Re: Does dual slit interference cause distribution of probability or E/M wave?

The energy in each photon is not changed at all, energy is conserved.  
Opening the second slit changes the probability of a photon arriving at a 
given point on the screen, according to the interference equations for an 
EM wave.  

If you integrate energy arriving at a given point over a period of time, 
the total energy (energy/time=power) ranges from twice to zero, not 
because the energy of the photons is different but because there are more 
or fewer of them arriving in total. Because the probability has has 
changed due to coherent interference between 2 slits.

Its easy to see how Coherent Interference happens with waves, because some 
of the wave passes thru each slit.  

The mystery:  How can a single photon or even a particle like an electron 
or a silver nucleus can interfere with itself?  Apparently it goes with 
50% probability thru both slits.  Like Schroedinger's cat, the photon 
passing thru each slit is 50% real, and the "superposition" of two half-
real photons make up a whole one.  At the screen they collapse to form a 
real photon.  Each slit does NOT receive half a photon, each slit receives 
a 50% probability of a whole photon containing ALL the energy.  

If you detect the photons at each slit, each photon appears 100% real but 
only at one slit, and the interference disappears.  When you 
quit "watching" the slits, each photon "goes thru both slits" and the 
coherent interference happens.

Well said that this simple experiment contains the paradox that physics 
has stubbed its toe on for over 70 years.  If you are intrepid in 
mathematics, read "Interpreting the Quantum World" by Dr. J. Bubb, else I 
recommend "In Search of Schroedinger's Cat" by a popular author.  See 
www.altair.org for constructing a working model.