Symmetry Sectors in Many-Body Numerics

A symmetry sector is an invariant subspace selected by exact conserved quantum numbers or by an irreducible representation of the finite problem’s symmetry group. When a Hamiltonian preserves such a subspace, one may construct and solve its block without storing states that can never mix with it.

For a compatible family of sector labels $\mathbf q$,

$$\mathcal H = \bigoplus_{\mathbf q} \mathcal H_{\mathbf q}, \qquad H = \bigoplus_{\mathbf q} H_{\mathbf q}.$$

This decomposition reduces memory and runtime, but its scientific value is broader. It identifies quantum numbers, separates symmetry-protected crossings from avoided crossings, exposes selection rules, prevents unrelated spectra from being mixed, and supplies strong validation identities.

The reduction is exact only if the implemented finite Hamiltonian has the claimed symmetry. A sector label inherited from an idealized infinite model may fail after choosing a boundary condition, cluster shape, gauge, local cutoff, disorder realization, or numerical convention.

Required background. Finite diagonalization supplies the basis construction; group representations explains irreducible sectors.

Invariant subspaces and reduced Hamiltonians

Let $P_{\mathbf q}$ be the orthogonal projector onto a sector. Exact invariance means

$$[H,P_{\mathbf q}] = 0.$$

Equivalently,

$$(I-P_{\mathbf q}) H P_{\mathbf q} = 0.$$

The second equation is the most direct numerical statement: a vector initialized in the sector has no Hamiltonian component outside it.

If $V_{\mathbf q}$ has orthonormal columns spanning the sector, then

$$\begin{aligned} V_{\mathbf q}^{\dagger} V_{\mathbf q} &= I_{\mathbf q}, \\ P_{\mathbf q} &= V_{\mathbf q} V_{\mathbf q}^{\dagger}, \\ H_{\mathbf q} &= V_{\mathbf q}^{\dagger} H V_{\mathbf q}. \end{aligned}$$

The full operator can be reconstructed as

$$H = \sum_{\mathbf q} V_{\mathbf q} H_{\mathbf q} V_{\mathbf q}^{\dagger}.$$

In a large calculation one usually does not form the full $H$, the full $P_{\mathbf q}$, or the dense matrix $V_{\mathbf q}$. Representatives, phases, normalization factors, and lookup rules implement the same map directly.

Symmetry Is a Property of the Declared Finite Problem

A claimed unitary symmetry $U_g$ must satisfy two conditions:

$$U_g^{\dagger}U_g=I, \qquad U_gHU_g^{\dagger}=H.$$

The first tests the representation. The second tests the Hamiltonian. Both can fail in code even when the continuum or thermodynamic model is symmetric.

For example, a periodic uniform chain may admit one-site translation. An open chain does not. A periodic chain with one modified bond also does not. A twisted boundary may preserve a modified translation only after a gauge phase is included. The finite operator, not the intended model name, decides.

Compatible Labels

For Hermitian conserved operators $Q_a$, a simple simultaneous labeling requires

$$[H,Q_a]=0, \qquad [Q_a,Q_b]=0.$$

Then

$$\mathcal H = \bigoplus_{\mathbf q} \mathcal H_{\mathbf q}, \qquad \mathbf q = (q_1,q_2,\ldots).$$

Pairwise commutation is not merely formal. It determines whether a basis can carry all proposed labels at once.

Several distinctions matter:

• Two symmetries may each commute with $H$ but fail to commute with each other.
• A symmetry may preserve one sector while exchanging two other sectors.
• A non-Abelian symmetry generally requires irreducible-representation labels, not simultaneous eigenvalues of every group element.
• An antiunitary symmetry constrains matrix structure and degeneracies but is not inserted into an ordinary complex-linear projector in the same way as a unitary symmetry.
• A fixed eigenvalue of one generator, such as $S^z_{\mathrm{tot}}$, need not specify a full irreducible multiplet of a larger symmetry such as $SU(2)$.

