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.

If you could just direct me to some materials for more information about this, I will appreciate that too.


1 Answer 1


There are several ways to think about this, but keep in mind that the preservation of the in-plane component of momentum is a very general concept in physics, so feel free to use your intuition on this one.

What might be regarded as the final word on this is Noether’s Theorem, which derives conserved quantities from symmetries in the system (such as translation invariance). Thus, since you have in-plane translation symmetry, you’ll have in-plane momentum conservation. But this is also a rather abstract approach.

You can also think about it intuitively: If you throw an (ideal) ball down at an angle at the (ideal) floor, when it bounces it’s vertical velocity changes sign (there’s acceleration), but it’s horizontal velocity stays the same. Why? Because there’s nothing to “push against” in the horizontal direction, only the vertical one. So the force can only be normal to the floor. If there’s nothing to absorb momentum in any given direction, then momentum cannot change.

In the case of a Bloch electron escaping through an interface, the only force it can experience is in the direction normal to the interface, and the parallel momentum will remain unchanged (up to a reciprocal lattice vector, as usual).

  • $\begingroup$ 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. $\endgroup$
    – Andrej
    Commented May 3, 2020 at 14:58
  • 1
    $\begingroup$ @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. $\endgroup$
    – Gilbert
    Commented May 3, 2020 at 18:16

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