Lecture 16 found the ground states. Physics happens above them. So the question is

The answer: dressing (Definition 17.1) gives an excitation’s energy and the derivative of its momentum (Proposition 17.2), which gives Lieb’s two branches; at $T>0$ the Yang–Yang equation (Theorem 17.7) gives the pressure. The worked example is the gas at $T=0$.

Every question asked of a Bethe state in the thermodynamic limit has the same shape. Adding one particle to the sea moves every other root a little, so the change in any extensive quantity is the particle’s own contribution plus the backflow. One linear operator does the bookkeeping for all of them.

The definition is worth its own name because (16.5) is already an instance of it. It is a response, not a total: the dressed bare momentum $k_0^{\mathrm{dr}}$ is not the excitation’s momentum; the momentum comes from integrating $(k_0')^{\mathrm{dr}}$.

Two things fix the object in place. Equation (17.1) is linear with the same kernel for every $f$, so dressing is one operator $(1-\tfrac1{2\pi}K\vartheta)^{-1}$: a single numerical solve answers every question. And $\vartheta$ carries all the state dependence — sharp at $T=0$, a smooth Fermi function at finite temperature (below), position- and time-dependent out of equilibrium (Lecture 20) — with (17.2) the dressed velocity throughout. What does not carry over is the route to it: at $T>0$ the identity $\varepsilon'=(\varepsilon_0')^{\mathrm{dr}}$ comes from differentiating Yang–Yang’s nonlinear equation for the pseudo-energy (Corollary 17.8).

The gas has a Fermi sea with edges, so it has two ways to be excited; Lieb found both in 1963, and both come off one pair of functions. The $T=0$ dressed energy and momentum are $$\varepsilon(k)-\frac1{2\pi}\int_{-q}^{q}K(k-k')\varepsilon(k')\dd k' =k^2-\mu, \qquad p'(k)=2\pi\rho_t(k),$$(17.3) the first being (17.1) for the bare energy $k^2-\mu$ and the second Proposition 17.2; the constant $\mu$ — the chemical potential, and in this lecture there is no spectral parameter to confuse it with — is fixed by $\varepsilon(\pm q)=0$. That condition does two things at once: it puts both branches at zero energy where they begin, and it is exactly the hypothesis of Proposition 17.2, so the velocity formula applies. Now add a particle just above the right edge, $k>q$: energy $\varepsilon(k)>0$ and momentum $p(k)-p(q)$ trace the Type I branch, the Bogoliubov mode (a particle below $-q$ gives the mirror branch, referred to $-q$). Move a particle from inside the sea to its edge, leaving a hole at $\abs k<q$: energy $-\varepsilon(k)>0$ and momentum $p(q)-p(k)$ trace Type II, which rises, turns over and returns to zero at momentum $2\pi n$ sketched: the branches are computed here from (17.3); that they exhaust the elementary excitations is cited (Lieb 1963).

Type II has no counterpart in Bogoliubov theory — it is the exact version of a dark soliton cited: Kulish–Manakov–Faddeev (1976) — and both branches are linear at small momentum with the same slope, the sound velocity that makes the gas a Luttinger liquid.

Everything in this lecture rests on one sentence of Lecture 16: a Bethe state is not a list of rapidities but a choice of quantum numbers from the list of vacancies. Choose a subset, solve (16.1), and the roots come out. So counting states is counting subsets — and that is a problem with an answer. For the chain the list is finite, of length $N-M$, and on the branch where $\zeta$ increases the roots occupy $M$ distinct members of it (Proposition 16.1, whose second sentence Remark 16.2 showed to be a hypothesis); for the gas the list is all of $\Z+\tfrac{N-1}2$ and is not finite at all. What the count below uses is neither, but the local density of vacancies $\rho_t$, which both models have.