Use only compatible labels supported by the actual finite Hamiltonian.

Finite-Group Projectors

Let a finite unitary group $G$ act through $U_g$. The projector onto the isotypic component associated with an irreducible representation $\lambda$ is

$$P_{\lambda} = \frac{d_{\lambda}}{|G|} \sum_{g\in G} \chi_{\lambda}(g)^{*} U_g,$$

where $d_{\lambda}$ and $\chi_{\lambda}$ are the representation dimension and character.

For an Abelian group, $d_{\lambda}=1$, every irreducible representation is one-dimensional, and the expression directly selects a quantum-number sector. For a non-Abelian group, $P_{\lambda}$ selects all copies of the irrep. Further row or multiplicity labels are required if one wants the smallest repeated block.

The projector identities are

$$\begin{aligned} P_{\lambda}^{\dagger} &= P_{\lambda}, & P_{\lambda}^{2} &= P_{\lambda}, \\ P_{\lambda}P_{\mu} &= 0 \quad (\lambda\neq\mu), & \sum_{\lambda}P_{\lambda} &= I. \end{aligned}$$

They are also numerical tests.

For several compatible Abelian symmetries, the joint projector is the product

$$P_{\mathbf q} = \prod_a P_{q_a}^{(a)}.$$

This formula fails if the factors do not commute.

Particle-Number Sectors

For $L$ modes,

$$\widehat N = \sum_{i=1}^{L} n_i.$$

If

$$[H,\widehat N]=0,$$

the Fock space decomposes as

$$\mathcal F = \bigoplus_N \mathcal H_N.$$

In an occupation basis, fixed-$N$ construction is direct: enumerate only configurations whose occupations sum to $N$.

For spinless fermions on $L$ orbitals,

$$\dim\mathcal H_N = \binom{L}{N}.$$

For bosons in $L$ modes with fixed total number and no additional local cutoff,

$$\dim\mathcal H_N = \binom{N+L-1}{N}.$$

For spinful fermions with separately conserved populations,

$$\dim \mathcal H_{N_{\uparrow},N_{\downarrow}} = \binom{L}{N_{\uparrow}} \, \binom{L}{N_{\downarrow}}.$$

These formulas count different model spaces. A local bosonic cutoff, forbidden double occupancy, gauge constraint, or orbital restriction changes the count and must be included before a dimension is used as a validation target.

Number conservation versus number parity

A pairing term such as

$$H_{\Delta} = \sum_{ij} \left( \Delta_{ij} c_i^{\dagger}c_j^{\dagger} + \Delta_{ij}^{*} c_jc_i \right)$$

changes particle number by two. Therefore

$$[H_{\Delta},\widehat N] \neq 0.$$

It can nevertheless preserve fermion-number parity,

$$\Pi_N = (-1)^{\widehat N}, \qquad [H_{\Delta},\Pi_N] = 0.$$

The correct blocks are then even and odd Fock-space sectors, not fixed-$N$ sectors. Confusing these two symmetries can remove physically required anomalous couplings.

Total Spin z-Component

For $L$ spin-$1/2$ sites,

$$S^z_{\mathrm{tot}} = \sum_{j=1}^{L} S_j^z.$$

If $N_{\uparrow}$ sites are up,

$$S^z_{\mathrm{tot}} = \hbar \left( N_{\uparrow} - \frac{L}{2} \right).$$

Thus a fixed-magnetization basis has dimension

$$D_{N_{\uparrow}} = \binom{L}{N_{\uparrow}}.$$

Exchange terms preserve this number because they move an up spin rather than create or destroy net magnetization:

$$S_i^{+}S_j^{-} : \quad N_{\uparrow} \longmapsto N_{\uparrow}.$$

The Heisenberg and XXZ chains conserve $S^z_{\mathrm{tot}}$. A transverse field generally does not:

$$\left[ \sum_j h_xS_j^x, S^z_{\mathrm{tot}} \right] \neq 0.$$

The fixed-$S^z_{\mathrm{tot}}$ basis is therefore model-dependent even when the local degrees of freedom are the same.

