# Why is the parallel component of the electron momentum preserved when passing through the solid-vacuum interface? Why is the perpendicular not?

I am reading Surface Science: An Introduction and in the chapter about angle resolved photoemission spectroscopy (page 106) it says

To consider the wave vector of an electron inside the solid $$k^{in}$$, one should recall that, when passing through the solid-vacuum interface, only the parallel component of the electron momentum is preserved as $$k^{ex}_\parallel = k^{in}_\parallel + G_{hk},$$ where $$G_{hk}$$ is a vector of the 2D surface reciprocal lattice. In contrast, the perpendicular component $$k_\perp$$ is not preserved and, thus, does not bear any particular relationship to $$k^{in}_\perp$$.

My question is: Why?

I found the same information in several other books, but no explanation. Sometimes they just mention translational invariance.

• So does this mean that both $k^{in}$ and $k^{ex}$ are wavevectors that decribe electron after the photoemission? I though that $k^{in}$ was the one before the interaction with photon. Then it didn't make sense to me how could kinetic energy of photon in $k^{ex}$ determine electron's wavevector in the past, before interaction happened. – Andrej May 3 '20 at 14:58
• @Andrej The way I’m reading that notation, $k^{in}$ is the wavevector in the interior of the surface, $k^{ex}$ is the wavevector in the exterior. You can sort of think of both momenta as representing the time after the interaction with the photon, when it has all this energy and is leaving the surface. Whenever the electron would leave the surface (such as when it is kicked out by photoemission), then the equation applies. – Gilbert May 3 '20 at 18:16