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.

With outgoing physical photon polarization vectors, the Feynman rule givesThe overall phase does not affect the decay width. Replacing either polarization vector by its own four-momentum gives zero by antisymmetry. Thus the Ward identity permits replacement of each physical photon polarization sum by $-{\eta }^{\mu \nu }$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-{\eta }_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.

The contraction of the two Levi-Civita symbols givessince $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero axion. The result is independent of direction in its rest frame. Including the identical-particle factor explained in solution therefore givesOmitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.^[ref]

Evaluate the two-body Lorentz-invariant phase space in the parent's rest frame, $p=(m,0)$. The spatial Dirac delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. For massless daughters $E_1=E_2=q$, soThe factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the decay width gives, for labelled or distinct daughters,The angular domain is one sphere: $0\le \phi <2\pi$ and $-1\le \cos \theta \le 1$, equivalently $0\le \theta \le \pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.

For identical daughters, integration over the full labelled two-body phase space counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical photons in solution.

The axion variation gives $\partial \mathcal{L}/\partial a=-m_a^{2}a+g_a{F}_{\mu \nu }{\tilde{F}}^{\mu \nu }/4$, soFor the photon variation, the dual electromagnetic field tensor obeys $\delta (F_{\rho \sigma }{\tilde{F}}^{\rho \sigma })=2{\tilde{F}}^{\rho \sigma }\delta F_{\rho \sigma }$. Antisymmetry then givesThere is no undifferentiated $A_{\nu }$, henceThe last equality uses the electromagnetic Bianchi identity ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$. These signs follow from the positive $aF\tilde{F}$ interaction in the source; reversing that interaction reverses both source terms. The induced electromagnetic current is identically conserved: its divergence contracts a symmetric second derivative of $a$ with an antisymmetric tensor, while the remaining term vanishes by the same Bianchi identity.

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.

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.

The Lagrangian density is Lorentz invariant, so a scalar-valued static solution can be composed with a Lorentz transformation. Let $X=\gamma (x-vt-x_0)$, with $\gamma =(1-v^2{)}^{-1/2}$ and $|v|<1$. If $\varphi (x,t)=f(X)$, the chain rule givesThe boosted equation of motion is therefore exactly the static one already verified in solution, andis a solution. The center moves as $x=x_0+vt$: $v$ is the physical velocity in units where the limiting speed is one. The kink width contracts from $k^{-1}$ to $(\gamma k{)}^{-1}$, while its topological charge remains unchanged.

Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed Sine-Gordon kink givesThe angle identity givesSince $k^2/b=a$, these formulas show $\boxed{{\varphi }^{\prime \prime }=a\sin (b\varphi )}$, precisely the static equation of motion.

For $x\to -\infty$ and $x\to +\infty$, the field approaches $0$ and $2\pi /b$ respectively. These are neighboring minima of the scalar potential when $ab>0$, and ${\varphi }^{\prime }$ vanishes exponentially at either end. The energy is therefore finite and the topological charge $Q=b[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.

Varying the real scalar field gives the Euler-Lagrange field equation ${\partial }_t^2\varphi -{\partial }_x^2\varphi +V^{\prime }(\varphi )=0$. Since $V^{\prime }(\varphi )=a\sin (b\varphi )$, the answer isFor $b\ne 0$, introducing $u=b\varphi$ gives $u_{tt}-u_{xx}+ab\sin u=0$, the Sine-Gordon equation. The proposed real localized kink requires $ab>0$; arbitrary real $a,b$ do not guarantee that condition. In what follows put $k=\sqrt{ab}>0$. The displayed potential itself requires $b\ne 0$, unless a limiting prescription is specified.

For $t\ll D_r^{-1},|\omega {|}^{-1}$, expand the exact expression:At asymptotically short time, translational diffusion dominates if $D>0$. A ballistic window occurs when $4D/v^2\ll t\ll \min (D_r^{-1},|\omega {|}^{-1})$, provided these scales are separated. The swimmer then retains nearly constant direction.

For $D_r>0$ and $t\gg D_r^{-1}$, the oscillatory exponential decays andAt intermediate times, the angular drift bends persistent trajectories into arcs. If $|\omega |\gg D_r$, several rotations occur before orientational decorrelation, producing oscillatory shoulders rather than a long straight ballistic regime. In the limit $D_r=0$, $\omega \ne 0$, the active contribution is bounded and periodic: $2v^2(1-\cos \omega t)/{\omega }^2$. At $D_r=\omega =0$, it remains ballistic forever, so ordinary late-time active diffusion does not follow.

