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.

  • $\begingroup$ 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 $E\propto k^2$ $\endgroup$
    – Prongs
    Nov 21, 2020 at 17:31
  • $\begingroup$ @Prongs I definitely agree with you. $\endgroup$
    – Laplacian
    Nov 22, 2020 at 8:49
  • $\begingroup$ 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. $\endgroup$ Apr 21, 2021 at 16:00

1 Answer 1


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:

$$\epsilon\left(\vec{k}\right) = \frac{\hbar^2k^2}{2m}$$


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

(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.


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