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ODE/IM TBA from TQ and Y-System Analyticity

Chapter 12 ended with exact functional relations and a warning: algebra does not choose an analytic solution. A TQTQ equation knows how shifted functions fit together, and a YY-system knows a finite-difference operator, but neither specifies which logarithm to take, which zeros a contour encloses, which asymptotic solution is physical, or which homogeneous term survives Fourier inversion.

Those are precisely the data that turn a functional relation into a nonlinear integral equation. The central principle of this chapter is therefore:

This page states that contract, keeps the TQ and fused-YY routes separate, and proves one scalar calibration completely. The detailed strip lemmas, matrix kernels, driving-term calculations, source terms, and numerical algorithms belong to the pages that follow.

Three integral-equation languages have different unknowns

Section titled “Three integral-equation languages have different unknowns”

The phrases “TBA equation” and “NLIE” are often used too loosely. Three constructions occur in this book, and a formula from one cannot be moved to another merely because all three contain logarithms and convolutions.

RoutePrimary unknownDiscrete dataWhat closes the derivation
TQ/Destri–de VegaAuxiliary ratio aa or a counting functionZeros of 1+a1+a, separated into roots, holes, and transfer zerosA contour around the intended divisor, plus large-rapidity data
Fused YY-system TBANode functions YaY_a or pseudoenergies εa\varepsilon_aZeros and poles of YaY_a and 1+Ya1+Y_aA finite fusion graph, analytic strips, branches, and a matrix inverse
Exact-WKB/GMN Riemann–Hilbert equationResummed periods or Darboux coordinatesBPS charges, intersection pairings, and Stokes raysA Riemann–Hilbert jump problem and its asymptotic normalization

The first two are ODE/IM inversion routes. The third starts from exact-WKB jump data and will enter on Page 3. Any equivalence between them is an additional theorem for a specified example, not a generic identity.

Two ODE/IM lanes begin with a TQ relation and determinant divisor or a fused Y-system, pass through a shared passport containing the rapidity cover, strip, divisor, logarithm, asymptotic, contour, and state data, and end in a DDV-type NLIE or a multi-node TBA; a separate exact-WKB lane leads from BPS and Stokes data to a Riemann–Hilbert equation, with comparison requiring an explicit map.

The three inversion routes use different unknowns and different discrete data. The two ODE/IM lanes share an analytic passport but not necessarily the same kernel or contour. The exact-WKB lane is separate until an explicit map of variables, jumps, asymptotics, and normalizations is proved.

A rapidity coordinate turns rotations into translations

Section titled “A rapidity coordinate turns rotations into translations”

Retain the Chapter 12 homogeneous-oscillator conventions

q=exp ⁣(πiM+1),ρM=M+12M.q=\exp\!\left(\frac{\pi\ii}{M+1}\right), \qquad \rho_M=\frac{M+1}{2M}.

Choose a positive scale ss_\star and the growth-adapted logarithmic coordinate

θ=ρMLog ⁣(ss),s=sexp ⁣(2MM+1θ).\theta =\rho_M\Log\!\left(\frac{s}{s_\star}\right), \qquad s=s_\star \exp\!\left(\frac{2M}{M+1}\theta\right).

The logarithm is part of the parameter passport. On this cover,

q±1sθ±πi2M,q±2sθ±πiM.\begin{aligned} q^{\pm1}s &\longleftrightarrow \theta\pm\frac{\pi\ii}{2M}, \\ q^{\pm2}s &\longleftrightarrow \theta\pm\frac{\pi\ii}{M}. \end{aligned}

This is why the shift ledger from Chapter 12 matters: fusion relations use q±1sq^{\pm1}s, while the elementary Baxter equation and the auxiliary root ratios use q±2sq^{\pm2}s. The two additive shifts differ by a factor of two.

