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A beam transfer matrix is a linear small-angle approximation to transverse transport between two positions along a beamline:Here $x$ is transverse displacement and $x^{\prime }=dx/ds$ is trajectory slope. A drift of length $d$ and a focusing thin quadrupole magnet have respective matricesSuccessive elements compose by matrix multiplication, with the first element acting on the right. At fixed longitudinal momentum, ideal uncoupled transport governed by Hamiltonian mechanics preserves transverse phase space area and gives $\det M=1$. During acceleration, the slope coordinate is not a fixed multiple of canonical transverse momentum; the corresponding area in $(x,x^{\prime })$ can decrease through adiabatic damping of geometric emittance.

For a real one-cell matrix with $\det M=1$, its eigenvalues obey ${\lambda }^2-(\mathrm{tr}M)\lambda +1=0$. Thus $|\mathrm{tr}M|<2$ gives a conjugate pair of distinct unit-modulus eigenvalues and bounded repeated-cell motion. If $|\mathrm{tr}M|>2$, one real eigenvalue has magnitude greater than one, producing an unstable direction. At equality, a nontrivial Jordan block can produce linear growth, so unit-modulus eigenvalues alone do not establish boundedness. This is a map for beam optics; the transfer matrix used in combinatorics has a different application.

For Ising spins $s,t=\pm 1$ separated by a decorating spin $\tau =0,\pm 1$, summing its Boltzmann weights removes that spin exactly. If its two exchange interactions are $-\tau (J_{1}s+J_{2}t)$, the effective transfer matrix has entries $1+2\cosh [\beta (J_{1}s+J_{2}t)]$. Its eigenvalues are ${\lambda }_{+}=2+4\cosh (\beta J_1)\cosh (\beta J_2)$ and ${\lambda }_{-}=4\sinh (\beta J_1)\sinh (\beta J_2)$. A transfer step contains two physical sites, so the limiting Helmholtz free energy per site is $-(2\beta {)}^{-1}\log {\lambda }_{+}$.

Keep every $b$th spin in a ferromagnetic nearest-neighbour Ising model at zero field, with coupling $K=J/(k_{B}T)$. Multiplying its two-state transfer matrix $b$ times yields the displayed recursion. For $b=2$, direct summation of one intermediate spin gives $2\cosh [K(s+s^{\prime })]=A{e}^{K^{\prime }s{s}^{\prime }}$, with $K^{\prime }=\frac{1}{2}\log \cosh 2K$. The new spacing is $ba$ before coordinate rescaling. All finite positive couplings flow toward the disordered fixed point $K=0$; the ordered zero-temperature endpoint is unstable to a nonzero domain-wall number per bond.

Use the symmetric transfer matrix $T_{s{s}^{\prime }}=\exp [Ks{s}^{\prime }+h(s+s^{\prime })/2]$ with dimensionless field $h$. For blocking length two set $W=T^2$, then write $W_{s{s}^{\prime }}=A\exp [K^{\prime }s{s}^{\prime }+h^{\prime }(s+s^{\prime })/2]$. Matching its three independent entries proves the title relations and gives $A=(W_{++}{W}_{--}{W}_{+-}^2{)}^{1/4}$. Explicitly $W_{\pm \pm }=e^{2K\pm 2h}+e^{-2K}$ and $W_{+-}=2\cosh h$. At zero field the exact chain recursion is recovered; at $K=0$ the independent-spin field stays unchanged. Splitting the field equally between neighboring bonds is essential for this normalization.

For positive equal exchange constants and $2L$ symmetric Ising spins on a ring, a minimum-energy triangle has precisely one spin different from its other two. Given adjacent base spins, there is one permitted apex choice if they agree and two if they disagree. The counting transfer matrix is $C=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, with eigenvalues $3,-1$, giving ${\Omega }_L=\mathrm{Tr}{C}^L=3^L+(-1{)}^L$. Thus the residual entropy per spin is $(k_B/2)\log 3$. If the two apex bonds are strictly stronger than the base bond, only the two globally reversed configurations minimize all triangles and this residual entropy vanishes.

