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Absorption in Indirect Bandgaps

  1. Apr 29, 2015 #1
    Hi guys I just wanted to make sure I was thinking about this the right way. So in indirect bandgap semiconductors absorption needs a phonon to assist the electron to the conduction band from the valence band because a photon has virtually no momentum and only contributes energy to the electron while a phonon can supply the momentum needed to sort of go horizontally (k axis) on the band structure diagram. Im just confused with what the phonon actually does with the electron. Does it collide with the electron to transfer its momentum? Also was wondering what effect does the actual placement of the minimum of the conduction band and maximum of the valence band have on the phonon needed to make the jump. Im assuming if less momentum is needed to make the jump (min on left max on right) the phonon needed would have to have a negative momentum relative to the maximum point to assist the electron? Thanks for the help!
     
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  3. Apr 30, 2015 #2

    DrDu

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    It is probably helpful to think of the absorption process to be localized in space. Basically, you are transferring an electron from a bonding orbital between two atoms into an anti-bonding one. This will change the equilibrium bond length between the two atoms and therefore classically, they will start to oscillate. In a QM treatment, you get a distribution of the amount of phonons released instead which is given by the Franck Condon factor.
     
  4. Apr 30, 2015 #3

    DrDu

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    It is probably helpful to think of the absorption process to be localized in space. Basically, you are transferring an electron from a bonding orbital between two atoms into an anti-bonding one. This will change the equilibrium bond length between the two atoms and therefore classically, they will start to oscillate. In a QM treatment, you get a distribution of the amount of phonons released instead which is given by the Franck Condon factor.
     
  5. Apr 30, 2015 #4
    Correct me if I'm wrong, but I assumed that when making the jump, the electron could either absorb or emit a phonon.

    Meaning that when emitting a phonon,

    ħωphoton = Eg + ħωphonon

    And absorbing

    ħωphoton = Eg - ħωphonon

    Please wait for someone else to respond to my comment before you take it as truth however, I'm still new to this topic.
     
  6. Apr 30, 2015 #5

    DrDu

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    I just realized that my previous explanation is not correct. Here is a (hopefully) better one.
    In the dipole approximation (which corresponds to the neglect of the momentum of the photon), the intensity is proportional to the square of the dipole matrix element
    ##\langle i| d|f \rangle##, where i and f are the initial and final state, both electronic and vibrational. These states depend on the nuclear displacements e.g. for f as
    ## | f(k)\rangle =|e_f(k)\rangle |0_v\rangle+ |e_f(0)\rangle \langle e_f(0)| \partial H/\partial Q(k) |e_f(k)\rangle/(E_f(0)+E_v(k)-E_f(k)) Q(k)|0_v \rangle## in first order of perturbation theory.
    Here ##|e_f(k)\rangle## is the electronic wavefunction for the undisplaced lattice and ##|0_v\rangle## the vibrational ground state wavefunction of the lattice.
    The operator Q of the nuclear displacement either creates or destroys one phonon, so that ##Q(q)|0_v\rangle\propto a^+_q |0_v\rangle=|1_v(q) \rangle##.
    Similarly, the ground state reads
    ## | i(0)\rangle =|e_i(0)\rangle |0_v\rangle+ |e_i(k)\rangle \langle e_i(k)| \partial H/\partial Q(-k) |e_i(0)\rangle/(E_i(k)+E_v(-k)-E_i(0)) Q(-k)|0_v \rangle##.
    Taking into account that only electronic dipole matrix elements between states with the same k are non-vanishing, you see that one phonon will be generated or distroyed in the indirect transition.
     
  7. May 17, 2015 #6
    You need phonons to be present to enable an indirect absorption process.
    At absolute zero only direct absorption occurs and indirect transitions are forbidden.
    The indirect transitions at high T occur between vibronic states and these are not occupied at T~0 K.
     
  8. May 17, 2015 #7

    DrDu

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    No, because phonons need not be present beforehand. They can be emitted in the transition.
     
  9. May 17, 2015 #8
    At T near zero only direct bandgap transitions are allowed. That means that in transitions from the ground state no phonons are emitted. Of course you can argue about high T. A transition from one vibronic state to another in a sense involves the absorption or emission of one or more phonons.
    In any case phonons need to be present beforehand.
     
  10. May 17, 2015 #9

    DrDu

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    I did derive in post #5 that an indirect transirion is even possible starting from the vibronic ground state containing no phonons. Do you have any proof for your claim?
     
  11. May 17, 2015 #10
    I have sources that at low T silicon becomes transparent for red light and longer wavelengths according to this link.
    "For example, silicon is opaque to visible light at room temperature, but transparent to red light at liquid helium temperatures, because red photons can only be absorbed in an indirect transition."
    http://en.wikipedia.org/wiki/Direct_and_indirect_band_gaps#Implications_for_light_absorption
    A reference to a scientific paper is not attached there, however.
    I expect this behaviour also from the k-values at 20 C at
    http://refractiveindex.info/?shelf=main&book=Si&page=Vuye-20C
    If the temperature goes to 0 K, the absorption at wavelengths longer than say 500 nm will vanish.
    So experiment says that a transition at or near the indirect bandgap energy is forbidden at 0 K, when there are no phonons present.
    Clearly visible is the absorption at 354 nm corresponding to the direct band gap, which is at 3.5 eV (http://arxiv.org/pdf/1211.0591.pdf).
    Only the direct transition is allowed if there are no phonons.
    Or is your claim that ecen in the direct transition phonons are emitted ?
     
  12. May 17, 2015 #11

    DrDu

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    Ok, clearly the likelhood for an indirect transitions increases when there are phonons present. Nevertheless, it does not vanish completely at T=0.
     
  13. May 17, 2015 #12
    Last edited: May 17, 2015
  14. May 17, 2015 #13

    DrDu

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    Interesting, thank you. I would't have guessed that these indirect transitions give rise to so strong thermochromic effect.
     
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