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Excited States, Contour Deformation, and Wall Crossing

A ground-state TBA is not discarded when its parameters are continued. Its contour is transported with the analytic solution. If a logarithmic singularity crosses that contour, returning the contour to its reference position creates an explicit source and a new root condition. If instead a kernel pole crosses because two period phases rotate through π/2\pi/2, a residue changes the equation and can force a larger set of unknowns. In exact-WKB/BPS examples that enlargement can encode a genuine wall crossing, but the three mechanisms are not synonyms.

This page derives each operation with its orientation visible. The excited-state prototype is the Dorey–Tateo analytic continuation of the scaling Lee–Yang TBA. The wall-crossing prototype is the A2A_2 polynomial Schrödinger problem of Ito–Mariño–Shu: after the first phase wall, the minimal two-cycle equation closes as a three-charge system containing γ12=γ1+γ2\gamma_{12}=\gamma_1+\gamma_2. The final section reconciles that residue calculation with the inverse-jump Kontsevich–Soibelman pentagon used in Chapter 9.

The following distinctions prevent most sign and scope errors.

EventWhat moves?What changes in the equation?What need not change?
Solution monodromyParameters circle a branch pointPossibly only the chosen solution branchContour and formula can remain unchanged
Excited-state crossingA zero or pole of 1+eεa1+\ee^{-\varepsilon_a} crosses an integration contourA source primitive and a root quantization conditionFusion graph and BPS indices
Kernel-pole crossingA pole of KabK_{ab} crosses while a phase or strip is continuedA residue evaluation of LbL_bIt does not alone prove a new BPS state exists
BPS wall crossingCentral-charge rays reorder and the active spectrum mutatesCharge set, indices, and a refactorized RH/TBA descriptionThe ordered sector automorphism and continued physical coordinate

An important fifth possibility is a pinch: singularities approach a contour from opposite sides so that no local deformation avoids both. A pinch is typically a branch point in parameter space. Circling it can produce solution monodromy before any singularity crosses the final reference contour.

Keep four ledgers during any continuation:

  1. the divisor ledger of zeros, poles, multiplicities, and branch integers;
  2. the contour ledger of orientations, banks, kernel poles, and winding numbers;
  3. the chamber ledger of charges, phases, BPS indices, and ordered products;
  4. the observable ledger of determinant, period, energy, and boundary normalization.

Changing one ledger does not silently update the others.

Three panels distinguish a logarithmic zero crossing a rapidity contour and producing a source, a phase-shifted kernel pole crossing and producing a composite-charge node, and a pair of BPS rays swapping order while the ordered sector product remains invariant.

Three continuation events. A crossed nonlinear divisor changes the state passport; a crossed kernel pole changes the contour representation; a BPS wall changes the charge factorization while preserving the ordered sector automorphism. In an established exact-WKB model the last two can describe the same continuation, but that identification is additional structure.

A crossed logarithmic zero produces a source primitive

Section titled “A crossed logarithmic zero produces a source primitive”

Write a multi-node equation on transported contours Cb\mathcal C_b as

εa(θ)=da(θ)bCbKab(θθ)Lb(θ) ⁣dθ,\varepsilon_a(\theta) = d_a(\theta) - \sum_b \int_{\mathcal C_b} K_{ab}(\theta-\theta') L_b(\theta')\,\dd\theta',

where

Lb(θ)=Logzb(θ),zb(θ)=1+eεb(θ).L_b(\theta) = \Log z_b(\theta), \qquad z_b(\theta) = 1+\ee^{-\varepsilon_b(\theta)}.

Assume that the kernel has a scattering primitive in the chosen strip,

Kab(u)=12πi ⁣d ⁣duLogSab(u).K_{ab}(u) = \frac{1}{2\pi\ii} \frac{\dd}{\dd u} \Log S_{ab}(u).

This is a convention, not a consequence of the word “TBA.” A matrix Fourier kernel can be treated similarly by first defining a local primitive χab\chi_{ab} with uχab=2πiKab\partial_u\chi_{ab}=2\pi\ii K_{ab}.

Suppose zbz_b has isolated zeros and poles θbj\theta_{bj} between Cb\mathcal C_b and the reference real line. Let

νbj=ordθbjzb\nu_{bj} = \operatorname{ord}_{\theta_{bj}}z_b

be positive for a zero and negative for a pole. Define wbjw_{bj} as the winding number of CbR\mathcal C_b-\mathbb R around θbj\theta_{bj}, positive for a counterclockwise loop. Since

θLb(θ)νbjθθbj,\partial_{\theta'}L_b(\theta') \sim \frac{\nu_{bj}}{\theta'-\theta_{bj}},

integration by parts and the residue theorem give

εa(θ)=da(θ)bRKab(θθ)Lb(θ) ⁣dθb,jνbjwbjLogSab(θθbj).\boxed{ \begin{aligned} \varepsilon_a(\theta) ={}& d_a(\theta) - \sum_b \int_{\mathbb R} K_{ab}(\theta-\theta') L_b(\theta')\,\dd\theta' \\ &- \sum_{b,j} \nu_{bj}w_{bj} \Log S_{ab}(\theta-\theta_{bj}). \end{aligned} }

The formula assumes that endpoint terms vanish or are already included in dad_a, and that no singularity of SabS_{ab} is crossed during the contour homotopy. It is deliberately orientation-complete: reversing the path changes wbjw_{bj} and hence the source sign. Memorizing an unsigned “+logS+\log S per root” rule is unsafe.