The same coordinate also linearizes the leading growth. Since a homogeneous determinant has

logD(s)csρM,\log D(s)\sim c\,s^{\rho_M},

its rapidity asymptotic is proportional to eθe^\theta. The coordinate has therefore matched two independent structures: the rotation angle and the WKB growth order.

For the particularly simple polynomial closure used here, take l=0l=0 and integer M>1M>1, and put

h=2M.h=2M.

The finite Ah1A_{h-1} system of Chapter 12 then has the standard shift θθ±πi/h\theta\mapsto\theta\pm\pi\ii/h. For the quartic oscillator, M=2M=2 and h=4h=4: a qq-shift is ±πi/4\pm\pi\ii/4, while a q2q^2-shift is ±πi/2\pm\pi\ii/2.

The TQ and fused-Y routes carry different divisor ledgers

Section titled “The TQ and fused-Y routes carry different divisor ledgers”

The TQ route begins from the Page 6 auxiliary functions

a+(s)=e2iϑQ+(q2s)Q+(q2s),a(s)=e2iϑQ(q2s)Q(q2s).\begin{aligned} a_+(s) &=\ee^{2\ii\vartheta} \frac{Q_+(q^2s)}{Q_+(q^{-2}s)}, \\ a_-(s) &=\ee^{-2\ii\vartheta} \frac{Q_-(q^2s)}{Q_-(q^{-2}s)}. \end{aligned}

Their factorized TQ equations imply

1+a±(s)=e±iϑT(s)Q±(s)Q±(q2s).1+a_\pm(s) =\ee^{\pm\ii\vartheta} \frac{T(s)Q_\pm(s)}{Q_\pm(q^{-2}s)}.

Thus 1+a±=01+a_\pm=0 does not label one spectrum by itself. Subject to the displayed denominator, it can detect a radial determinant zero, a transfer-function zero, or—after a contour has been deformed—a hole.

A useful model-specific separation theorem is available for the regular radial oscillator when

M>1,12<l<M2.M>1, \qquad -\frac12<l<\frac M2.

In this window the radial Q+Q_+ zeros are positive and simple, while the associated lateral-transfer zeros are negative. A contour around the positive ray can therefore select exactly the radial divisor. This is spectral input from self-adjointness and the PT-symmetric lateral problem—not a consequence of the TQTQ identity—and Page 2 will state the contour formula in that passport.

The DDV-type route replaces the logarithmic derivative of a product over the selected roots by a Cauchy integral of log(1+a)\log(1+a). Its characteristic shape is

loga=twist+driving term+one boundary-value convolutionthe opposite boundary-value convolution+declared source terms.\begin{aligned} \log a ={}&\text{twist}+\text{driving term} \\ &+\text{one boundary-value convolution} \\ &-\text{the opposite boundary-value convolution} \\ &+\text{declared source terms}. \end{aligned}

This line is structural, not a contour formula. The contour orientation, kernel, boundary values, and sources are model data and will be derived on Page 2.

The fused route starts instead from the finite nearest-neighbor system

Ya ⁣(θ+πih)Ya ⁣(θπih)=b=1h1(1+Yb(θ))Iab,a=1,,h1,\begin{aligned} &Y_a\!\left(\theta+\frac{\pi\ii}{h}\right) Y_a\!\left(\theta-\frac{\pi\ii}{h}\right) \\ &\qquad= \prod_{b=1}^{h-1} \bigl(1+Y_b(\theta)\bigr)^{I_{ab}}, \qquad a=1,\ldots,h-1, \end{aligned}

where II is the Ah1A_{h-1} incidence matrix. This route exists as a finite TBA only when the fusion orbit actually truncates. For generic MM the one-function TQ/NLIE route may still exist even though no finite YY-graph closes.

Explicitly,

Iab=δa,b+1+δa,b1,Y0=Yh=0.I_{ab}=\delta_{a,b+1}+\delta_{a,b-1}, \qquad Y_0=Y_h=0.

