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Let $K_i$ generate Lorentz boosts with the same exponential convention as the rotation generators of rotations $J_i$. The Lie algebra of the Lorentz group isConsequently non-collinear Lorentz boosts do not close among themselves: the product contains a rotation, the Wigner rotation. For an explicit counterexample, in coordinates $(ct,x,y)$ write $c_a=\cosh a$, $s_a=\sinh a$ and similarly for $b$. ThenA pure rotationless Lorentz boost has a symmetric matrix, whereas this product has spatial entries $(xy)=0$ and $(yx)=s_a{s}_b\ne 0$ for $a,b\ne 0$. It is therefore a concrete product of two pure Lorentz boosts that is not a pure Lorentz boost. Collinear Lorentz boosts are the special case that do form a one-parameter group, with additive rapidities.

Allowing all spatial directions and closure generates both the $K_i$ and, through their commutators, all $J_i$. The smallest connected group containing all these Lorentz boosts is the proper orthochronous Lorentz groupIt contains arbitrary compositions of spatial rotations and Lorentz boosts continuously connected to the identity. It has six continuous parameters. Here “smallest” refers to the subgroup generated by all continuous pure Lorentz boosts, not to an arbitrary collection already including discrete reversals.

The full Lorentz group $O(1,3)$ has four connected components, distinguished by the sign of the determinant and the direction of time. Representatives are the identity, parity $P=\mathrm{diag}(1,-1,-1,-1)$, time reversal $\mathsf{T}=\mathrm{diag}(-1,1,1,1)$, and $P\mathsf{T}=-I$. ThusMultiplication by these discrete transformations relates the components algebraically; no continuous path within the Lorentz group joins different components. Translations are not required for closure here; adding them gives the larger Poincaré group.

Set $C(\beta )=[A,e^{-\beta B}]$. The product rule givesMultiplying by the integrating factor $e^{\beta B}$ and integrating yields the commutator of an operator with an exponential:Equivalently, differentiate $e^{\beta B}A{e}^{-\beta B}-A$ directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.

The given real symmetric matrix is Hermitian and has trace one. A spectral decomposition supplies its positivity without relying on diagonal entries alone:The remaining normalized eigenvector is $(-1,1,1{)}^{\mathsf{T}}/\sqrt{3}$ with eigenvalue zero. Consequently the spectrum is $\{3/4,1/4,0\}$, every eigenvalue is nonnegative, and the matrix is permissible. Its purity is $(3/4{)}^2+(1/4{)}^2=\boxed{5/8}$, so it is mixed.

Use normalized instantaneous modes ${\phi }_p(x,t)=\sqrt{2/L}\sin (p\pi x/L)$ and expand $\Psi ={\sum }_p{a}_p{e}^{-i{\vartheta }_p}{\phi }_p$. Differentiating at fixed $x$ givesFor $n\ne p$, integration usesso the moving-wall derivative coupling isFor $n=p$ the normalization derivative and the remaining integral cancel exactly, giving zero. The moving integration endpoint contributes no boundary term because every mode vanishes there. ThereforeThe real coupling matrix is antisymmetric, as required by conservation of total probability.

Set $P=a^{†}{b}^{†}$ and $Q=ab$. Independent bosonic canonical commutation relations give $[Q,P]=N_a+N_b+1=2K_3$, while $[K_3,P]=P$ and $[K_3,Q]=-Q$. Substituting $K_1=(P+Q)/2$ and $K_2=(P-Q)/(2i)$ yieldsThereforewhich matches the matrix Lie algebra. This boson-pair realization of SU(1,1) differs from the number-conserving Schwinger boson representation of $SU(2)$.

For general spin $S$, use dimensionless generators $S^j$ obeying $[S^i,S^j]=i{ϵ}_{ijk}{S}^k$. The two exponentials represent the same group elements as for spin $1/2$, now in the $(2S+1)$-dimensional representation of $SU(2)$. A representation preserves products, sowith exactly the same axis and angle found in (c); after $N$ periods the angle is $N\vartheta$. One must not substitute higher-spin matrices into the two-term Pauli exponential in (b), which used a property special to spin $1/2$.