For $D_r>0$, the central-limit coarse graining of the position gives the late-time Fokker-Planck equationThere is no surviving drift after orientation averaging. This is an effective large-time, large-distance equation, not the exact closed equation for position at all times.

Integrate the two velocity factors over the time square $0\le s,s^{\prime }\le t$. The white noise contribution is $4Dt$. The orientation contribution depends only on the time separation, givingThis double-integral derivation fixes the delta contribution without ambiguity about a half-weight at an endpoint. Put $z=D_r-i\omega$. Elementary integration givesUsing $\mathrm{Re}(1/z)=D_r/(D_r^2+{\omega }^2)$ and multiplying out the second term yieldsThe constant and decaying terms cancel at $t=0$, as required. The apparently singular case $D_r=\omega =0$ is evaluated from the integral, giving $M=4Dt+v^2{t}^2$.

Use the unwrapped angle on the real line. Integrating its Langevin equation gives $\theta (t)-\theta (0)=\omega t+\eta (t)$, where $\eta ={\int }_0^t\beta (s)ds$ is a centered Gaussian random variable with variance $2D_{r}t$. ThereforeA wrapped angular difference is bounded and would not satisfy this formula.

For the orientation correlation, $\mathbf{n}(t)\cdot \mathbf{n}(0)=\cos [\theta (t)-\theta (0)]$. The characteristic function gives $⟨e^{i\eta }⟩=e^{-⟨{\eta }^2⟩/2}=e^{-D_{r}t}$, soThe translational noise is independent of the orientation, hence the cross terms vanish. Summing its two component correlations gives $⟨\mathbf{u}(t)\cdot \mathbf{u}(0)⟩=4D\delta (t)$. Consequently, for $t\ge 0$,The extension to negative time uses $e^{-D_r|t|}$. Because the position contains Brownian motion, the instantaneous noise velocity and its correlation are distributions; the delta contribution is not a finite ordinary velocity at $t=0$.

The term $v\mathbf{n}$ is directed self-propulsion at fixed speed, while the two components of $\mathbf{u}$ generate translational Brownian motion with diffusion coefficient $D$. The angle has systematic angular velocity $\omega$ and rotational diffusion coefficient $D_r$. With independent standard Brownian motions, the precise stochastic differential equations areIndependent translational and rotational noises are the usual interpretation; the supplied autocorrelations alone would not specify arbitrary cross-correlated noises.

The complete state $(\mathbf{r},\theta \mathrm{mod}2\pi )$ is a Markov process: its future increments depend only on its present state and independent future noise. Its generator is $v\mathbf{n}\cdot \nabla +\omega {\partial }_{\theta }+D{\nabla }^2+D_r{\partial }_{\theta }^2$. Position alone generally lacks the Markov property, since an unobserved current orientation carries memory into future displacements. At long scales orientation averaging produces the effective positional diffusion equation in solution.

Write $G=G^{(0)}(\tau -{\tau }^{\prime })$ and $g_0=G^{(0)}(0)$. With two identical degree-$n$ vertices, let $r$ lines join them. The remaining $n-r$ half-edges on each vertex form self-loops. Both $n$ and $r$ are even. The number of Wick contractions issince one chooses and matches the joining half-edges and then pairs the rest independently. Dividing by the expansion denominator $2!(n!{)}^2$ gives every symmetry factor weight without guessing graph multiplicities.

For two quartic vertices, $r=0,2,4$ givesThe graphs are respectively the product of two first-order two-loop graphs; two vertices joined by two lines with one self-loop on each; and two vertices joined by four lines with no self-loops.

For two sextic vertices, $r=0,2,4,6$ givesThe corresponding numbers of self-loops on each vertex are $3,2,1,0$. All seven second-order graphs are drawn in solution; in each series the $r=0$ graph is disconnected and the others are connected.

For completeness, a mixed quartic-sextic pair, not needed for the two pure second-order coefficients, contributesTaking $\log Z$ removes the products of first-order graphs, that is, the $r=0$ terms. The requested partition function retains them.

Set the external source to zero in solution and apply Wick's theorem. At a single quartic vertex, the four position insertions have three pairings; at a single sextic vertex, six insertions have $5!!=15$ pairings. ThusThermal time-translation invariance makes this $1-\lambda \beta g_0^2/8-\mu \beta g_0^3/48+\cdots$, with $g_0$ from solution. If desired the absolute corrections follow by multiplying by the free partition function $Z_0=[2\sinh (\beta \Omega /2){]}^{-1}$.