Logarithms create a strip boundary-value problem

Section titled “Logarithms create a strip boundary-value problem”

Assume first that the ground-state YaY_a are positive on the real line and that both YaY_a and 1+Ya1+Y_a are nonzero on a simply connected domain containing the shifted contours. Equivalently, one may assume directly that the required single-valued logarithms have been chosen with zero winding. Define the Dorey–Dunning–Tateo pseudoenergy convention

εa=LogYa,La=log ⁣(1+eεa).\varepsilon_a=\Log Y_a, \qquad L_a=\log\!\left(1+\ee^{-\varepsilon_a}\right).

Because

log(1+Yb)=εb+Lb,\log(1+Y_b)=\varepsilon_b+L_b,

the multiplicative system becomes

εa ⁣(θ+πih)+εa ⁣(θπih)bIabεb(θ)=bIabLb(θ)+2πiNa.\begin{aligned} &\varepsilon_a\!\left(\theta+\frac{\pi\ii}{h}\right) +\varepsilon_a\!\left(\theta-\frac{\pi\ii}{h}\right) -\sum_b I_{ab}\varepsilon_b(\theta) \\ &\qquad= \sum_b I_{ab}L_b(\theta)+2\pi\ii N_a. \end{aligned}

The integer NaN_a is not decorative. On a connected region where all logarithm arguments are nonzero and the branches vary continuously, it is locally constant. A zero or pole of YaY_a, a zero of 1+Ya1+Y_a, or a crossed cut can change the branch ledger and introduce a source.

The leading ODE asymptotic supplies a homogeneous solution

da(θ)=κsin ⁣(πah)eθ.d_a(\theta) =\kappa\sin\!\left(\frac{\pi a}{h}\right)\ee^\theta.

Indeed, the sine vector is the Perron–Frobenius eigenvector of the incidence matrix:

bIabsin ⁣(πbh)=2cos ⁣(πh)sin ⁣(πah),\sum_b I_{ab}\sin\!\left(\frac{\pi b}{h}\right) =2\cos\!\left(\frac{\pi}{h}\right) \sin\!\left(\frac{\pi a}{h}\right),

while shifting eθe^\theta by ±πi/h\pm\pi\ii/h produces the same factor. Thus the difference equation cannot determine the coefficient κ\kappa. That coefficient is fixed by determinant/WKB asymptotics and is part of the state passport.

For the ground-state branch considered here, choose the continuous logarithms with Na=0N_a=0. A nonzero NaN_a would contribute at zero Fourier momentum and must be retained as a constant particular solution or an explicit source.

There is one more endpoint issue. In the finite ODE system, fa=εadaf_a=\varepsilon_a-d_a and LaL_a generally approach nonzero zero-spectral constants as θ\theta\to-\infty. Their Fourier transforms are therefore not ordinary L1L^1 transforms. After a boundary regulator or an equivalent distributional prescription has been declared, the nonzero-momentum multiplier to invert is

bAab(k)f^b(k)=bIabL^b(k),k0,Aab(k)=2cosh ⁣(πkh)δabIab.\begin{aligned} \sum_b \mathcal A_{ab}(k)\widehat f_b(k) &= \sum_b I_{ab}\widehat L_b(k), \qquad k\ne0, \\ \mathcal A_{ab}(k) &= 2\cosh\!\left(\frac{\pi k}{h}\right)\delta_{ab}-I_{ab}. \end{aligned}

Inverting this matrix produces the nonzero-momentum part of the multi-node TBA kernel. This compact formula records the algebraic step; Page 2 will restore the regulated endpoint constants and justify the contour shifts, invertibility, and real-space kernels.

The common shorthand ANZC means analytic, nonzero, and constant-asymptotic in a declared strip. It is rarely the raw pseudoenergy that has this property. In the finite fused system, εada\varepsilon_a\sim d_a as θ+\Re\theta\to+\infty, so the useful object is

exp ⁣(εada)=edaYa.\exp\!\left(\varepsilon_a-d_a\right) =\ee^{-d_a}Y_a.

