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Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.

For a constant real parameter, ${\partial }_{\mu }(e^{i\alpha }\varphi )=e^{i\alpha }{\partial }_{\mu }\varphi$. The phases therefore cancel separately in the kinetic term and in ${\varphi }^{\ast }\varphi$, leaving any $f({\varphi }^{\ast }\varphi )$ unchanged. This is the global phase symmetry of a complex scalar field, with circle group $U(1)$.

Use infinitesimal variations per unit parameter, $\delta \varphi =i\varphi$, $\delta {\varphi }^{\ast }=-i{\varphi }^{\ast }$. The Lagrangian density is exactly invariant, so no total-derivative correction is needed in Noether's theorem. The Noether current isIt depends only on derivatives of the Lagrangian density with respect to field derivatives, so the derivative-free scalar potential cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:Thus it is conserved on solutions of the Euler-Lagrange equations. Its spatial integral is a conserved Noether charge when the flux through the boundary vanishes. Its overall sign is fixed here by $\delta \varphi =+i\varphi$; reversing the orientation of the phase transformation reverses the current.

In $d$ spacetime dimensions, $[d^{d}x]=-d$ and a dimensionless action requires $[\mathcal{L}]=d$. Since $[{\partial }_{\mu }]=1$, the canonical kinetic term fixes $2+2[\varphi ]=d$. The cubic and quartic terms then giveThese brackets denote mass dimensions. The cubic coupling is dimensionless at $d=6$, whereas the quartic coupling is dimensionless at $d=4$. Hence there is no single dimension in which both nonzero couplings are dimensionless. For example, at $d=4$ the cubic coupling still has mass dimension one.^[ref]

Divide the infinitesimal variation by its constant parameter $\alpha$, so the transformation generator is $(\delta {\varphi }_1,\delta {\varphi }_2)=({\varphi }_2,-{\varphi }_1)$. The variation of the kinetic term isand that of ${\varphi }_1^2+{\varphi }_2^2$ is $2{\varphi }_1{\varphi }_2-2{\varphi }_2{\varphi }_1=0$. Thus the Lagrangian density is invariant pointwise, without a nonzero boundary current. The finite transformation is a rotation of the two real scalar fields, giving an SO(2) global symmetry in field theory.

Noether's theorem supplies the parameter-independent Noether currentIts divergence is ${\partial }_{\mu }{j}^{\mu }={\varphi }_2\square {\varphi }_1-{\varphi }_1\square {\varphi }_2$, since the first-derivative terms cancel. Both fields obey the same Klein-Gordon equation, so this vanishes on solutions. With spatial boundary conditions giving zero flux at infinity, the Noether charge isA finite box with periodic boundary conditions works as well. The spatial integral is conserved because the local continuity equation converts its time derivative into a boundary flux; neither vanishing local current nor constant individual fields is required. Equivalently, $\Phi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ transforms as $\Phi \mapsto e^{-i\alpha }\Phi$, exhibiting the equivalent global U(1) symmetry and fixing the sign of the generator.

Use mass dimensions with $[x]=-1$, $[{\partial }_{\mu }]=1$ and dimensionless action. The Lagrangian density in Euclidean space has dimension $D$, and the kinetic term fixes $2+2[{\varphi }_0]=D$. ThusA dimensionless coupling constant is thereforeHere $\mu$ is a positive regularization scale with mass dimension one. The interaction is marginal at its upper critical dimension $D_c=2k/(k-1)$, relevant below it and irrelevant above it by this engineering-dimension test. In particular, $d_2=4-D$, while $d_k=4-2k+2(k-1)ϵ$ near $D=4-2ϵ$. At this stage $g_{2k}$ is the dimensionless version of the displayed bare coupling; after introducing counterterms one may use the same scale factor for the renormalized coupling.

