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The supplied matrices are the chiral gamma-matrix representation. The Pauli matrix multiplication law implies $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication gives $({\gamma }^0{)}^2=I_4$, $\{{\gamma }^0,{\gamma }^i\}=0$, and $\{{\gamma }^i,{\gamma }^j\}=-2{\delta }^{ij}{I}_4$. Hence their Clifford algebra relation isSince partial derivatives commute, multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ givesThus every component of the Dirac spinor satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation with the same mass.

For a Lorentz transformation $x^{\prime }=\Lambda x$, set ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$. The derivative transforms as ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. With the given spinor-matrix identity,This proves the Lorentz covariance of the Dirac operator explicitly.

For parity, let $x_P=(x^0,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. Spatial differentiation brings a minus sign, which is canceled by ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$. ThereforeSince ${\gamma }^0$ is Hermitian and squares to one, the Dirac adjoint transforms asThe chirality matrix ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with ${\gamma }^0$. The parity transformation of a Dirac bilinear consequently givesThe first bilinear is a Lorentz scalar even under parity, and the second is a pseudoscalar.

In this representation, ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$. Its square is one, so the chiral projectors $P_{\pm }=(1\pm {\gamma }^5)/2$ are idempotent, mutually orthogonal and sum to the identity. They select ${\psi }_L=P_{-}\psi$ and ${\psi }_R=P_{+}\psi$, the two Weyl spinors. Because ${\gamma }^{\mu }{P}_{\pm }=P_{\mp }{\gamma }^{\mu }$, a solution of the massless Dirac equation has each chiral component solving that equation separately. In two-component form,The two chirality sectors are thus independent under massless propagation and proper Lorentz transformations. A nonzero Dirac mass term couples them, while parity exchanges them. This explains why the projectors are particularly useful in the massless theory.

Keep the printed coupling $g$ symbolic and use the Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$ and ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Varying the action with respect to the independent adjoint variable gives the Euler-Lagrange field equationVarying with respect to $\psi$ and integrating its derivative term by parts gives the corresponding left-acting adjoint Dirac equationNow differentiate the axial current $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. The first equation supplies ${\gamma }^{\mu }{\partial }_{\mu }\psi =-i(m+g\varphi {\gamma }^5)\psi$, and the second supplies $({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }=i\bar{\psi }(m+g\varphi {\gamma }^5)$. Since $\{{\gamma }^{\mu },{\gamma }^5\}=0$ and $({\gamma }^5{)}^2=1$,Therefore the axial-current divergence for a pseudoscalar Yukawa interaction isBoth the fermion mass and the interaction can break the axial symmetry. This is the classical field-equation identity being asked for, not a claim that quantum composite operators need no renormalization.

There is a convention issue in the original PDF: its interaction is written without an explicit $i$. The Hermiticity of a pseudoscalar Dirac bilinear follows directly fromFor real $\varphi$, the printed interaction is Hermitian if $g^{\ast }=-g$, rather than if $g$ is real. The PDF does not specify that $g$ is real, so a consistent physical convention is $g=i{g}_P$ with real $g_P$. Its interaction is $-i{g}_P\varphi \bar{\psi }{\gamma }^5\psi$, and the divergence becomes $2im\bar{\psi }{\gamma }^5\psi -2g_P\varphi \bar{\psi }\psi$. If $g$ were instead required to be real, the printed action would be non-Hermitian; the variational formulas above would still be formal calculations, but the variational adjoint equation would not be its Hermitian-conjugate equation. No missing factor is silently inserted into the solution.

Continue with metric signature $+---$. The Pauli matrices obey $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication in the chiral gamma-matrix representation givesandHence $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$, the required Clifford algebra. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ yieldsThe antisymmetric part of ${\gamma }^{\nu }{\gamma }^{\mu }$ drops out because partial derivatives commute. Therefore every component of the Dirac spinor satisfies the Klein-Gordon equation. The converse is not true: four arbitrary scalar solutions need not obey the first-order Dirac equation.

The Lorentz group consists of real invertible matrices $\Lambda$ preserving the Minkowski metric: ${\Lambda }^{T}g\Lambda =g$. Writing $\Lambda =I+\omega +O({\omega }^2)$ gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, so there are six continuous parameters: three spatial rotation angles and three boost rapidities. The proper orthochronous Lorentz group is the connected component with determinant $+1$ that preserves time orientation. Parity and time reversal are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.

For the real-generator convention used here, vector generators can be writtenTake ${\Omega }_{\rho \sigma }=-{\Omega }_{\sigma \rho }$ and use the same parameters in the spinor representation. The Lorentz-spinor generators from a Clifford algebra areTo verify the Lorentz algebra, first commute a generator with one gamma matrix:For example, this follows by moving ${\gamma }^{\tau }$ through each product using the Clifford anticommutator. Applying the commutator derivation rule to $\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$ then givesNo extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.

The same gamma commutator provesIndeed, to first order $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }+\frac{1}{2}{\Omega }_{\rho \sigma }[{\gamma }^{\mu },M_{\mathrm{sp}}^{\rho \sigma }]$, exactly the infinitesimal vector action defined above. Exponentiating proves the finite relation. The spinor field transforms as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $x^{\prime }=\Lambda x$; combining this relation with ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$ preserves the Dirac equation. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover spin representation, rather than a single-valued ordinary representation of the Lorentz group itself.