In the scalar convention used below, Y=eεY=\ee^{-\varepsilon} tends to zero and the stripped object is instead edY\ee^dY. The sign changes with the pseudoenergy convention; the principle does not.

A sufficient inversion passport records all of the following:

  1. the logarithmic cover and the exact strip width;
  2. analyticity and nonvanishing of every logarithm argument;
  3. controlled boundary values on both shifted lines;
  4. the branch integers and Schwarz-reflection convention;
  5. the complete growing asymptotic to subtract;
  6. integrability or Hardy-class control of the remainder;
  7. the zeros, poles, holes, and contour indentations assigned to the chosen state.

Checking only the real axis is not enough. Nor does the entirety of a function in ss prove nonvanishing after the exponential pullback to a rapidity strip.

A scalar zero-free strip yields a prototype TBA equation

Section titled “A scalar zero-free strip yields a prototype TBA equation”

The matrix system above is the actual finite ODE/IM route. To calibrate the analytic inversion without graph notation, consider the one-component prototype

Y(u+iη)Y(uiη)=1+Y(u),Y=eϵ.Y(u+\ii\eta)Y(u-\ii\eta)=1+Y(u), \qquad Y=\ee^{-\epsilon}.

This is a normalization model, not the physical quartic-oscillator TBA, which has the A3A_3 graph. Assume:

  • η>0\eta>0 and R>0R>0;
  • YY and 1+Y1+Y are analytic and nonzero for u<η|\Im u|<\eta and have controlled boundary values;
  • Y(uˉ)=Y(u)Y(\bar u)=\overline{Y(u)} and Y(u)>0Y(u)>0 for real uu;
  • the continuous logarithms are chosen with branch integer N=0N=0;
  • with d(u)=Rcosh ⁣(πu2η),d(u)=R\cosh\!\left(\frac{\pi u}{2\eta}\right), the remainder v=ϵdv=\epsilon-d is analytic in the strip, has L1L^1 boundary values, and obeys the Hardy-type bound supy<ηRv(x+iy) ⁣dx<;\sup_{|y|<\eta} \int_{\mathbb R}|v(x+\ii y)|\,\dd x<\infty;
  • L=log(1+eϵ)L=\log(1+\ee^{-\epsilon}) belongs to L1(R)L^1(\mathbb R), and the vertical-edge integrals vanish in the boundary contour shifts.

The logarithmic equation is

ϵ(u+iη)+ϵ(uiη)=L(u).\epsilon(u+\ii\eta)+\epsilon(u-\ii\eta)=-L(u).

The driving term is invisible because

d(u+iη)+d(uiη)=0.d(u+\ii\eta)+d(u-\ii\eta)=0.

Therefore v++v=Lv^++v^-=-L. Fix the Fourier convention

f^(k)=Reikuf(u) ⁣du,f(u)=12πReikuf^(k) ⁣dk.\begin{aligned} \widehat f(k) &=\int_{\mathbb R}\ee^{\ii ku}f(u)\,\dd u, \\ f(u) &=\frac{1}{2\pi} \int_{\mathbb R}\ee^{-\ii ku}\widehat f(k)\,\dd k. \end{aligned}

Analytic contour shifts give

v(+iη)^(k)=eηkv^(k),v(iη)^(k)=eηkv^(k).\widehat{v(\,\cdot+\ii\eta\,)}(k) =\ee^{\eta k}\widehat v(k), \qquad \widehat{v(\,\cdot-\ii\eta\,)}(k) =\ee^{-\eta k}\widehat v(k).

Hence

2cosh(ηk)v^(k)=L^(k).2\cosh(\eta k)\widehat v(k)=-\widehat L(k).