A continuous Lorentz transformation in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0{)}^{\mathsf{T}}=\gamma (1,v{)}^{\mathsf{T}}$. Preservation of the Minkowski norm gives ${\gamma }^2(1-v^2)=1$, so $\gamma =(1-v^2{)}^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared Minkowski norm $-1$; continuity fixes it to $\gamma (v,1{)}^{\mathsf{T}}$. ThusHere $x=(t,x{)}^{\mathsf{T}}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.

Define the rapidity $\alpha =\mathrm{artanh}v$. The hyperbolic function identities giveIn light-cone coordinates, $t^{\prime }+x^{\prime }=e^{\alpha }(t+x)$ and $t^{\prime }-x^{\prime }=e^{-\alpha }(t-x)$. Therefore the derivatives transform oppositely:Use the field-transformation convention ${\psi }^{\prime }(x^{\prime })=S(\alpha )\psi (x)$. The component action in c is invariant underIndeed, the right-component kinetic term gains $e^{\alpha /2}{e}^{-\alpha }{e}^{\alpha /2}=1$, and the left-component kinetic term gains $e^{-\alpha /2}{e}^{\alpha }{e}^{-\alpha /2}=1$. The two factors in each Dirac mass term cancel as well. Finally $\det B=1$, so $d{t}^{\prime }d{x}^{\prime }=dtdx$ and the whole action is unchanged. These are the chiral boost weights in two spacetime dimensions.

In spinor notation, $S(\alpha )=\exp (\alpha {\gamma }^5/2)$ satisfies $S^{-1}{\gamma }^{\mu }S=B^{\mu }{}_{\nu }{\gamma }^{\nu }$ and $S^{†}{\gamma }^{0}S={\gamma }^0$, so ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$. These identities directly verify covariance of the Dirac equation and the Lorentz invariance of its action. The matrices $S$ are not unitary with respect to the ordinary component norm; the Dirac adjoint supplies the invariant bilinear.

The statement about there being only Lorentz boosts applies to the identity component of the One-dimensional Lorentz group. The full Lorentz group also has discrete transformations: the concrete spatial reflection $\mathrm{diag}(1,-1)$ preserves the Minkowski metric but has determinant $-1$, so it cannot equal any displayed $B(\alpha )$, whose determinant is $+1$.

For a connected amputated Feynman diagram, write $L$ for the number of independent loop momenta. Each loop momentum integration scales as ${\lambda }^4$, each internal scalar propagator as ${\lambda }^{-2}$, and each internal Dirac propagator as ${\lambda }^{-1}$ when both four-momenta and mass are scaled. The Yukawa coupling has mass dimension zero. ThusCounting the two fermion ends and one scalar end at each interaction vertex, and using connectedness, givesConsequently $n_f=n_v$, $L=n_s+1$, andThe requested exponent requires amputation and removal of the overall momentum delta function. An ordinary unamputated connected Green function has an additional $n$ external scalar propagators, giving mass dimension $4-3n$ after stripping that delta function, or $-3n$ if it is retained. For a concrete check, the free two-point scalar propagator scales as ${\lambda }^{-2}$, not ${\lambda }^2$. The exponent $4-n$ is the engineering result for the amputated object. A fixed renormalization scale introduces logarithms; the displayed homogeneity is a naive power counting statement rather than an exact quantum scaling law at fixed subtraction scale.

More generally, with $E_s$ external scalar ends and $E_f$ external fermion ends, the same counting gives the superficial degree of divergenceThis is an upper bound on the overall ultraviolet divergence. Numerator cancellations and Lorentz invariance may lower it; an apparently convergent whole Feynman diagram can still have ultraviolet subdivergences. Local subtraction of those ultraviolet subdivergences is required first.

For pure scalar ends, $E_s=0,1,2,3,4$ can require counterterms. Lorentz invariance reduces the two-point terms to a mass counterterm and wave-function renormalization; derivatives of a single scalar field are total derivatives, and a one-derivative cubic scalar term cannot be a Lorentz scalar. With two external fermions, $E_s=0$ requires their kinetic term and mass counterterm, while $E_s=1$ requires Yukawa coupling renormalization. Four external fermions give $D=-2$, so no independent four-fermion counterterm is required.

