Figures — Singlet

Figure 01 / Conditional recouplingExact · Four qubits

Line weight

Singlet probability · 0 → 1

Measure pair

Prepared two singlets. Both outcomes retain total spin zero.

Selected pair 1–3

Two singlets have total spin zero. Measurements can rearrange their correlations within that sector.

Measurement history

Prepared two singlets on 1–2 and 3–4

Read the mathematics & all six probabilities

The ideal instrument applies $P_s=(I-\mathrm{SWAP})/2$ or $P_t=(I+\mathrm{SWAP})/2$ and normalizes. Inspecting a branch conditions on that outcome; sampling draws an outcome according to the Born rule.

For $A=|s_{12}\rangle|s_{34}\rangle$ and $B=|s_{13}\rangle|s_{24}\rangle$ , measuring 1–3 gives $B$ with probability ¼, or $(2A-B)/\sqrt{3}$ with probability ¾. The pairings have overlap ½; they are not an orthogonal basis.

Singlet probabilities for the current state

Pair 1–2100%

Pair 1–325%

Pair 1–425%

Pair 2–325%

Pair 2–425%

Pair 3–4100%

What does “all triplet” mean?

The second input is $|t_0\rangle_{12}|t_0\rangle_{34}$ , where $|t_0\rangle=(|01\rangle+|10\rangle)/\sqrt2$ . Pairs 1–2 and 3–4 are certainly triplet, but each cross-pair has singlet probability ¼. Two triplet pairs are not a fully symmetric four-qubit state.

The fully symmetric input is the entangled Dicke state $|D_4^{(2)}\rangle$ : the equal superposition of all six bit strings containing two ones. Every SWAP leaves it unchanged, so every pair gives triplet with certainty. Every STP sequence on these four qubits leaves the state unchanged. This holds for any state supported in the fully symmetric subspace.

For the two-triplet input, only J = 0 and J = 2 have support. The displayed percentages refer to the current conditional state. These prepared pure examples differ from the mixed triplet outcomes in Figure 5.

Lines visualize six possible tests of the current state. They are not simultaneous definite spin labels or literal bonds. Selecting an edge changes the next pair, not the quantum state. These prepared inputs illustrate recoupling and symmetry, not the full universality construction from mixed input.

Why the SWAP test measures singlet or triplet

Q01 / Research directionWhich algorithmic primitives does subsystem spin suggest?Can filtering, recoupling, or state comparison give an STP algorithm for a specified task? Identify its interference mechanism and compare its resource cost with a circuit implementation.

Figure 02 / Schur–Weyl decompositionA change of coordinates

Count the coordinates in each spin sector

Each cell represents one basis coordinate, not a particle. For three qubits, the two spin sectors contain four coordinates each. For four qubits, the spin-zero sector has two: these are the two basis states used in Figure 3.

Number of qubits 3

28

J = 1/24 dim.

2 × 2

4/8 of mixed input

J = 3/24 dim.

4 × 1

4/8 of mixed input

4 + 4 = 8Total dimension

Columns: spin basis states. Collective rotations act on the $V_J$ factor, with basis labels $m=-J,\ldots,J$ and dimension $2J+1$ .

Rows: multiplicity. Permutations act on $M_J$ . Multiple coupling paths can reach the same total spin.

The decomposition, multiplicities & Young shapes

(\mathbb C^2)^{\otimes n}\cong\bigoplus_J V_J\otimes M_J

\dim M_J=\binom n{n/2-J}-\binom n{n/2-J-1}

The two-row Young shape is $(n/2+J,n/2-J)$ . Each rectangle above contains one cell per dimension, arranged as multiplicity × spin dimension. The small two-row diagram identifies its representation. Measuring J, sampling a Schur label, and performing a coherent Schur transform are different operations.

Read the explanation of Schur–Weyl duality

Q09 / QuestionWhat is the simulation cost of STP at fixed total spin?With a fixed register in sector

J=N/2-k

, determine how STP computation and classical simulation scale with the spin deficit

k

and the multiplicity dimension.

Figure 03 / Relative phase at fixed total spinFour qubits · J = 0

This figure asks whether total spin determines what a pair measurement will do. It does not. We compare superpositions of two orthogonal four-qubit states, both with total spin zero.

$|e_0\rangle$ · two singlets

|e_0\rangle=|s\rangle_{12}|s\rangle_{34}

Pairs 1–2 and 3–4 each have spin 0.