Fix a coarse-graining: cut the rapidity line into cells of width $\dd\lambda$, small on the scale over which $\rho$ varies but wide enough to hold many vacancies. In the cell at $\lambda$ there are $L\rho_t(\lambda)\dd\lambda$ slots and $L\rho(\lambda)\dd\lambda$ of them carry roots, where $L$ is the number of sites $N$ or the length of the box, as in (16.5). The cells are independent, so

proved as a count, from the hypotheses named below

At temperature $T$ and chemical potential $\mu$ the equilibrium state minimises the grand free energy per unit length, $$g[\rho]=\int\bigl(\varepsilon_0(\lambda)-\mu\bigr)\rho(\lambda)\dd\lambda-Ts ,$$(17.5) where $\varepsilon_0$ is the bare energy of one root — $\varepsilon_0=k^2$ for the gas of Lecture 11, $\varepsilon_0=-\pi Ja_1$ for the chain (the reading at the end). Here $\mu$ is the chemical potential, as in Lieb’s branches.

proved from Proposition 17.4; that the minimiser gives the true free energy is Yang and Yang’s saddle-point argument, sketched after the proof

Why the minimum is the free energy. The argument is a saddle point. Write the partition function as a sum over sets of quantum numbers, group the sets by their coarse-grained densities, and the sum becomes $Z=\sum_{\{\rho\}}e^{-L g[\rho]/T}$ over the coarse-grained configurations, with $g$ of (17.5): the number of terms in each group is $e^{Ls}$ by Proposition 17.4 and each contributes $e^{-L(e-\mu n)/T}$. As $L\to\infty$ the sum is dominated by its largest term and $-T\ln Z/L\to\min g$, provided the number of groups grows sub-exponentially — which is the usual requirement, and the usual thing that is waved through.

Equation (17.7) is nonlinear — the unknown sits inside a logarithm inside the integral — and that is the price of the temperature. Everything else about it is a gift: one scalar equation, one scalar unknown, and no reference to $N$ or to any particular state. And what the second half of (17.8) says is that the minimisation is a Legendre transform. The minimum value of $g=e-\mu n-Ts$ is the grand potential density, $-P(\mu,T)$; so $P$ is the Legendre transform of the energy density in both $n$ and $s$, its partners come back by differentiating, $$n=\frac{\partial P}{\partial\mu},\qquad s=\frac{\partial P}{\partial T},\qquad \frac EL=Ts+\mu n-P ,$$(17.9) and one scalar function of two variables carries the whole thermodynamics. Students lose the last of these by trying to compute $E/L$ from $\int\varepsilon\rho$, which is not it: $\varepsilon$ is not an energy.

A new equation earns trust by reproducing the two answers already known.

Part (1) carries a sentence of physics, and it is the opposite of the one that suggests itself. The shift raises $\mu$, so the gas has more pressure than the impenetrable one, while hard rods of length $a$ give $P=P_{\mathrm F}(\mu-aP)$ and less: matching, $a=-2/c$. At large finite $c$ the gas is free fermions in a box lengthened by $2/c$ per particle. The Bethe equations say so directly — $\theta(k)\approx2k/c$ turns (11.11) into $k_j(L+2N/c)=2\pi I_j$ in the rest frame, $\sum_lk_l=0$ — and so does Lecture 16’s $E/N=n^2e=\tfrac{\pi^2n^2}3(1-4/\gamma+\dots)$, which to this order is $\pi^2n_{\mathrm{eff}}^2/3$ with $n_{\mathrm{eff}}=n/(1+2n/c)$. At $c=100$, $T=\mu=1$ the shift accounts for the whole of the $0.6\%$ by which the interacting pressure exceeds the free-fermion one.

Equation (17.7) has no closed-form solution and does not need one: it is a fixed-point equation, and the obvious iteration works. It also has one fewer parameter than it looks: substituting $k=\sqrt T\,\kappa$ and $c=\sqrt T\,\hat c$ makes every $\sqrt T$ cancel between $K$ and $\dd k'$, so that $\varepsilon/T$ and $P/T^{3/2}$ depend on $(\mu/T,\hat c)$ alone. One plot is the whole equation of state.

The linearisation being differentiated here is nothing other than the dressing operator $\tfrac1{2\pi}K\vartheta$ of (17.1), so the equation that answers every question about the state is also the one that says how fast the solver converges.

Two features of Figure 17.4 deserve naming. On the dilute side every curve collapses onto $P/T^{3/2}=e^{\mu/T}/2\sqrt\pi$: a gas dilute enough is classical, and the coupling cannot be seen. On the degenerate side the curves fan out, and the free-fermion curve is the floor: since $K>0$, (17.7) gives $\varepsilon\le k^2-\mu$ pointwise and hence $P\ge P_{\mathrm F}$ at the same $\mu$ and $T$. The impenetrable gas has the least pressure there is, which is the statement of Proposition 17.9(1) seen from the other side.

