Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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 $\hat{v}$.

share|cite|improve this answer
I write a comment at the end of the original question, could you help me to check why I got this Contradictory result? – Laurent Feb 19 '13 at 21:59
  1. In principle, your formulas are correct in the framework of the tight-binding approximation.
  2. If you do a basis expansion, the operators never act on the coefficients ($\psi_i$, in your notation).
  3. All operators you use, also $\mathbf r$, must have two indices because the position operator is not diagonal in the tight-binding basis.
  4. 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.

share|cite|improve this answer
sorry, still do not understand why Wannier representation have continuum real space representation while atomic orbit representation doesn't. – Laurent Feb 7 '13 at 17:57

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.

share|cite|improve this answer
Note that we do have Mathjax here, which allows for Latex-like syntax to get prettified equations. – Kyle Kanos Mar 8 at 12:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.