$|e_1\rangle$ · two triplets, coupled to spin 0

|e_1\rangle=\frac{|t_+t_-\rangle-|t_0t_0\rangle+|t_-t_+\rangle}{\sqrt3}

|\psi_\varphi\rangle=\frac{|e_0\rangle+e^{i\varphi}|e_1\rangle}{\sqrt2}

Change only the relative phase

Both components have probability ½. Moving the slider changes their interference, not their weights. Watch the cross-pair bars: 1–3 and 1–4 exchange their singlet probabilities between φ = 0 and φ = π.

This slider selects a mathematical state; it does not implement an STP gate.

Relative phase φ 0

0π2π

Coherent superposition · equal amplitudes

Probability of a singlet outcome

1–2

50%

1–3

93.3%

1–4

6.7%

2–3

6.7%

2–4

93.3%

3–4

50%

Total spin 0Sum over pairs 3

Which coherence? Which resource?

|\psi_\varphi\rangle=\frac{|e_0\rangle+e^{i\varphi}|e_1\rangle}{\sqrt{2}}

The two states above form an orthonormal basis of the four-qubit J = 0 multiplicity space. The mixture is $\tfrac12|e_0\rangle\langle e_0|+\tfrac12|e_1\rangle\langle e_1|$ . Both remain in the same spin sector.

\sum_{i<j}P_s^{(ij)}=\frac{n(n+2)}8 I-\frac12\mathbf J^2

Thus total singlet weight is fixed by total spin. This is why a resource theory must look beyond that single number. This figure studies coherence within a sector; it does not assume free access to coherence between different global-spin sectors.

The phase slider compares mathematical states; it is not an STP phase gate. Starting from singlets or maximally mixed input, bare STP has real projectors and preserves real density matrices. At φ = π/2 or 3π/2, all six probabilities are ½, just as for the mixture. Even adaptive bare-STP sequences cannot detect that imaginary coherence: their measurement effects remain real.

Q11 / QuestionDoes multiplicity-space coherence improve STP protocols?Within one total-spin sector, compare a state with its dephased version in a fixed coupling basis. Seek a separation in task success probability or measurement cost.

Figure 04 / A complexity phase transitionMarwaha & Sud · 2026

The singlet in the local spectrum

h(s)=-s|\Psi^-\rangle\langle\Psi^-|-|\Psi^+\rangle\langle\Psi^+|

Positive-weight sums of this two-qubit interaction on arbitrary graphs; the inverse-polynomial precision ground-energy decision problem.

Local energy levels

Proved classification

QMA-complete

Past the crossing, the singlet is the unique local ground state.

Singlet coefficient s 1.50

−0.52

∈ BQPvia EPREPR∈ BQPStoqMANPQMA

Local energies are exactly computed; complexity labels are theorem annotations, not inferred by the simulation.

EPR and EPR* are in BQP. Rayudu and Takahashi’s Lee–Yang gap theorem gives the quantum upper bound, following the framework of Wong, Bravyi, Gosset, and Liu. Whether EPR* is in BPP remains open. [16][17]

Lee–Yang bound, classification, and singlet conjecture

This one-parameter slice has energies −s, −1, 0, 0. The labels refer to worst-case many-body promise problems, not the difficulty of diagonalizing this four-dimensional matrix. At s = 0 the interaction is locally unitarily equivalent to the EPR interaction. [5]

Why the Lee–Yang theorem applies to EPR*

For the EPR* family in Q14, direct expansion gives

K_b=\frac b2 I+\frac12(XX+bYY-ZZ),\qquad -1\le b\le1.

Removing the known scalar, each positive-weight edge satisfies the Lee–Yang condition $w^z\ge\max(|w^x|,|w^y|)$ for $-w^zZZ+w^xXX+w^yYY$ . This identification is our application of their theorem.

Rayudu–Takahashi use a uniform field $-h\sum_i Z_i$ with gap at least $h/4$ . Adiabatic preparation and energy estimation, with inverse-polynomial $h$ and energy perturbation at most $nh$ , give BQP containment. This does not require a zero-field gap or prepare a prescribed exact zero-field ground state.

Marwaha and Sud’s singlet conjecture says that, for a symmetric two-local interaction K and any triplet state τ, replacing K by K − |τ⟩⟨τ| gives a positive-weight Hamiltonian problem that is at most as hard. This is a conjecture about making a triplet more favorable. It is stronger than the plotted classification, and its consequences include StoqMA = NP.