With (Kf)(u)=RK(uu)f(u) ⁣du(K*f)(u)=\int_{\mathbb R}K(u-u')f(u')\,\dd u', the inverse kernel is

K^η(k)=12cosh(ηk),Kη(u)=14ηsech ⁣(πu2η).\begin{aligned} \widehat K_\eta(k) &=\frac{1}{2\cosh(\eta k)}, \\ K_\eta(u) &=\frac{1}{4\eta} \operatorname{sech}\!\left(\frac{\pi u}{2\eta}\right). \end{aligned}

The prototype TBA equation is therefore

ϵ(u)=Rcosh ⁣(πu2η)RKη(uu)log ⁣(1+eϵ(u)) ⁣du.\begin{aligned} \epsilon(u) ={}&R\cosh\!\left(\frac{\pi u}{2\eta}\right) \\ &- \int_{\mathbb R} K_\eta(u-u') \log\!\left(1+\ee^{-\epsilon(u')}\right) \,\dd u'. \end{aligned}

At the standard calibration η=π/2\eta=\pi/2,

ϵ(u)=Rcoshulog ⁣(1+eϵ(u))2πcosh(uu) ⁣du.\begin{aligned} \epsilon(u) ={}&R\cosh u \\ &- \int_{-\infty}^{\infty} \frac{\log\!\left(1+\ee^{-\epsilon(u')}\right)} {2\pi\cosh(u-u')} \,\dd u'. \end{aligned}

Two normalization checks are immediate. First,

RKη(u) ⁣du=K^η(0)=12,\int_{\mathbb R}K_\eta(u)\,\dd u =\widehat K_\eta(0)=\frac12,

The zero-drive case R=0R=0 lies outside the L1L^1 hypotheses used in the Fourier derivation. The same equation nevertheless has a bounded-convolution extension because KηL1K_\eta\in L^1. In that extension, the positive constant solution is

Y=1+52.Y=\frac{1+\sqrt5}{2}.

Indeed Y2=1+YY^2=1+Y, while the integral equation gives logY=12log(1+Y)-\log Y=-\tfrac12\log(1+Y). The two calculations agree only with the kernel normalization shown above.

For this real scalar calibration with R0R\geq0, the fixed-point map for the bounded remainder is a contraction because  ⁣dlog(1+ex)/ ⁣dx1|\dd\log(1+\ee^{-x})/\dd x|\le1 and Kη1=1/2\lVert K_\eta\rVert_1=1/2. This proves uniqueness in the toy class. It does not prove uniqueness for complex twists, matrix systems, excited-state contours, or kernels after analytic continuation.

Solving an NLIE or TBA determines the analytic function named in its passport on the integration contour. Recovering ODE spectral data then requires the declared dictionary back to TT, QQ, or a determinant, analytic continuation to the quantization locus, and one absolute normalization or WKB asymptotic.

For a TQ auxiliary ratio, simple roots obey

a(θj)=1,Loga(θj)=πi(2Ij+1),a(\theta_j)=-1, \qquad \Log a(\theta_j)=\pi\ii(2I_j+1),

where the integers IjI_j are branch and quantum-number data. The ratio form additionally assumes that the shifted denominator is nonzero. If both shifted QQ-values vanish, one must return to the denominator-free TQ equation.

A solved YY-system TBA can similarly reconstruct selected transfer or determinant combinations, but not the full ODE connection matrix. It also does not retroactively prove the zero locations or strip nonvanishing assumed in its derivation.

The rest of Chapter 13 supplies the missing layers: Page 2 proves the strip, kernel, driving, and contour machinery; Pages 3 and 4 introduce the exact-WKB/GMN construction and audit possible equivalences; Page 5 adds excited-state sources and wall crossing; Page 6 treats numerical convergence and error; Page 7 compares WKB and NS periods; and Page 8 performs a full spectral computation.

Taking logarithms after checking only YY. The argument 1+Y1+Y must also be nonzero on the chosen domain. Its zeros are precisely where a new logarithmic source can enter.

