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In Modern Semiconductor Devices (available online from the author), it's stated that "any energy above $E_C$ is the electron kinetic energy", accompanied by the figure below. Additionally, in this lecture at 1:15:01, it's stated that even if there were some number of electrons moved up to the conduction band and they were right at $E_C$, and a voltage were applied across the material, they would still be unable to conduct current due to 0 kinetic energy and thus 0 velocity and 0 current. This is a bit confusing to me.

band diagram

A seemingly plausible line of thought to me is that if two conduction band electrons are at the same location $x_1$ in the band diagram, and one electron is at $E_1>E_C$ and the other $E_2>E_1>E_C$, then since they have the same electrostatic potential $V(x_1)$, the difference in their energy $E_2-E_1$ must be from kinetic energy.

It's not clear to me from high-level explanations of the band diagram's origin why any given energy level $E$ should necessarily correspond to some amount of KE and some amount of PE.

Is saying that electrons at $E_C$ have 0 KE and 0 velocity an arbitrary choice? If a voltage were applied to the material to slant the band diagram, why couldn't electrons that move up to $E=E_C$ move in response to the potential difference and carry current?

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Bloch electrons and their velocity

By Bloch's theorem, the single-electron energy eigenstates in a crystal (with periodic potential) are Bloch states of the form $$\psi_{\mathbf k}(\mathbf r)=u(\mathbf r)e^{i\mathbf k\cdot\mathbf r}$$ where $u(\mathbf r)$ is a periodic function with the periodicity of the crystal and $\mathbf k$ is the wave vector. The Bloch states $\psi_{\mathbf k}(\mathbf r)$ are the single-particle states whose energy eigenvalues make up the nearly-continuous conduction band. If you have studied any quantum mechanics, ignoring the $u(\mathbf r)$ factor for a moment, you will recognize $\psi_{\mathbf k}(\mathbf r)$ as a momentum (and hence velocity) eigenfunction with momentum eigenvalue $\hbar \mathbf k$. However, Bloch states are not momentum (or velocity) eigenfunctions and $\hbar \mathbf k$ is not their momentum (this is called crystal momentum). Nonetheless, we can still compute the expected value of velocity in a Bloch state, and it turns out to be $$\mathbf v(\mathbf k) =\frac{1}{\hbar}\nabla_\mathbf k E(\mathbf k)$$ where $E(\mathbf k)$ is the energy eigenvalue. You can verify that this also gives the correct velocity for a free particle. It might also look familiar from group velocity of waves, which is calculated from the dispersion relation as $\mathbf v(\mathbf k) =\nabla_\mathbf k \omega(\mathbf k)$.

To answer your question: since $E_C $ is the minimum of $E(\mathbf k)$ in the conduction band, and since we must have $\nabla_\mathbf k E(\mathbf k) = 0$ at the minimum, the Bloch electron with $E(\mathbf k)=E_C$ has zero mean velocity and thus cannot contribute to current. It is therefore reasonable to say that such an electron has zero kinetic energy (i.e. energy due to motion). However, it is potentially misleading to say that an electron with energy $E_C$ cannot contribute to current because such an electron can still accelerate under the influence of an external E-field, moving up in the conduction band and acquiring a non-zero velocity.

Parabolic bands

In many (most?) semiconductors, it turns out that the conduction band is parabolic. Mathematically, this means that for energies close to $E_C$, $$E(\mathbf k) = E_C +\frac{\hbar^2}{2}(\mathbf k - \mathbf k_0)^T\mathbf M^{-1}(\mathbf k - \mathbf k_0).$$ Here, $\mathbf M^{-1}$ is called the inverse effective mass tensor (basically a symmetric $3\times 3$ matrix), and $\mathbf k$ and the constant $\mathbf k_0$ are now thought of as column vectors. Purely out of convenience I will assume $\mathbf k_0 = 0$, but you can just substitute $\hat{\mathbf{k}} = \mathbf k - \mathbf k_0$ for $\mathbf k$ in what follows to proceed without this assumption.

With an appropriate choice of coordinates, we can write the energy as $$E(\mathbf k) = E_C +\frac{\hbar^2}{2}\left(\frac{k_x^2}{m^*_x}+\frac{k_y^2}{m^*_y}+\frac{k_z^2}{m^*_z}\right)$$ which explains the term "parabolic": the energy varies parabolically as you move in a straight line in $\mathbf k$-space. In an isotropic band where $m^*_x = m^*_y = m^*_z =m^*$, we have simply $$E(\mathbf k) = E_C +\frac{\hbar^2k^2}{2m^*}$$ where the second term might be familiar as the (kinetic) energy eigenvalue of a free particle with mass $m^*$.

Velocity and kinetic energy in a parabolic band

The mean (group) velocity in a parabolic band is $\mathbf v(\mathbf k) =\frac{1}{\hbar}\nabla_\mathbf k E(\mathbf k) = \hbar \mathbf M^{-1}_{ij}\mathbf k$, and the electron energy can be written as $$E(\mathbf k) = E_C + \frac{1}{2}\mathbf v^T \mathbf M \mathbf v.$$ The second term is now easy to think of as kinetic energy (the energy due to motion). It is entirely analogous to the formula $\frac12mv^2$, and in fact reduces to it in an isotropic band.

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  • $\begingroup$ Some region around a maxima or minima will be 'parabolic' in the Taylor series expansion, since if there was a linear slope that point would not be a maxima or minima. $\endgroup$
    – Jon Custer
    Commented May 24, 2023 at 20:36
  • $\begingroup$ @JonCuster There are bands (most frequently valence bands I believe) in which the energy isn't a "smooth" function of $\mathbf k$ near its minimum or maximum, so the parabolic band approximation doesn't work very well even for small $\mathbf k$, at least not without some corrections. $\endgroup$
    – Puk
    Commented May 24, 2023 at 20:39
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    $\begingroup$ Thanks very much for the detailed answer - I'll need to spend some time to truly understand it in its entirety, but it's reassuring to see where the justification for these claims comes from. $\endgroup$
    – Halleff
    Commented May 24, 2023 at 21:28

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