Why can't valence electrons conduct electricity?

Question

The simplified $E-k$ diagram obtained from Kronnig-Penney model looks like that:

The bottom curve corresponds to the so-called valence electrons, and the upper curve - to conduction electrons.

But these are simply the allowed states. Kronnig-Penney model doesn't (at least, immediately) tell us that only the bottom states are for the electrons that can move freely and cause electricity to flow.

Is the fact that the bottom electrons are bound to nuclei and the upper electrons are free to move just an assumption? Or does this fact have theoretical derivation (if so, I would like to see one)?

Answer 1

To be clear, valence band electrons (or lack thereof, i.e. holes) can and do carry current, you might find this answer useful. A band that is completely filled doesn't carry current. Here is a proof, following Ashcroft & Mermin. We start by proving an integral identity. Suppose that $f(\mathbf r)$ is a periodic function with the periodicity of a Bravais lattice, meaning in 3D, $$f(\mathbf r + n_1 \mathbf a_1 + n_2 \mathbf a_2 + n_3 \mathbf a_3) = f(\mathbf r)$$ where $\mathbf a_j$ are the primitive vectors spanning the lattice, and $n_j$ are arbitrary integers. Define $$I(\mathbf r')=\int\limits_Cf(\mathbf r+\mathbf r')\ d\mathbf r,$$ a volume integral over primitive cell $C$. This integral is just $\int\limits_{C'}f(\mathbf r)\ d\mathbf r$ where $C'$ is $C$ shifted by $\mathbf r'$, which is another primitive cell. Due to the periodicity of $f$, its integral over any primitive cell is the same, meaning $I(\mathbf r')$ is independent of $\mathbf r'$: $$\mathbf \nabla'I(\mathbf r')=\int\limits_C\mathbf \nabla'f(\mathbf r+\mathbf r')\ d\mathbf r=\int\limits_C\mathbf \nabla f(\mathbf r+\mathbf r')\ d\mathbf r = 0$$ Plugging in $\mathbf r'=0$, $$\int\limits_C\mathbf \nabla f(\mathbf r)\ d\mathbf r = 0.$$ This is the result we were looking for: a Green's theorem for periodic functions. It says that the integral of the gradient of a periodic function over a primitive cell is zero.

Now, the current density contribution of an energy band is

$$\mathbf J =\int\limits_B\frac{d\mathbf k}{4\pi^3}(-q)g(\mathbf k)\mathbf v(\mathbf k).$$ Here, $B$ is the first Brillouin zone, $g(\mathbf k)$ is the probability of occupancy of the state with wave vector $\mathbf k$, and $\mathbf v(\mathbf k)=\frac 1 \hbar\mathbf \nabla_{\mathbf k} E(\mathbf k)$ is the group velocity where $E (\mathbf k)$ is the energy of the state with wave vector $\mathbf k$. For a completely filled band, $g(\mathbf k)=1$, so $$\mathbf J =\frac{-q}{4\pi^3\hbar}\int\limits_Bd\mathbf k\nabla_{\mathbf k} E(\mathbf k).$$ Since $E(\mathbf k)$ is a periodic function with the periodicity of the reciprocal lattice, and the first Brillouin zone is a Wigner-Seitz primitive cell, the integral identity we proved above applies, and the current density is zero.

Answer 2

Usually bands are symmetric, i.e., for any state with quasimomentum $\mathbf{k}$, we have a state with quasimomentum $-\mathbf{k}$. Electrons occupying these states have velocities $$ v(\mathbf{k})=\frac{1}{\hbar}\nabla_\mathbf{k}E(\mathbf{k}) $$ and $$-v(\mathbf{k}).$$ Thus, the contribution of every electron to current is cancelled out by the electron with the opposite momentum.

When applying electric field, we shift the chemical potentials for electrons moving in the directions along and against the field, that is the states in conduction band with positive momenta have higher/lower occupancy than the states with negative momenta. This is not the case with in the valence band, where all the states are already occupied and nothing changes... unless the field is so strong that it transfers some electrons to the conduction band (Zener tunneling.)

See also Filled band cannot generate current