- #1
Johan0001
- 108
- 4
My limited understanding of a photon is a pulsating/vibrating energy packet which oscillates in the x, y plane and propagates in the z direction of motion.
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?