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For two identical spin-one-half fermions with no additional antisymmetric internal factor, the antisymmetric spin-one-half singlet state combines with an even-$\ell$ symmetric orbital state, while the symmetric spin-one-half triplet state combines with an odd-$\ell$ antisymmetric orbital state.

For spin $1/2$, $\mathbf{S}=\hslash \mathbf{\sigma }/2$. The Pauli matrices satisfy ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric term vanishes on contraction with $n_i{n}_j$, and $|\mathbf{n}|=1$, so $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Separating the even and odd terms of the exponential series givesFor the single-particle Hamiltonian, unitary time evolution consequently acts asFor two spins the joint eigenstates of total angular momentum are the spin-one-half triplet statewith $J^2=2{\hslash }^2$ and $J_3=m\hslash$, and the spin-one-half singlet stateThese follow by adding the two spin-$1/2$ angular momenta, or by direct action of the Pauli matrices.

The two terms of the specified Hamiltonian commute, so its evolution factorizes into rotations with opposite angles on the two tensor factors. The unique initial $J_3=\hslash$ state is $|\uparrow \uparrow ⟩$. With $c=\cos (\lambda t/2)$, $s=\sin (\lambda t/2)$, its evolved state isThe singlet amplitude is $i\sqrt{2}cs$, giving the requested probability

The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. ThenThus $|\psi ⟩$ is unchanged. It is the spin-one-half singlet state, so joint rotations leave it invariant. Equivalently, direct use of the Pauli matrices showsIt is therefore an eigenstate of every Cartesian component of the combined spin, with eigenvalue zero.

The Affleck--Kennedy--Lieb--Tasaki state places two virtual spin-one-half degrees of freedom at every site, puts neighboring virtual spins into singlets, and projects the two virtual spins at each site onto their symmetric spin-one triplet. Its local tensor is equivalently proportional to ${\sigma }_x,{\sigma }_y,{\sigma }_z$ in the spin-one Cartesian basis, and its Affleck--Kennedy--Lieb--Tasaki parent Hamiltonian is

The one-dimensional cluster state is the simultaneous $+1$ eigenstate of the commuting stabilizersOn a periodic chain it is obtained by applying a controlled-$Z$ gate on every neighboring pair of the product state $|+{⟩}^{\otimes N}$.

After suitable blocking, the two states have the same nontrivial projective virtual symmetry of a matrix product state for the protecting group ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$. The two virtual symmetry generators can be represented by anticommuting Pauli matrices, so they realize the nontrivial projective class in group cohomology. Consequently the AKLT and cluster states can be connected by a symmetry-preserving gapped path, or equivalently by a symmetry-preserving finite-depth local circuit: this is the Symmetry-protected equivalence of the Affleck--Kennedy--Lieb--Tasaki state and cluster state.

Both states therefore exhibit one-dimensional symmetry-protected topological order. On an open chain their nontrivial virtual representation produces protected edge degrees of freedom and a characteristic degeneracy in the entanglement spectrum of a matrix product state. They do not have intrinsic topological order: if the protecting symmetry is discarded, either state can be connected to a product state by a finite-depth local circuit.

Let $|s⟩=|{\psi }^{-}⟩$ be the spin-one-half singlet state, complete it to an orthonormal basis $\{|s⟩,|e_1⟩,|e_2⟩,|e_3⟩\}$, and chooseEach excited population decays into the singlet. On the matrix-unit basis, $|e_j⟩⟨e_k|$ has decay eigenvalue $-\gamma$ after separating off its contribution to the stationary population, while $|s⟩⟨e_j|$ and its adjoint have eigenvalue $-\gamma /2$. The sole zero mode is $|s⟩⟨s|$, so every initial state converges toThe nonzero eigenvalue closest to the origin is therefore

Taking the sum once over each unordered pair, write the All-to-all Heisenberg model asFor even $N$, the minimum total spin is $S=0$, henceFor odd $N$, $S=1/2$ and $E_0=3/2-3N/2$. If the paper's ${\sum }_{ij}$ counts ordered pairs or includes $i=j$, the corresponding harmless factors and additive constant change, but the minimizing total-spin sector is the same.

Now partition an even number of spins into disjoint pairs and put every pair in a spin-one-half singlet state. Each pair has total spin zero, so their tensor product also has $S=0$ and is itself a ground state. Its internal pair contributes $-3$, while correlations between different singlets vanish, giving $-3N/2$ in total. ThereforeThe large degeneracy is special to equal all-to-all coupling: the Hamiltonian sees only total spin and cannot distinguish different singlet coverings.

The two-spin singlet isSpin measurements along axes $\mathbf{a}$ and $\mathbf{b}$ have correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$, and equal axes give perfect anticorrelation.

For two spin-one-half particles,has two eigenvalues $\alpha {\hslash }^2/4$ and two further eigenvaluesFor positive $\alpha$, the zero-field limit has a unique spin-one-half singlet state ground state and a triply degenerate spin-one-half triplet state excited level.