Take $2M$ base Ising spins on a ring and add one apex Ising spin over every other base bond. Give all base and apex bonds the antiferromagnetic energy $Js{s}^{\prime }$ with $J>0$. There are $3M$ sites, and adjacent triangles are connected by a single undecorated bridge rather than sharing a corner. With $K=J/(k_{B}T)$, summing the apex spins yields alternating spin-chain transfer matricesTheir common symmetric eigenvector gives the leading cell eigenvalue $3+e^{-4K}+e^{-2K}+3e^{2K}$. A minimum-energy triangle has one frustrated bond and energy $-J$; its connecting bridge can have energy $-J$. Requiring the bridge to be antiparallel leaves the counting transfer matrix $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ on successive first base spins. Its eigenvalues $3,1$ give ground-state degeneracy $g_M=3^M+1$ and ground-state energy $-2JM$. The residual entropy per site is $k_B\ln 3/3$.

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 $-J{\sum }_j{\tau }_j({\sigma }_j+{\sigma }_{j+1})$. The subsequently mentioned $h$ has no term in that energy, so no magnetic field or single-ion anisotropy is added. Such an extra term would define another model and change the answer.

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.

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.

Let $M=L/2$, so that the total number of Ising spins is $3M$. Write $K=\beta J$ and group each triangle's two base spins and its apex into one cell. The Ising chain with alternating triangular decorations has a triangle bond followed by an undecorated connecting bond. Summing an apex Ising spin first gives a decorated-bond Boltzmann weightThe undecorated bridge has Boltzmann weight $B_{t{s}^{\prime }}=e^{-Kt{s}^{\prime }}$. In the basis $+1,-1$,Each cell contributes $T=AB$, so periodic boundary conditions giveThe two matrices have common normalized eigenvectors $(1,1{)}^{\mathsf{T}}/\sqrt{2}$ and $(1,-1{)}^{\mathsf{T}}/\sqrt{2}$, expressing spin inversion symmetry. Their respective eigenvalues areConsequently the cell transfer matrix has eigenvaluesand the finite-size partition function isFor $K>0$, ${\lambda }_{-}\ge 0$ and ${\lambda }_{+}-{\lambda }_{-}=4+2e^{-2K}+2e^{2K}>0$, so only ${\lambda }_{+}$ survives in the thermodynamic limit. Divide the Helmholtz free energy by $3M$, the number of sites rather than the number of cells or base spins:As a counting check, $K=0$ gives ${\lambda }_{+}=8$, ${\lambda }_{-}=0$, hence $Z_L=8^M=2^{3M}$, exactly the number of independent Ising spin configurations. This model has separated decorated bonds and should not be confused with the corner-sharing sawtooth Ising chain, which has a different site count.

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

Continue the transmission amplitude to complex wavenumber $q=i\kappa$ with $\kappa >0$, equivalently complex momentum $k=i\hslash \kappa$. Then $E=-{\hslash }^2{\kappa }^2/(2m)<0$. In the left exterior region the $A$ term grows at negative infinity and the $B$ term decays; in the right exterior region the $D$ term grows and the $C$ term decays. A bound state thus requires $A=D=0$, with nonzero $B,C$. The transfer matrix relation gives $M_{22}(i\kappa )=0$. Equivalently, a zero of the incoming coefficient is a bound-state pole of $t(q)$, whose numerator is nonzero at $q=i\kappa$.

The denominator found in part (c) givesFactorization produces two branches. To identify their parity, use an even interior wavefunction $A\cosh (\kappa x)$ or an odd interior wavefunction $A\sinh (\kappa x)$ for $|x|<a$, joined to a decaying exterior wavefunction at $a$. Continuity and the derivative jump give, respectively,Rearrangement gives the implicit energy eigenvalue equationsOnly strictly positive equation roots $\kappa$ are retained. With $\nu =-\mu >0$, the even equation root lies above $\nu$ and the odd equation root, when it exists, lies below $\nu$, so the even bound state has lower energy. The formal zero equation root in the odd equation does not itself establish a normalizable wavefunction.

For a ferromagnetic nearest-neighbour Ising model on a chain let $s_i=\pm 1$, $K=J/(k_{B}T)>0$ and $h=H/(k_{B}T)$. The dimensionless Hamiltonian is $-K{\sum }_i{s}_i{s}_{i+1}-h{\sum }_i{s}_i$. In one dimension, decimating every other spin closes exactly within the nearest-neighbour family, including its additive constant. This is the exact chain building block of the Migdal-Kadanoff approach; bond moving becomes an additional approximation in higher-dimensional lattices.