Equation (17.7) is the equation of state of a real gas: a density profile measured on an atom chip in 2008 is reproduced with no free interaction parameter (the reading at the end).

• Dressing, one linear operator for every quantity, and the dressed velocity (Proposition 17.2 and Corollary 17.8): the objects Lecture 20 carries out of equilibrium.

• Lieb’s two branches and the speed of sound (Figure 17.1).

• The exact equation of state of an interacting quantum gas at every temperature and coupling (Theorem 17.7), checked in both limits (Proposition 17.9) and measured. Goal G3 is reached.

• Status. Proposition 17.2, Theorem 17.7, Corollary 17.8, Proposition 17.9 and Proposition 17.10 are proved, the first, fourth and fifth in readings; Theorem 17.7 rests on the count, whose assumptions (Remark 17.5) are not proved. That the minimum is the free energy is a saddle-point argument, sketched; that the Yang–Yang equation has one solution is cited, as is that Lieb’s two branches are all the elementary excitations.

• Two hypotheses in Proposition 17.2. Its second half needs $T=0$ and $\varepsilon(\pm q)=0$; at $T>0$ the identity is derived again, from the Yang–Yang equation (Corollary 17.8).

• The effective velocity at $T>0$ is a definition (17.10); that it is a propagation speed is not proved here.

• The chain needs strings, and the string hypothesis is false as a statement about individual solutions; its thermodynamics is a reading.

Return to the antiferromagnet and make the cheapest excitation there is: remove one root. The sector is $M=N/2-1$, which by Proposition 16.1 has $N/2+1$ vacancies for $N/2-1$ roots — two holes, at $\lambda_{h_1},\lambda_{h_2}$ — while the magnetisation has gone up by one, $\Delta S^z=1$. Two objects, one unit of $S^z$ between them: by itself that would also allow a spin $1$ and a spin $0$. What settles it is the multiplet structure — at fixed $\lambda_{h_1},\lambda_{h_2}$ the states form one triplet and one singlet, exactly as two spin-$\tfrac12$ objects do cited: Faddeev–Takhtajan (1981). They are the spinons, and that is the most surprising sentence about the chain: the elementary excitation of a spin chain is not a magnon of spin $1$ but half of one, and it cannot be made alone.

proved from the first-order form of the assumption noted after the proof; the dispersion is that of des Cloizeaux and Pearson (1962)

The proof is a backflow computation. With the bare magnon energy $\varepsilon_0=-2J/(4\lambda^2+1)=-\pi J a_1$, putting $r$ delta-function holes into (16.4) and inverting $1+a_2*$ dresses the energy of the removed magnon, $-\varepsilon_0=\pi Ja_1$ — and that equation is (16.7) again, so $\varepsilon=(-\varepsilon_0)^{\mathrm{dr}}=\pi J\rho$. It runs on one order finer information than Theorem 16.3’s assumption supplies: at fixed $r$ the hole density tends to zero, and (17.11) needs $\lim N(\rho_N-\rho_{(0),N})$. The careful route subtracts the two finite systems and linearises before letting $N\to\infty$ cited: Takahashi §3.4; it gives the same $\varepsilon$, and finite chains agree (the check script, $N=64$ to $1024$).

A two-spinon state has $\omega=\varepsilon(p_1)+\varepsilon(p_2)$, $k=p_1+p_2$, with $p_1,p_2\in(0,\pi)$ distinct (the holes sit at distinct vacancies). At fixed $k$, $\sin p_1+\sin p_2=2\sin\tfrac k2\cos\tfrac{p_1-p_2}2$ grows as $p_1-p_2$ shrinks and falls as one spinon reaches an end of its band.

The lower boundary is the des Cloizeaux–Pearson dispersion of 1962, read for thirty years as the spin-wave branch; the upper one shows it is only the edge of a continuum. That is the measurable difference between a spin chain and a magnet: a three-dimensional antiferromagnet scatters neutrons into a sharp magnon line, the spin-$\tfrac12$ chain into a broad continuum, because a neutron flips one spin and therefore makes spinons in pairs — two at least, any even number in principle — which share the energy as they like. Inelastic scattering on KCuF$_3$ sees such a continuum, and the 1995 measurement was fitted with an approximate two-spinon cross-section, the Müller ansatz. The exact two-spinon contribution, computed later, exhausts about $73\%$ of the integrated intensity; four spinons and more carry the rest cited: Tennant, Cowley, Nagler and Tsvelik, Phys. Rev. B 52 (1995) 13368; Müller, Thomas, Beck and Bonner, ibid. 24 (1981) 1429; Lake et al., Phys. Rev. Lett. 111 (2013) 137205.

