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The action is dimensionless, so in four dimensions the Lagrangian density has mass dimension four. Its kinetic term fixes $[\varphi ]=1$ and $[{\partial }_{\mu }]=1$. An interaction $g_D{\mathcal{O}}_D$ with $[{\mathcal{O}}_D]=D$ has $[g_D]=4-D$. ThereforeIn an effective field theory with heavy scale $\Lambda$, a higher-dimensional local operator enters as $c_D{\mathcal{O}}_D/{\Lambda }^{D-4}$. For characteristic scales $E,m\ll \Lambda$ and coefficients compatible with this expansion, its effects are suppressed by powers of the small ratios. Omitting these terms defines the leading approximation and its truncation error in effective field theory. It is not a fundamental prohibition: such terms must be retained if that accuracy is required. A continuum renormalizable quantum field theory can instead be defined by the closed set of counterterms of dimension at most four.

Up to total derivatives, all Lorentz scalar terms of dimension at most four built from one real scalar field areThe term $\varphi \square \varphi$ is equivalent to the kinetic term by integration by parts; $\square \varphi$ and $\square ({\varphi }^2)$ contribute only at the boundary. The displayed Lagrangian density omits the linear and cubic interactions and a field-independent constant. The exact Z2 symmetry $\varphi \mapsto -\varphi$ forbids both odd interactions, and radiative corrections preserving this symmetry cannot generate them. The constant shifts the vacuum energy and the normalization of $Z$ and can be dropped when gravity and absolute vacuum energy are not being studied. A constant vacuum counterterm may still be used when renormalizing unnormalized vacuum diagrams. The kinetic coefficient has already been fixed by field normalization.

A useful model of spontaneous breaking of a Z2 scalar symmetry is the four-dimensional real scalar theoryThe action is invariant under the Z2 symmetry $\varphi \mapsto -\varphi$. Its classical vacuum manifold consists of the two values $\varphi =\pm v$. In the broken quantum phase, choose a pure vacuum with $⟨\varphi ⟩=v$; the symmetry maps it to a different vacuum with opposite expectation value, although it leaves the action unchanged. To distinguish quantum symmetry breaking from simply minimizing a classical potential, introduce a small source $J\varphi$ and take the infinite-volume limit before $J\to 0^{+}$. A finite-volume symmetry-preserving ground state can remain an even superposition because of tunneling, whereas the selected infinite-volume pure vacua have nonzero vacuum expectation values. The parameter $v$ below is the tree-level expectation, with its quantum value determined by renormalized parameters.

Writing $\varphi =v+h$ givesThe original sign symmetry relates expansions about the two vacua; in one expansion it acts as $h\mapsto -2v-h$. It is not a symmetry that fixes the chosen vacuum. Since the broken group is discrete, there is no continuous broken generator and no required Goldstone boson.

With relativistically normalized external states, the general two-body partial decay width isHere the bar sums final spins and averages initial spins only if the initial ensemble is unpolarized. The identical final-state symmetry factor is $\mathcal{S}=2!$ for two identical daughters and one for distinguishable daughters. Integrating the three-momentum delta function in the parent rest frame leaves ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes $k=|{\mathbf{p}}_1|$ and has radial Jacobian $k(1/E_1+1/E_2)$. Hence the two-body decay phase space isConsequentlywhen the spin-summed amplitude is angle independent.

For Higgs decay to two W bosons, the scalar parent has no spin average and the amplitude, up to an irrelevant phase, is $\mathcal{M}=g{M}_W{\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$. Use the massive vector polarization sum with $p_i^2=M_W^2$:Momentum conservation gives $p_1\cdot p_2=(M_H^2-2M_W^2)/2$. With the PDF's ratio $x=M_W/M_H$, this becomesThe charged daughters are distinguishable. Combining the last two equations with $g^2/M_W^2=4\sqrt{2}{G}_F$ givesFor Higgs decay to two Z bosons, the Higgs boson coupling to Z bosons is $2i{M}_Z^2{g}_{\mu \nu }/v=ig{M}_Z^2{g}_{\mu \nu }/M_W$, using $v=2M_W/g$. This compensates the changed powers of $M_Z$ in the polarization contraction, giving the same normalization of the squared amplitude with $x$ replaced by $y=M_Z/M_H$. The two Z bosons are identical, so the phase-space symmetry factor supplies an additional half:Both expressions have mass dimension one, vanish at their two-body thresholds and approach the ratio $2:1$ for $M_H\gg M_Z$. The cubic large-mass behavior comes from longitudinal-vector polarizations. These are on-shell tree-level widths under the stated heavy-Higgs hypothesis; below threshold the physical off-shell multi-particle decays require a different calculation.