Using the same rapidity shift for qq and q2q^2. In the convention above they give πi/(2M)\pi\ii/(2M) and πi/M\pi\ii/M. Confusing them changes the Fourier multiplier and the kernel.

Fourier transforming the growing pseudoenergy. The driving term is not integrable. Subtract it first and transform only the controlled remainder.

Deriving the drive from the YY-system. The drive is a homogeneous mode invisible to the difference equation. ODE/WKB asymptotics fix its coefficient.

Calling every nonlinear equation TBA. A DDV counting-function NLIE, a fused-node TBA, and a GMN Riemann–Hilbert equation have different unknowns and divisor data.

Inferring a strip from a real-axis plot. A numerical scan can miss a nearby complex zero or pole. The contour shift needs control on the whole strip and its boundary values.

Promoting one convergent iteration to uniqueness. The scalar toy model has a contraction proof. General ODE/IM equations require their own existence and uniqueness analysis.

1. Audit the multiplicative-to-additive shift

Section titled “1. Audit the multiplicative-to-additive shift”

Starting from

q=exp ⁣(πiM+1),θ=M+12MLog ⁣(ss),q=\exp\!\left(\frac{\pi\ii}{M+1}\right), \qquad \theta=\frac{M+1}{2M}\Log\!\left(\frac{s}{s_\star}\right),

derive the rapidity displacement produced by sqjss\mapsto q^js. Then specialize to the qq- and q2q^2-shifts for the quartic oscillator.

Solution

On the chosen logarithmic sheet,

Log ⁣(qjss)=Log ⁣(ss)+jπiM+1.\Log\!\left(\frac{q^js}{s_\star}\right) =\Log\!\left(\frac{s}{s_\star}\right) +\frac{j\pi\ii}{M+1}.

Multiplication by (M+1)/(2M)(M+1)/(2M) gives

sqjsθθ+jπi2M.s\mapsto q^js \quad\Longleftrightarrow\quad \theta\mapsto\theta+\frac{j\pi\ii}{2M}.

For M=2M=2, a qq-shift gives πi/4\pi\ii/4 and a q2q^2-shift gives πi/2\pi\ii/2. These are respectively the fusion and elementary TQ shifts in the conventions of this chapter.

2. Prove that the logarithmic defect is locally constant

Section titled “2. Prove that the logarithmic defect is locally constant”

Suppose all factors in a multiplicative YY-system are nonzero on a connected domain UU, and continuous logarithms have been selected. Show that the discrepancy between the two logarithmic sides is 2πiN2\pi\ii N with NN locally constant. State how NN can change.

Solution

Let FF be the logarithm of the left-hand side minus the logarithm of the right-hand side. The multiplicative identity implies

eF(u)=1.\ee^{F(u)}=1.

Consequently F(u)/(2πi)F(u)/(2\pi\ii) is integer-valued. It is also continuous on UU, because every chosen logarithm is continuous there. A continuous integer-valued function on a connected set is constant, so F=2πiNF=2\pi\ii N for one integer NN on each connected component.

The argument fails when a factor YaY_a or 1+Ya1+Y_a vanishes or has a pole, when a shifted contour crosses a cut, or when the domain is split. Crossing such an obstruction can change NN and can also generate an explicit source term.

3. Separate what ANZC does and does not prove

Section titled “3. Separate what ANZC does and does not prove”

Let dd and gg be analytic in a strip, with gg non-singular there and g(u)0g(u)\to0 at both asymptotic ends. Set

Y(u)=ed(u)+g(u).Y(u)=\ee^{d(u)+g(u)}.

Which ANZC properties follow for edY\ee^{-d}Y? Does this information alone justify taking Log(1+Y)\Log(1+Y) and shifting its contour?

Solution

The stripped function is

ed(u)Y(u)=eg(u).\ee^{-d(u)}Y(u)=\ee^{g(u)}.

It is analytic and nonzero wherever gg is analytic, and it tends to 11 at both stated ends. Thus it has the advertised analytic, nonzero, constant-asymptotic behavior in the open strip.