Fixed magnetization is not fixed total spin

For an $SU(2)$-invariant Hamiltonian,

$$[H,\mathbf S_{\mathrm{tot}}^2] = 0, \qquad [H,S^z_{\mathrm{tot}}] = 0.$$

A fixed-$M$ sector with

$$S^z_{\mathrm{tot}}|S,M\rangle = \hbar M|S,M\rangle$$

contains every total-spin multiplet satisfying $S\geq |M|$. Restricting to $M=0$ does not restrict to singlets. One may identify $S$ afterward from $\mathbf S_{\mathrm{tot}}^2$, construct a fully $SU(2)$-adapted basis, or use highest-weight counting. These routes have different implementation complexity.

Translation Momentum on a Periodic Cluster

Consider a periodic chain with $L$ sites and one-site translation $T$ satisfying

$$T^L=I.$$

Choose the convention

$$T|\psi_k\rangle = e^{-ik} |\psi_k\rangle.$$

The allowed dimensionless crystal momenta are

$$k_n = \frac{2\pi n}{L}, \qquad n=0,1,\ldots,L-1.$$

The momentum projector is

$$P_{k} = \frac{1}{L} \sum_{r=0}^{L-1} e^{ikr} T^r.$$

Translation is available only if $T$ maps the declared finite cluster, boundary conditions, local basis, and Hamiltonian coefficients onto themselves. A periodic drawing is not enough; the implemented bond list must be invariant.

Translation Orbits and Representatives

First consider a spin or boson configuration basis on which translation is an unsigned permutation. Start from a configuration $|a\rangle$ in a fixed additive-charge sector. Its translation orbit is

$$\mathcal O_a = \left\{ |a\rangle, T|a\rangle, \ldots \right\}.$$

Let $R_a$ be the smallest positive integer satisfying

$$T^{R_a}|a\rangle = |a\rangle.$$

The orbit has $R_a$ distinct configurations. A deterministic convention, such as the smallest encoded integer, chooses one representative.

The normalized momentum state built from that orbit is

$$|a;k\rangle = \frac{1}{\sqrt{R_a}} \sum_{r=0}^{R_a-1} e^{ikr} T^r|a\rangle,$$

provided

$$e^{ikR_a} = 1.$$

If this compatibility condition fails, the projected state vanishes. This is the simplest stabilizer constraint: configurations with a shorter spatial period do not contribute to every momentum.

The momentum-block dimension can be checked independently from the character trace,

$$D_k = \operatorname{Tr}P_k = \frac{1}{L} \sum_{r=0}^{L-1} e^{ikr} \operatorname{Tr}_{\mathcal S} T^r,$$

where $\mathcal S$ is the prefiltered configuration space. For an unsigned permutation, the trace counts fixed configurations.

Fermionic translations also reorder creation operators and can attach a sign. If the shortest return of the occupation pattern satisfies $T^R|a\rangle=s_a|a\rangle$, the compatibility condition becomes $e^{ikR}s_a=1$. Sum the $R$ distinct translated patterns with their actual signed amplitudes and normalization $1/\sqrt R$. The character trace remains valid, but counts fixed configurations with their phases rather than only their number. A two-site fully occupied fermion state illustrates this: translation exchanges its two creators, so its pattern has period one but its translation eigenvalue is $-1$.

Worked Orbit Count for Six Spins

Take a six-site spin-$1/2$ ring at

$$N_{\uparrow}=3, \qquad S^z_{\mathrm{tot}}=0.$$

The fixed-magnetization dimension is

$$\binom{6}{3} = 20.$$

Eighteen configurations form three translation orbits of length six. The two Néel configurations

$$|\uparrow\downarrow\uparrow\downarrow \uparrow\downarrow\rangle, \qquad |\downarrow\uparrow\downarrow\uparrow \downarrow\uparrow\rangle$$

form one orbit of length two.

