Why can we treat the Bloch hamiltonian as an effective hamiltonian?

Often when looking at topological insulators the hamiltonian is broken down into the Bloch hamiltonian and then analysed ignoring the creation/annhilation operators of the Bloch waves. Why is it okay to just ignore these?

For example for an n-band Bloch Hamiltonian in 2D we have, for each Brillouin zone the effective hamiltonian:

$$H(\mathbf k)=\left[a^\dagger_{\mathbf k, 1}\qquad a^\dagger_{\mathbf k, 2} ....a^\dagger_{\mathbf k, n}\right]\mathcal H(\mathbf k) \begin{bmatrix} a^\dagger_{\mathbf k, 1}\\ a^\dagger_{\mathbf k, 2} \\ a^\dagger_{\mathbf k, n}\end{bmatrix}$$

It is then given that for some $\mathbf k$ we have the eigenstates

$$\mathcal H (\mathbf k)\lvert m(\mathbf k)\rangle=E_m(\mathbf k )\lvert m(\mathbf k)\rangle$$

But what do these eigenstates mean? I don't understand why it means anything since we have redefined what our hamiltonian is.

The first thing to note is that the Hamiltonian you are presenting describes non-interacting electrons and it conserves the respective occupation numbers. So there is no real need for the whole machinery of second quantization.

In principle, the single-particle states and energies provide you with all the information you need and this information will be sufficient to deduce all many body properties. So why are some models "effective"?

The exact solution has an infinite number of eigenstates in the thermodynamic limit. But usually you are interested in things which happen close to the energy of the Fermi level. Since the single-electron states don't talk to each other via the Hamiltonian, you simply need to take a snapshot of your band-structure across the energy window you are interested in.

Take e.g. the Rashba Hamiltonian (wiki) $$H({\bf k})= \frac{\hbar^2k^2}{2m^*} + \alpha(k_y\sigma_x - k_x\sigma_y)$$ which is written in the basis of Bloch states and spin state, $|{\bf k},\sigma\rangle$, $\sigma=\uparrow\downarrow$ and yields to bands. As one can easily see, the spin quantum number is not a good quantum number and $|{\bf k},\sigma\rangle$ is not an eigenstate. Thus, you have to be clear about the representation of the Hamiltonian. In most cases it will not only be $\langle r|{\bf k}\rangle =e^{-i {\bf k}\cdot{\bf r}}$.

The important point here is that your Hamiltonian is the one of a Linear system, ie there is only quadratic term in it. This allow you, when your system is periodic in space, to rewrite you Hamiltonian as a sum of quadratic form: $$H = \sum_k \vec{a}_k^\dagger h(k) \vec{a}_k$$

When you write this equation, what you are saying is that every wave vectors is not coupled to others k. You can treat each wavevectors $k$ separetely. In the subspace of a fixed $k$, you then only have to diagonlise the matrix $h(k)$. You gets the eigenvectors of $h(k)$: $\vec{v}(k)$. This gives you the eigenmode of your initial Hamiltonian: $\vec{v}_\vec{k}\cdot \vec{a}_\vec{k}$

• Linear, i.e. quadratic? Jun 3, 2018 at 19:17
• Sorry for the confusion... It's because quadratic Hamiltonian leads to linear equation of motion. When i say linear i refer to the Equation of motion. when i say quadratic i refer to the Hamiltonian Jun 3, 2018 at 20:45
• Agreed! Maybe you could edit this into your answer Jun 3, 2018 at 21:01