Symmetry Sector Decomposition

Symmetry decomposition reduces a Hamiltonian to independent blocks when the selected symmetry is exact. It also changes how observables act: an operator can connect different sectors even when the Hamiltonian does not. Numerical reduction therefore needs both a basis projector and an operator-selection rule.

Helpful background. Spin-chain bases introduces fixed magnetization; sparse assembly provides the unreduced reference.

Isometric sector embeddings

Let columns of $P_q$ be orthonormal full-space states spanning sector $q$. Then

$$P_q^\dagger P_q=I_q,\qquad H_q=P_q^\dagger HP_q.$$

Exact invariance requires $(I-P_qP_q^\dagger)HP_q=0$. A small commutator with a symmetry generator is helpful but does not replace checking the actual implemented projector.

For particle number or magnetization, $P_q$ selects occupation basis states. For parity or momentum it forms linear combinations. If all sectors are included, $\sum_qP_qP_q^\dagger=I$ and $\sum_qD_q=D$. A missing orbit or duplicated representative breaks these completeness identities.

Translation orbits

Let $T^L=I$ translate a periodic spin chain. An occupation state may have orbit length $\ell$ dividing $L$. For $k=2\pi m/L$, define

$$|s;k\rangle=\frac1{\sqrt\ell} \sum_{r=0}^{\ell-1}e^{-ikr}T^r|s\rangle.$$

This formula defines a momentum eigenstate only when $e^{ik\ell}=1$, and then satisfies $T|s;k\rangle=e^{ik}|s;k\rangle$. The compatibility condition follows because translating the final term wraps the orbit. The full group projector is

$$\Pi_k=\frac1L\sum_{r=0}^{L-1}e^{-ikr}T^r.$$

For an incompatible short orbit, $\Pi_k|s\rangle=0$: successive traversals cancel. In contrast, blindly evaluating the displayed $\ell$-term sum at that momentum generally produces a nonzero normalized vector which is not a translation eigenstate. Reject incompatible orbits before constructing that shortened sum.

Choose one canonical representative per orbit, for example the smallest integer label. Track the shift needed to map each translated state back to its representative; it supplies the phase in matrix elements. Fermionic translation must also include the sign of permuting occupied modes.

Four-site example

In the one-up-spin sector of a four-site ring, all four configurations form one orbit of length four. There is one state at each $k=0,\pi/2,\pi,3\pi/2$.

In the two-up-spin sector, the alternating configurations form an orbit of length two. They contribute only at $k=0,\pi$. The remaining four configurations form a length-four orbit. Hence the momentum dimensions are $(2,1,2,1)$ and sum to $\binom42=6$.

This dimension test catches missing or duplicated short-orbit contributions before any Hamiltonian is built. It does not test column normalization, which must be checked separately. Comparing the union of block eigenvalues with the full six-dimensional spectrum checks the phases and matrix elements.

For example, the alternating orbit has $\ell=2$. At $k=\pi/2$, the shortened sum is $(|s\rangle-iT|s\rangle)/\sqrt2$, which has norm one but does not satisfy $T|\psi\rangle=i|\psi\rangle$. The four-term group projection vanishes. Test the eigenvalue equation for each retained basis column as well as orthonormality; normalization alone cannot detect this mistake.

Observables between sectors

For an observable $A$, the relevant reduced matrix is $A_{q'q}=P_{q'}^\dagger AP_q$. A spin-raising operator changes magnetization; a Fourier component carrying momentum $p$ changes momentum according to the chosen translation convention. Restricting every observable to $A_{qq}$ can erase a physically nonzero response.

At degeneracies, an eigensolver can return arbitrary mixtures within an unresolved symmetry multiplet. Compare invariant subspaces or diagonalize commuting symmetry operators within that multiplet. Individual eigenvector components are not a stable cross-solver diagnostic.

Validity and convergence

Check the actual boundary conditions, disorder, fields, and driving terms before enabling a symmetry. A periodic clean chain may conserve momentum while its open or disordered counterpart does not. Symmetry reduction is exact only for the model that commutes with it.

Use the full Hilbert space for the smallest tractable size and compare spectra, traces, and selected matrix elements. Enlarging system size does not repair an incorrect sector construction.

Exercise

Why does a fully polarized spin configuration contribute only at zero momentum?

Solution

Its translation orbit has length one. The compatibility condition is $e^{ik}=1$, so only $k=0$ modulo $2\pi$ survives. Assigning it to every momentum sector duplicates the same physical state and violates completeness.

References

• Weinberg, P., and Bukov, M., “QuSpin: a Python package for dynamics and exact diagonalisation of quantum many body systems. Part I: spin chains,” SciPost Physics 2, 003 (2017), arXiv:1610.03042.
• Weinberg, P., and Bukov, M., “QuSpin. Part II: bosons, fermions and higher spins,” SciPost Physics 7, 020 (2019), arXiv:1804.06782.