi minus. The measurement is based on devices you know well. Firstly, a beam splitter insensitive to polarization couples the input modes, alpha and beta, to the output mode gamma and delta. We have the usual relations between the input and the output written here for the creation operators a-dagger. These expressions are valid for each polarization x or y. For all channels, we call x the polarization in the plane of the figure and y the orthogonal polarization. In each output of the beam splitter, we have a polarization splitter followed by two detectors, which allows us to measure the polarization in the x-y basis. We need to know what will be detected at the various detectors if we have at the input one of the Bell states. I will do the detailed calculation for the state Psi minus. Let us find the expression of Psi minus in the output space. We express it in the input space with the creation operators and we transform them as seen in Quantum Optics 1. Let us do it for the first term expressed with creation operators applied to the vacuum and transformed. Do it yourself for the second term. This is the result. Notice that you obtain the same set of four terms since the various creation operators commute, but the signs are different. When you add the two lines to get Psi minus, four terms cancel, and you are left with two terms only. Applying them to the vacuum yields the expression of Psi minus in the output channels of the two beam splitters. Note that if you have taken into account the various coefficients, the state is normalized as it should. You can now find what signal you get at each of the four detectors for a state Psi minus at the input. It suffices to read the expession of the state in the output space to get the answer. You have two possibilities, each with one-half probability. The first one is a joint detection of x gamma and y delta, the other one is a joint detection of y gamma and x delta. In both cases, there is one photon detected in channel gamma and one photon detected in channel delta. In fact, we could as well remove the polarization splitters and have one detector in channel gamma and one detector in channel delta, and we would have a joint detection in the two channels, gamma and delta. It is remarkable that among the four Bell states, it is the only case where there is a joint detection in channels gamma and delta. Indeed, for the Psi plus state, you can make a similar reasoning to show that you have either a joint detection of both polarizations of channel gamma or a joint detection of both polarization of channel delta. I leave it to you to demonstrate that for the states Phi plus or Phi minus at the input, one has two photons at one of the four detectors, each case happening with the probability one-fourth. To summarize, it is possible to detect, without ambiguity, the Psi minus component or the Psi plus component of the input state. The first case is characterized by a joint detection between the two channels gamma and delta, while the second case is characterized by a joint detection between the two output channels of one of the polarization splitters. For components Phi plus or Phi minus in the input alpha-beta, only one detector fires and shows the detection of two photons. One knows that it is either Phi plus or Phi minus, but one cannot distinguish the two cases. The case of Psi minus is remarkable. It is the only case corresponding to a joint detection by one detector gamma and one detector delta. Check that no other Bell state at the input yields such a response. If one is interested only in the detection of the Psi minus component at the input, the apparatus can thus be significantly simplified. In fact, to detect the Psi minus component in the input space, it suffices to use one detector in each output channel of the beam splitter. A joint detection with these two detectors means that one has detected a Psi minus component in the input channel and that the input state has been projected onto the corresponding eigenspace. You may wonder why I have not shown you how to perform a full Bell state measurement, that is to say, showing a device that would have four outputs, each associated with one of the four Bell states. Such a measuring apparatus can be built in principle, but it must incorporate a quantum two-bit gate, able to entangle individual qubits. Such a two-bit gate demands a non-linear interaction, which is not yet possible in general for polarized single photons to my knowledge. There is, however, one existing non-linear operation that we have already used. It is photodetection, more precisely, joint photodetection. It was a remarkable discovery of Emmanuel Knil, Raymond Laflamme and Gerald Milburn, that one can achieve efficient quantum computation with linear optics and single-photon detectors. The possibility to identify the two-state Psi minus and Psi plus is an example of application of the KLM method. Unfortunately, there does not seem to be a way to use a KLM method to realize a complete measurement of the four Bell states in the case of single-photon polarization. But realizing in such a simple way an efficient partial measurement of the Psi minus and the Psi plus cases is very useful and has allowed quantum optics experimentalists to demonstrate many quantum information operations, in particular, quantum teleportation.