Which way does $p$ run? Theorem 17.11 could as well have printed $p=2\pi\int_\lambda^\infty\rho=\tfrac\pi2-\arctan\sinh\pi\lambda$: it satisfies $\varepsilon=\tfrac{\pi J}2\sin p$ just as well, and for a pair of holes at $\pm\lambda$ gives the same total, $\pi$. A symmetric test cannot tell the two apart — which is exactly why one must test an asymmetric pair. Against finite chains with the holes placed asymmetrically the printed orientation is right to $10^{-4}$ at $N=256$ and the other is wrong by $1.13$. The counterexample lives in the check script.

Equation (17.7) is not a model equation of state; it is the equation of state of a real gas. Ultracold $^{87}$Rb on an atom chip, confined so tightly in two directions that only the lowest transverse mode is populated, is the Lieb–Liniger gas of Lecture 11 with $c$ set by the transverse confinement. Measuring the in situ density profile of such a cloud, and reading the trap’s slowly varying potential as a local chemical potential $\mu(x)=\mu_0-V(x)$, turns one image into a curve of $n$ against $\mu$ at fixed $T$. Van Amerongen and co-workers did this in 2008 for a weakly interacting gas on an atom chip, fitting only $\mu_0$ and $T$: no interaction parameter is free, since $c$ follows from the transverse confinement. One complication is instructive — at their temperatures the higher transverse modes are not empty, so what is fitted is $n_{\mathrm{YY}}(\mu(x),T)+\sum_{j\ge1}(j+1) n_{\mathrm{ideal}}(\mu(x)-j\hbar\omega_\perp,T)$: Yang–Yang for the one-dimensional gas, an ideal gas for everything above it. With that, the profiles are reproduced across the crossover from the quasi-condensate to the nearly ideal gas — where mean-field theory has nothing to say cited: van Amerongen et al., Phys. Rev. Lett. 100 (2008) 090402.

Everything so far used only real roots, and for the gas that is a theorem (Proposition 11.4). The chain is different. Lecture 10 already met a solution with $\lambda_{1,2}=x\pm\tfrac\ii2+O(\delta)$ and $\delta$ exponentially small in $N$ (10.23), and its argument — an imaginary part forces a nearly vanishing denominator — applies to any number of roots.

cited, and false in detail: Essler, Korepin and Schoutens (1992)

The hypothesis is known to fail. There are “wide pairs” whose imaginary parts are not $\pm\tfrac12$, and $2$-strings that collapse onto the real axis as $N$ grows; Lecture 16 met the tail of the same phenomenon in the pair of real roots sharing a quantum number at $N=22$ (Remark 16.2). What survives — and what the thermodynamics needs — is the counting: the string configurations reproduce the right number of states even where individual solutions are wrong. That is less than it sounds, and this reading returns to it: a partition function weights states by energy, so equal cardinality is not equal free energy. This is an uncomfortable position to be in and it is the honest one.

Put the string of Definition 17.14 into the Bethe equations and multiply the $n$ equations of its members together. The exponentially small deviations cancel between a diverging and a vanishing factor, and what is left is one equation per string centre: $$N\theta_n(\lambda^n_\alpha) -\sum_{m,\beta}\Theta_{nm}(\lambda^n_\alpha-\lambda^m_\beta)=2\pi I^n_\alpha , \qquad \theta_n(\lambda)=2\arctan\frac{2\lambda}n ,$$(17.14) with the scattering phase between an $n$-string and an $m$-string $$\Theta_{nm}=\theta_{\abs{n-m}}+2\theta_{\abs{n-m}+2}+\dots +2\theta_{n+m-2}+\theta_{n+m}, \qquad \theta_0\equiv0 .$$(17.15)

sketched: the cancellation of the deviations is the content, and it is cited (Takahashi 1971)