A complete counterterm Lagrangian consistent with the species symmetry can therefore be writtenThe vacuum energy density term matters for an unnormalized vacuum functional and cancels from normalized Green functions. The tadpole counterterm can impose $⟨\varphi ⟩=0$. There is no $\varphi \mapsto -\varphi$ symmetry of the massive Yukawa interaction, so linear, cubic and quartic scalar terms cannot be excluded. For instance, in a constant background the divergent fermion loop contribution contains $(m+g\varphi {)}^4$, exhibiting all these powers. This is counterterm closure of a massive Yukawa theory.

The two species have identical couplings, so their common unitary flavour rotations preserve the Lagrangian density and their counterterms can be chosen species independent. Equality of the scalar and fermion masses, however, is not protected: $\delta m_s^2$ and $\delta m_f$ are independent. Including the displayed finite set of operators yields a renormalizable quantum field theory. Restricting the bare Lagrangian density to the terms printed in the question is not closed under renormalization.

Here three means the total spacetime dimension. A dimensionless action requires $[\mathcal{L}]=3$ in mass dimensions. The kinetic term gives $2+2[\varphi ]=3$, so $[\varphi ]=1/2$. Matching the mass dimensions of both field interaction terms givesThus $g_1$ has units of mass and $g_2$ is dimensionless. The quartic interaction is superrenormalizable by power counting, and $g_2$ is a marginal coupling by power counting.

A dimensionless action in six spacetime dimensions requires the Lagrangian density to have mass dimension six. The kinetic term then fixes $2+2[\varphi ]=6$, and each field interaction term must have the same mass dimension. ThereforeEquivalently, in powers of a length unit $L$, $\varphi$ has units $L^{-2}$, $g_1$ has units $L^0$, and $g_2$ has units $L^2$. Thus literal length exponents would be $(-2,0,2)$; the $d$'s in the supplied superficial degree of divergence formula must instead mean mass dimensions, namely $(2,0,-2)$. This distinction resolves the wording's convention mismatch.

To check the convention independently, let $I,V_3,V_4,\ell$ count internal lines, cubic interaction vertices, quartic interaction vertices, and independent loop momenta. A connected Feynman diagram obeys $2I+n=3V_3+4V_4$ and $\ell =I-V_3-V_4+1$. Hence its superficial degree of divergence is $6\ell -2I=6-2n+2V_4$, agreeing with the supplied formula only with the mass dimensions above.

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.

For nonzero real $g$, the continuous internal symmetry of a classical field theory is the global U(1) symmetryThe kinetic terms and mass terms are invariant separately, while the phases in ${\varphi }^2{\Phi }^{†}$ cancel. Independent phase rotations would instead require $2{\alpha }_{\varphi }-{\alpha }_{\Phi }=0$ modulo $2\pi$, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives $U(1)⋊{\mathbb{Z}}_2\cong O(2)$ for generic parameters. At $g=0$ the continuous symmetry enlarges to $U(1)\times U(1)$.

The Noether current with particles assigned positive charges isFor example, put $A={\varphi }^2{\Phi }^{†}$ and $B={\varphi }^{†2}\Phi$. The Euler-Lagrange equations give ${\partial }_{\mu }{j}_{\varphi }^{\mu }=ig(A-B)$ and ${\partial }_{\mu }{j}_{\Phi }^{\mu }=ig(B-A)$, with the factor $2$ already included in $j_{\Phi }^{\mu }$. Their sum vanishes. The Noether charge $Q=\int d^{3}x{j}^0$ is conserved and, after normal ordering, countsThus a $\Phi$ particle carries two units of the same conserved charge carried by a $\varphi$ particle; their antiparticles carry the negatives. The interaction permits conversion of one $\Phi$ particle into two $\varphi$ particles when allowed by conservation of energy, while preserving this weighted number.