Use the metric signature $(-,+,\dots ,+)$, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the weight $e^{iS}$. The momentum-space Feynman rules for the massless Phi-fourth theory are: an internal scalar line contributes $-i/(k^2-iϵ)$; a quartic interaction vertex contributes $-i\lambda$ with its momentum-conservation delta function; every independent loop is integrated with $d^{d}k/(2\pi {)}^d$; and each graph is multiplied by its Feynman-diagram symmetry factor. All external momenta can be taken incoming. These rules follow by expanding the interaction exponential and contracting free fields with the Wick theorem.

For a connected graph with $I$ internal lines, $V$ interaction vertices and $L$ loops,Here $D$ is the superficial degree of divergence. In four dimensions,Odd $n$ vanish by Z2 symmetry. Thus only the nonvacuum two-point and four-point one-particle-irreducible Feynman diagrams can have overall ultraviolet divergences: degree two for $n=2$, degree zero for $n=4$. Vacuum graphs with $n=0$ may also diverge but cancel from normalized correlators. This is an overall power-counting statement; subgraphs must be subtracted before using it to conclude finiteness for $n\ge 6$.

For the full one-particle-irreducible vertices, the tree-level quadratic kernel and quartic interaction giveIf ${\hat{\tau }}_2$ is instead reserved for interaction-generated self-energy insertions, its tree value is zero and $-p^2$ is displayed separately as the free inverse kernel. The distinction is only that convention; the loop insertions below are unchanged.

The requested one-loop diagrams are the two-point tadpole diagram and the three four-point bubble diagrams. Each has symmetry factor $1/2$. The three bubble channels correspond to $P_s=p_1+p_2$, $P_t=p_1+p_3$, $P_u=p_1+p_4$.

The tadpole insertion obeysIn massless dimensional regularization this scaleless integral is zero. A mass or cutoff regulator exposes the ultraviolet divergence allowed by power counting; its vanishing in the scaleless convention does not remove the need to include this diagram.

For a bubble, the vertex and quantum field theory propagator factors give $(-i\lambda {)}^2(-i{)}^2/2={\lambda }^2/2$. The PDF defines $f$ with the factor $1/i$ outside the Minkowski integral, so the integral itself is $if(P^2)$. ThereforeThe converted TeX misplaces this $i$ as though it were part of the exponent of $(2\pi )$; keeping the printed $1/i$ is essential to this real pole coefficient.

For nonexceptional Euclidean external momentum, Wick rotation cancels that factor $1/i$. Use a Feynman parameter and translate the integration variable:For example, represent the inverse square as ${\int }_0^{\infty }d\alpha \alpha e^{-\alpha ({\ell }^2+\Delta )}$, do the Gaussian momentum integral, and then use the Gamma function. This also derives the formula rather than assuming a loop-integration table. At $d=4-\epsilon$, $\Gamma (\epsilon /2)=2/\epsilon +O(1)$ and the parameter integral tends to one, givingTake $P^2>0$ before continuation back to Minkowski momenta. At $P=0$, a massless scaleless integral mixes ultraviolet and infrared issues, so it cannot be used to read off this pole.

Introduce a renormalization scale $\mu$ and a quartic counterterm. With $\lambda$ denoting a dimensionless renormalized coupling constant,After factoring out ${\mu }^{\epsilon }$ from the vertex, each loop contains ${\mu }^{\epsilon }{f}_d$. The counterterm contributes $-\delta \lambda$ to ${\hat{\tau }}_4$ and cancels the pole from all three channels. More explicitly,In modified minimal subtraction, replace the pole-only counterterm bywhich subtracts the first three terms inside the brackets. The finite renormalized quartic scalar vertex is thenAnother renormalization condition can change the finite constant, but not the pole cancellation or the momentum-dependent logarithms. The corresponding timelike expressions follow by the same Feynman continuation.

Let the shell-restricted scalar propagator beWrite $\Delta S=S_0[\hat{\varphi }]+V[\varphi ,\hat{\varphi }]$ and average with the normalized Gaussian shell measure. The cumulant expansion givesThe subtraction removes disconnected Feynman diagrams; the logarithm retains connected Feynman diagrams made from shell contractions. This is the connected shell-contraction expansion. An external line denotes the background low field, not an additional low-momentum propagator.