This does not imply that 1+Y1+Y is nonzero: the value Y=1Y=-1 is compatible with Y0Y\ne0. Nor does open-strip analyticity by itself give the boundary estimates needed to move a Fourier contour. One must check 1+Y1+Y, boundary values, branch continuity, and suitable decay or Hardy-class control separately.

For the Ah1A_{h-1} incidence matrix, show that

da(θ)=κsin ⁣(πah)eθd_a(\theta) =\kappa\sin\!\left(\frac{\pi a}{h}\right)\ee^\theta

solves the homogeneous logarithmic YY-system operator.

Solution

With the boundary convention d0=dh=0d_0=d_h=0, the incidence matrix gives

bIabdb(θ)=da1(θ)+da+1(θ)=2cos ⁣(πh)da(θ).\begin{aligned} \sum_b I_{ab}d_b(\theta) &=d_{a-1}(\theta)+d_{a+1}(\theta) \\ &=2\cos\!\left(\frac{\pi}{h}\right)d_a(\theta). \end{aligned}

On the other hand,

da ⁣(θ+πih)+da ⁣(θπih)=(eπi/h+eπi/h)da(θ)=2cos ⁣(πh)da(θ).\begin{aligned} d_a\!\left(\theta+\frac{\pi\ii}{h}\right) +d_a\!\left(\theta-\frac{\pi\ii}{h}\right) &=\left(\ee^{\pi\ii/h}+\ee^{-\pi\ii/h}\right)d_a(\theta) \\ &=2\cos\!\left(\frac{\pi}{h}\right)d_a(\theta). \end{aligned}

Their difference is zero. Hence the functional relation cannot fix κ\kappa; the ODE/WKB asymptotic must do so.

5. Reconstruct the scalar inversion kernel

Section titled “5. Reconstruct the scalar inversion kernel”

Using the Fourier convention on this page, derive the multiplier and real-space kernel that invert

v(u+iη)+v(uiη)=L(u).v(u+\ii\eta)+v(u-\ii\eta)=-L(u).

Check the total mass of the kernel.

Solution

Analytic contour displacement gives

v(+iη)^=eηkv^(k),v(iη)^=eηkv^(k).\widehat{v(\,\cdot+\ii\eta\,)} =\ee^{\eta k}\widehat v(k), \qquad \widehat{v(\,\cdot-\ii\eta\,)} =\ee^{-\eta k}\widehat v(k).

Therefore

v^(k)=L^(k)2cosh(ηk).\widehat v(k) =-\frac{\widehat L(k)}{2\cosh(\eta k)}.

The standard transform pair yields

K^η(k)=12cosh(ηk),Kη(u)=14ηsech ⁣(πu2η).\widehat K_\eta(k)=\frac{1}{2\cosh(\eta k)}, \qquad K_\eta(u)=\frac{1}{4\eta} \operatorname{sech}\!\left(\frac{\pi u}{2\eta}\right).

Thus v=KηLv=-K_\eta*L. Evaluating the multiplier at zero, or integrating the hyperbolic secant directly, gives

RKη(u) ⁣du=K^η(0)=12.\int_{\mathbb R}K_\eta(u)\,\dd u =\widehat K_\eta(0)=\frac12.

6. Audit the TQ divisor before drawing a contour

Section titled “6. Audit the TQ divisor before drawing a contour”

Use

1+a±(s)=e±iϑT(s)Q±(s)Q±(q2s)1+a_\pm(s) =\ee^{\pm\ii\vartheta} \frac{T(s)Q_\pm(s)}{Q_\pm(q^{-2}s)}

to classify a zero of 1+a±1+a_\pm. What can be concluded when the zero comes from Q±Q_\pm, from TT, or at a point where the shifted denominator vanishes? Where do holes enter this ledger?

