Symmetry of Bloch Hamiltonian If a crystal system preserve a symmetry C, why its Bloch Hamiltonian satisfy
$H(C\vec k)=CH(\vec k)C^{-1} $
 A: That a Hamiltonian preserves a symmetry means
$$ [H, C] = 0  \Rightarrow CHC^\dagger = CHC^{-1} = H$$
For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal).
The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is:
$$ H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k) $$
Where $$\psi(\vec k) = \begin{pmatrix} \vdots \\ c_i(\vec k) \\ \vdots \end{pmatrix}.$$
And $c_i(\vec k)$ are the annihilators for electrons at $\vec k$ in state $i$ (which is a combined index of spin and band). As $C$ is applied to $\vec k$, it must be a spatial symmetry (that is a lattice symmetry).
The transformation of $\psi(\vec k)$ under $C$ is easily obtained by argument.
$C$ is a spatial transformation, thus $C\psi(\vec k)C^{-1}$ does the following:


*

*the coordinates are transformed.

*a particle with momentum $\vec k$ is destroyed in the new coordinates.

*the space rotation is undone leaving a particle with momentum $C \vec k$ destroyed in the original coordinates.


I effect this means that $C\psi(\vec k)C^{-1} = \psi(C\vec k)$. 
Using the above results:
$$ H = CHC^{-1} = \sum_{\vec k} C\psi^\dagger(\vec k)C^{-1}C H(\vec k) C^{-1}C \psi(\vec k)C^{-1}. $$
As $C\psi(\vec k)C^{-1} = \psi(C\vec k)$, $C H(\vec k) C^{-1} = H(C\vec k)$ must hold, in order to assure the equality to $H$ (which is then shown by relabelling the summation variable $\vec k' = C\vec k$).
