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Indirect absorption processes - need help!

  1. May 22, 2007 #1
    Indirect absorption processes -- need help!

    On page 188-190 in Introduction to Solid State Physics by C. Kittel, I read

    I don't get it. I thought it was as simple as this: the highest point in the valence band and the lowest point in the conduction band have two different energies, and they therefore have two different wavevectors (this is because different energies gives different wavenumbers). The energy difference is E, and therefore the least energetic photon possible of exciting an electron from the valence band to the conduction band must have this energy. But according to the above quote, the energy of the photon is enough, but not the magnitude of the wavevector. This leads me to think that the photon wavenumber is different from a matter wave wavenumber, which is 2pi/lambda. Is this correct?
  2. jcsd
  3. May 23, 2007 #2
    The different states in a band may have different energies but also other differences like the components of the wavevector.

    The picture below shows the band diagram for Si: si_banddiagram.gif
    For full explanations see: http://www.tf.uni-kiel.de/matwis/amat/semi_en/kap_2/backbone/r2_1_5.html.

    Just observe that the x-axis is the wave vector.
    The various curves correspond to different directions of the wavector (this is a 3D topic!).
    The picture indicates the band gap, and you can guess the two bands shown ont heis because above and below the band gap.
    It is clear that these two bands are made of states with different energy but also different wavevectors.

    You shoul read the details to understand why and how e-states in solids are characterised by more than 1 quantum number. Basically this is related to the crystaline symmetries: the hamiltonian for an electron does not change for translations compatible with the cristaline symmetry.

    Sometimes the bands are represented without the x-axis. In such a representation, some information is lost. This information is not always useful, but for the analysis of photons emissions it is necessary.

    Don't ask me for much more detail, I have forgotten 99% of what I have learned long ago.
    But I could still go a little bit further.
    Last edited: May 23, 2007
  4. May 23, 2007 #3


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    If you have a look at free electrons, the wavevector is an essential quantity describing the momentum of an electron ([tex] \vec p= \hbar \vec k[/tex]).

    Bloch waves are not eigenstates of the momentum operator, so there is just a quasimomentum for crystal electrons, but in principle everything stays the same.
    So the dispersion relation (band structure) shows the combinations of energy and momentum, which are allowed for electrons. If you have a look at the photon dispersion relation, you will see, that it is linear ([tex] \omega=c k[/tex]). Compared to the electron dispersion relation, the photon dispersion is almost a vertical line at k=0. Therefore transitions to electron states where k is not 0 need some other mechanism to transfer the needed momentum, but almost no energy. Usually these are phonon processes.

    So it's just the usual conservation of (crystal) momentum and energy, but it should be clear, that crystal momentum is not equal to usual momentum.
    The change in usual momentum depends on the total force acting on an electron. The change in crystal momentum depends only on external forces, but not on the effects, which arise due to the periodic lattice structure.
    Last edited: May 23, 2007
  5. May 23, 2007 #4
    I think that was helpful, thanks.

    So an electron with a particular energy doesn't have the same wavevector (and hence not the same wavenumber) as a photon with the same energy then?
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