I have a question about the principle of momentum conservation the modeling of ARPES:


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We split the initial momentum of the electron ${\displaystyle \hbar k_{i}}$ into the component $ \hbar k_{i\parallel }$ parallel to the surface and $\hbar k_{i\perp }$ perpendicular to it.

Similary we split the momentum of the outgoing electron $\hbar k$ also into the component $ \hbar k_{\parallel }$ parallel to the surface and $\hbar k_{\perp }$ perpendicular.

The photon with energy $\hbar \omega$ hits under abitrary angle on the surface and ionizates an atom and therefore induces the outgoing electron.

My question is why concretely the parallel component of the momentum is conserved while the perperdicular not. I don't see any assumption that the the photon comes perpendiculary to the surface so naively I don't se any reason for momentum conservation for parallel momentum component.

One argument that I often found was "symmetry breaking" although I don't understand how the concept of sb proides here the desired conservation, therefore why should

$ \hbar k_{i\parallel }=\hbar k_{\parallel }$ hold?


1 Answer 1


The kind of classical idea is that the electrons need to gain a certain velocity away from the surface in order to break free of the crystal. This is the 'work function' of the material.

In terms of symmetry, you can think like this: inside the crystal the potential is periodic (i.e. like an array of valleys) and if the electron moves between symmetric points in the potential then it will have the same potential energy. If there was no scattering (assumed for ARPES) then the total energy is conserved and so is we can see that the kinetic energy (and momentum) will also be conserved. But when you get near the edge of the crystal the potential begins to change as you transfer to free space. The momentum has to change to account for the changing potential.

  • $\begingroup$ So the angle of the oncoming photon doesn't matter? Only the scalar "portion" of extra energy is transfered to the electron such that - in order to conserve the momentum - has to change it's trajectory to compensate the added energy? $\endgroup$
    – user267839
    Dec 7, 2018 at 3:49
  • $\begingroup$ So the change or perpendicular momentum comes just with exclusion principle $\endgroup$
    – user267839
    Dec 7, 2018 at 3:53
  • $\begingroup$ The angle of the photon does not matter (notice it's not even drawn in your picture!). Although it can change the absoprtivity and thus the quantum efficiency. We often think of a 3 step process: The photon is absorbed (within a skin depth), an electron is transported to the surface (hopefully without scattering), and finally it leaves the material (provided it still has enough perpendicular momentum). $\endgroup$ Dec 7, 2018 at 5:04

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