The dimension and action on spin states change, but the spin-independent axis-angle composition does not. In particular,Integer-spin states return with positive sign and half-integer-spin states with negative sign. spin expectation vectors transform as spatial vectors in either case, while a general higher-spin state also carries higher multipole information.

Choose orthonormal, flux-normalized incoming and outgoing channels and include all channels of this lossless scattering problem. Conservation of probability requires $\Vert \beta {\Vert }^2=\Vert \alpha {\Vert }^2$ for every input vector. With $\beta =S\alpha$,For a square channel matrix this also gives $S{S}^{†}=I$. A submatrix obtained by omitting loss channels need not be unitary, but that is not the stated two-channel model.

The given real Hadamard matrix is symmetric, unitary and its own inverse. HenceThis same transformation preserves the canonical commutation relations for bosons and anticommutation relations for fermions. To extract probabilities one must use normalized number states, including their factorial factors, even when the supplied input expression is unnormalized.

The defining condition isThe entries are complex and the matrices are invertible; products and inverses retain determinant one. There is no unitarity condition $S^{†}S=I$ in general. Four complex entries subject to one complex determinant equation give three complex, or six real, local group parameters.

The association between four real coordinates and Hermitian two-by-two matrices is a real-linear bijection. Congruence by $S$ preserves Hermiticity, so $X^{\prime }=SX{S}^{†}$ defines another real four-vector depending linearly on the original. Its determinant isThus the induced invertible linear map preserves the Minkowski quadratic form:The maps obtained continuously from the identity are proper and orthochronous. For example, a future timelike vector gives a positive-definite $X$, which remains positive definite under invertible congruence; its time direction is preserved. The matrices $S$ and $-S$ induce the same four-vector transformation, illustrating the Lorentz spinor double cover.

The source uses $x+iy$ in the upper-right entry, corresponding to $X=ctI+x{\sigma }_x-y{\sigma }_y+z{\sigma }_z$ for the standard Pauli matrices. This is a permissible basis convention, but its minus sign in the $y$ basis must also be retained in the spinor identification.

For nonzero $\chi$, the outer product $X=\chi {\chi }^{†}$ is rank-one Hermitian positive semidefinite, so $\det X=0$ and $ct={\chi }^{†}\chi /2>0$. It therefore represents a future-directed null vector. Conversely any nonzero future-null Hermitian $X$ has eigenvalues $2ct$ and zero, so its positive eigenvector multiplied by $\sqrt{2ct}$ supplies $\chi$. The phase $\chi \mapsto e^{i\zeta }\chi$ leaves $X$ unchanged. A past-directed null vector needs $X=-\chi {\chi }^{†}$, rather than the positive outer product alone.

For the printed normalized spinor, direct multiplication yieldsComparing with the printed $x+iy$ convention givesIts usual spatial azimuth is $-\varphi$, not $\varphi$. The source's asserted usual-angle identification has a sign inconsistency. Either conjugating the two spinor phases or replacing the upper-right matrix entry by $x-iy$ restores the usual azimuth $\varphi$. Neither alteration is made silently here. A further positive scale factor sets the magnitude of an arbitrary future-null vector; the displayed normalized spinor fixes $ct=1/2$ in the chosen coordinate units.

The spinor transformation is ${\chi }^{\prime }=S\chi$, with $S=\mathrm{diag}(e^{\kappa /2},e^{-\kappa /2})$. Its squared norm becomesDividing by $\sqrt{\mathcal{R}}$ isolates the new direction without discarding the physical scale factor in $X^{\prime }$. The lower-to-upper component ratio givesEquivalently,This is relativistic aberration for the active positive-$z$ boost specified in the question. At the poles, azimuth is undefined rather than a measurable transformed variable. Under the literal matrix convention from (c), the actual azimuth $-\varphi$ is likewise unchanged. The time coordinate is multiplied by $\mathcal{R}$; normalizing a spinor for its direction must not erase that boost-induced change of null-vector scale.