Solution

Assume first that Q±(q2s)0Q_\pm(q^{-2}s_\star)\ne0. If Q±(s)=0Q_\pm(s_\star)=0 and T(s)0T(s_\star)\ne0, then ss_\star is a radial determinant zero. If T(s)=0T(s_\star)=0 and Q±(s)0Q_\pm(s_\star)\ne0, it is a lateral or transfer-function zero. If both numerator factors vanish, the equation alone does not assign the zero to one spectral problem; the declared divisors and multiplicities must be inspected.

A hole is not a third factor in the displayed identity. It is a state-and-contour label for a solution of 1+a=01+a=0 that is not occupied by the selected QQ-root set; in the ODE factorization it is typically tracked through the transfer divisor after a contour deformation.

If Q±(q2s)=0Q_\pm(q^{-2}s_\star)=0, the ratio defining a±a_\pm is singular or indeterminate, so the factorized equation is not a safe quantization test. One must return to the denominator-free TQ relation and test for cancellation before assigning any spectral meaning.

7. Check the quartic zero-spectral A₃ solution

Section titled “7. Check the quartic zero-spectral A₃ solution”

The quartic fusion values at zero spectral parameter are

(T0,T1/2,T1,T3/2,T2,T5/2)(0)=(1,3,2,3,1,0).(T_0,T_{1/2},T_1,T_{3/2},T_2,T_{5/2}) (0)=(1,\sqrt3,2,\sqrt3,1,0).

Using

Yn=T(n1)/2T(n+1)/2,Y_n=T_{(n-1)/2}T_{(n+1)/2},

compute (Y1,Y2,Y3)(Y_1,Y_2,Y_3) and verify the constant A3A_3 YY-system.

Solution

The three node values are

(Y1,Y2,Y3)=(2,3,2).(Y_1,Y_2,Y_3)=(2,3,2).

With Y0=Y4=0Y_0=Y_4=0, the constant nearest-neighbor equations are

Ya2=(1+Ya1)(1+Ya+1).Y_a^2=(1+Y_{a-1})(1+Y_{a+1}).

For the end nodes, 22=(1+0)(1+3)=42^2=(1+0)(1+3)=4. For the middle node, 32=(1+2)(1+2)=93^2=(1+2)(1+2)=9. This also shows why the scalar golden-ratio calibration is not the quartic oscillator: the latter has three coupled nodes and different constant data.

8. Audit a claimed equivalence of nonlinear equations

Section titled “8. Audit a claimed equivalence of nonlinear equations”

Two authors begin from the same multiplicative YY-system but obtain integral equations with different driving terms. Explain why the common functional relation does not prove that the equations are equivalent. Give a minimal comparison checklist.

Solution

The linear difference operator cannot determine a homogeneous drive dad_a, so an inversion must import its coefficient from asymptotics. This is not an additive symmetry of the nonlinear equation: although DH=0\mathcal D H=0, one generally has

log ⁣(1+eεaHa)log ⁣(1+eεa).\log\!\left(1+\ee^{-\varepsilon_a-H_a}\right) \ne \log\!\left(1+\ee^{-\varepsilon_a}\right).

Thus adding HH to a known solution need not produce another solution. Different driving asymptotics define different nonlinear boundary-value problems, and each candidate integral equation must be checked against the functional relation independently. The scalar family dR=Rcosh(πu/(2η))d_R=R\cosh(\pi u/(2\eta)) is the prototype: every dRd_R is homogeneous, but the value of RR must be supplied before solving the fixed-point problem.

At minimum, one must match:

  1. the unknown functions and rapidity variables;
  2. the shift and Fourier conventions;
  3. the analytic strip and its boundary values;
  4. zeros, poles, holes, branch integers, and contour prescriptions;
  5. the complete driving asymptotic and its normalization;
  6. kernel normalizations and declared source terms;
  7. the dictionary from the solution back to the same determinant or spectral observable.

Without these identifications, the two equations may be different inversions of the same algebraic relation rather than equivalent descriptions of the same state.

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