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Wave experiences refraction when it propagates into another medium which has different refraction index. Lights surely does experience refraction at the border of mediums which have different refraction index because of its wave-like property. But I am wonder if high energy photons, whose energy exceeds MeV or GeV, also experiences refraction.

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A photon is light. – Kvothe Mar 20 '14 at 9:29

The individual photons interact with the electrons of the material. The absorption and spontaneous re-emission of these propagating photons takes a finite amount of time. This time difference results in our perceived slowing of light in the medium. It also induces a phase difference in the photons which is responsible for the light bending in the medium, due to the requirement that the phase be continuous at the boundary (or off by a factor of Pi, depending on the boundary). Thus, as the density of the medium changes, or the medium has more or less electrons available for interaction, the index changes, resulting in a change in bending and effective "speed" through the medium. As you increase the photon energy, the likelihood of interaction with electrons decreases, thus rendering the medium more "transparent" to the photons. There is a lot more to deal with the index and how various photon energies correspond to the index based on different energy interactions. If you would like more on this, consider reading Feynmans's lectures on Physics, predominantly the second volume.

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You should add a paragraph about electron interaction in addition to electron density. The photon-electron interaction is actually more dependent on the photon energies and the electron probability clouds and that coupling than on the density of electrons themselves. You are correct that as photons reach very high energies most materials begin to "appear more transparent." – daaxix Jul 25 '14 at 22:58
@daaxix I consider the electron probability clouds as a density of electron states for interaction. As the wavelength shortens (or frequency increases), the effective relative density decreases, thus the interaction probability drops. How this drop occurs depends on the initial distribution of electron states (i.e. metals behave differently than gasses) – Chris Jul 26 '14 at 11:16
since this came up to front again, I have to note that the photons are not absorbed and re-emitted. Absorption is a quantum effect for a photon. It either gives up all its energy and disappears, and if a photon is re emitted it will have a 2 pi angle to decay into, so the phases will be lost, or it scatters with the various ways photons can scatter elastically or inelastically. – anna v Aug 30 '14 at 7:18

All photons change speed in any medium. $ v = \frac{c}{n} $ It may well be that $n$ is essentially equal to 1 for certain wavelength ranges (for a given medium).

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An electromagnetic wave is composed of zillions of photons with the same frequency as the wave. The waves emerge by the coherent addition of all the individual photon behaviors. Photons, as elementary particles, will scatter when meeting fields. This means that entering a medium their individual paths will change. In an opaque medium the scattering is drastic, the ones going forward have been absorbed and the reflected ones rebuild the wave. In a transparent medium individually they will scatter off the fields of the medium but they reorder coherently so as to present the wavefront of the classical electromagnetic equation since both the quantum and classical framework merge in light. See this answer to a similar query.

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