For exact zero-field Ising-chain decimation, sum over an intermediate spin $\sigma$ between two retained spins:Parallel and antiparallel end spins give $A{e}^{K^{\prime }}=2\cosh 2K$ and $A{e}^{-K^{\prime }}=2$. HenceFor an even periodic chain, $Z_N(K)=A^{N/2}{Z}_{N/2}(K^{\prime })$. Rescaling the lattice spacing $2a$ back to $a$ completes the renormalization step. Decimation of $b$ successive links gives $\tanh K^{\prime }=(\tanh K{)}^b$, as follows by multiplying the transfer matrix $b$ times.

Let $v=\tanh K$. Then $v^{\prime }=v^2$ on $0\le v\le 1$. The infinite-temperature renormalization-group fixed point $v=0$ attracts every $0\le v<1$; the zero-temperature renormalization-group fixed point $v=1$ is unstable to heating, with slope two. There is no intermediate ferromagnetic renormalization-group fixed point. Equivalently $K^{\prime }\simeq K^2$ for small $K$, while $K^{\prime }\simeq K-\frac{1}{2}\log 2$ for large finite $K$. Thus there is no finite-temperature ordered thermodynamic phase in the short-range chain.

The domain-wall scaling coordinate of an Ising chain near $T=0$ is the domain-wall Boltzmann factor $w=e^{-2K}$. Its recursion and the zero-field correlation length areSince $\xi (K^{\prime })/a=[\xi (K)/a]/2$, the recursion has the correct length scaling. Its exponent is one with respect to $w$; it is not a finite power of physical temperature about $T_c=0$. In the thermodynamic limit, a bond is a domain wall with probability $(1-\tanh K)/2\sim e^{-2K}$, giving the same exponentially large length scale. Their finite energy allows entropy to proliferate walls at any positive temperature.

The Ising-chain decimation constant matters for thermodynamics. With dimensionless free energy per spin $f=-\lim N^{-1}\log Z_N$,The second expression follows independently from transfer eigenvalues $2\cosh K$ and $2\sinh K$ and satisfies the recursion. Multiplying it by $k_{B}T$ restores the physical free energy. Dropping $A$ would preserve normalized retained-spin probabilities but lose the eliminated-spin entropy.

Finite-field Ising-chain decimation is also exact. Use the symmetric transfer matrixand read the new couplings fromHere $W_{++}=e^{2K+2h}+e^{-2K}$, $W_{--}=e^{2K-2h}+e^{-2K}$ and $W_{+-}=2\cosh h$. The latter off-diagonal entry, the field convention split between adjacent bonds and the constant must all be tracked consistently. Near the zero-temperature zero-field renormalization-group fixed point $h^{\prime }\simeq 2h$, while at $K=0$ the independent-spin field is unchanged.

In the higher-dimensional Migdal-Kadanoff approximation, moving bonds creates chains that can be decimated. With bond moving first on a $d$-dimensional hypercubic lattice and length factor $b$, the strengthened links have $b^{d-1}K$, giving the conventionFor $d=1$ no transverse bonds move, so this reduces to the exact chain result. In higher dimensions it discards geometric correlations and is generally approximate on the original lattice; moving and decimating in another order yields another approximation, not a contradictory exact chain formula.^[ref]

For the sawtooth Ising chain, each apex spin occurs in only one triangle. Put $K=\beta J$. Summing that spin first gives the effective transfer matrixIn the ordered basis $s=+1,-1$ this isIts symmetric and antisymmetric eigenvectors are $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$, with eigenvaluesConsequentlyAt $J=0$, the result is $Z_L=4^L$, correctly counting $2L$ independent Ising spins. As $T\to 0$, $f_{\infty }\to -3J/2$, the energy of three satisfied ferromagnetic bonds per two sites.

The printed summand lacks grouping brackets. The result above uses the translation-invariant common-$J$ interpretation. If base and apex bonds instead have independent couplings $J_b,J_a$, the same derivation gives ${\lambda }_{\pm }=2e^{\beta J_b}\cosh (2\beta J_a)\pm 2e^{-\beta J_b}$; specifying those couplings is essential before giving a different answer.

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.

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.

