So when it encounters a linear polariser, and is transmitted through , what actually happens to the photon. Has it stopped vibrating in the plane perpendicular to the polariser, when it emerges? This would be my obvious conclusion. Since putting a second perpendicular polariser after the first reveals no photons ever emerging. Consider my analogy for simplicity. If I hold a rope on one end, and tie the other end to a pole in the distance. Between myself and the pole the rope runs through a gate( having vertical beams evenly spaced) placed between myself and the pole. 1.If I induce a vertical vibration (Y- pulse) on the rope . Then the pulse goes through the gate without any , disturbance. The energy of the pulse is totally transmitted through to the pole. 2.If I induce a horizontal vibration ( X- pulse) on the rope. The energy of the pulse is totally absorbed by the gate. No pulse energy is transmitted to the pole. 3. If I induce a diagonal vibration 45% to the horizontal and vertical. From simple vector analysis , I can calculate that a certain % of the energy will be transmitted to the pole. We would find a substantially smaller pulse being transmitted to the pole, and the rest being absorbed/reflected at the gate. All this can be visualized in a classical , mathematical way. No problem there. But now our original pulsating /vibrating ( in x-y plane) photon is not quite like this. If it passes the polariser , why does it not lose most of its energy( or does it)? If it is now only vibrating in 1 plane then it should be in the order of 1/360 its original energy? Since it was vibrating in 360 degrees (unpolarised), its now down to one degree of freedom. My understanding of the photon passing through a polariser must be incomplete. OR Is the vector superposition of states not then incorrect in describing the polarization of a photon?