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For a qubit density matrix $\rho =(I+n\cdot \sigma )/2$ with nonzero Bloch vector, $\hat{n}=n/\Vert n\Vert$ and the multiplication rule for the Pauli matrices give these orthogonal rank-one spectral projectors, with eigenvalues $(1\pm \Vert n\Vert )/2$. Their Bloch vectors are opposite unit vectors $\pm \hat{n}$. At $n=0$ the state is maximally mixed and the direction is undefined; any orthonormal eigenbasis gives an antipodal pair, but no pair is distinguished by the state.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

For signature $+---$, use ${\gamma }^0=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ and ${\gamma }^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$, where ${\sigma }^i$ are Pauli matrices. Their Clifford algebra follows directly from the Pauli anticommutator. The chirality matrix is ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$, separating the two chiral components of a Dirac spinor.

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

The four Pauli matrices form an orthogonal basis of $2\times 2$ matrices for the Hilbert-Schmidt inner product, since $\mathrm{Tr}({\sigma }_{\mu }{\sigma }_{\nu })=2{\delta }_{\mu \nu }$. A Hermitian density matrix therefore has real coefficients in this basis, and its trace-one condition fixes the identity coefficient. ThusThe multiplication rule for the Pauli matrices ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ implies $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I$. Its trace is zero, so for $n\ne 0$ its eigenvalues are $\pm \Vert n\Vert$; for $n=0$ it vanishes. HencePositivity is equivalent to $\Vert n\Vert \le 1$. Conversely every real vector in this unit ball defines a positive trace-one matrix by the boxed formula. Therefore $\Phi (\rho )=n$ is a one-to-one correspondence with the Bloch ball, with the displayed inverse. Each coordinate is linear in $\rho$, proving the affine property under any convex mixture.

A density matrix is pure precisely when its eigenvalues are $1,0$, equivalently when ${\rho }^2=\rho$. The eigenvalue formula makes this equivalent to $\Vert n\Vert =1$. Alternatively $\mathrm{Tr}{\rho }^2=(1+\Vert n{\Vert }^2)/2$. Thus the pure states correspond exactly to the Bloch sphere.

For $n\ne 0$, put $\hat{n}=n/\Vert n\Vert$. The spectral projectors for the larger and smaller eigenvalues areThey are orthogonal rank-one projections, have sum $I$, and give $\rho ={\lambda }_{+}{P}_{+}+{\lambda }_{-}{P}_{-}$. For normalized eigenvectors in this nondegenerate case they equal $|{\varphi }_1⟩⟨{\varphi }_1|$ and $|{\varphi }_2⟩⟨{\varphi }_2|$. Their Bloch vectors are therefore $\boxed{n_1=n/\Vert n\Vert ,n_2=-n/\Vert n\Vert }$, the antipodal eigenprojectors of a qubit density matrix.

At $n=0$, the state is $I/2$ and both eigenvalues coincide. There is no distinguished direction and the printed directional expression is undefined. Any orthonormal eigenbasis gives an antipodal pair of pure-state Bloch vectors, since its two projectors sum to $I$, but that pair can point in any direction. The nonzero-$n$ qualification is essential. The PDF denominators are $\Vert n\Vert$, not the converted TeX denominators involving $n_1,n_2$.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

The Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

The resource here is one copy of a two-qubit Werner state, with the physical range $1/2<F\le 1$. The impossibility must include successful conditional branches of local operations and classical communication; average monotonicity alone would not exclude probabilistic enhancement.

For a normalized two-qubit pure state $|\chi ⟩={\sum }_{i,j}{M}_{ij}|ij⟩$, define its concurrence as $C(\chi )=2|\det M|$. Its mixed-state concurrence of two qubits is the convex roof extensionCombining near-optimal decompositions proves convexity. A separable quantum state has zero concurrence, because it has a decomposition into product vectors with zero coefficient determinant.

First establish the concurrence of a two-qubit Werner state. A pure state with squared Schmidt coefficients $\lambda ,1-\lambda$ has $C=2\sqrt{\lambda (1-\lambda )}$. In this Schmidt basis a maximally entangled state has coefficient matrix $V/\sqrt{2}$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$, because $|V_{ii}|\le 1$. Thus the singlet overlap satisfiesAveraging any decomposition of $W_F$ gives $C(W_F)\ge 2F-1$.

For the reverse bound use the orthonormal triplet vectorsEach is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C({\chi }_{j,\pm })=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence

Every fully specified record of an LOCC protocol gives a product Kraus operator $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output asFor invertible $A,B$, the concurrence scaling under invertible local filters isHere is a derivation of that scaling. Each pure coefficient matrix transforms to $AM{B}^{\mathsf{T}}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized concurrence is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_j{p}_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of ${\rho }^{\prime }$, so taking both infima proves equality. If either local filter has rank one, the output is a separable quantum state and the desired bound is immediate.

It remains to show that the filter normalization cannot make this ratio exceed one. Use the Pauli matrices representationWrite the positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, so $a_0\ge |a|$, $b_0\ge |b|$. Taking traces givesFor completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$, and both sides are nonnegative. This proves the single-copy Werner filtering normalization bound, and hence

Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F^{\prime }}$ would therefore obey $2F^{\prime }-1\le 2F-1$, contradicting $F^{\prime }>F$. Thus even heralded single-copy LOCC cannot produce the specified improved Werner state. This proves that single-copy Werner filtering cannot increase concurrence. It does not forbid collective entanglement distillation from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.