Typical operators in tight binding Let the tight-binding Hamiltonian be $\sum\limits_{ij} {{t_{ij}}\left| i \right\rangle \left\langle j \right|}$. Where ${\left| i \right\rangle }$ is the atomic orbit at lattice site $i$.
My question is, is it correct to express the usual operators in this representation like this:
Position operator:
$\mathop {\bf{r}}\limits^ \wedge   = \sum\limits_i {{{\bf{r}}_i}\left| i \right\rangle \left\langle i \right|} $, where ${{\bf{r}}_i} = \left\langle i \right|\mathop {\bf{r}}\limits^ \wedge  \left| i \right\rangle$ is just the lattice position of  ${\left| i \right\rangle }$ if the atomic orbit is symmetric.
Velocity operator:
$\mathop {\bf{v}}\limits^ \wedge   = \sum\limits_{ij} {{{\bf{v}}_{ij}}\left| i \right\rangle \left\langle j \right|} $
${{\bf{v}}_{ij}} = \left\langle i \right|\mathop {\bf{v}}\limits^ \wedge  \left| j \right\rangle  = \frac{{ - i}}{\hbar }\left\langle i \right|\left[ {\mathop {\bf{r}}\limits^ \wedge  ,H} \right]\left| j \right\rangle  = \frac{{ - i}}{\hbar }\left( {{{\bf{r}}_i} - {{\bf{r}}_j}} \right){H_{ij}}$
If the wavefunction of the Hamiltonian is 
$\begin{array}{l}
 \left| \psi  \right\rangle  = \sum\limits_i {{\psi _i}\left| i \right\rangle }  \\ 
 {\psi _i} = \left\langle i \right|\left. \psi  \right\rangle  \\ 
 \end{array}$
Is it right to write:
$\begin{array}{l}
 \left\langle \psi  \right|\mathop {\bf{r}}\limits^ \wedge   \times \mathop {\bf{v}}\limits^ \wedge  \left| \psi  \right\rangle  = \sum\limits_{ij} {\left\langle i \right|\psi _i^*\left( {\mathop {\bf{r}}\limits^ \wedge   \times \mathop {\bf{v}}\limits^ \wedge  } \right){\psi _j}\left| j \right\rangle }  \\ 
  = \sum\limits_{ij} {\psi _i^*{\psi _j}\left\langle i \right|\left( {\mathop {\bf{r}}\limits^ \wedge   \times \mathop {\bf{v}}\limits^ \wedge  } \right)\left| j \right\rangle }  = \sum\limits_{ij} {\psi _i^*{\psi _j}{{\bf{r}}_i} \times {{\bf{v}}_{ij}}}  \\ 
 \end{array}$, 
where ${\psi _i^*{\psi _j}}$ is not operated by ${\left( {\mathop {\bf{r}}\limits^ \wedge   \times \mathop {\bf{v}}\limits^ \wedge  } \right)}$?
I ask this because if this is done in continuum real space, the wavefunction ${\psi _i^*{\psi _j}}$ should also be operated by ${\left( {\mathop {\bf{r}}\limits^ \wedge   \times \mathop {\bf{v}}\limits^ \wedge  } \right)}$ since it's position dependent.
Comment 1:
To calculate $\widehat{\bf{r}} \times \widehat{\bf{v}}$ at lattice site $i$, according to my formalism, in the tight-binding formalism it should be $\left\langle i \right|\widehat{\bf{r}} \times \widehat{\bf{v}}\left| i \right\rangle  = \sum\limits_k {\left\langle i \right|\widehat{\bf{r}}\left| k \right\rangle  \times \left\langle k \right|\widehat{\bf{v}}\left| i \right\rangle  = } \frac{{ - i}}{\hbar }\sum\limits_{k,l} {{{\bf{r}}_i} \times \left( {{{\bf{r}}_i}{h_{ii}} - {h_{ii}}{{\bf{r}}_{i}}} \right)}  = 0$, while it should be proportional to angular momentum at site $i$, for $p_z$ orbit, it should be nonzero, why I got this Contradictory result?
 A: *

*In principle, your formulas are correct in the framework of the tight-binding approximation.

*If you do a basis expansion, the operators never act on the coefficients ($\psi_i$, in your notation).

