With $n$ individuals, each of the $n$ can reproduce at rate $\lambda$ and independently die at rate $\lambda$. Thus the total birth and death rates are $q_n=r_n=\lambda n$. The gain into state $n$ comes from a birth in state $n-1$ and a death in state $n+1$, while its loss rate is $2\lambda n$. ThereforeAt $n=0$, the birth inflow from negative states is absent and ${\dot{P}}_0=\lambda P_1$. The empty state is absorbing: once the population reaches zero, neither process can restart it. Equal birth and death rates make this a critical branching process.

For the Poisson process, $q_n=g$ and $r_n=0$, soThis is an ordinary differential equation in time for each fixed $z$, with solutionConsequentlyThe initial condition is interpreted as $P_0(0)=1$, with all other initial probabilities zero. Normalization follows from $F(1,t)=1$. The mean and variance are both $gt$, as expected for independent events arriving at constant rate.

Multiply the master equation by $z^n$ and sum over $n$. Shifting indices producesThe apparent negative power causes no problem at $z=0$ when $r_0=0$. Let $D=z{\partial }_z$, so $D{z}^n=n{z}^n$. If the rates are polynomials in $n$, replace their arguments by $D$ to obtain a finite-order partial differential equation:For arbitrary rates the weighted sums still give an exact generating-function evolution, but need not reduce to a finite-order differential operator. The probability generating function obeyswhich is the normalization condition. The evolution has ${\partial }_{t}F(1,t)=0$ when the rate sums converge, so an initially normalized distribution stays normalized. Its derivatives at $z=1$ also give the factorial moments, and expansion around $z=0$ recovers individual probabilities.

This is a birth-death master equation: $q_n$ is the transition rate $n\to n+1$ and $r_n$ is the transition rate $n\to n-1$. The first two terms are incoming probability flows from the adjacent states; the last removes probability at the total escape rate $q_n+r_n$. Rates must be nonnegative. For a population on the nonnegative integers, take $r_0=0$ and no incoming flux from negative states.

Summing over all states and shifting the two incoming indices givesThus an initially normalized probability distribution remains normalized. The cancellation assumes no boundary flux and legitimate summation; on an infinite state space this means the nonexplosive process or a boundary prescription that conserves total probability. All rates used in the later parts satisfy this requirement.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

Choose the local positive $\tilde{z}$ axis along the classical spin. For nonzero field, a proper rotation isIt preserves the Spin commutation relations and turns the field term into $-H_{f}S{\sum }_j{\tilde{S}}_j^z$. At zero field choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this rotation would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.

To the required order,The longitudinal exchange gives $-J{S}^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange givesFinally the field contributes its classical constant and $H_{f}S{n}_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,These are the order-$S$ terms; the discarded products of number operators and the next square-root terms are order $S^0$ for occupations held fixed as $S\to \infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the spectrum, but the sign above follows from the stated rotation and oscillator convention.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne 0$, $s=\mathrm{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either field direction.

For classical vectors of length $S$, define ${\mathbf{q}}_j=(S_j^x,S_j^y)$. The Cauchy-Schwarz inequality gives ${\mathbf{q}}_j\cdot {\mathbf{q}}_{j+1}\le |{\mathbf{q}}_j||{\mathbf{q}}_{j+1}|\le S^2$, and $h{S}_j^x\le H_{f}S$. Each term is therefore bounded below, soUniform alignment along the field saturates every bound:Equality requires all planar vectors to be parallel with length $S$, and the nonzero field then fixes their direction. At $h=0$, every uniformly oriented vector in the $xy$ plane is a classical ground state:A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

Insert the decorating-spin observable into the one bond that contains it. The decorated-spin occupation insertion uses the matrix:The partition function still contains $L$ decorating-spin sums; one of them has been replaced by this weighted sum. HenceTaking the dominant eigenvector givesThis is the probability that the decorating site is in either nonzero state. It tends to $2/3$ at high temperature, when its three states are equally likely, and to one at low temperature, when the nonzero aligned states gain bond energy. Unlike the magnetization, this even observable is not forced to vanish by spin reversal.

Insert $D=\mathrm{diag}(1,-1)$ at the first undecorated site. In the eigenbasis of the transfer matrix, $D$ exchanges $v_{+}$ and $v_{-}$, and both diagonal matrix elements vanish. Explicitly,Thus its large-$L$ limit is zero at every fixed $T>0$. The result is exact even before taking that limit. Equivalently, simultaneous reversal of all spins pairs configurations of equal energy and opposite ${\sigma }_1$. Positivity of the transfer matrix at fixed positive temperature prevents a degenerate dominant eigenvalue and spontaneous magnetization in this one-dimensional zero-field chain. Selecting one aligned state at strictly zero temperature is a different order of limits.

The sum needs a stated interpretation. The original equation leaves the second bond outside explicit parentheses, with its index then formally free. The natural translation-invariant, equal-bond interpretation used here is the energy $-J{\sum }_j{\tau }_j({\sigma }_j+{\sigma }_{j+1})$.