Each length-six orbit contributes one state to every $k_n$. The length-two orbit contributes only when

$$e^{i2k_n}=1,$$

which occurs for $n=0$ and $n=3$. Therefore

$$\left( D_0,D_1,D_2,D_3,D_4,D_5 \right) = \left( 4,3,3,4,3,3 \right).$$

The dimensions add to the original count:

$$\sum_{n=0}^{5}D_n = 20.$$

Assuming every momentum block has dimension $20/6$ would have failed because orbit stabilizers make the blocks unequal.

The three length-six orbits each supply one state at every momentum. The single length-two orbit adds a fourth state only at $k=0,\pi$, giving dimensions $(4,3,3,4,3,3)$.

Reflection and Parity

Let $\mathcal R$ be a reflection with

$$\mathcal R^2=I.$$

If

$$[H,\mathcal R]=0,$$

its eigenvalues are

$$p=\pm1.$$

In a basis where reflection is compatible with all previous labels, the projectors are

$$P_p = \frac{1}{2} \left( I+p\mathcal R \right).$$

For a one-dimensional periodic lattice, reflection reverses translation:

$$\mathcal R T \mathcal R^{-1} = T^{-1}.$$

Consequently,

$$\mathcal R : \quad \mathcal H_k \longrightarrow \mathcal H_{-k}.$$

Momentum and reflection parity can be imposed simultaneously only when

$$k \equiv -k \pmod{2\pi}.$$

Thus:

• $k=0$ always supports parity labels;
• $k=\pi$ supports parity labels when $L$ is even;
• generic $k$ and $-k$ are exchanged, so a single complex momentum block has no independent reflection-parity label.

At generic momentum one may retain separate complex $k$ and $-k$ blocks, combine them into a real two-component representation, or use a basis adapted to the appropriate dihedral-group irrep. One must not simply add a $p=\pm1$ label to each $k$ block.

For the six-site example and the site-centered reflection $j\mapsto -j$, each special momentum block splits as

$$\begin{aligned} k=0: &\qquad D_{+}=3, \quad D_{-}=1, \\ k=\pi: &\qquad D_{+}=3, \quad D_{-}=1. \end{aligned}$$

The split depends on the precisely defined reflection operation. Even rings have site-centered and bond-centered reflections related by a translation, and their labels can differ by a momentum-dependent phase.

Several meanings of parity

The word parity is overloaded in many-body work. Distinguish:

• spatial inversion or reflection;
• fermion-number parity $(-1)^{\widehat N}$;
• spin inversion, which flips every $S_j^z$;
• sublattice or chiral operations, which may anticommute rather than commute with $H$.

These operators have different algebras and different sectors. Name the operator, not only the eigenvalue.

Building a Reduced Hamiltonian Directly

Suppose $|a;\mathbf q\rangle$ are normalized symmetry-adapted states indexed by representatives. The block matrix is

$$\left( H_{\mathbf q} \right)_{ba} = \langle b;\mathbf q| H |a;\mathbf q\rangle.$$

A practical matrix-element kernel does the following:

1. take representative $a$;
2. apply one local Hamiltonian term to its underlying configuration;
3. verify that the complete Hamiltonian has no output outside the assumed additive-charge sector; unexpected leakage is an error, not a transition to silently discard;
4. map each surviving output to its symmetry representative $b$;
5. record the group element used in that map;
6. multiply by its character or momentum phase;
7. include the orbit-normalization ratio;
8. accumulate the contribution in row $b$.

The phase convention and orbit normalization must be derived together. Reusing a phase from one translation convention with normalization from another can preserve plausible eigenvalues in special sectors while corrupting generic-momentum matrix elements.

