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?