Define $\alpha =\hslash T/(2I)$. The free propagator acts diagonally on each angular harmonic:The kick multiplies by the Jacobi-Anger seriesMultiply the two series and collect the output harmonic $n=m+\ell$. The one-step coefficient update iswhere both sampling times are just after their respective kicks. Free evolution supplies the input-index phase before the kick mixes angular momenta.

The source uses the unnormalized basis functions $e^{im\theta }$. Thus a unit-norm wavefunction satisfies $2\pi {\sum }_m|c_m{|}^2=1$. If the basis is instead $e^{im\theta }/\sqrt{2\pi }$, its amplitudes have sum of squared moduli one and obey the same matrix update. Mixing these normalization conventions would change an energy expectation incorrectly.

Assume the composite pure state has unit norm. Put its amplitudes into the matrixThe partial trace contracts the same $B$ index between ket and bra:ThusIt is Hermitian, positive semidefinite and has trace one. Its determinant is particularly useful:This follows from multiplicativity of the determinant, or by expanding the two-by-two expression. If the supplied amplitudes were not normalized, the physical reduced state would be $C{C}^{†}/\mathrm{tr}(C{C}^{†})$ instead.

A matrix $R$ belongs to the special orthogonal group in four dimensions precisely whenDifferentiating the orthogonality relation along a path $R(ϵ)=I+ϵX+O({ϵ}^2)$ gives $X^{\mathsf{T}}+X=0$. Thus its Lie algebra consists of real antisymmetric four-by-four matrices. Such a matrix has one free entry for each unordered coordinate pair, giving dimension $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$.

Each displayed generator has a single two-coordinate antisymmetric block. The $K$ matrices cover planes $(1,2),(1,3),(1,4)$; the $L$ matrices cover $(3,4),(2,4),(2,3)$, with their printed orientations. These are all six distinct planes. Therefore the generators are linearly independent and span the full antisymmetric space: they form a basis of $\mathfrak{so}(4)$.

Conversely, exponentiating any real antisymmetric $X$ gives $(e^X{)}^{\mathsf{T}}{e}^X=I$ and $\det (e^X)=e^{\mathrm{tr}X}=1$. This verifies that the infinitesimal matrices generate proper rotations rather than reflections.

In the printed phased Bell basis, the actions on its four kets areEach matrix is purely imaginary and antisymmetric. Let $T$ have these four basis kets as columns in the computational basis. The infinitesimal generators of the standard rotation $U_A=\exp [-i\theta \mathbf{n}\cdot {\mathbf{\sigma }}_A/2]$ areUsing the displayed actions gives the phase-convention-specific identitiesThey are real antisymmetric matrices. A finite $SU(2)$ rotation on $A$ therefore acts on the magic-basis coefficient vector as $b\mapsto Ob$, with $O\in SO(4)$. ConsequentlyIt preserves concurrence and hence preserves the reduced eigenvalues and entanglement entropy. A general $U(2)$ operation adds a common phase to $b$; that changes $q$ by a phase of twice the angle but leaves $|q|$ unchanged. The real-rotation description of local transformations is a characteristic property of the magic basis.^[ref]

As an independent direct check, a local unitary sends ${\rho }_A$ to $U_A{\rho }_A{U}_A^{†}$, which is a unitary similarity transformation and cannot change its spectrum. This local-unitary invariance of entanglement entropy also explains why merely rotating one subsystem cannot create entanglement.

Put $X={\sum }_s{a}_s^{†}{b}_s$ and $Y=X^{†}$, and take real hopping amplitude $t$ and real $\mathcal{E}$. Write the anti-Hermitian ansatz with its ordering visible:Using $Yf(\Delta )=f(\Delta +2)Y$, its commutator with $\mathcal{E}$ isBoth terms equal their required hopping coefficients $-t$ ifThe shifted expression is then $f(\Delta +2)=t/[\mathcal{E}(\Delta +2)-\mathcal{E}(\Delta )]$. For a transition from difference $d$ to $d+2$, the left-ordered factor evaluates as $f(d+2)=t/[\mathcal{E}(d+2)-\mathcal{E}(d)]$, the final-minus-initial energy denominator.