For a matrix-free solver, the same representative map implements

$$x \longmapsto H_{\mathbf q}x$$

without storing the sparse block. The computational saving comes from never visiting equivalent configurations as independent basis vectors.

Complex arithmetic is sometimes physical labels

For generic momentum,

$$e^{ikr} \notin \mathbb R.$$

The reduced Hamiltonian may therefore be complex Hermitian even when the original configuration-basis Hamiltonian is real. Discarding imaginary parts does not make the calculation more exact; it changes the representation. Real formulations exist by pairing $k$ and $-k$, but they require the corresponding two-component basis.

Observables Can Change Sectors

A Hamiltonian-preserving label need not be preserved by every observable.

If an additive charge satisfies

$$[Q,O] = \Delta q\,O,$$

then

$$P_{q'} O P_q = 0 \quad \text{unless} \quad q'=q+\Delta q.$$

An annihilation operator connects $N$ to $N-1$. A spin-raising operator connects magnetization sectors differing by $\hbar$. A momentum-resolved operator satisfying

$$T O_p T^{-1} = e^{-ip}O_p$$

maps

$$\mathcal H_k \longrightarrow \mathcal H_{k+p}.$$

An operator odd under reflection connects opposite parity sectors at compatible momentum.

This matters for spectral functions. A ground state may be computed in one sector, while every intermediate state contributing to a response lies in another. Restricting both sides of the matrix element to the ground-state block can force the desired signal to zero by construction.

Ground States Require a Cross-Sector Search

Let

$$E_0^{(\mathbf q)} = \min \operatorname{spec} H_{\mathbf q}.$$

The unconstrained finite-system ground-state energy is

$$E_0 = \min_{\mathbf q} E_0^{(\mathbf q)}.$$

Solving one convenient block establishes only the ground state within that block. It does not establish the global ground state unless a theorem or an explicit cross-sector comparison identifies the winning label.

This distinction is especially important when:

• a coupling drives a level crossing between different symmetry sectors;
• the first excitation lies at momentum different from the ground state;
• boundary conditions reorder low-energy levels;
• a magnetic field changes the favored magnetization;
• a pairing problem requires comparing even and odd number parity;
• finite clusters in a sequence admit different momentum grids.

Crossings between distinct exact sectors can be sharp because the states cannot hybridize. Levels with identical complete symmetry labels generically repel unless another conservation law or integrable structure intervenes.

Finite Symmetry and Spontaneous Symmetry Breaking

A finite Hamiltonian with an exact symmetry can be diagonalized in symmetry sectors even when its thermodynamic phase breaks that symmetry.

Finite eigenstates normally carry definite symmetry labels. A symmetry-broken configuration is then represented by a superposition of low-lying states from different sectors. The thermodynamic pattern is inferred from:

• near-degenerate parity doublets;
• towers of states in several spin sectors;
• low-energy levels at ordering momenta;
• order-parameter correlations and structure factors;
• systematic scaling with size and geometry.

Selecting one broken-symmetry product state and calling it the exact finite ground state discards the finite-system symmetry structure that reveals the approach to the thermodynamic limit.

Thermal Traces Across Sectors

For a complete orthogonal decomposition,

$$Z(\beta) = \operatorname{Tr} e^{-\beta H} = \sum_{\mathbf q} \operatorname{Tr}_{\mathcal H_{\mathbf q}} e^{-\beta H_{\mathbf q}}.$$

For any observable, set $O_{\mathbf q}=V_{\mathbf q}^\dagger O V_{\mathbf q}$. Only these diagonal blocks contribute to a one-point thermal trace, whether or not $O$ preserves sectors:

$$\langle O\rangle_{\beta} = \frac{1}{Z} \sum_{\mathbf q} \operatorname{Tr}_{\mathcal H_{\mathbf q}} \left( e^{-\beta H_{\mathbf q}} O_{\mathbf q} \right).$$

A trace over one sector gives a constrained ensemble. Sector-changing intermediate states must still be included in products such as $O^\dagger(t)O$; multiplying only the diagonal blocks is generally wrong.