For $n=m=1$ this is $\Theta_{11}=\theta_2$ and (17.14) is Lecture 16’s (16.1). Two consequences follow as they did there, by reading off the values at $\pm\infty$: since $\Theta_{nm}(\pm\infty)=\pm\pi t_{nm}$ with $t_{nm}=2\min(n,m)-\delta_{nm}$, the number of vacancies for $n$-strings is $$\#\{\text{vacancies for }n\text{-strings}\}=N-\sum_m t_{nm}M_m , \qquad \rho_n+\rho^h_n=a_n-\sum_m A_{nm}*\rho_m , \qquad A_{nm}=\frac{\Theta_{nm}'}{2\pi},$$(17.16) with $M_m$ the number of $m$-strings and $a_n=\theta_n'/2\pi$ as in Theorem 16.3. The energy follows from the identity $2\sum_{a=1}^{n}(4\lambda_a^2+1)^{-1}=\pi a_n(\lambda)$ for an exact string, together with Lecture 10’s $E=-\sum_j2J/(4\lambda_j^2+1)$ (10.16): an $n$-string of centre $\lambda$ carries energy $-\pi Ja_n(\lambda)$ and $S^z$ lowered by $n$. For $n=1$ that is $\varepsilon_0=-\pi Ja_1$, the bare energy §17.5 asked for.

Repeat §17.5 with one density per string length. Put the chain in a field, $H\mapsto H-h\sum_nS^z_n$. An $n$-string lowers $S^z$ by $n$, so the Zeeman energy is $-hN/2+h\sum_nnM_n$: the magnons’ chemical potential is $-h$, and it is $+h$ that takes the place of the gas’s $-\mu$. Nothing else changes except the bookkeeping, and the answer is an infinite coupled system: with $\eta_n=e^{\varepsilon_n/T}=\rho^h_n/\rho_n$, $$\varepsilon_n=-\pi Ja_n+hn +T\sum_m A_{nm}*\ln\bigl(1+\eta_m^{-1}\bigr), \qquad f=-\frac h2-T\sum_n\int a_n\ln\bigl(1+\eta_n^{-1}\bigr)\dd\lambda .$$(17.17)

sketched: this is Theorem 17.7’s variation with one density per string length, not redone here

The sign in front of the sum has flipped because $K_{\mathrm{XXX}}/2\pi=-a_2$ (Lecture 16): at $n=m=1$, (17.17) is (17.7). Infinitely many equations, each with infinitely many terms, is not progress. It becomes progress because the matrix $1+A$ has an inverse that is almost diagonal: in Fourier variables $(1+A)^{-1}=1-\hat s\,(\text{shift in }n)$ with $\hat s(\omega)=\bigl(2\cosh(\omega/2)\bigr)^{-1}$, i.e. with the kernel $s(\lambda)=\bigl(2\cosh\pi\lambda\bigr)^{-1}$. Applying it collapses (17.17) to

sketched: the Fourier inversion is checked in the script; the convergence of the truncation is numerical

Three things fall out at once. The driving term now sits at $n=1$ alone, and every other equation couples only to its two neighbours — a chain in the index $n$. The constant $-J\ln2$ in $f$ is Lecture 16’s Hulthén ground-state energy (16.9), and it arrives here as $\int s\,\varepsilon_0=-J\int s\,\pi a_1=-J\ln2$, not by assumption. And at $T\to\infty$ the driving vanishes and (17.18) has the constant solution $\eta_n=n(n+2)$, whence $f\to-T\int s\ln4=-T\ln2$ — the free energy of $N$ free spins, by a telescoping product.

In practice one truncates at some $n_{\max}$ and, at zero field, imposes $\eta_n\to n(n+2)$ beyond it. At $T=J$ the truncation error falls below $10^{-5}$ by $n_{\max}=20$, and the free energy so computed agrees with exact diagonalisation of the finite chain — with a finite chain, where no string ever appears — to five decimals at $T=2J$, $J$ and $J/2$. The check script does exactly that comparison, and it is the strongest evidence this reading offers that the string hypothesis, false in detail, counts correctly — not the strongest there is, since the same equations follow algebraically from the quantum transfer matrix with no string hypothesis at all cited: Essler et al. (2005). Note what it settles. That the string configurations preserve the number of states does not by itself give the free energy, since $Z$ weights states by energy; it is the agreement of the energies that is the evidence.

Two things §17.12 asserted are worth doing properly: the field’s normalisation, which no check of the equations can see, and the promise Lecture 15 made about the T–Q relation, which is worth keeping exactly as far as it is true.

