Detecting a photon without changing it: Does it break conservation laws? This is about an article published on ScienceMag: Nondestructive Detection of an Optical Photon. I don't have access to full text, but you can see a brief transcription in this link.
Basically, it says that a photon causes a phase shift in another system. This phase shift can be detected, and it does not change photon properties, such as frequency (pulse shape) and polarization.
How can that be true? I thought that for a photon to cause any change on a system, it must lose some energy, which is transferred to the detector. What am I missing?
 A: The abstract of that paper mentions the annihilation of the previously measure photons, where now, they aren't totally conserved, but not totally annihilated. Further down, they mention survival probability, and without access to the whole article, I can only conjecture, but I would go so far as to say that that is them implying that not ALL of their photons are conserved. Only 66% so far. They then mention tweaking that number.
Their measurement apparatus is basically a single atom in a superposed stated. It will require a fairly small amount of energy to shift it into a definite state. I'm thinking potentially orders below the photon energy that they launch at the aforementioned atom. If this is truly how they're doing their experiment, then they aren't truly leaving that photon EXACTLY the same. It does interact with the reflective surface. It does interact with the atom, and it does have the direction of it's momentum change if it's being reflected. All of these process do not happen with no energy transfer. It seems this team is merely exaggerating what they can do towards the head of the abstract, then throw in the limitations later.
A: I dont now about this system but i have had a lecture about a similar non destructive measurement aperatus. they trap a photon in a cavity and have a atom move throw the cavity. right throw the destructive interference spot (there where the photon has low change of being so low change of destroying it) They trick lies in the atom. When that atom moves throw a field wit a certain frequencies. it will start switching between a ground an a exited state. however if such a time period is exactly timed it will experience a so called half pie shift where it just switches phase and not state. Moving throw the cavity if there is a photon the field of that photon can cause that same half pie shift neither giving nor taking away energy from the atom. this shift can then be measured. However whether polarization remains the same I dont now nor do I know if secretly the photon is simple absorbed and then emited straight away by the atom
A: There exists elastic scattering where only directions are changed , momentum is conserved and energy is conserved.
There exists elastic scattering of photons:

In Thomson scattering a photon interacts with electrons.
Thomson scattering is the elastic scattering of electromagnetic radiation by a free charged particle, as described by classical electromagnetism. It is just the low-energy limit of Compton scattering: the particle kinetic energy and photon frequency are the same before and after the scattering. This limit is valid as long as the photon energy is much less than the mass energy of the particle: nu very much smaller than mc^2/h .
In Rayleigh scattering a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon. In this scattering process, the energy (and therefore the wavelength) of the incident photon is conserved and only its direction is changed. In this case, the scattering intensity is proportional to the fourth power of the reciprocal wavelength of the incident photon.

Without having read the experiment I would consider it a utilization of elastic scattering of photons, which should involve phases since directions of individual photons change in elastic scattering.
A: the simple answer is that a quantum state has several variables (degrees of freedom), so if you measure only one of them and leave the others unchanged, then you detect the photon , change its state but do not destroy it completely. this is what they say in introduction 

Second, nondestructive detection can serve as a herald that signals
  the presence of a photon without aecting its other degrees of
  freedom, like its temporal shape or its polarization.

look at the full article in arxiv
