diff --git a/main.tex b/main.tex index 2178e41794d0daf15d4c74f852b30b2505e760a8..9712fe479f5b47700bddb317285a84b96a2ef66f 100644 --- a/main.tex +++ b/main.tex @@ -387,14 +387,15 @@ presented in the following. The divergent eigenstate will belong to one of the irreps of the point symmetry group of the lattice (\textit{i.e.} \sym C {6v}). -When the Hamiltonian has an \su 2 spin symmetry, the point symmetry irreps -can be identified by matching the symmetry behavior of the spatial pairing with -one of the spatial irrep basis functions shown +When the Hamiltonian has an \su spin rotation symmetry, the spin component of +of the state decouples from the spatial pairing, and the point symmetry irreps +can be solely identified by matching the symmetry behavior of the spatial pairing +with one of the spatial irrep basis functions shown in \cref{tab:irrep_basis_fns}, second column. -If the \su 2 symmetry is broken by the introduction of SOC, the transformation -behaviour of the spins under symmetry group operations must be -considered\,\cite{kaba2019}. +If the \su symmetry is broken by the introduction of SOC, the spins are `frozen' +into the lattice, and must transform under the point symmetry group operations +\,\cite{sigrist1987, kaba2019}. Here, the irrep basis functions of the total superconducting state, hereafter referred to as the \textit{total irrep}, can be understood only as a combination of the spatial pairing symmetry and the symmetry behavior of the two spins, @@ -420,8 +421,8 @@ representation, in the same way as described for the spin-pair representation described in \cref{app:spin_irreps}, or by a group theory calculation exploiting the orthogonality of the group characters \todo{source?}. Either results in: -\begin{equation} - \sym E 1 \otimes \sym E 1 = \sym A 1 \oplus \sym B 2 \oplus \sym E 2 \, . +\begin{equation} \label{eqn:irrep-breakdown} + \sym E 1 \otimes \sym E 1 = \sym A 1 \oplus \sym A 2 \oplus \sym E 2 \, . \end{equation} We show all possible basis functions for each irrep in \cref{tab:irrep_basis_fns}, fourth column. @@ -434,25 +435,26 @@ We show all possible basis functions for each irrep in \label{sec:nn_results} We analyze the simplest case of the nearest neighbor Rashba Hubbard model on -the triangular lattice, i.e., $t_1=1, t_{n \neq 1}=0, \alpha_1=\alpha, -\alpha_{n \neq 1}=0$ where we choose the -- rather strong -- interaction $U_0 = +the triangular lattice, i.e., $t_1=1, \, t_{n \neq 1}=0, \, \alpha_1=\alpha, +\, \alpha_{n \neq 1}=0$ where we choose the -- rather strong -- interaction $U_0 = 8, U_{n \neq 0}=0$. -By virtue of FRG we are able to resolve both particle-particle and +By the virtue of FRG we are able to resolve both particle-particle and particle-hole phases in an unbiased way. We therefore provide the phase diagram in \cref{fig:phases}, showing the critical scale as well as phase transition. We postpone analysis of the particle-hole instabilities, noting only that the dominant instabilities are spin-density waves (SDW). -As already shown in Ref.~\onlinecite{beyer2022a}, the admixing of charge +As previously shown in Ref.~\onlinecite{beyer2022a}, the admixing of charge density waves into SDWs is prohibited by symmetry in the absence of relative -momentum dependencies in the eigenstate. +momentum dependencies in the eigenstate. \todo{I don't understand what this +sentence is trying to tell me.} The frayed boundaries between the instabilities are a product of the finite resolution of the calculations. As shown on the right side of \cref{fig:phases}, the points of seeming inconsistencies are the onsets of new pockets. -The accurate capture of these features of vanishing size is impossible with the -resolution of calculation. +These features are vanishing in size, making capturing them impossible +with the resolution of calculation. We therefore disregard these points and focus our attention instead on the extended superconducting regions around half-filling and around $\nu=0.2, @@ -493,6 +495,8 @@ the formfactors irreducible representation of the singlet subspace. With the singlet corresponding to \sym A 1, the total irreducible representation will equal that of the momentum component. We thus identify the \sym E 2 irreducible representation. +\todo{I think it would be cool to talk about the form of the triplet +component that you see, i.e. which E2 basis function is it?} Because we would naively expect singlet-triplet mixing under SOC, it is surprising that the -- initially triplet -- superconducting @@ -500,23 +504,22 @@ region at low filling we find negligible singlet-like contributions up to $\alpha=0.4$ (where a phase transition to a SDW occurs). Determining the irreducible representation is more involved as it necessitates a decomposition of the spin-component as presented above. -For the case of $\alpha=0$, we find \todo{Matt}. -When we now add SOC, the spin-degeneracy is broken and the system enters the -$d_x, d_y$ spin subspaces, which follow the $E_1$ irreducible representation. - -If we determine the momentum irreducible representation to be $E_1$ as well, we -are left with a total space of possibilities following -\begin{equation} - \sym A 1 \oplus \sym B 2 \oplus \sym E 2\,. -\end{equation} +For the case of $\alpha=0$ --- no SOC --- \su 2 spin rotation symmetry enforces +the degeneracy of the the spin triplet states, which are decoupled from the point +symmetry group of the crystal field\,\cite{sigrist1987}. +\todo{the state we get is...} + +When we now add SOC, the overall symmetry of the Hamiltonian is reduced and the +spin-triplet state degeneracy is broken. The resulting superconducting state is +doubly degenerate, fixed in the $d_x, d_y$ spin subspaces with $p$-wave spatial +pairing, as shown on the right hand side of \cref{fig:sing}. +This is however not enough to identify the irrep of the state, as we are left +with a total space of possibilities in \cref{eqn:irrep-breakdown}. In our case, this is disambiguated by the degeneracy of the eigenstates, i.e., because we find a pair of eigenstates we must be in the \sym E 2 irreducible -representation's subspace. -For other disambiguations one could have considered the transformation behavior -of the eigenstates of the system as shown in the right side of \cref{fig:sing}. -Their transformation behavior clearly does not follow \sym A 1 or \sym B 2 -representations, we therefore conclude that we remain in the \sym E 2 -irreducible representation. +representation's subspace. This can be confirmed by identifying the state +plotted on the right hand side of figure \cref{fig:sing} with the basis +functions as listed in Table\,\ref{tab:irrep_combinations}. \begin{figure*} \includegraphics{plots/sc_inst.pdf}