Determining representative states

In SymBasis.jl, each symmetry-related orbit of basis states is represented by a single representative state. Starting from a state $\lvert a'(q)\rangle$, you generate all symmetry images under the symmetry operator $\hat{S}$ and choose the one with the smallest hash value as the representative:

\[\lvert a(q)\rangle = \hat{S}^{\ell_{\mathrm{rep}}}\lvert a'(q)\rangle, \qquad \ell_{\mathrm{rep}}=\arg\min_{\ell}\Big\{\mathrm{hash}\!\big[\hat{S}^\ell\lvert a'(q)\rangle\big]\Big\}~.\]

Here, $q$ labels the symmetry sector, and $\ell_{\mathrm{rep}}$ is the shift needed to map $\lvert a'(q)\rangle$ onto its representative $\lvert a(q)\rangle$. This is handled automatically when you construct a basis via basis from SymBasis.Bases, ensuring that each symmetry sector contains only one representative per orbit.

Operator action and mapping back to representatives

Consider an operator decomposed into local contributions,

\[\hat{F} \coloneqq \sum_{j=0}^{N_F}\hat{F}_j~,\]

with $\hat{F}_0$ the diagonal part and $\hat{F}_{j>0}$ off-diagonal parts. Acting with one term on a representative basis state generally produces a (symmetry-related) state that is not itself in representative form:

\[\hat{F}_j\lvert a(q)\rangle = f_j[a(q)]\,\lvert b'_j(q)\rangle~.\]

To express the result in the representative basis, map $\lvert b'_j(q)\rangle$ to its representative $\lvert b_j(q)\rangle$ using the symmetry operation:

\[\lvert b_j(q)\rangle = \hat{S}^{\ell_j}\lvert b'_j(q)\rangle~,\]

where $\ell_j$ is the number of applications needed to reach the representative. Equivalently,

\[\hat{F}_j\lvert a(q)\rangle = f_j[a(q)]\,\hat{S}^{-\ell_j}\lvert b_j(q)\rangle~.\]

Including the symmetry phase and normalization factors, this becomes

\[\hat{F}_j\lvert a(q)\rangle = f_j[a(q)]\,\eta(q,\ell_j)\,\sqrt{\frac{N_{b_j(q)}}{N_{a(q)}}}\,\lvert b_j(q)\rangle~,\]

where $\eta(q,\ell_j)$ is the phase accrued when relating $\lvert b'_j(q)\rangle$ to $\lvert b_j(q)\rangle$, and $N_{a(q)}$, $N_{b_j(q)}$ are the state normalization factors.

Matrix elements

• Diagonal term:

\[\langle a(q)\lvert \hat{F}_0\rvert a(q)\rangle = f_0[a(q)]~.\]

• Off-diagonal term (after mapping to representatives):

\[\langle b_j(q)\lvert \hat{F}_{j>0}\rvert a(q)\rangle = f_{j>0}[a(q)]\,\eta(q,\ell_j)\,\sqrt{\frac{N_{b_j(q)}}{N_{a(q)}}}~.\]

Implementation in SymBasis.jl

To obtain the representative for an arbitrary state (and associated normalization information), use representative from SymBasis.Bases. It takes a state and a symmetry group and returns the corresponding representative data used to build symmetry-resolved matrix elements.

For example, if you have a state $\lvert \downarrow \uparrow \uparrow \downarrow \rangle$ in a spin-1/2 chain with total magnetization symmetry ($S^z=0$ sector) and translational symmetry ($k=0$ sector), you can find its representative state and normalization factor as follows:

using SymBasis.DigitBase
using SymBasis.DoFObjects
using SymBasis.SymGroups
using SymBasis.Bases

N = 4 # number of sites
dofo = dof_object(Spin(1 // 2)) # define a DoF-object for spin-1/2

Sz = 0 # total magnetization quantum number
# define the symmetry group for total magnetization symmetry
sg_Sz = sym(TotalMagnetization(Sz, N), dofo)

perm = mod1.((1:N) .+ 1, N) # permutation for translational symmetry
k = 0 # momentum quantum number
# define the symmetry group for translational symmetry
sg_translational = sym(Translational(k, perm), dofo)

# combine the total magnetization symmetry and the translational symmetry
csg = sg_Sz ∘ sg_translational

state = bi"0110"2 # state |↓↑↑↓⟩ in binary representation

# get the representative state and normalization
rep_state, norm_factor = representative(state, csg)

((1100)₂, 1.0 - 0.0im)

This will return the representative state in the specified symmetry sector, along with its normalization factor, which can be used to compute matrix elements of operators in the basis.