# Birotons and \"Dark\" Quantum Hall Hierarchies
## 2 Materials and Methods
### Reciprocal Vectors and Partial Translations  
For an operator to constitute an observable, it should commute both with the Landau Hamiltonian and magnetic translations \(\hat{t}_{m}(\tau_{1,2})\). A remarkable observation due to [PERSON] [15] was that there exists center mass magnetic translation operators \(\hat{T}(a)=\prod_{i=1}^{N_{c}}\hat{t}_{m,i}(a)\), where \(\hat{t}_{m,i}(a)\) acts on the \(i\)-th electron, to nontrivial vectors \(a\) satisfying both of these properties and also commuting with each other, thus constituting a vector quantum number similar to momentum.  
Indeed, it follows from (4) that \(\hat{T}(a)\hat{T}(b)=\exp\left(\frac{iN_{c}(a_{x}b_{y}-b_{x}a_{y})}{l_{B}^{2}} \right)\hat{T}(b)\hat{T}(a)\). Consider \(N=\gcd(N_{e},N_{s})\), then for \(Na=\tau_{1},\ Nb=\tau_{2}\ [\hat{T}(a),\hat{T}(b)]=0\):  
\[\hat{T}\left(\frac{\tau_{1}}{N}\right)\hat{T}\left(\frac{\tau_{2}}{N}\right)= \exp\left(\frac{i2\pi N_{e}N_{s}}{N^{2}}\right)\hat{T}\left(\frac{\tau_{2}}{N }\right)\hat{T}\left(\frac{\tau_{1}}{N}\right)\]  
What is the spectrum of these operators? Because each \(\prod_{i=1}^{N_{c}}a_{j_{i}}^{\dagger}|0\rangle\) is an eigenstate for \(\hat{T}\left(\frac{\tau_{1}}{N}\right)^{N}\), the \(\hat{T}\left(\frac{\tau_{1}}{N}\right)\) eigenvalues are \(\exp\left(\frac{2\pi il}{N}\right)\). Two translation operators commute; hence, they have a common eigenstate base, the eigenvalues being \(\exp\left(\frac{2\pi il_{i}}{N}\right)\) for \(\hat{T}(L_{i})\). Consider \(q\) such that \((q,L_{1})=2\pi l_{1},\ (q,L_{2})=2\pi l_{2}\) (i.e., \(q\) belongs to a reciprocal lattice), then the spectrum of \(\hat{T}(L)\) is \(\exp(\frac{i(q,L)}{N})\). This \(q\) is a reciprocal vector quantum number.  
Direct calculation shows that the one-electron states deduced in the previous section satisfy \(\hat{T}\left(\frac{\tau_{1}}{N}\right)\psi_{r}=\exp\left(\frac{2\pi ir}{N} \right)\psi_{r}\), \(\hat{T}\left(\frac{\tau_{2}}{N}\right)\psi_{r}=\exp\left(i\frac{(2\tau+1)}{N} \cdot\pi\frac{\tau_{2}|\cos\theta}{\tau_{1}}\right)\psi_{r+\frac{N_{s}}{N}}\). Using this, we constructed a base out of reciprocal momentum eigenstates, and to diagonalize (1), it is only needed to handle blocks corresponding to each eigenvalue.  
We will now focus on the allowed values of \(q\) because these will be the eigenvalues for the blocks, and their absolute values correspond to the points in the dispersion curve. As we mentioned, they lie in the reciprocal lattice \(\{n_{1}\tau_{1}+n_{2}\tau_{2}\}^{-1}\) and are defined up to a translation by an element of \(\{\frac{n_{1}\tau_{1}}{N}+\frac{n_{2}\tau_{2}}{N}\}^{-1}\). Therefore, the number of allowed values (\(N^{2}\)) of the reciprocal number at some fixed filling factor can be increased by increasing \(\gcd(N_{e},N_{s})\), which makes the computational complexity of our problem dramatically increase. Keeping \(N_{s}\) fixed, we can still increase the number of different absolute values of allowed \(q\) by considering a non-square cell (\"lifting degeneracy\" of two points, which are reflections in a diagonal) and the maximum absolute value of \(q\) by choosing a non-rectangular cell (maximizing the diagonal of a parallelogram while keeping its area). The latter may be crucial in constructing roton minima for some fractions.