Ballistic Transport and Bloch Oscillations Contradiction

Question

My question is as follows, below this I have included derivations of both effects:

The derivation of Bloch Oscillations implies that in a perfect crystal we will not have net current flow when an electric field is applied.

However, the ballistic transport derivation shows that, regardless of crystal quality, we will have a clear, quantised current. More accurately, the conductance is quantised.

I don't understand this apparent contradiction, what am I missing here?

Furthermore, assuming a perfect crystal and no scattering events, under what conditions would I observe Bloch oscillations and under what conditions would I observe ballistic transport, and why? By Ballistic transport I mean the quantisation of conductance, whereby conductance is an integer multiple of $\frac{2e^2}{h}$.

Bloch Oscillations

The Hamiltonian of a 1D system describing an electron in a periodic potential as well as a constant electric field is as follows:

$H=\frac{p^2}{2m} + V(x)+eEx$,

where the momentum and position operators are as standard and $E$ is a constant electric field. $V(x+a)=V(x)$. Let us evolve a state which is initially a bloch state. This state has projection in position space, $\phi(k_0,x)$, where $k_0$ is its wavenumber and is within the first Brillioun Zone. We don't bother noting the band index. Let $\psi(x,t)$ be the wavefunction after time t. Note that $\psi(x,0)=\phi(k_0,x)$. We have that,

$\psi(x,t)=\exp\bigg [\frac{-it}{\hbar}(\frac{p^2}{2m} + V(x)+eEx)\bigg ]\phi(k_0,x)$

We translate the $x$ variable by $a$, the lattice period. We obtain,

$\psi(x+a,t)=\exp\bigg [\frac{-it}{\hbar}(\frac{p^2}{2m} + V(x)+eE(x+a)\bigg ]\exp(ik_0a)\phi(k_0,x)$.

Where we have used the periodicity of $V(x)$ and the fact that, for a Bloch state, $\phi(k_0,x+a)=\exp(ik_0a)\phi(k_0,x)$.

Thus,

$\psi(x+a,t)=\exp\bigg [\frac{-it}{\hbar}(\frac{p^2}{2m} + V(x)+eEx\bigg ]\exp(ia[k_0 -\frac{Eet}{\hbar}])\phi(k_0,x)$.

This means that

$\psi(x+a,t)=\exp(ia[k_0 -\frac{Eet}{\hbar}])\psi(x,t)$.

This in turn means that our wavefunction at time $t$ is a Bloch state with wavevector $k_0 -\frac{Eet}{\hbar}$. We cannot necessarily say that the band index remains the same, but what we can say is that the state at time $t$ is a linear combination of bloch states, from potentially many bands, all with wavevector $k_0 -\frac{Eet}{\hbar}$. If interband tunnelling is rare then the state will simply move along the same band repeatedly and we have Bloch Oscillations with no net flow of current. Scattering events, if frequent enough, will prevent these oscillations.

Ballistic Transport

Here we say that we have two electron resevoirs which are physically at opposite ends of the material, the left and the right. As a potential difference is applied across the material, the chemical potentials of the reservoirs vary. The left reservoir has a chemical potential of $\mu_L$ and the right has $\mu_R$. Thus the possible energies of the electrons which are injected from the left and subsequently travel to the right differ to those which travel to the left after being injected from the right. We also use the fact that

$v(k)=\langle\frac{p}{m}\rangle=\frac{1}{\hbar}\frac{dE(k)}{k}$.

Accounting for spin degeneracy and counting of states, $I=\frac{-2e}{2\pi}\int^{k_L}_{k_R} \frac{1}{\hbar}\frac{dE(k)}{k} dk=\frac{-2e}{h}(u_L-u_R)=\frac{2e^2 V}{h} $.

This means, regardless of the quality of the crystal, we see current flow. This is apparently at odds with Bloch oscillations.

Answer 1

The oscillation length in Block oscillations is usually much longer than the mean free path. Indeed an electron usually scatters off an imputy or phonon long before ${\bf k}$ gets any significant distance away from the Fermi surface. It is therefore very hard to observe Bloch oscillations even in very good quality crystals. Consequently ballistic propagation is a reaonable approximation for motion up to the mean free path.

In the case of long mean free path quantum wires, in which the Landauer formula gives conductivity $\sigma=N(2 e^2/h)(\mu_R-\mu_L)$, there is not even an electric field ($V$ is the difference of electrochemical potentials) so the ${\bf k}$'s stay the same throughout the process, so again no oscillations.

Answer 2

The connection between the electrostatic (Hartree in QM problems) potential and chemical potential (or node voltage in electronics) is one that always causes a great deal of confusion. The short answer is that, to get it right, one has to solve a nonlinear electrostatic screening problem, which has no helpful symmetries, and in cases like the one under discussion, is not even well-defined because all of the boundary conditions are not known. In a uniform sample of conductive material that is large enough to be considered a reservoir, there will usually be a connection between these quantities defined by something like a work function. In the spaces between such regions, which always include the "device" whose current response is desired, one is generally reduced to invoking the existence of "fringing electric fields," which make the electrostatic potential satisfy the boundary conditions at the contacts, but which are otherwise not well-defined.

The strength, and also a fundamental limitation, of the Landauer formula is that these effects are ignored, except as they might affect the transmission coefficient.