# Group Velocity Formalism vs. Current Operator Formalism in band theory

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.

• I have not come up with a clear answer but maybe the following thoughts could be useful: (1). The first formalism is a single-body formalism while the second is a many-body one since the field operators are encountered. Thus if you are dealing with a strongly interacting system where bands or dispersion relations for single particles do not even make sence, the first formalism won’t be useful. (2). A simple Fourier transform gives $\nabla \rightarrow i\vec{k}$ so the second formalism seems to be about momentum, which could be neither phase velocity or group velocity unless $E\propto k^2$ Nov 21, 2020 at 17:31
• @Prongs I definitely agree with you. Nov 22, 2020 at 8:49
• Your current $\mathbf{J}$ is only valid for the dispersion $E = k^2/2m$, i.e. in continuum. I would regard it as less general than your first equation which is generally true. In general the current operator is defined by $\mathbf{J}(\mathbf{x},t) = -\delta H / \delta \mathbf{A}(\mathbf{x},t)$ where $\mathbf{A}$ is included by Peierls substitution/minimal coupling. See e.g. physics.stackexchange.com/questions/70613/… for current operators in lattice models. Apr 21, 2021 at 16:00

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.

$$\epsilon\left(\vec{k}\right) = \frac{\hbar^2k^2}{2m}$$
$$\psi\left(\vec{x}, t\right) = e^{i\left(\vec{k}\cdot\vec{x}-\omega t\right)}$$