Two photons are initially prepared at a source and move along the y-direction in opposite directions toward polarisers at A and B. They are initially in the state VZ [IVA VB) + [HAHB)], where IV,) and [ Hi) are the two linear polarisation states, vertical and horizontal, of the photon moving to the polariser at I = A or B. To be explicit, we will fix the vertical polarisation direction here as the z-direction. In particular, the V state is in the direction of 2 and the H state is in the x-direction, thus perpendicular to both the 2-direction and the direction of motion for the photon. Let n, be other directions at polarisers I = A or B that are perpendicular to the direction of motion of the photons, so in the x-2 plane, and at an angle GAZ to the z-direction. If photon I is measured through a polariser oriented with respect to the n-direction: IVI,I) = cos OLzIV) - sin OIz |H), [HI,I) = sin OLz(V) + cos OLZ |H). The two photons initially in the state |P) are then measured respectively at A and B, with polarisers oriented along the nA and n,-directions, respectively. Let Pvv(nA, B) be the probability that the vertical polarisation of the photon at A is measured along direction n for polarisers rotated by angle GAZ and for the photon at B the vertical polarisation is measured along direction ng for polarisers rotated by angle OBz. Show that Pvv (nA, B) = KVA,A VB,BIT)12 = 2 COS? (OBI - 0AL). Also compute PHH(nA, B), PVH (nA, B) and PHV (nA, B).Show that E(nA, B) = Pvv(nA, nB) + PHH(nA, nB) - PVH (nA, B) - PHV (nA, B) = cos[2(0AZ - OBZ)]. For the expression S = E(n , nB) - E(nA, ng) + E(n'A, nB) + E(n'A, ng), a realistic local theory gives the Bell inequality -2