Let $K=\beta J$. Sum over each decorating spin before summing over the remaining spins. The decimation of an alternating mixed-spin chain gives the effective transfer matrix:No on-site entropy has been omitted: the term $1$ is the weight of the zero decorating spin. The normalized eigenvectors are $v_{\pm }=(1,\pm 1{)}^{\mathsf{T}}/\sqrt{2}$, with eigenvaluesThere are $L$ transfer steps but $2L$ physical sites, soThe free energy per two-site unit cell is twice this value. At $J=0$, the formula gives $Z_L=6^L$, as required by two states on one sublattice and three on the other. As $T\to 0$, the free energy per physical site tends to $-J$, the energy of an aligned chain with two satisfied bonds per unit cell. These checks also fix the factor of two in the site count.

Put $\beta =(k_{B}T{)}^{-1}$ and label the two spin values in the order $+1,-1$. Assign each bond its Boltzmann weight,The symmetry of the bond energy makes $V$ a real symmetric matrix. For a fixed configuration the product of successive matrix entries is its Boltzmann weight. Summing over all spins and identifying the last index with the first therefore giveswhereThis replaces a sum over $2^L$ configurations by the powers of a two-by-two matrix. For finite bond energies at $T>0$, every matrix entry is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>0$ and $|{\lambda }_{-}|<{\lambda }_{+}$. Hence the Helmholtz free energy per spin tends to $-k_{B}T\log {\lambda }_{+}$. A diagonal local observable $O$ can be inserted at one site: $⟨O_1⟩=\mathrm{tr}(O{V}^L)/Z_L$. Its limiting average is the expectation in the normalized eigenvector of ${\lambda }_{+}$.

Label the initial particle and antiparticle $p_1,p_2$ and the corresponding final momenta $p_3,p_4$, with $s=(p_1+p_2{)}^2$ and $t=(p_1-p_3{)}^2$. Charge flow permits an $s$-channel heavy-scalar annihilation graph and a $t$-channel heavy-scalar exchange graph. It excludes a third $u$-channel graph for these fixed charge assignments. The effective theory has one contact graph.

The two cubic Feynman vertices and the heavy propagator giveFor $|s|,|t|\ll M^2$,This independently checks the sign and the factor $1/2$ in solution. The next difference is the derivative correction of order $M^{-4}$ in the amplitude.

Let $E=\sqrt{s}$ denote the total center-of-mass energy, with $m\ll E\ll M$. The final particle and antiparticle are distinct, so no identical-particle factor is needed. The leading isotropic differential scattering cross-section and total cross section areCorrections are relatively of order $E^2/M^2$ and, in a high-energy expansion, $m^2/E^2$, in addition to loop corrections when they matter. If instead $E$ denotes each beam particle's energy, substitute $s=4E^2$ before comparing numerical prefactors.

Put $\rho ={\phi }^{†}\phi$. The heavy-field equation of motion isThis solution describes the low-energy sector without an independent incoming heavy-particle wave. It is valid when all external momentum invariants and light masses squared are small compared with $M^2$.

After an integration by parts, the terms involving the heavy field are $-\frac{1}{2}\varphi (\square +M^2)\varphi -g\rho \varphi$. Substituting the solution into both terms, rather than the source term alone, givesHence, with the question's sign convention for the local quartic interaction,This is integrating out a field followed by a derivative expansion. The negative quartic means attraction. It also makes the truncated potential unbounded at large fields; minimizing the original potential over $\varphi$ reveals the same negative quartic. Thus neither displayed model, without extra stabilizing interactions, supplies a globally stable nonperturbative vacuum. The matching describes perturbative low-energy scattering about the local vacuum. The simple equation-of-motion matching fixes the leading tree-level coefficient; precision quantum matching requires consistent loop renormalization and the additional effective operators at the requested accuracy.

The first-order Dyson series contains $-ig\int d^{4}w{\phi }^{†}\phi \varphi$. For an amputated three-point Feynman diagram, each of the three distinct field types contracts in exactly one way. Writing each external plane wave in momentum space producesThus the momentum-space Feynman rule, with all four-momenta incoming, isAfter stripping the conservation delta factor, it is simply $-ig$. There is no $3!$ combinatorial factor because the interaction has three distinct field slots. The charged lines carry opposite charge-flow orientations and the real heavy line is neutral.

Let ${\Delta }_M$ be the free real-scalar Feynman propagator and $D_m(x-y)=⟨0|T\phi (x){\phi }^{†}(y)|0{⟩}_0$ the free complex-scalar one. Expanding the interaction exponential once and applying Wick's theorem gives the connected decay Green's functionThere is one allowed contraction of each external field with the appropriate vertex field, so no factorial remains. Fourier transformation gives the momentum-conserving delta factor times $(-ig){\tilde{\Delta }}_M(p){\tilde{D}}_m(q_1){\tilde{D}}_m(q_2)$, and amputation of external propagators gives the decay vertex $-ig$. An on-shell physical decay additionally requires $M>2m$.