The formula describes occupation-dependent virtual hopping on actual transitions with nonzero energy gaps. Perturbation theory also requires the hopping matrix element, including its bosonic occupation factors, to be small compared with that gap. The resonance obstruction to perturbative elimination at a zero denominator on a nonzero hop requires degenerate perturbation theory. Values on unreachable states whose hopping amplitude vanishes need not determine the generator.

The fixed-particle-number sector is spanned by $|1⟩=c_1^{†}|0⟩$ and $|2⟩=c_2^{†}|0⟩$. In this basis,The relevant quantum Gibbs state is normalized within this two-dimensional sector, not over all particle numbers. With $\beta =(k_{B}T{)}^{-1}$ and $x=\beta t$, the exponential of a Pauli vector givesThereforeIts eigenstates are $|\pm ⟩=(|1⟩\pm |2⟩)/\sqrt{2}$, with energies $\mp t$ and probabilities $p_{\pm }=e^{\pm x}/(2\cosh x)$. These are nonnegative and sum to one, so the matrix is Hermitian and has trace one. Its purity of a density operator isFor $t\ne 0$, the zero-temperature limit is the pure ground-state projector,with purity one. As $T\to \infty$, $x\to 0$ and $\rho \to I/2$, with purity $1/2$. At every strictly positive finite temperature and finite $t$, both eigenvalues are nonzero, so the state is mixed. If $t=0$, the two levels are degenerate and the Gibbs state is $I/2$ at all temperatures, including its zero-temperature limit.

Using dimensionless Von Neumann entropy, $S=-\mathrm{Tr}(\rho \ln \rho )$, givesFor nonzero $t$, $S\to 0$ as $T\to 0$, while $S\to \ln 2$ as $T\to \infty$. More precisely, the high-temperature expansion is $S=\ln 2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim (1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.

Fix the mode ordering with the $b^{†}$ operator before the $c^{†}$ operator, as in the given state. In the sector with one fermion in each subsystem, use the ordered basisThe state vector is $v=(1/2,1/\sqrt{2},0,1/2{)}^T$, whose squared norm is one. Thus the global density matrix is the rank-one projectorIt has trace one and satisfies ${\rho }_{BC}^2={\rho }_{BC}$, so the global state is pure.

Take the partial trace over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt{2}{)}^T$ for $B_1$ and $v_2=(0,1/2{)}^T$ for $B_2$. Therefore the reduced density matrix isIts determinant is $1/16$, giving eigenvaluesBoth are positive; in particular, $\mathrm{Tr}{\rho }_C^2=7/8<1$, so the reduced state is mixed despite global purity. The entanglement entropy is obtained by diagonalizing this reduced matrix and evaluatingFor a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.

Put $\beta =(k_{B}T{)}^{-1}$ and label the two spin values in the order $+1,-1$. Assign each bond its Boltzmann weight,The symmetry of the bond energy makes $V$ a real symmetric matrix. For a fixed configuration the product of successive matrix entries is its Boltzmann weight. Summing over all spins and identifying the last index with the first therefore giveswhereThis replaces a sum over $2^L$ configurations by the powers of a two-by-two matrix. For finite bond energies at $T>0$, every matrix entry is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>0$ and $|{\lambda }_{-}|<{\lambda }_{+}$. Hence the Helmholtz free energy per spin tends to $-k_{B}T\log {\lambda }_{+}$. A diagonal local observable $O$ can be inserted at one site: $⟨O_1⟩=\mathrm{tr}(O{V}^L)/Z_L$. Its limiting average is the expectation in the normalized eigenvector of ${\lambda }_{+}$.

Insert the decorating-spin observable into the one bond that contains it. The decorated-spin occupation insertion uses the matrix:The partition function still contains $L$ decorating-spin sums; one of them has been replaced by this weighted sum. HenceTaking the dominant eigenvector givesThis is the probability that the decorating site is in either nonzero state. It tends to $2/3$ at high temperature, when its three states are equally likely, and to one at low temperature, when the nonzero aligned states gain bond energy. Unlike the magnetization, this even observable is not forced to vanish by spin reversal.