Put $x_P=(t,-x,y)$ and $U_P={\gamma }^1$. Since $U_P^{†}=-{\gamma }^1$ and ${\gamma }^0{\gamma }^1{\gamma }^0=-{\gamma }^1$, the Dirac adjoint transforms asThe chain rule gives ${\partial }_{\mu }\psi (x_P)=r_{\mu }({\partial }_{\mu }\psi )(x_P)$, with $(r_0,r_1,r_2)=(1,-1,1)$ and no sum in this formula. The Clifford algebra givesMultiplication by $r_{\mu }$ therefore restores each kinetic term:However, the Dirac scalar bilinear changes sign because $({\gamma }^1{)}^2=-I_2$:Changing variables in the action has unit absolute Jacobian determinant. HenceA nonzero Dirac mass term for this single two-component field breaks parity symmetry in quantum field theory with the specified transformation.

The specified time reversal includes complex conjugating $i{\gamma }^{\mu }{\partial }_{\mu }$; omitting that operation changes the answer. Put $x_T=(-t,x,y)$, $U_T={\gamma }^2$, and $(r_0,r_1,r_2)=(-1,1,1)$. As in c, the Dirac adjoint in the source's transformed bilinear is ${\bar{\psi }}^{\prime }(x)=\bar{\psi }(x_T){\gamma }^2$. The conjugated Dirac operator has coefficient matricesCombining these coefficients with the chain rule signs and the Clifford algebra gives, without summing over $\mu$,For example the time coefficient is ${\gamma }^2(-i{\gamma }^0){\gamma }^2(-1)=i{\gamma }^0$, and the second spatial coefficient is ${\gamma }^2(-i{\gamma }^2){\gamma }^2=i{\gamma }^2$. Thus the kinetic term is invariant. The real Dirac mass term again reverses sign since ${\bar{\psi }}^{\prime }{\psi }^{\prime }=\bar{\psi }({\gamma }^2{)}^2\psi =-\bar{\psi }\psi$. The spacetime integration has unit absolute Jacobian determinant, soEach operation reverses the sign of the mass, while their composition reverses it twice and preserves the kinetic term. In an operator formulation the complex conjugation of coefficients expresses the antiunitary operator character of time reversal; the transformation must not be treated as an ordinary unitary substitution with all $i$'s held fixed. These statements concern the stated free action.

First conjugate the Dirac equation: $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. A matrix $\Gamma$ taking ${\psi }^{\ast }$ into another solution of the same Dirac equation must obeyBecause ${\gamma }^1$ anticommutes with the real ${\gamma }^0,{\gamma }^2$ and commutes with the purely imaginary ${\gamma }^1$, a unitary choice isAny constant overall phase is equally valid. To verify invariance of the action, rather than only of its equation, keep the fermion exchange signs. Define the charge-conjugation matrix $C=\Gamma {\gamma }^0={\sigma }_2$. It obeysFor a classical Grassmann field, interchanging two components contributes a minus sign; equivalently use normal ordering for quantum fermionic operators. ThusandThe last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the kinetic term and the Dirac mass term therefore leave $S$ invariant under charge conjugation. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the fermion statistics is essential.

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Use natural units $\hslash =c=1$ and the mostly-minus Minkowski metric. Question 3 has two spatial dimensions: its gamma matrices are the specified two-by-two matrices. The printed ${\sigma }^{\mu }$ in its kinetic term is interpreted as those ${\gamma }^{\mu }$, rather than as three unmodified Pauli matrices. All spin sums in question 2 instead use four-component Dirac spinors in four spacetime dimensions.

In $2+1$ spacetime dimensions, the dimensionless action gives $[\mathcal{L}]=3$ and $[{\partial }_{\mu }]=1$. The scalar kinetic term requires $2[\varphi ]+2=3$, whereas the fermion kinetic term requires $2[\psi ]+1=3$. Thus the engineering dimensions of relativistic fields areMatching each interaction to mass dimension three givesHere the brackets mean exponents of mass in natural units. The coefficients $g_1,g_3$ are marginal couplings by classical power counting, and $g_2$ is a relevant coupling.