Let $M=L/2$, so that the total number of Ising spins is $3M$. Write $K=\beta J$ and group each triangle's two base spins and its apex into one cell. The Ising chain with alternating triangular decorations has a triangle bond followed by an undecorated connecting bond. Summing an apex Ising spin first gives a decorated-bond Boltzmann weightThe undecorated bridge has Boltzmann weight $B_{t{s}^{\prime }}=e^{-Kt{s}^{\prime }}$. In the basis $+1,-1$,Each cell contributes $T=AB$, so periodic boundary conditions giveThe two matrices have common normalized eigenvectors $(1,1{)}^{\mathsf{T}}/\sqrt{2}$ and $(1,-1{)}^{\mathsf{T}}/\sqrt{2}$, expressing spin inversion symmetry. Their respective eigenvalues areConsequently the cell transfer matrix has eigenvaluesand the finite-size partition function isFor $K>0$, ${\lambda }_{-}\ge 0$ and ${\lambda }_{+}-{\lambda }_{-}=4+2e^{-2K}+2e^{2K}>0$, so only ${\lambda }_{+}$ survives in the thermodynamic limit. Divide the Helmholtz free energy by $3M$, the number of sites rather than the number of cells or base spins:As a counting check, $K=0$ gives ${\lambda }_{+}=8$, ${\lambda }_{-}=0$, hence $Z_L=8^M=2^{3M}$, exactly the number of independent Ising spin configurations. This model has separated decorated bonds and should not be confused with the corner-sharing sawtooth Ising chain, which has a different site count.

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

For the sawtooth Ising chain, there is a source inconsistency: the printed Ising spin values are $\{-1,r+1\}$, with no definition of $r$. The requested transfer matrix follows for $r=0$, namely the usual Ising model with values $\{-1,+1\}$. For arbitrary $r$, summing an apex spin actually givesFor example, $r=1$, $s=s^{\prime }=-1$ and $\beta \bar{J}=1$ give the bracket $e^{-2}+e^4$, whereas the claimed hyperbolic cosine gives $e^{-2}+e^2$. Thus the printed identity is false literally. The rest of Question 1 uses the intended symmetric Ising spin values, $r=0$.

For a fixed configuration of base spins, each apex spin appears in only one local Boltzmann factor. The apex spins are therefore conditionally independent, and their finite sums factorize:Only the base spins carry information from one triangle to the next. The apex spins have already been summed out, so no apex index remains in the transfer matrix; the final trace implements the periodic boundary condition.

In the $s=+1,-1$ basis the transfer matrix and its eigenvalues areFor finite positive temperature, $a,b>0$ and ${\lambda }_{+}>|{\lambda }_{-}|$. Consequently $Z={\lambda }_{+}^L+{\lambda }_{-}^L$ and $L^{-1}\log Z\to \log {\lambda }_{+}$ in the thermodynamic limit. There are $2L$ spins, so the Helmholtz free energy per spin isThe factor $1/2$ is essential: one transfer matrix step counts two spins. For completeness, if the undefined $r$ is retained, set $a_r=A_{-1,-1}^{(r)}$, $d_r=A_{r+1,r+1}^{(r)}$ and $b_r=A_{-1,r+1}^{(r)}$. The same thermodynamic limit gives $f_r=-(2\beta {)}^{-1}\log [(a_r+d_r+\sqrt{(a_r-d_r{)}^2+4b_r^2})/2]$ whenever these are two distinct spin values.

For one triangle, equal base spins $s=s^{\prime }=\sigma$ give energy $J+2\bar{J}\sigma \bar{s}$. Their minimum occurs at $\bar{s}=-\sigma$ and is $J-2\bar{J}$. Opposite base spins give energy $-J$, independently of the apex spin. The all-equal configurations have energy $J+2\bar{J}$.

When $\bar{J}=J$, the six configurations with precisely one spin different from the other two all have minimum energy $-J$. For a fixed first base spin, there is one minimum choice if the next base spin is the same, and two choices of apex if it is opposite. The ground-configuration counting transfer matrix is thereforeThis ground-state counting of the frustrated sawtooth Ising chain proves the extensive geometric frustration and the residual entropy $k_B{\lim }_{L\to \infty }\log {\Omega }_L/(2L)=(k_B/2)\log 3$.

When $\bar{J}>J$, $J-2\bar{J}<-J$. Every triangle must have equal base spins and the opposite apex spin. Shared base spins propagate that choice around the chain, leaving just two global ground states, of energy $L(J-2\bar{J})$. Their entropy per spin is $k_B\log 2/(2L)\to 0$.

For couplings favoring ferromagnetism, write $J=-K$, $\bar{J}=-\bar{K}$ with $K,\bar{K}>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base spins then make the whole chain uniform. There are two global ground states for either $K=\bar{K}$ or $K\ne \bar{K}$, so no corresponding distinction in residual entropy occurs. This statement assumes all bonds remain nonzero.