The field. Everything in (17.18) except the boundary condition is blind to $h$: the operator that produced it annihilates the sequence $hn$ (check script). So the normalisation of $h$ survives in exactly one place, and it is the one place no check of the equations can see. Set $J=0$, where the answer is known. The recurrence has the constant solution $1+\eta_n=\bigl(\sinh((n+1)x)/\sinh x\bigr)^2$, for which $(\ln\eta_n)/n\to2x$, and (17.18) gives $f=-T\ln(2\cosh x)$. A free spin in $H-h\sum S^z$ has $f=-T\ln\bigl(2\cosh(h/2T)\bigr)$, so $x=h/2T$ and the boundary condition is $h/T$. Takahashi’s is $2h_{\mathrm T}/T$ because his field term is $-2h_{\mathrm T}\sum S^z$: taking his boundary condition with our Hamiltonian multiplies the zero-field susceptibility by four. As $x\to0$ the hyperbolic solution becomes $\eta_n=n(n+2)$, which is why that is the zero-field truncation.

Where $Q$ reappears, and where it does not. Fusing the auxiliary space of Lecture 14 to spin $\tfrac n2$ gives a tower of transfer matrices $T_n$, and their eigenvalues obey a T-system: a relation of the shape $$T_n^{[1]}T_n^{[-1]}=T_{n-1}T_{n+1}+f_n , \qquad g^{[m]}(\lambda)=g\bigl(\lambda+\tfrac{\ii m}2\bigr),$$(17.20) with $f_n$ a known scalar built from the vacuum factors. Put $Y_n=T_{n-1}T_{n+1}/f_n$; then (17.20) at $n\mp1$ gives $1+Y_{n\mp1}=T_{n\mp1}^{[1]}T_{n\mp1}^{[-1]}/f_{n\mp1}$, and the $Y_n$ satisfy (17.19) provided $f_n^{[1]}f_n^{[-1]}=f_{n-1}f_{n+1}$ — a condition on the scalar alone, which the standard $f_n$ satisfies identically (check script). And every fused eigenvalue is generated from a single Baxter $Q$ together with those vacuum factors: that is the sense in which Lecture 15’s T–Q relation (15.20) generalises cited: Kuniba, Nakanishi and Suzuki (2011) §§6, 8.

Now the warning, which is the reason this is a reading and not a theorem. The $Y_n$ so constructed are not the thermal $\eta_n$. They cannot be: an ordinary transfer matrix of a finite chain has no temperature in it. At $N=2$ on the reference state the ratio $T_0T_2/f_1$ is a non-constant rational function of $\lambda$, negative on an interval, where the $J=h=0$ thermal solution is the constant $\eta_1=3$. A functional equation does not determine its solution; the temperature enters through analyticity and asymptotics, and the object whose ratios are the thermal $\eta_n$ is the quantum transfer matrix, a Trotter-limit construction this course does not build cited: Essler et al. (2005), §13.2 and appendix 13.A. What Lecture 15 promised is delivered here at the level of the hierarchy — the Y-system of (17.19) is the T–Q relation’s, seen from the other side — and not at the level of the solutions.

Theorem 17.7 minimises a functional. What entitles anyone to believe that the minimum is the free energy?

The saddle-point argument is in §17.5.

Yang and Yang proved what can be proved cleanly: that (17.7) has a unique solution for every $c>0$, $T>0$, that $\varepsilon$ is analytic in $\mu$ and $T$ and increasing in $\abs k$, and that the resulting $P(\mu,T)$ is convex — so that the thermodynamics it defines is consistent cited: Yang and Yang (1969). What their argument does not establish is the step before: that the set of Bethe states really is in bijection with the sets of quantum numbers, and that the coarse-graining loses nothing. For the gas, half of the first is a theorem: quantum numbers and solutions are in bijection (Proposition 11.5); that the states so built are all of them is still cited. For the chain even that much rests on the string hypothesis, which is false as a statement about individual solutions. The subject’s position is that the counting is right even so. Part of the evidence is numerical — §17.12’s exact diagonalisation, at the three temperatures it reports, and no further, since a ring of fourteen sites is not the thermodynamic limit at low temperature. The rest is theoretical, and independent: the same equations come out of the quantum transfer matrix without any string hypothesis cited: Essler et al. (2005), §13.2 and appendix 13.A.