*All operators you use, also $\mathbf r$, must have two indices because the position operator is not diagonal in the tight-binding basis.

*The problem is that you usually don't have a real space representation of the tight-binding orbitals and thus can't calculate the matrix elements of $\mathbf r$ (by the way, this is the dipole matrix element) and $\mathbf v$. This is different for Wannier orbitals, because they have a real space representation. I'm actually working on that.


Edit in response to the comment:
Wannier orbitals are atomic orbitals. The point is that not every tight binding scheme is generated in a way that includes a real space representation.


*

*Unitary transformation of a Hartree-Fock calculation: One obtains tight-binding parameters, and the real-space representation is the basis of the Hartree-Fock calculation (i.e. a Gaussian orbital basis, like 6-31G*) 

*DFT bandstructure fit: only energy results from the DFT calculation are used, and therefore, the tight-binding parameters only describe the energetic structure.

*Wannier orbitals: Unitary transformation of the DFT Kohn-Sham orbitals. Tight-binding parameters as well as a real-space representation of the orbitals are obtained.


Summarizing: Tight-binding always uses atomic orbitals, but the existence of a real space representation of the orbitals depends on the method.
A: Yes, your deduction is correct. In fact, your $v_{ij}$ is the matrix representation of the differential operator $i\partial$ on the discrete basis. The wave function $\psi_i^*\psi_j$ has beed operated in your formula. Acting the differential operator to the wave function in the continuum limit just becomes multiplying the matrix to the wave vector in the discrete case. In your formula, the summation over $j$ is actually multiplying the matrix $v_{ij}$ to the vector $\psi_j$, so in this way $\psi_j$ has been operated by $\hat{v}$. 
A: I think your deduction about angular momentum operator in tight binding model is partially right. You only calculate the moving part of the wave function but missing the intrinsic part (I'm not quiet sure about the statement).
We can choose the situation in atom for analogy. When we want to identify the total angular momentum in an atom with given electron configuration, the orbital and spin angular momentum of all the electrons are combined separately. The orbital angular momentum is
$${\bf L} = \sum_i {\bf L}_i$$
And the spin angular momentum is
$${\bf S} = \sum_i {\bf S}_i$$
The total angular momentum is
$${\bf J}_{\text{total}} = {\bf L} + {\bf S}$$
So returning to the tight binding model, your deduction posted in the question is similar to $\bf L$ in above example. As you commented once we want to calculate the angular momentum in a given site $\left|i\right\rangle$, there might be some troubles if we missed $\bf S$ in our tight binding model. Actually I don't mean there is spin contribution, but we need to add the intrinsic angular momentum contribution from the orbitals we choose.
For example, we can project the angular momentum operator $\bf L$ to the p-character sub-Hilbert space. When the space is represented by the three basis orbitals $\left|p_x\right\rangle, \left|p_y\right\rangle, \left|p_z\right\rangle$, the operator is represented by the following elements
$$\left({\bf L}^{(p)}_{\alpha}\right)_{\beta\gamma} = -i\hbar\epsilon_{\alpha\beta\gamma}$$
Where the superscript $(p)$ labels the sub-Hilbert space, $\epsilon$ is the Levi-Civita symbol and $\alpha,\beta,\gamma$ are directions. In every specific tight binding model, we can constructed the intrinsic angular momentum separately.


*

*8.9: The Allowed Values of J - the Total Angular Momentum Quantum Number - Chemistry LibreTexts

*Go, D.; Jo, D.; Kim, C.; Lee, H.-W. Intrinsic Spin and Orbital Hall Effects from Orbital Texture. Phys. Rev. Lett. 2018, 121 (8), 086602. https://doi.org/10.1103/PhysRevLett.121.086602.


A: There is a comment above which states that the calculation of the expectation value of the angular momentum operator defined in that way in |i> should be nonzero.
The above expectation can be examined by considering, for in the continuous real space, the operator, r cross p, where p is the momentum operator and r is the position operator. If one calculates the expectation value of that operator on a wavefunction which is a delta function at position r_0, then one finds it is also zero. Therefore, the result of defining velocity operator by such tight-binding language, in terms of this consistency of angular momentum projection, looks reasonable.  