At order $g$, panel A is the existing four-field interaction vertex. The new shell contractions are B, a tadpole diagram contributing a two-field vertex, and C, a Vacuum Feynman diagram contributing a constant. The Gaussian functional determinant in panel L is another field-independent term, of order $g^0$; the original quadratic kinetic term and mass term are retained as well.

For completeness, all two-vertex topologies from Wick contractions at order $g^2$ are shown. Let $r$ be the number of shell lines joining the vertices and $t_1,t_2$ their numbers of self-contractions. The external-field counts areInterchanging the two vertices identifies the same topology. Enumerating these conditions gives exactly the following eight possibilities:

The support qualification is important. In E or F, a vertex with one external low field and one bridge also has a tadpole whose two momenta cancel. Conservation forces the bridge momentum to equal that single low momentum, outside the shell. Equivalently, convolution by $C_{>}$ annihilates $\varphi$. This is momentum-support exclusion for Wilsonian bridge diagrams. D is different: its bridge carries the sum of three low momenta, which can lie in the shell. Its kernel is proportional to ${\varphi }^3(x)C_{>}(x-y){\varphi }^3(y)$ and need not vanish. Thus one-particle-reducible Feynman diagrams must not be excluded merely because the object being computed is an effective action.

These kernels need not be local before a derivative expansion. Their momentum dependence generates derivative interactions where such an expansion is valid. If all external momenta are sufficiently far below ${\Lambda }^{\prime }$, D vanishes too; in particular it does not produce a zero-momentum local ${\varphi }^6$ coupling at order $g^2$. A local six-field term is allowed and is generated at higher orders, for example by a three-vertex shell triangle.

To all orders, expect every scalar interaction allowed by the original symmetries: arbitrary even powers of the field and their allowed derivative couplings, together with vacuum terms. The Z2 symmetry $\varphi \mapsto -\varphi$ excludes odd-field vertices. The original quartic form is therefore not closed under exact Wilsonian integration.

Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.

The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension $2n-(n-1)d$, positive below $d_c$. Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At $d=d_c$, equality of the powers allows logarithmic corrections.

Write $r_i={\mu }_i^2$ and assume $g>0$, as required for quartic stability. In the mean-field approximation, the two-component radial quartic Landau potential isSet $x={\varphi }_1^2\ge 0$, $y={\varphi }_2^2\ge 0$. At fixed $\rho =x+y$, the quartic term is fixed and the quadratic term is minimized by assigning all of $\rho$ to the component with smaller $r_i$. Minimizing over $\rho$ then givesThe resulting phase diagram has these regions:

On $r_1=r_2=r<0$, the symmetry is enhanced to the orthogonal group $O(2)$, and the minima form the circleMixed-component minima occur on this line, rather than in an open coexistence region. Crossing the negative diagonal swaps the selected component discontinuously: this is a first-order phase transition, also called a spin-flop transition. For example, holding $r_2=r<0$ fixed and moving $r_1$ across $r$, the derivative $\partial f_{\min }/\partial r_1$ jumps from $-r/(8g)$ to zero. The angular degeneracy exactly on the line does not remove this discontinuity.

The positive half-axis $r_1=0$, $r_2>0$ is a continuous phase transition, as is $r_2=0$, $r_1>0$. Their endpoint at $(0,0)$ is a bicritical point, where the two continuous boundaries meet the first-order negative diagonal. Approaching the origin radially makes the ordered amplitude vanish continuously. The negative half-axes lie inside ordered phases and are not additional transition lines. There are two continuous half-axis boundaries and one first-order negative diagonal, meeting at the origin.

No. With only ${\lambda }_0{\varphi }^6$, the free energy has an exact Z2 symmetry $\varphi \mapsto -\varphi$. Integrating out fast modes and rescaling preserve that symmetry, whereas ${\varphi }^5$ is odd. Therefore no five-point vertex and no correction to $\gamma (\zeta )$ can be generated at any order in ${\lambda }_0$.

No. Even when the two masses differ, the free energy retains an independent Z2 symmetry for each component: ${\varphi }_1\mapsto -{\varphi }_1$ and ${\varphi }_2\mapsto -{\varphi }_2$. The bilinear ${\varphi }_1{\varphi }_2$ is odd under either transformation, so integrating out modes cannot generate it.

Let $L$ be the number of independent loop momenta in a connected Feynman diagram. Every scalar propagator falls as $k^{-2}$, and every loop momentum integration contributes three powers, so $\omega =3L-2e$. Counting line ends and using connectedness givesEliminating $L$ and $e$ yieldsThus increasing the sextic interaction vertex count does not worsen the superficial degree of divergence; adding a quartic interaction vertex improves it. Only finitely many external-line structures can have an overall ultraviolet divergence, even though arbitrarily high loop order is possible.