For non-Abelian reductions, record whether the numerical block retains every irrep row or only a multiplicity space. If only one repeated block is solved, the irrep dimension must be restored in full traces.

Spectral Statistics Must Be Desymmetrized

Level-spacing statistics compare neighboring eigenvalues that are allowed to repel. Combining independent exact sectors superposes unrelated spectra. Cross-sector crossings then produce many small spacings and can mimic Poisson-like behavior even when each irreducible block has random-matrix correlations.

A defensible level-statistics calculation therefore fixes:

• particle number or magnetization;
• translation momentum;
• every compatible point-group or reflection label;
• number parity or spin inversion where applicable;
• any remaining exact conservation law;
• an energy window and unfolding or spacing-ratio convention.

The Many-Body Quantum Chaos Preview explains the physical interpretation. Here the key numerical rule is simple: compare levels only after all exact unitary symmetries have been resolved as far as the analysis requires.

Dimension and Closure Checks

A complete set of sectors satisfies:

$$\sum_{\mathbf q} D_{\mathbf q} = D_{\mathrm{parent}}.$$

For nested reductions, check the equality at each stage:

$$D_{\mathrm{full}} \longrightarrow \sum_N D_N \longrightarrow \sum_{N,k}D_{N,k} \longrightarrow \sum_{N,k,p}D_{N,k,p}.$$

Only compatible parity labels belong in the final sum.

Projector closure can be tested by random vectors:

$$\epsilon_{\mathrm{leak}}(x) = \frac{ \left\| (I-P_{\mathbf q}) H P_{\mathbf q}x \right\| }{ \left\| H P_{\mathbf q}x \right\| + \epsilon }.$$

A symmetry-action test is

$$\epsilon_g(x) = \frac{ \left\| HU_gx-U_gHx \right\| }{ \left\| HU_gx \right\| + \left\| U_gHx \right\| + \epsilon }.$$

Both should scale with numerical roundoff for an exact implemented symmetry.

Spectral and Trace Reconstruction

On a small system where the unreduced calculation is feasible, the union of block spectra must equal the full spectrum with multiplicity:

$$\operatorname{spec}H = \biguplus_{\mathbf q} \operatorname{spec}H_{\mathbf q}.$$

Low-order trace moments provide basis-independent tests:

$$\begin{aligned} \operatorname{Tr}H &= \sum_{\mathbf q} \operatorname{Tr}H_{\mathbf q}, \\ \operatorname{Tr}H^2 &= \sum_{\mathbf q} \operatorname{Tr}H_{\mathbf q}^2. \end{aligned}$$

The first can reveal missing diagonal contributions; the second checks the sum of squared matrix-element magnitudes for a Hermitian matrix. Neither is a complete test: $\operatorname{Tr}H^2=\sum_{ab}|H_{ab}|^2$ is insensitive to changing only phases. Loop-sensitive higher moments, direct block comparison, or the full small-system spectrum are needed to detect such errors.

Additional checks include:

• Hermiticity of every block;
• representative uniqueness;
• group multiplication on encoded configurations;
• orbit length dividing the group order;
• vanishing of incompatible projected states;
• expected $k\leftrightarrow -k$ spectral pairing when a reflection or suitable antiunitary symmetry is present;
• agreement of symmetry-resolved observables reconstructed from the full eigenvectors.

Exercises

2. Why generic momentum has no reflection parity

Let

$$\mathcal R T \mathcal R^{-1} = T^{-1}.$$

Assume a nonzero state is simultaneously an eigenstate of $T$ and $\mathcal R$:

$$T|\psi\rangle = e^{-ik}|\psi\rangle, \qquad \mathcal R|\psi\rangle = p|\psi\rangle.$$

Determine the allowed momenta.