For a boost in direction $i$, $M_{\mathrm{sp}}^{0i}=\frac{1}{2}{\gamma }^0{\gamma }^i$ is Hermitian, with eigenvalues $\pm \frac{1}{2}$. A real rapidity $\eta$ therefore produces eigenvalues $e^{\pm \eta /2}$ in $S$, whose moduli are not one. The finite-component spinor representation is not unitary for boosts, although the rotation matrices are unitary. The Lorentz group is noncompact; this does not forbid the unitary infinite-dimensional action on the physical Hilbert space of states.

Nevertheless, since $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$,This Dirac spinor pseudo-unitarity implies ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$, and consequentlyThe Dirac adjoint is precisely the adjoint needed to form this Lorentz scalar; ordinary ${\psi }^{†}\psi$ alone is not a scalar.

Let ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\mathrm{diag}(-I_2,I_2)$. It anticommutes with every gamma matrix and commutes with every $M_{\mathrm{sp}}^{\rho \sigma }$, so $S^{-1}{\gamma }^{5}S={\gamma }^5$. Thus the axial current transforms as a four-vector under proper Lorentz transformations:Under parity it is an axial vector, acquiring the additional pseudovector sign. Contracting two axial currents cancels that sign and uses the invariant metric. Therefore $j_{5\mu }{j}_5^{\mu }$ is compatible with Lorentz invariance, including parity invariance of this contraction. Lorentz invariance of the interaction does not require the axial current to be conserved.

Use signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

Variation with respect to the Dirac adjoint, the Dirac field, and the real scalar field respectively givesThe derivative in the adjoint equation acts to the left; its sign follows by integrating $i\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\delta \psi$ by parts. The Yukawa interaction supplies a spacetime-dependent effective fermion mass and a scalar source.

Use $/p={\gamma }^{\mu }{p}_{\mu }$ and the convention $S_{fi}^{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p_f-p_i)\mathcal{M}$. The momentum-space Feynman rules are: a scalar internal line contributes $i/(p^2-m^2+i0)$; an oriented fermion line contributes $i(/p+m)/(p^2-m^2+i0)$; and each scalar-fermion vertex contributes $-ig$ times the identity in spinor space. Impose four-momentum conservation at each vertex. Incoming and outgoing fermions supply $u$ and $\bar{u}$, while incoming and outgoing antifermions supply $\bar{v}$ and $v$; scalar external legs supply one. Loop momenta are integrated with $d^{4}p/(2\pi {)}^4$, a closed fermion loop contributes a minus sign, and graph symmetry factors are included. Relative signs between distinct contractions of identical external fermions follow from their anticommutation relations. These specify the Feynman rules also beyond the tree approximation.

Label incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, with all external particles on shell. Write the Mandelstam variables as $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, and abbreviate $D_r=r-m^2+i0$. Spin indices on $u_i,v_i$ are implicit. The six required tree-level Feynman diagrams are:

For two incoming fermions, the two diagrams exchange a scalar in the $t$ and $u$ channels. With external state ordering $b_1^{†}{b}_2^{†}|0⟩$ and $b_3^{†}{b}_4^{†}|0⟩$, the result isEach scalar-exchange contraction has two factors $-ig$ and one scalar Feynman propagator. Exchanging the final fermions reverses the sign, as required by their identical-particle statistics.

For a fermion and antifermion, there is $t$-channel scalar exchange and $s$-channel annihilation. Take both initial and final states ordered as fermion creator followed by antifermion creator. ThenThe relative minus is not optional. One way to track it is the antifermion sign of a normal-ordered bilinear: the antifermion scattering part of $:\bar{\psi }\psi :$ is $-d^{†}d\bar{v}v$, whereas its annihilation part is $db\bar{v}u$. Thus the exchange contraction has the opposite fermionic sign to the annihilation contraction before multiplying by $(-ig{)}^{2}i/D_r$. A different overall phase convention for external states changes the common sign of this amplitude, but cannot change the relative sign.

For fermion-scalar scattering, the fermion can absorb the incoming scalar before emitting the outgoing one, or emit first and absorb afterwards. The internal momenta are $p_1+p_2$ and $p_1-p_4$ respectively, givingBoth orderings have the same sign: there is only one open fermion line and no exchange of identical external fermions. There is no scalar-exchange $t$ diagram because this Yukawa interaction has no three-scalar vertex. These amplitudes describe tree scattering in Yukawa theory.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

The sources are independent odd Grassmann variables and anticommute with the Dirac field. Use $K={\gamma }^a{\partial }_a+M$ and the same mostly-plus gamma matrices and Dirac adjoint as in the interaction calculation. The inverse $K_F^{-1}$ is selected with vacuum Feynman i-epsilon prescription boundary conditions. With integral kernels and spinor contractions understood, the exponent can be completed to a square:Translations preserve the Berezin integral, so the Gaussian generating functional for a Dirac field isThe positions of the sources matter. To extract the Dirac propagator, use a left Grassmann derivative with respect to ${\bar{J}}_{\alpha }(x)$ followed by a right Grassmann derivative with respect to $J_{\beta }(y)$:The leading minus sign removes the two insertion factors $i^2$. It can also be checked by differentiating the quadratic source exponential: its ordered second derivative is $-iZ[0,0]K_F^{-1}$.

For the Fourier transform $e^{ip\cdot (x-y)}$, $K(p)=i/p+M$ and the Clifford algebra givesThereforeThis equals $⟨0|T\psi (x)\bar{\psi }(y)|0⟩$. The fermionic time ordering is explicitlywith the minus sign supplied by exchanging odd fields. The poles put positive energy forward in time and negative energy backward, which is the antiparticle contribution. Finally, $K_x{S}_F(x,y)=iI{\delta }^{(4)}(x-y)$ checks the numerator and overall sign. All formulas are distributional limits with the indicated boundary prescription.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.