If one literally asks for the full three-point correlation function without a connected restriction or a tadpole condition, it also containsthe product of the induced heavy-field one-point function and the free light-field two-point function. This disconnected tadpole diagram does not describe a three-particle decay and drops out of its connected amputated amplitude.

For two equally charged daughters the corresponding Green's function vanishes to every perturbative order:Every interaction contains equal numbers of $\phi$ and ${\phi }^{†}$, so the two excess conjugate fields can never be paired. Equivalently the exact global U(1) symmetry $\phi \mapsto e^{i\alpha }\phi$, ${\phi }^{†}\mapsto e^{-i\alpha }{\phi }^{†}$ leaves the vacuum and action invariant, whereas this insertion has charge $-2$. Charge conservation forbids the decay of a neutral particle into two particles of the same nonzero charge.

Use a Hermitian gauge field $A_{\mu }=A_{\mu }^A{T}^A$ and $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. With fundamental covariant derivative ${\partial }_{\mu }-ig{A}_{\mu }$, the adjoint derivative isIt transforms as $D_{\mu }\Phi \mapsto U(D_{\mu }\Phi )U^{†}$, provided $A_{\mu }^{\prime }=U{A}_{\mu }{U}^{†}-(i/g)({\partial }_{\mu }U)U^{†}$.

The infinitesimal parameter in $U=1+iϵ+O({ϵ}^2)$ must be Hermitian and traceless, from unitarity and unit determinant. It is not generally antisymmetric as the hint states. The valid generator $ϵ=\mathrm{diag}(1,-1,0,0,0)$ is a concrete counterexample to that assertion.

Choose canonical adjoint kinetic normalization ${\mathcal{L}}_{\Phi }=\mathrm{tr}(D_{\mu }\Phi D^{\mu }\Phi )$, which gives $\frac{1}{2}\partial {\Phi }^A\partial {\Phi }^A$. For the diagonal vacuum expectation value $D=v\mathrm{diag}(2,2,2,-3,-3)$, the matrix unit $E_{ij}$ obeys $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the mass of the gauge boson associated with either normalized Hermitian generator joining $i,j$ is $g|d_i-d_j|$.

Within the three-dimensional and two-dimensional blocks, or for diagonal generators, the commutator vanishes. There are therefore $8+3+1=12$ massless gauge bosons. The $3\times 2$ interblock pairs supply two real generators each, hence 12 equally massive gauge bosons:The absolute mass formula assumes the stated kinetic normalization, which the question leaves implicit. Writing instead $\frac{1}{2}\mathrm{tr}(D\Phi {)}^2$ without rescaling $\Phi$ would multiply these squared masses by $1/2$.

This is the gauge group of the Standard Model, up to its global quotient, arising from one adjoint Higgs mechanism in a simple unified group. The unbroken abelian generator can be chosen as $Y=\mathrm{diag}(-1/3,-1/3,-1/3,1/2,1/2)$, so $D=-6vY$. This is the characteristic first breaking step of an SU(5) grand unified theory, rather than electroweak breaking all the way to electromagnetism.

Write $R^2={\sum }_i{x}_i^2$ and impose tracelessness by a Lagrange multiplier $\zeta$. Differentiating $V-\zeta {\sum }_i{x}_i$ givesFor $b\ne 0$, each eigenvalue is a real root of the same nonzero cubic polynomial, so at most three distinct values occur at any stationary point, in particular at any existing minimum. If the multiplicities are $n_1,\dots ,n_k$, the unbroken subgroup is the block centralizerIts connected Lie algebra is ${⨁}_{\alpha }\mathfrak{su}(n_{\alpha })\oplus \mathfrak{u}(1{)}^{k-1}$. The partitions allowed by this upper bound are $5$, $4+1$, $3+2$, $3+1+1$, and $2+2+1$. These are candidate patterns, not a proof that every one is a stable minimum for given couplings. For three distinct roots, the absent quadratic coefficient also implies that their unweighted sum is zero; tracelessness imposes an additional multiplicity-weighted relation.

The unrestricted printed claim has an exception. If $b=0$, $a>0$ and $m>0$, every traceless matrix with $R^2=m^2/(2a)$ is a global minimum. For examplehas five distinct eigenvalues and attains that minimum. Its unbroken subgroup is the maximal torus $U(1{)}^4$. The cubic argument therefore requires a nonzero $b$, or another condition excluding the identically vanishing stationarity polynomial. Couplings for which the potential is unbounded below have no global minimum to classify.

The spectral theorem for a Hermitian matrix gives a unitary diagonalizer $V$ with $V\Phi V^{†}=\mathrm{diag}(x_1,\dots ,x_5)$ and real $x_i$. Similarity preserves the trace, so ${\sum }_i{x}_i=0$.

To ensure that the diagonalizer belongs to SU(5) rather than only the unitary group $U(5)$, write $\det V=e^{i\alpha }$ and put $U=e^{-i\alpha /5}V$. Its determinant is one, and the scalar phase cancels from its conjugation action. ThusThe same argument shows why no additional reality condition beyond Hermiticity is needed. Permuting the real eigenvalues merely changes their order in the diagonal form.