After subtracting ultraviolet subdivergences, locality and Lorentz invariance restrict the divergent counterterms to a constant vacuum energy density, a mass counterterm, a wave-function renormalization, and quartic and sextic field interaction terms. Odd powers are forbidden by the $\varphi \mapsto -\varphi$ Z2 symmetry. In particular, $n=2$ permits a constant and $p^2$, $n=4$ permits only a constant, and $n=6$ permits only a constant. This is perturbative renormalizability, not a guarantee that every superficially allowed divergence actually occurs or that every unrenormalized graph with $\omega <0$ lacks divergent subgraphs.

The expansion is a local phenomenological description in a small scalar order parameter near a transition. Its coefficients summarize microscopic physics and generally depend smoothly on temperature. At zero field, spin-reversal Z2 symmetry requires even powers of magnetization; the conjugate field supplies the term $-hm$. The additive $F_0$ does not affect minimization with respect to $m$.

For the ordinary continuous-transition quartic model, $b>0$ ensures stability, and the quadratic coefficient changes sign, $a=a_{0}t+O(t^2)$ with $a_0>0$. If $b$ vanishes or is negative, stabilizing higher powers are required and the phase structure can change. Treating one uniform magnetization neglects spatial gradients, fluctuations and domains; the resulting exponents are predictions of mean-field theory.

Assume $b>0$ and temperature-independent $d$ for the local model. The zero-field stationary points obeyFor $d\ne 0$, the favoured ordered sign is opposite to $d$. The nonzero extrema first appear at $a=d^2/(4b)$; their existence is not yet a coexistence condition. To find the equilibrium transition, require both $F^{\prime }(m_{\ast })=0$ and $F(m_{\ast })=F(0)$. Substituting $a=-d{m}_{\ast }-b{m}_{\ast }^2$ into the free energy givesHenceAt coexistence the entire potential factorizes:It has equal minima at zero and $m_{\ast }$ separated by a barrier. Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model, with a finite magnetization jump before the quadratic coefficient reaches zero.

For $d^2/(4b)>a>a_{\ast }$, the ordered minimum is metastable and zero is globally stable. For $a_{\ast }>a>0$, the ordered state is globally stable and zero is metastable. At $a=0$ the zero minimum loses local stability. Thus the spinodals are $a=d^2/(4b)$ and $a=0$, distinct from $a_{\ast }$. For $a<0$, the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign $-d$.

If $a=a_0(T-T_c)/T_c$, the transition is shifted to $T_{\ast }=T_c[1+2d^2/(9b{a}_0)]$. With $b,d$ constant and $a^{\prime }(T)>0$, the entropy jump from the disordered to ordered phase is $-a^{\prime }(T_{\ast })m_{\ast }^2/2$, giving a latent heat of magnitude $T_{\ast }{a}^{\prime }(T_{\ast })m_{\ast }^2/2$. As $d\to 0$, the jump and shift vanish and the original continuous transition is recovered.

A genuine zero-field uniaxial ferromagnet with exact spin-reversal Z2 symmetry forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original ferromagnet.

Impose the discrete Z2 symmetry $\boxed{\Phi \mapsto -\Phi }$. The quadratic and quartic trace invariants are unchanged, while $\mathrm{tr}{\Phi }^3$ changes sign. Invariance therefore forbids a nonzero coefficient of that cubic operator. This is an additional symmetry of the scalar potential, not a consequence of SU(5) conjugation alone; cubic traces are permitted by the latter.

A real scalar field with the displayed potential is invariant under the Z2 symmetry $\varphi \mapsto -\varphi$, but a chosen broken-phase pure vacuum has nonzero vacuum expectation value of one sign. Expanding about $\varphi =v$ gives $V=\lambda v^2{h}^2+\lambda v{h}^3+\lambda h^4/4$. The excitation is massive, and no continuous broken generator requires a Goldstone boson. In a finite-volume theory with symmetry-preserving conditions, tunneling can produce a symmetric ground state. A broken pure vacuum is selected by taking the infinite-volume limit before removing an infinitesimal source, or by specifying appropriate boundary conditions. The two signs are related by the symmetry, although the individual pure vacua are not invariant under it.

A Z2 symmetry is generated by an operation whose square is the identity. For a real scalar field, the sign symmetry $\varphi \mapsto -\varphi$ forbids every odd power of $\varphi$.