Solution

Act with $T$ on the reflected state:

$$\begin{aligned} T\mathcal R|\psi\rangle &= \mathcal R T^{-1}|\psi\rangle \\ &= e^{ik} \mathcal R|\psi\rangle. \end{aligned}$$

But $\mathcal R|\psi\rangle=p|\psi\rangle$, so the same nonzero vector would have translation eigenvalues $e^{-ik}$ and $e^{ik}$. Therefore

$$e^{-ik} = e^{ik},$$

or

$$e^{i2k}=1.$$

Hence

$$k=0 \quad \text{or} \quad k=\pi \pmod{2\pi}.$$

The $k=\pi$ value belongs to the finite momentum grid only when $L$ is even.

3. Pairing preserves parity but not number

Consider

$$H_{\Delta} = \Delta c_1^{\dagger}c_2^{\dagger} + \Delta^{*} c_2c_1.$$

Show that $H_{\Delta}$ does not conserve $\widehat N$ but does conserve $\Pi_N=(-1)^{\widehat N}$.

Solution

The number-operator commutators are

$$\begin{aligned} \left[ \widehat N, c_1^{\dagger}c_2^{\dagger} \right] &= 2c_1^{\dagger}c_2^{\dagger}, \\ \left[ \widehat N, c_2c_1 \right] &= -2c_2c_1. \end{aligned}$$

Therefore

$$\left[ \widehat N, H_{\Delta} \right] \neq 0$$

for nonzero $\Delta$.

Number parity acts on a single fermion operator as

$$\Pi_N c_i^{\dagger}\Pi_N^{-1} = -c_i^{\dagger}, \qquad \Pi_N c_i\Pi_N^{-1} = -c_i.$$

Every pairing monomial contains two such operators, so the two minus signs cancel:

$$\Pi_N H_{\Delta}\Pi_N^{-1} = H_{\Delta}.$$

Thus the Hamiltonian mixes $N$ with $N\pm2$ but never mixes even and odd $N$.

4. Momentum carried by an observable

Let local operators translate according to

$$T O_j T^{-1} = O_{j+1},$$

and define

$$O_q = \frac{1}{\sqrt L} \sum_{j=0}^{L-1} e^{iqj} O_j.$$

Show that $O_q$ maps momentum $k$ to momentum $k+q$.

Solution

Translate the Fourier component:

$$\begin{aligned} T O_q T^{-1} &= \frac{1}{\sqrt L} \sum_j e^{iqj} O_{j+1} \\ &= e^{-iq} \frac{1}{\sqrt L} \sum_{\ell} e^{iq\ell} O_{\ell} \\ &= e^{-iq}O_q. \end{aligned}$$

For

$$T|\psi_k\rangle = e^{-ik}|\psi_k\rangle,$$

one obtains

$$\begin{aligned} T O_q|\psi_k\rangle &= \left( T O_qT^{-1} \right) T|\psi_k\rangle \\ &= e^{-i(k+q)} O_q|\psi_k\rangle. \end{aligned}$$

Therefore

$$O_q : \quad \mathcal H_k \longrightarrow \mathcal H_{k+q}.$$

5. Magnetization does not determine total spin

Four spin-$1/2$ degrees of freedom decompose into two spin-zero multiplets, three spin-one multiplets, and one spin-two multiplet. Determine the dimension and total-spin content of the $M=0$ sector.

Solution

Every integer-spin multiplet contains exactly one state with $M=0$. Therefore the $M=0$ sector contains:

• two states from the two singlet multiplets;
• three states from the three triplet multiplets;
• one state from the quintet multiplet.

Its dimension is

$$D_{M=0} = 2+3+1 = 6.$$

This agrees with direct fixed-magnetization counting:

$$D_{M=0} = \binom{4}{2} = 6.$$

The equality confirms the count, while the decomposition shows why $M=0$ cannot be used as a synonym for $S=0$.

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