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Group Velocity Formalism vs. Current Operator Formalism in band theory

Physics Asked on May 19, 2021

There are at least two ways to argue about the velocity (or current) in band theory.

The first one is the group-velocity formalism
$$mathbf v_g = frac{1}{hbar} nabla_{mathbf k} epsilon_{mathbf k}$$
and the second one is the current operator formalism
$$mathbf J = frac{hbar}{2mi} psi^dagger nabla psi + h.c.$$
Here $psi$ is the field operator. In many condensed matter textbooks, transport properties in band theory is derived by the first formalism, rather than the more microscopic second formalism. I wonder whether all of the well-known properties can be derived by the second formalism. Here, the well-known properties could be a conductance in integer quantum hall effect, Landauer-Buttiker formula, etc.

For this purpose, it would be helpful to analyze how much these two formalisms are similar and different in general. Below I summarize the similarities and differences of two formalisms that I found.

  • The group velocity formalism is only valid for narrowly-peaked wavepacket in $mathbf k$-space, but the current operator formalism is valid in general situations.
  • In the current operator formalism, we obtain $mathbf J(mathbf r)$ as a function of $mathbf r$, but in the group velocity formalism we only obtain the single quantity $mathbf v_g$. I am also confused how to relate these two quantities.
  • Considering the derivation of the group velocity, the explicit time-dependence $psi(mathbf r, t)$ is important. On the other hand, in the second formalism, we don’t need $psi(mathbf r, t)$ as a function of $t$. Rather, we only need the wavefunction at an instant time $t_0$ and we can argue the probability current at
    $t_0$. Considering the fact that conductance is associated to the time-dependent behavior, more or less the first formalism could be more natrual, but I am not sure about this.

Any ideas would be appreciated a lot.

One Answer

Can the "well-known properties" be developed using free particles? (I think that the answer should be "yes".) Because the two equations are the same for free particles.

For free particles:

$$epsilonleft(vec{k}right) = frac{hbar^2k^2}{2m}$$

and

$$psileft(vec{x}, tright) = e^{ileft(vec{k}cdotvec{x}-omega tright)}$$

(If it works for single-particle wavefunctions, it should work for field operators too.)

Things get a little more tricky for non-free particles. IIRC, you need to make some approximation to get the equivalence for Bloch wave functions. However, I may be misremembering.

Answered by lnmaurer on May 19, 2021

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