Time-of-flight vs. which-path information for single photon interference I wonder how different path length and therefore different times required for each path lead to interference. Wouldn't it be in principle be possible to tell which path was taken by looking carefully at the time required for the photon to pass the interferometer?
Is this apparent contradiction completely solved by the fact that which-path information is not a binary thing and the differences in time needed are too small?
 A: If we want to see interference in white light beams, the two paths must have a very small difference of path length. With monochromatic light, the path length difference can be longer.
Let us call $\lambda$ the (average) wavelength, $\delta \lambda$ the wavelength interval (roughly, the range of wavelengths is between $\lambda-\delta\lambda$ and $\lambda+\delta\lambda$), and $\delta L$ the difference of path length between the two paths. In order to see the interference, the phase difference of the various wavelengths along $\delta L$ must be small. Very roughly:
$$ \frac{\delta L}{\lambda-\delta\lambda} - \frac{\delta L}{\lambda+\delta \lambda} \ll 1 \tag{1}$$
Assuming $\delta\lambda\ll\lambda$:
$$ \delta\lambda \ll \frac{\lambda^2}{2\delta L} \tag{2}$$
I remind again the reader that this is a very rough calculation. In order to have the requested $\delta\lambda$ and $\lambda$, the wave packet must be long enough; roughly, it should contain $\frac{\lambda}{\delta\lambda}$ oscillations (waves). So the wave packet should have at least a length $\frac{\lambda^2}{\delta\lambda}$. Its duration is thus at least $\delta T$: 
$$ \delta T = \frac{\lambda^2}{c\delta\lambda} \tag{3}$$
Using the inequality 2:
$$ \delta T \gg 2\frac{\delta L}{c} \tag{4} $$
So, the time duration of the pulse must be much longer than the time delay between the two paths. It is thus not possible to find which path the photon followed.
A: The photon is better described in quantum optics as a "photon wave function" (PWF). Photon interference is a temporary thing, photons never interact to cancel each other out, this would violate conservation of energy.  For example in the double slit experiment, the dark areas are referred to as cancellation areas but it's not possible. A modern explanation based on PWF says that a photon wants to travel an integer number of its wavelength, no photons can travel to the dark spots because the path length is not ideal, the bright spots correspond to n lambda wavelength paths (Feynman), the multiple spots are all n lamba and caused by photon interaction with the slit (as an aperture).  Take a laser for example ... very bright on a piece of paper (rough surface) .... now stick a mirror at exactly n lambda plus 0.5 lambda ... the laser stops lasing!  But you really can't observe the laser not lasing so we call it interference but really it is the lack of an ideal PWF.
If you did TOF in the double slit, you would need single photons, but single photons cannot be generated at a certain time, you apply energy to your light source and they appear randomly.  But if you could there would be no mystery, a photon that went thru the left side could indeed appear over on one of the right hand side peaks and yes it would take more time!