Writing $x_P=(t,-\mathbf{x})$, the standard parity transformation of a Dirac field isSince ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, the fermion bilinear $\bar{\psi }{\gamma }^5\psi$ changes sign. Invariance of the pseudoscalar Yukawa interaction therefore requiresThus $\varphi$ is a pseudoscalar, despite having the usual scalar-form kinetic term and mass term. In this convention $({\gamma }^5{)}^{†}=-{\gamma }^5$, but $\bar{\psi }{\gamma }^5\psi$ is Hermitian because of the additional ${\gamma }^0$ in the Dirac adjoint. A real $g$ is consequently consistent with a Hermitian Lagrangian density.

In perturbation theory about the parity-invariant vacuum state, renormalization can preserve this non-anomalous parity symmetry in quantum field theory. Every local counterterm must then be parity even, at every loop order. The allowed terms are precisely ${\mathcal{L}}_{\mathrm{ct}}^{(+)}$ in b; all coefficients in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ are zero. In particular, no $\varphi$ or ${\varphi }^3$ counterterm is generated, whereas ${\varphi }^4$ is allowed and required. The one-point and three-point pure-boson vertices also vanish to all orders: a parity-odd Lorentz scalar cannot be formed from their at most two independent external four-momenta. This last reasoning is specific to these low-point vertices; it does not assert that parity symmetry in quantum field theory is an independent $\varphi \mapsto -\varphi$ internal symmetry with the fermions held fixed.

The continuous spacetime symmetries are translations, spatial rotations and Lorentz boosts, together the connected Poincaré group. The theory also has local electromagnetic U(1) gauge symmetry, which is an internal symmetry rather than a spacetime transformation. A nonzero mass and a dimensionful interaction do not give an additional exact scale invariance.

In four dimensions the action is dimensionless and $[\partial ]=1$. The two kinetic terms giveThe axion-photon operator has mass dimension five, so this is an effective field theory. Its dimensionless interaction strength grows with energy as $|g_a|E$. Without information about an ultraviolet completion, a conservative parametric validity condition is $E\ll \Lambda$ with $\Lambda \sim |g_a{|}^{-1}$, and $m_a$ must also lie below that scale for the axion to be a degree of freedom in the effective theory. Numerical factors such as $4\pi$ depend on the ultraviolet dynamics and on the chosen perturbative criterion; the Lagrangian alone does not fix them.

The Proca action is invariant under spacetime translations and proper continuous Lorentz transformations, together forming the connected Poincaré group. The fixed nonzero mass excludes scale invariance. In four dimensions, a dimensionless action and the canonical kinetic term implyThese are mass dimensions. The field-strength term is invariant under $A_{\mu }\to A_{\mu }+{\partial }_{\mu }\chi$, but the mass term is not: its change contains $m_A^2{A}^{\mu }{\partial }_{\mu }\chi +\frac{1}{2}{m}_A^2(\partial \chi {)}^2$. Thus the displayed massive-vector theory has no Maxwell-type local gauge invariance. The massless limit has a different constraint structure.

Introduce three independent local parameters ${\alpha }_j(x)$. A consistent convention for the three Abelian gauge transformations isIndeed, $g(B_{\mu }^k-q{B}_{\mu }^{k+1})$ shifts by $-{\partial }_{\mu }({\alpha }_k-q{\alpha }_{k+1})$, cancelling the derivative of the scalar phase. Thus $D_{\mu }{\varphi }_k$ transforms by the same phase as ${\varphi }_k$. The two kinetic terms, the absolute-value potentials and all Abelian field strengths are invariant.

In units where the gauge coupling is $g$, the charge vectors under the three factors areLocally this is the gauge algebra of $U(1{)}^3$. To interpret it globally as three compact circle groups, the charges must belong to a compatible charge lattice: in the stated unit-charge convention integer $q$ is sufficient, while arbitrary irrational $q$ is not a single-valued compact-group representation. The local transformations and mass calculation remain meaningful for general positive real $q$ as a Lie-algebra calculation, or with noncompact Abelian factors.