I'm currently doing an introductory quantum mechanics and solid state physics course.

I understand why the density of states in the valence band of an intrinsic semiconductor increases with increasing energy; at higher energy levels there are more degeneracies and thus more states per energy level that an electron can inhabit. What I don't understand is why the density of states increases in the valence band as energy decreases.

I've attached an image of a DOS plot below. I thought it was possible that considering the zero of energy to be at the Fermi energy may have altered the graph but I no longer think so; why would choosing an arbitrary point to designate as zero physically alter the available states for an electron? It's also possible that this is two plots in one; the top half of the plot could be the DOS of electrons in the conduction band, and the bottom half could be the DOS of holes in the valence band. This possibility doesn't make sense to me either; holes are just the absence of electrons, if there are few states for electrons at a certain energy then how could there be a whole lot of states for holes? That would seem to imply that there are more missing electrons than the number of electrons that could be there in the first place.

enter image description here

  • $\begingroup$ Of course there are lots of states for electrons in the valence band - it is full of them after all. And those can be occupied by holes instead of electrons. Same states. $\endgroup$
    – Jon Custer
    Commented May 25, 2021 at 12:27
  • $\begingroup$ But why does the number of states trend up as energy trends down in the valence band? Shouldn’t the number of states increase with energy? $\endgroup$
    – Cdizzle
    Commented May 25, 2021 at 22:22

3 Answers 3


One concept to correct before the math derivation: They are the density of states for "electrons" not "hole". The concept of hole will come in later, as one starting to filling the states, determining by Fermi-Dirac function:

\begin{align} f_n (E) =& \frac{1}{e^{\frac{E-\mu}{KT}} + 1},\,\, \text{ for } E > E_c\\ f_h(E) =& 1 - f_n(E) = \frac{1}{e^{-\frac{E-\mu}{KT}} + 1},\,\, \text{ for } E < E_v \end{align}

Come back to the problem of density of states, We may adopt simple parabolic band for conduction band and valence band:

\begin{align} E(\mathbf{k}) =& E_c + \frac{\hbar^2 k^2}{2 m_c}, \,\, \text{ for conduction band} E > E_c \tag{1}\\ & \\ E(\mathbf{k}) =& E_v - \frac{\hbar^2 k^2}{2 m_v},\,\, \text{ for valence band } E < E_v \tag{2} \end{align}

The density of states is found by consider to total number of states between $n(E)$ per volume between energy $E$ abd $E+dE$: $$ n(E) = D(E)dE = D(k) dk = 2\frac{1}{8\pi^3}4\pi k^2 dk. $$ Which renders the density of state in energy: $$ D(E) = 2\frac{1}{8\pi^3}4\pi k^2 \vert\frac{dk}{dE}\vert = \frac{1}{\pi^2} \frac{2m^*E}{\hbar^2} \vert\frac{dk}{dE}\vert. \tag{3} $$

Apply Eq.(1) and Eq.(2) to formular Eq.(3):

For the valence band:

$$ D_v(E) = \frac{1}{2\pi^2} \left(\frac{2m_v}{\hbar^2}\right)^{3/2} \sqrt{E_v-E}.\,\, \text{ for } E < E_v. $$

  • $\begingroup$ I think you are assuming the answer to the question in your answer. Sure, if you assume your E(k) relations then you get the DOS as described, but I think the real answer is in why the E(k) relation is that way. Especially because your E(k) relations as written are approximations for all real materials. $\endgroup$
    – Matt
    Commented May 26, 2021 at 16:48
  • $\begingroup$ I don't understand your comment about the "real answer". For arbitrary band structure, the D(E) is an integral with $\vec{\nabla}_kE(\vec{k})\cdot d\vec{A}$ in the denominator. $\endgroup$
    – ytlu
    Commented May 29, 2021 at 3:53
  • $\begingroup$ I don't understand why we start counting the density of states from the top of the valence band down, instead of from the bottom of the valence band up. Why do we use $E_v - E$ instead of $E$ in equation (2)? $\endgroup$
    – Cdizzle
    Commented May 29, 2021 at 4:24
  • $\begingroup$ (1) There has no energy states between $E_v < E < E_c$, This region is called energy gap. Therefore, you can only count the number if states for $E>E_c$ or $E<E_v$. (2) counting state from $E_v$ downward or from the band bottom upward will give the some density of state, by my Eq.(3), only the $dE/dK$ is involved. $\endgroup$
    – ytlu
    Commented May 29, 2021 at 4:32
  • $\begingroup$ The term $E-E_v$ is coming from the band staructure Eq.(2). $\endgroup$
    – ytlu
    Commented May 29, 2021 at 4:34

The density of occupied states in the valence band increases with decreasing energy and the density of occupied states in the conduction band decreases with increasing energy. The opposite is true for both bands for the density of available states. It seems like you are confusing them.

  • No carrier electrons exist in the valence band, all valence band electrons make up the covalent bonds.
  • All conduction band electrons are those which had enough energy to escape the crystal covalent bonds and are present in the crystal lattice.

When a covalent bond is formed energy is released, so the electrons which make it up have low energy. Since higher energy electrons cannot make up covalent bonds this would be equivalent to saying that fewer electron states exist as the energy of a covalent bond increases or equivalently high energy states in a covalent bond do not exist.

Since the diagram gives only electron DOS, I think this analogy can suffice. I only had an introductory course in Semiconductor Physics, I am unaware of the quantum mechanical process which results in such a DOS.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Feb 24 at 2:14

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