scilover89 said:
1. Observation means interaction in QM.
Theoretically considered
Observation in QM mechanics means that your state vector \Psi
collapses to one of the eigenvectors \Phi_{n} of the observable (operator).
For example if you measure the energy of an electron in the hydrogen atom,
your state \Psi collapses into one of the energy eigenstates
of the Hamilton-operator (H \Phi_{n} = E_{n} \Phi_{n})
(That's an axiom of QM)
Ok the above example is not about the photon's superposition but it shows
you what measurement means in QM.
So here an example for photons:
An example would be polarizers for light. Say we have a polarizer at 45 degree to the original light polarization. Then there's a certain probability that the wavefunction will collaps into the state that can pass the polarizer.
scilover89 said:
2. Photon doesn't interact with gravity or electromagnetic force.
I am not sure about that. Physicists believe that photons are for example attracted by a black hole due to its great gravity.
scilover89 said:
3. Thus, light wave is never observed and will therefore maintain superposition.
Light is obviously observed because we can see it with our eyes..hmm..
or look at this site: http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm
You can see the photon 'dots' on the screen.
(see my answer to 1)
I think that you are asking yourself how exactly the photon is interacting
with matter, for example photons are scattered by electrons, they are absorbed and emitted by an electron in an atom and so on.
I think someone else can explain that to you with QED (unfortunately I can't tell you anything about QED).
-Edgardo