The plotted family is field-free and Bell-diagonal. Exchange symmetry alone does not make every interaction Bell-diagonal. The named conjecture is attributed in its source’s broader wording.

Q13 / ConjectureThe singlet conjectureFor an exchange-symmetric interaction, does subtracting a triplet projector ever increase the complexity of its positive-weight local-Hamiltonian problem?

Figure 05 / Combining resourcesThree qubits

What can a cross-pair measurement prepare?

Start with independent pieces. Measure a pair that joins them, and compare the two possible outputs. Click another edge to continue with the resulting state.

P_s^{12}\otimes I_3/2

Line width: probability that this pair measures singlet

Measure pair

Available now · a singlet and a mixed qubit

Probabilities are conditional on the current state. Choosing a branch retains its state; sampling chooses according to the probabilities shown.

The input resources and the measurement

A mixed qubit has state $I/2$ . A singlet has state $P_s$ , the rank-one projector onto the antisymmetric subspace. A mixed triplet has state $P_t/3$ , uniform on the symmetric subspace. The mixed triplet is separable; an arbitrary pure triplet need not be.

A fully symmetric triple here means $P_{\mathrm{sym}}^{123}/4$ : the uniform mixture on the four-dimensional spin-3/2 subspace. Every internal pair is triplet. As with the mixed triplet, this state is separable; an arbitrary pure symmetric state need not be.

The blocks are supplied independently. Singlets and mixed triplets can be prepared by measuring mixed pairs and retaining the corresponding outcome. The symmetric triple can be prepared by measuring the total spin of three mixed qubits and retaining spin 3/2, which occurs with probability ½. The displayed probabilities begin after the input blocks have been prepared.

P_{s,t}^{ab}=\frac{I\mp\mathrm{SWAP}_{ab}}{2},\qquad p_o=\mathrm{Tr}(P_o^{ab}\rho),\qquad \rho_o=\frac{P_o^{ab}\rho P_o^{ab}}{p_o}.

For each prepared input, the first measurement across independent blocks gives singlet with probability ¼ and triplet with probability ¾. Each selected qubit is maximally mixed, so their joint reduced state before this measurement is $I/4$ . Correlations with the other qubits determine the output on the full register.

Two singlets

A singlet outcome on 2–3 leaves $P_s^{23}\otimes P_s^{14}$ . A triplet outcome gives a pure state with total spin zero: the spin-one pairs 2–3 and 1–4 are entangled with each other.

Singlet + mixed qubit

A singlet outcome on 2–3 leaves $I_1/2\otimes P_s^{23}$ . A triplet outcome instead leaves a total-spin-½ state with singlet probability ¾ on each of 1–2 and 1–3.

Mixed triplet + mixed qubit

A singlet outcome on 2–3 leaves a singlet on that pair and a mixed qubit on 1. A triplet outcome leaves singlet probability $1/12$ on each of 1–2 and 1–3. Two successive triplet outcomes on overlapping pairs do not make the three qubits fully symmetric.

Symmetric triple + another resource

Start with a symmetric triple on 1–2–3 and measure 3–4. With a mixed qubit on 4, the singlet outcome leaves a mixed triplet on 1–2 and a singlet on 3–4. With a singlet on 4–5, it leaves a symmetric triple on 1–2–5 and a singlet on 3–4: qubit 5 replaces qubit 3 in the symmetric block.

With a mixed triplet on 4–5, the singlet outcome again factors off a singlet on 3–4, but the remaining triple is not fully symmetric. Its total-spin weights are $p_{1/2}=4/9$ and $p_{3/2}=5/9$ . In every case, pair 1–2 remains triplet because the measurement acts on neither qubit.

Two mixed triplets

A singlet outcome on 2–3 leaves $P_s^{23}\otimes\left(P_s^{14}/3+2P_t^{14}/9\right)$ . A triplet outcome creates entanglement across $2\mid134$ even though every reduced two-qubit state is separable. Pair probabilities describe predictions for pair measurements; they do not describe all the correlations in the state.

Q32 / Research directionCan intermediate state resources guide an STP compiler?Schedule a finite library of STP state conversions, reuse states from unsuccessful attempts, and bound the resulting compilation cost.