Let us have a 2D lattice model with three sites in one unit-cell (basically 2D Kagome lattice). In Fourier space, the Hamiltonian is written as

$$ H = \sum_{k_x,k_y} \begin{bmatrix} a^\dagger & b^\dagger & c^\dagger \end{bmatrix} \begin{bmatrix} h_{11}&h_{12}&h_{13}\\ h_{21}&h_{22}&h_{23}\\ h_{31}&h_{32}&h_{33} \end{bmatrix} \begin{bmatrix} a\\b\\c \end{bmatrix}\equiv\Psi^\dagger h({k_x,k_y}) \Psi $$ here $a,b,c$ are operators of particles on 3 sites of the unit-cell.


How to write the velocity operator for $a,b,c$ particles in x-direction and y-direction separately.

My attempt:

Using Ehrenfest theorem for particles $a$, we have:

$$\partial_t a^\dagger a = \frac{1}{i\hbar}[a^\dagger a,H]$$

(to not make it a shopping-question:) I think the Ehrenfest theorem is the right way to write velocity operators. But I don't see how exactly one can separate velocity of particles in x- and y-directions or $k_x,k_y$ directions using this theorem. How does quantum mechanics define velocity in different directions?

  • 1
    $\begingroup$ Have you tried considering the commutator of H and X/Y? $\endgroup$ Commented Jul 27, 2020 at 16:02
  • $\begingroup$ @JahanClaes yes, but how do I define X and Y operators here? (pardon me if this is a very stupid question) $\endgroup$ Commented Jul 27, 2020 at 16:11
  • $\begingroup$ Your question title needs to be a little more specific. $\endgroup$
    – Yejus
    Commented Jul 27, 2020 at 16:14

1 Answer 1


In the first quantization representation the velocity operator is obtained using $$\hat{\dot{x}} = \frac{1}{i\hbar}\left[\hat{x},\hat{H}\right]_-,$$ which defines the operator of the derivative or the derivative of the operator, depending on whether you work in Schrödinger or Heisenberg picture.

In second quantization one switches to the operators using the usual prescription - calculating the matrix element between the field operators $\psi^\dagger(x),\psi(x)$.

For trancated Hamiltonians, like it seems to be the case in the question, one is usually interested in the current operator, which is obtained using the continuity equation, as the derivative of the charge operator in the region of interest, e.g. $$\partial_t a^\dagger a = \frac{1}{i\hbar}\left[a^\dagger a,\hat{H}\right]_-.$$ A caveat is that one may have some challenges when switching the gauge, e.g., when trying to apply the Kubo formula (doable, but not straightforward).


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