Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absorption spectra of T = 0K semiconductor?

  1. Apr 30, 2014 #1
    I'm trying to understand the form of the absorption spectrum of a T=0K intrinsic semiconductor. The valence band/conduction band energy diagram looks like;


    As [itex] E = \frac{\hbar^2 k^2}{2m^*} [/itex]

    where [itex] m^* [/itex] denotes the effective mass. The image doesn't indicate too well that the effective mass in the valence band is MUCH larger than that in the conduction band.

    I'm trying to understand the absorption spectrum of photons sent in to the semiconductor. For photons with energy less than the band gap there is no absorption.

    When photons with energy more than the band gap are sent in, there is absorption. We assume that the momentum of the photons is low and so in the diagram above there can only be vertical transitions from the valence to the conduction band - so the transition energies are uniquely determined (i.e. an electron is promoted from [itex] E_2 [/itex] to [itex] E_1 [/itex] and [itex] E_1,E_2[/itex] are uniquely determined).

    I would hazard a guess that the absorption spectrum is proportional to the product of the density of states [itex] Z(E_1)Z(E_2) [/itex]. There are two ways I can interpret this...

    1) Examine the change of the absorption spectrum from a photon energy [itex] E [/itex] to a small increase [itex] E + \Delta E [/itex]. For a vertical transition, [itex] E_2 [/itex] barely changes because the effective mass in the valence band is so large, but [itex] E_1 [/itex] changes quite rapidly since it is so low in the conduction band. So [itex] Z(E_2) [/itex] is practically constant and so the absorption spectra depends only on [itex] Z(E_1) [/itex]. Since [itex] Z(E_1) \propto E_1^{1/2} [/itex] we see the absortion spectrum increase like [itex] E^{1/2} [/itex].

    2) Mathematically you can see, since the momentum is conserved

    [itex] E_1 = \frac{\hbar^2 k^2}{2m_c^*} [/itex]
    [itex] E_2 = \frac{\hbar^2 k^2}{2m_v^*} [/itex]

    [itex] E_1/E_2 = m_v^*/m_c^* [/itex]

    Then the absorption rate is [itex] \propto Z(E_1)Z(E_2) \propto (E_1 E_2)^{1/2} \propto E_1 [/itex]

    And so it varies linearly with [itex] E_1 [/itex] and hence the energy of the photon.

    I'm told actually it varies like [itex] E^{1/2} [/itex] which would imply the second way is wrong. But how does the maths lie?

    {Edit: I've included the image so that one doesn't have to click the link to see it - Zz}
    Last edited by a moderator: Apr 30, 2014
  2. jcsd
  3. Apr 30, 2014 #2


    User Avatar

    We have to be a bit more careful in definitions here. From your figure, you can see that valence band dispersion is given by [itex]E_v(k) = -\hbar^2 k^2/2m_v[/itex] and the conduction band is given by [itex]E_c(k) = E_g + \hbar^2 k^2/2m_c[/itex]. From this you can work out the density of states for each band (there is indeed a square root involved) - but importantly, the density of states for the valence band vanishes where the density of states for the conduction band is nonzero, and vice versa. This should convince you that absorption is not related to the product of the densities of states. Instead, think about the energy difference [itex]E_c(k)-E_v(k)[/itex] for each [itex]k[/itex] ... this should lead you to a quantity called the "joint density of states" which is indeed closely related to absorption.
  4. May 1, 2014 #3
    I apologise but I'm still lost. Even with the clean up of definitions I run in to the same issue.

    Using [itex]E_v(k) = -\hbar^2 k^2/2m_v[/itex], [itex]E_c(k) = E_g + \hbar^2 k^2/2m_c[/itex] as you said. One then gets [itex] E_c(k) - E_v(k) = E_{\gamma} [/itex] which, for a given [itex] E_{\gamma}, E_g [/itex], fixes the constant k. This, in turn, fixes [itex] E_c(k) [/itex] and [itex] E_v(k) [/itex].

    I get that the valence band DoS vanishes where the conduction band DoS is zero. But I don't get how this means the absorption magnitude is not related to the product of the two. I'm not using [itex] Z_v(E)Z_c(E) [/itex] where [itex] E [/itex] is a fixed energy. I'm using [itex] Z_v(E_2)Z_c(E_1) [/itex] where [itex] E_1 = E_c(k) [/itex], [itex] E_2 = E_v(k) [/itex].

    I imagine the absorption magnitude is a product of the number of electrons states that can take that photon energy (proportional to [itex] Z_v(E_2) [/itex]) and the number of electrons states they can be promoted to (proportional to [itex] Z_c(E_1) [/itex]). So the absorption magnitude would proportional to the product of [itex]Z_c(E_1)Z_v(E_2)[/itex]...?

    Ideally [itex] Z_c(E_c(k)) \propto (E_c(k) - E_g)^{1/2} \propto k [/itex]. [itex] Z_v(E_v(k)) \propto (-E_v(k))^{1/2} \propto k [/itex]

    Then [itex] Absorption~magnitude \propto k^2 \propto E_{\gamma} - E_G[/itex]. Where [itex] k^2 \propto E_{\gamma} - E_G[/itex] is seen by re-arranging [itex] E_c(k) - E_v(k) = E_{\gamma} [/itex]. So it varies linearly with [itex] E_{\gamma} [/itex]?
    Last edited: May 1, 2014
  5. May 4, 2014 #4


    User Avatar

    The problem with taking the product of the individual densities of states is that you are not enforcing that the transitions take place at a given value of the vector [itex]k[/itex]. Remember that the density of states is only a function of energy, not [itex]k[/itex]; it is only for the free electron (parabolic) dispersion can you invert to get the magnitude of [itex]k[/itex] given the energy (but you can not invert to get the vector [itex]k[/itex]). So in some sense you are overcounting by taking the product as you suggest, because you are counting transitions which are not vertical in the vector space of [itex]k[/itex]. What you want is the joint density of states which enforces that the transitions happen vertically in [itex]k[/itex] space,
    [tex] JDOS(E) = \int d^3k\ \delta(E_c(k)-E_v(k)-E) [/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook