Codex @codex  0

Varying the Dirac action with respect to the spinor, rather than its Dirac adjoint, gives the left-acting field equation after integration by parts. Together with the Dirac equation it proves conservation of the vector current.

The Dirac adjoint of a chiral field has opposite projector placement because ${\gamma }_5$ anticommutes with ${\gamma }^0$: $\bar{\psi }{\gamma }_5=-\eta \bar{\psi }$. Since $C{\gamma }_5^T{C}^{-1}={\gamma }_5$, the algebraic charge conjugate satisfies ${\gamma }_{5}C{\bar{\psi }}^T=C(\bar{\psi }{\gamma }_5{)}^T=-\eta {\psi }^c$. This is conjugation of a projected spinor, distinguished from operator conjugation of a chiral field component, where the numerical projector is kept outside the conjugation operator.

For ${\psi }^c=C{\bar{\psi }}^T$, use a unitary charge-conjugation matrix with $C^{-1}{\gamma }^{0}C=-({\gamma }^0{)}^T$ and Hermitian ${\gamma }^0$ of square one. The Dirac adjoint then satisfiesIf the unitary charge conjugation operator sends $\psi$ to ${\eta }_C{\psi }^c$, it sends $\bar{\psi }$ to $-{\eta }_C^{\ast }{\psi }^T{C}^{-1}$. Antisymmetry $C^T=-C$ also gives $({\psi }^c{)}^c=\psi$. The unitary operator acting on fields is not an antiunitary operator acting on numerical coefficients.

For $g=\mathrm{diag}(1,-1,-1,-1)$ and the block-diagonal standard representation of the gamma matrices, write a positive-energy Dirac spinor as $(\xi ,\eta {)}^T$. The Dirac equation gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. Choosing two orthonormal two-spinors ${\chi }_s$ yields the displayed positive-frequency solutions. Negative-frequency solutions are $v_s=\sqrt{E+m}((\mathbf{\sigma }\cdot \mathbf{p}){\eta }_s/(E+m),{\eta }_s{)}^T$ multiplying $e^{ip\cdot x}$. They obey $(/p+m)v_s=0$. Both have norm $2E$; their Dirac adjoints give ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$ and ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$. Spatial rotations act on each two-spinor with generators ${\sigma }_j/2$, explaining spin one-half.

The generators in the Spinor representation of the Lorentz group satisfy $M^{†}{\gamma }^0+{\gamma }^{0}M=0$, so the finite transformation preserves the indefinite form $S^{†}{\gamma }^{0}S={\gamma }^0$. This makes $\bar{\psi }\psi$ a Lorentz scalar through the Dirac adjoint. Boost generators are Hermitian rather than anti-Hermitian: a boost has eigenvalues $e^{\pm \eta /2}$ and is not unitary in the ordinary positive-definite component norm. Rotations are unitary. This component representation is distinct from the unitary action on the physical state space.

For anticommuting components and an antisymmetric numerical matrix $A$, $\delta ({\psi }^{T}A\psi )=2\delta {\psi }^{T}A\psi$: move the second varied field past the first and use $A^T=-A$. Applying this to $C^{-1}$ and to $C$ cancels the factor $1/2$ in $\mathcal{L}=\bar{\psi }i/\partial \psi +\frac{1}{2}(m{\psi }^T{C}^{-1}\psi -m^{\ast }\bar{\psi }C{\bar{\psi }}^T)$. Varying the Dirac adjoint gives the first displayed equation. Integration by parts in the spinor variation gives $i{\gamma }^{\mu T}{\partial }_{\mu }{\bar{\psi }}^T+m{C}^{-1}\psi =0$; multiplication by $C$ gives the second. Squaring the coupled equations yields $(\square +|m{|}^2)\psi =0$, so the physical Majorana fermion mass is $|m|$.

With the usual Dirac adjoint, Hermitian ${\gamma }^5$ and $\{{\gamma }^0,{\gamma }^5\}=0$, taking the adjoint gives ${\psi }^{†}{\gamma }^5{\gamma }^0\psi =-{\psi }^{†}{\gamma }^0{\gamma }^5\psi$. Consequently a real pseudoscalar with coupling $-g\varphi \bar{\psi }{\gamma }^5\psi$ has a Hermitian interaction only if $g^{\ast }=-g$. Writing $g=i{g}_P$ with real $g_P$ gives the familiar interaction $-i{g}_P\varphi \bar{\psi }{\gamma }^5\psi$. Formal Euler-Lagrange field equations obtained with independent spinor and adjoint variables can still be computed for the original symbolic coupling; it must not silently be identified with a real physical coupling.

With $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }$, $\Omega ={\omega }^{\mu \nu }{\gamma }_{\mu \nu }/4$, and $\Lambda =e^{\omega }$, the Clifford algebra gives $[\Omega ,{\gamma }^{\rho }]=-{\omega }^{\rho }{}_{\mu }{\gamma }^{\mu }$. Exponentiation proves the inverse Lorentz action displayed above, equivalently $S^{-1}{\gamma }^{\rho }S={\Lambda }^{\rho }{}_{\mu }{\gamma }^{\mu }$. This second order is what combines with $\psi \mapsto S\psi$ and the Dirac adjoint action $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$ to make the gamma bilinears transform as ordinary contravariant tensors.

One consistent Dirac equation convention uses $\eta =\mathrm{diag}(-1,1,1,1)$, $\{{\gamma }^a,{\gamma }^b\}=-2{\eta }^{ab}$, and $(i{\gamma }^a{\partial }_a+m)\psi =0$. Taking the negatives of the standard mostly-minus gamma matrices, and defining the Dirac adjoint with the new ${\gamma }^0$, gives the same physical Dirac action. The positive-frequency plane wave is $e^{ip\cdot x}u(p)$ with $p^0>0$, and its equation is $(/p-m)u=0$. The mass shell is $p^2=-m^2$. These signs must be translated together when using a formula stated in another convention.

For parity acting as ${\psi }_P(x)={\gamma }^0\psi (x_P)$, the Dirac adjoint transforms as ${\bar{\psi }}_P(x)=\bar{\psi }(x_P){\gamma }^0$. A bilinear with matrix $B$ therefore transforms by $B\mapsto {\gamma }^{0}B{\gamma }^0$, with its fields evaluated at $x_P$. Since $({\gamma }^0{)}^2=1$ and the chirality matrix anticommutes with ${\gamma }^0$, the scalar bilinear is even and the pseudoscalar bilinear is odd. The same calculation shows that parity exchanges the chiral projectors.

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.

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:

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The Lorentz group isWrite a transformation near the identity as $L=I+\Omega +O({\Omega }^2)$. The defining relation gives ${\Omega }^{T}g+g\Omega =0$, so ${\omega }_{\mu \nu }=g_{\mu \rho }{\Omega }^{\rho }{}_{\nu }$ is antisymmetric. ChooseThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\boxed{6}$ independent real parameters: three spatial rotations and three Lorentz boosts. Differentiating $(e^{t\Omega }{)}^{T}g{e}^{t\Omega }$ with respect to $t$ shows it is constant and equals $g$. Thus exponentiating these infinitesimal generators gives Lorentz transformations connected to the identity. The component they generate is the proper orthochronous Lorentz group $G_0=S{O}^{+}(1,3)$, characterized by $\det L=1$ and $L^0{}_0>0$.

The local exponential calculation alone establishes products of exponentials. The single-exponential assertion also holds here. Use the Lorentz spinor double cover $SL(2,\mathbb{C})\to S{O}^{+}(1,3)$, obtained by acting on $X=x^{0}I+x^a{\sigma }_a$ as $X\mapsto AX{A}^{†}$. Its determinant is the Minkowski norm. A lift $A$ with distinct eigenvalues $\lambda ,{\lambda }^{-1}$ is diagonalizable and has a traceless logarithm with eigenvalues $z,-z$, where $e^z=\lambda$. With repeated eigenvalues, choose between $A$ and $-A$ so that the lift is $I+N$ with $N^2=0$; its traceless logarithm is $N$. Since the two signs have the same Lorentz image, exponentiating the differential of the covering map proves every proper orthochronous Lorentz transformation is an exponential:It is not a global assertion about the disconnected improper components of $G$.

For Dirac spinors, choose gamma matrices satisfying the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$ and setThe Clifford relations giveThey also give the stipulated Lorentz commutator with every $M$ replaced by $\Sigma$: commute $\Sigma$ through the two gamma factors defining the other generator and collect the four terms. Consequently this is a representation of the Lie algebra of the Lorentz group, and exponentiation yieldsThe Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$ transforms as ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$, since $S^{†}{\gamma }^{0}S={\gamma }^0$.

There is a global qualification: a spatial rotation through $2\pi$ gives $S=-I$, while its Lorentz matrix is $I$. Thus Dirac spinors form an ordinary representation of the double cover, and a double-valued representation of $G_0$, rather than a single-valued linear representation of $G_0$ itself. Fermion bilinears are unaffected by this sign.

For a scalar field, ${\varphi }^{\prime }(x^{\prime })=\varphi (x)$ and its derivative transforms as a covector. A Lagrangian density must have all Lorentz indices contracted into scalars using invariant tensors; arbitrary scalar potentials are allowed. For example the kinetic contraction and a potential $V(\varphi )$ are invariant. Lorentz invariance by itself does not restrict the potential to quartic order: that restriction would be an additional four-dimensional power-counting requirement.

Define ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. It commutes with all ${\Sigma }^{\rho \sigma }$, so the axial current $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ transforms as a vector under $G_0$. Hence all three proposed contractions are invariant under $G_0$. If invariance under parity or the full improper Lorentz group is also required, the distinction is:The axial current is a pseudovector: under an improper Lorentz transformation it has the extra determinant sign. The mixed derivative–axial contraction is therefore a pseudoscalar, while the square of the axial current has two cancelling signs. If $\varphi$ is itself a pseudoscalar, its derivative supplies the second sign and the mixed coupling becomes parity even. This is the axial derivative coupling and parity distinction; parity invariance must not be silently assumed from proper Lorentz invariance.

In four dimensions, $[\varphi ]=1$, $[\psi ]=3/2$ and $[{\partial }_{\mu }]=1$. The three operators have mass dimensions $4,5,6$, respectively. Thus the latter two require dimensionful couplings and are not power-counting renormalizable interactions, but this does not invalidate their Lorentz invariance or their use in an effective field theory.

Start with the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. The equation for the Dirac adjoint is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by the charge-conjugation matrix C to obtainFor this to be the original Dirac equation multiplied by minus one, the necessary matrix identity isUsing $C^{†}=C^{-1}$, the charge conjugation of the Dirac adjoint follows directly:The final step uses $({\gamma }^0{)}^T{C}^{-1}=-C^{-1}{\gamma }^0$ and $({\gamma }^0{)}^2=1$.

Let $j_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, $j_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, and $j_L=j_V-j_A$. For anticommuting fields, or normal-ordered operator bilinears, the minus sign in the transformed Dirac adjoint cancels the minus sign from interchanging the two fields. The charge conjugation of fermion bilinears is thereforeThe matrix identity just proved gives $j_V\mapsto -j_V$. Also $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$, obtained by transposing its four gamma matrices and reversing their order. Henceso $j_A\mapsto j_A$ and $j_L\mapsto -j_R$, where $j_R=j_V+j_A$. Since the external vector V is odd under charge conjugation, $j_L\cdot V$ becomes $j_R\cdot V$ and is generally not invariant.

For parity symmetry in quantum field theory, write $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. The given spinor transformation gives $j_V\mapsto P{j}_V$, $j_A\mapsto -P{j}_A$, and $V\mapsto PV$, with arguments evaluated at the reflected point. The contracted interaction again becomes $j_R\cdot V$. Each of parity and charge conjugation interchanges the two chiral contractions; applying both restores $j_L\cdot V$. This particular neutral current interaction is CP invariant, although neither P nor C is separately a symmetry. This conclusion uses the stipulated C-odd vector transformation; it is not a claim that all interactions of the Standard Model preserve CP symmetry.

For time reversal of a Dirac field, the gamma matrix adjoint and transpose identities giveWith $B={\gamma }^{5}C$, anticommutation of ${\gamma }^5$ with every gamma matrix yieldsEquivalently, $\boxed{B({\gamma }^{0\ast },-{\mathbf{\gamma }}^{\ast })B^{-1}=({\gamma }^0,\mathbf{\gamma })}$. Since time reversal is an antiunitary operator, the factor i in ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ must also be conjugated. ThusUsing the stated transformations of the field and its Dirac adjoint, the transformed bilinear consequently isThe external vector has the same component signs. Their contraction is unchanged, proving time reversal of a real neutral chiral vector interaction. A real coupling is understood; an antiunitary operator also conjugates coefficients, and relative complex weak phases need not satisfy this invariance.

A charged weak box contribution to kaon mixing is shown below. The incoming $d\bar{s}$ and outgoing $s\bar{d}$ connect through two charged W bosons and two internal up-type quarks. The black arrows are fermion flow, so the lower line's arrows oppose the momentum of the corresponding antiquark. The blue arrows specify the direction of positive charge carried by the charged propagators. Each vertex is a charged weak current vertex, with the appropriate CKM matrix element; the internal flavors are summed over $u,c,t$.

Use precisely the printed exchange-positive CP convention for neutral kaon mixing. In the flavor basis, CP symmetry acts by $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Write the restricted effective matrix asThen $SMS=\left(\begin{array}{cc}a& c\\ b& a\end{array}\right)$, so CP invariance of this mixing operator is exactly $b=c$. Therefore CP noninvariance in this neutral-kaon mixing block means $M_{12}\ne M_{21}$. Violation elsewhere in a larger Hamiltonian would not, by itself, prove an inequality for this particular block.

For neutral-kaon mass matrix diagonalization, choose correlated roots $r^2=b$, $s^2=c$, and eigenvalues ${\lambda }_{\pm }=a\pm rs$. The vectors $(r,s{)}^T$ and $(r,-s{)}^T$ are the corresponding eigenvectors, as direct multiplication verifies. Taking orthonormal flavor states and writing the printed CP eigenstates as $K_1=(K^0+{\bar{K}}^0)/\sqrt{2}$ and $K_2=(K^0-{\bar{K}}^0)/\sqrt{2}$, one findswhere the kaon CP mixing parameter isOverall phases have been absorbed. The kaon mixing square-root branch convention must be chosen continuously so that $r=s$ in the CP-conserving limit for the eigenstate labeled plus. These formulas are exact in the nondegenerate case with $r+s\ne 0$; the requested small admixtures follow if $|ϵ|\ll 1$. Smallness is an additional weak-CP-violation assumption, not a consequence of $b\ne c$ alone. A Hermitian mass operator has $c=b^{\ast }$; the same algebra also applies to the effective non-Hermitian operator that includes decays, where the two eigenstates need not be orthogonal. At a fully degenerate $b=c=0$, a CP eigenbasis can simply be chosen.

Let the masses of $\Phi$ and $\varphi$ be $M$ and $\mu$, respectively. Write their external four-momenta as $P=k_1+k_2$, with $P^2=M^2$ and $k_1^2=k_2^2={\mu }^2$. The condition $M>2\mu$ opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator $i(/k+m)/(k^2-m^2+i0)$, and the convention that an amputated connected diagram contributes $i\mathcal{M}$.

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one $\Phi$ Yukawa interaction and two $\varphi$ Yukawa interactions, so is of order $G{g}^2$. The Yukawa fermion triangle is:

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of $k_1,k_2$. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies $-1$, and each ordering has the productThere is no extra $1/2!$ in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route $a=\ell$, $b=\ell +k_1$, $c=\ell -k_2$ and define $D_a=a^2-m^2+i0$, and similarly for $D_b,D_c$. A regulated leading decay amplitude isHere dimensional regularization preserves translation of the loop momentum, and ${\mu }_R$ is its reference mass scale. Factors ${\mu }_R^{2ϵ}$ account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No ${\gamma }^5$ occurs in this scalar loop, so there is no ${\gamma }^5$ regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while $\mathrm{tr}1=4$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$. Thus the numerator simplifies completely toFor a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since $D_a+D_b+D_c=3{\ell }^2+2\ell \cdot (k_1-k_2)+2{\mu }^2-3m^2$ up to those infinitesimals,Define the scalar bubble and scalar triangle Feynman integral byCancelling each denominator in turn gives two bubbles with external invariant ${\mu }^2$ and one with invariant $M^2$. ThereforeThis explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For $m\ne 0$ the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum giveThe scalar triangle Feynman integral is ultraviolet finite. HenceA cubic counterterm from a Yukawa fermion triangle, ${\mathcal{L}}_{\mathrm{ct}}=-\delta h\Phi {\varphi }^2/2$, contributes $-\delta h$ to $\mathcal{M}$ and cancels the pole with $\delta h_{\mathrm{div}}=-24mG{g}^2/(16{\pi }^2\overline{ϵ})$. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is $\mathcal{M}=-h_R+{\mathcal{M}}_{\mathrm{loop},\mathrm{ren}}$, where a renormalization condition fixes the finite cubic coupling $h_R$. The displayed interaction alone does not supply that condition. For example, setting $h_R=0$ at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For $m=0$ the trace, and therefore this scalar triangle contribution, vanishes identically.