The first Vacuum Feynman diagram is one quartic vertex carrying two self-loops, with weight $-\lambda /8$ before time integration. The second is one sextic vertex carrying three self-loops, with weight $-\mu /48$. No disconnected subtraction is appropriate for $Z$ itself. The top row of the following figure shows these first-order terms and the conventions for solution.

Keep the same measure normalization as in the free generating functional and defineHere $W$ is the source-dependent partition function, despite the common alternative convention in which that letter denotes its logarithm. A functional derivative of $e^{\int Jx}$ brings down one factor $x(\tau )$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta /\delta J(\tau )$ therefore givesCompleting the square in the free Gaussian integral gives $W_0[J]=W_0[0]\exp (\frac{1}{2}\int J{G}^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic boundary conditions. Restoring $\hslash$ replaces the upper limit by $\hslash \beta$ and the interaction coefficients by $\lambda /(4!\hslash )$ and $\mu /(6!\hslash )$, exactly as in the source.

The free Euclidean action has inverse propagator $D=-m{\partial }_{\tau }^2+\kappa$. Therefore the thermal Green's function obeyswhere ${\delta }_{\beta }$ is the periodic Dirac delta function. It is continuous, has continuous first derivative away from its source, and satisfies $G^{\prime }(0^{+})-G^{\prime }(0^{-})=-1/m$. These conditions follow by integrating the differential equation across the source and using the bosonic thermal trace.

For $\Omega =\sqrt{\kappa /m}$, solving the homogeneous equation on $0<\tau <\beta$ and imposing these conditions givesIn particular $g_0\equiv G^{(0)}(0)=\coth (\beta \Omega /2)/(2m\Omega )$. Equivalently $G^{(0)}={\beta }^{-1}{\sum }_n{e}^{-i{\omega }_n\tau }/(m{\omega }_n^2+\kappa )$ with bosonic Matsubara frequencies ${\omega }_n=2\pi n/\beta$.

Here the imaginary-time Heisenberg picture position operator is $\bar{x}(\tau )=e^{\tau H_0}\hat{x}{e}^{-\tau H_0}$, where $H_0=H(0,0)$. Bosonic time ordering $T_{\tau }$ places the larger imaginary time to the left without an exchange sign. The brackets denote the normalized thermal expectation valueThe trace is cyclic and $H_0$ is time independent, so the two-point correlation function depends only on the time difference and extends periodically with period $\beta$. This periodicity is a property of the thermal correlation function, not the operator identity $\bar{x}(\tau +\beta )=\bar{x}(\tau )$.

Use $\hslash =1$. Taking the trace of the imaginary-time time-evolution operator identifies the initial and final position, giving the path integralThe measure includes integration over the common endpoint. The paths are real periodic bosonic paths on the imaginary-time circle, not paths with fixed zero endpoints and not only classical trajectories. The kinetic measure is defined by time slicing; typical contributing paths need not be differentiable. For the free expansion assume $m,\kappa >0$. The interacting partition function requires a potential bounded below; otherwise the following formulas have the meaning of a formal perturbation series about the free harmonic oscillator.

Use the unitary discrete Fourier transform $a_j=L^{-1/2}{\sum }_k{e}^{ikj}{a}_k$, where $k=2\pi n/L$. Thenwith ${\alpha }_k=S[2J^x-(J^y+J^z)\cos k]$ and ${\beta }_k=-S(J^y-J^z)\cos k$. The factor $1/2$ accounts for the duplicated $k,-k$ pair in the full momentum sum.

Make a bosonic Bogoliubov transformation $a_k=\cosh {\theta }_k{b}_k+\sinh {\theta }_k{b}_{-k}^{†}$. Its coefficients preserve the canonical commutation relation because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The coefficient of the pair term vanishes when $\tanh 2{\theta }_k=-{\beta }_k/{\alpha }_k$. ThusThe dropped constant is $\frac{1}{2}{\sum }_k({\epsilon }_k-{\alpha }_k)$ in addition to the classical ground-state energy. Since $J^x>J^y>J^z>0$, ${\alpha }_k>|{\beta }_k|$ at every $k$, so the bosonic Bogoliubov diagonalization is stable. The minimum occurs at $k=0$ and gives the energy gap $2S\sqrt{(J^x-J^y)(J^x-J^z)}$. In the isotropic limit the pair coefficient vanishes and the familiar quadratic small-$k$ ferromagnetic dispersion relation is recovered.