If zbz_b has a simple zero at θbj\theta_{bj}, the source position is not a free parameter. It obeys

1+eεb(θbj)=0,1+\ee^{-\varepsilon_b(\theta_{bj})}=0,

or

εb(θbj)=(2Ibj+1)πi,IbjZ.\varepsilon_b(\theta_{bj}) = (2I_{bj}+1)\pi\ii, \qquad I_{bj}\in\mathbb Z.

The integer is branch and state data fixed by continuous tracking. Solving the sourced integral equation while holding a guessed root fixed does not solve the excited-state problem; the integral equation and all root conditions form one coupled nonlinear system.

One pair of roots calibrates the excited-state sign

Section titled “One pair of roots calibrates the excited-state sign”

For the scaling Lee–Yang model, Dorey and Tateo use

ε(θ)=rcoshθϕL(θ),\varepsilon(\theta) = r\cosh\theta - \phi*L(\theta),

with

L(θ)=log(1+eε(θ)),ϕ(θ)=iθlogS(θ),(ϕL)(θ)=R ⁣dθ2πϕ(θθ)L(θ).\begin{aligned} L(\theta) &= \log(1+\ee^{-\varepsilon(\theta)}), \\ \phi(\theta) &= -\ii\partial_\theta\log S(\theta), \\ (\phi*L)(\theta) &= \int_{\mathbb R} \frac{\dd\theta'}{2\pi} \phi(\theta-\theta')L(\theta'). \end{aligned}

After a particular analytic continuation, two logarithmic singularities lie at θ0-\theta_0 and +θ0+\theta_0. Their transported contour winds counterclockwise around θ0-\theta_0 and clockwise around +θ0+\theta_0. Thus

w(θ0)=1,w(+θ0)=1,w(-\theta_0)=1, \qquad w(+\theta_0)=-1,

and the sourced equation is

ε(θ)=rcoshθ+LogS(θθ0)S(θ+θ0)ϕL(θ).\boxed{ \begin{aligned} \varepsilon(\theta) ={}& r\cosh\theta + \Log\frac{ S(\theta-\theta_0) }{ S(\theta+\theta_0) } - \phi*L(\theta). \end{aligned} }

On the continuously selected one-particle branch,

ε(θ0)=πi.\varepsilon(\theta_0)=\pi\ii.

Evaluating the equation at the root gives the companion condition

0=rcoshθ0LogS(2θ0)ϕL(θ0),0 = r\cosh\theta_0 - \Log S(2\theta_0) - \phi*L(\theta_0),

with the logarithm branch inherited from the continuation. More general one-particle branches allow ε(θ0)=(2n+1)πi\varepsilon(\theta_0)=(2n+1)\pi\ii and, away from the spin-zero sector, conjugate rather than reflection-paired roots.

Straightening the contour must also be done in every observable built from LL. In the normalization of Dorey and Tateo, the effective scaling function changes from its ground-state integral to

cexc(r)=12riπsinhθ0+3π2RrcoshθL(θ) ⁣dθ.\begin{aligned} c_{\mathrm{exc}}(r) ={}& \frac{12r\ii}{\pi}\sinh\theta_0 + \frac{3}{\pi^2} \int_{\mathbb R} r\cosh\theta\,L(\theta)\,\dd\theta. \end{aligned}

The first term is the residue of the same crossed pair. Adding sources to the pseudoenergy while retaining the old energy or determinant formula would therefore mix two different contour passports.

This example makes two limitations concrete. First, a ground-state formula can acquire a physically excited solution by monodromy before any zero crosses the real line. Second, analytic continuation cannot jump between arbitrary superselection sectors. In the Lee–Yang example, a parity-symmetric seed does not generate nonzero-spin states without an appropriately asymmetric sourced system.

For the Page 2 ODE/IM equations, the same logic applies to whichever nonlinear factor the passport declares. In the fused convention the crossing divisor is 1+Ya1=01+Y_a^{-1}=0; in the DDV convention it is 1+a=01+a=0. The latter has

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

Which lip of the DDV keyhole is crossed fixes the source sign. A fused root, a DDV hole, and a zero of a spectral determinant are related only after the Page 1 unknown and divisor dictionaries are applied.

A mass phase drives a kernel pole through the contour

Section titled “A mass phase drives a kernel pole through the contour”

Return to the minimal polynomial-potential equation of Page 4. Write

ma=maeiϕa,m_a=|m_a|\ee^{\ii\phi_a},

and shift the real-axis functions by

ε~a(θ)=εa(θiϕa),L~a(θ)=La(θiϕa).\widetilde\varepsilon_a(\theta) = \varepsilon_a(\theta-\ii\phi_a), \qquad \widetilde L_a(\theta) = L_a(\theta-\ii\phi_a).

The phase-shifted kernel is

Ka,b(u)=12πcosh ⁣(u+i(ϕbϕa)).K_{a,b}(u) = \frac{1}{ 2\pi\cosh\!\left( u+\ii(\phi_b-\phi_a) \right) }.

The residue-free minimal equation is valid while

ϕbϕa<π2|\phi_b-\phi_a|<\frac{\pi}{2}

for every adjacent pair. The full pole set satisfies

u+i(ϕbϕa)=πi(n+12),nZ.u+\ii(\phi_b-\phi_a) = \pi\ii\left(n+\frac12\right), \qquad n\in\mathbb Z.

The nearest pair lies at ±πi/2\pm\pi\ii/2 before the phase shift. Thus a pole first reaches the real uu-axis at relative phase ±π/2\pm\pi/2. Either equality is a contour singularity, not a point at which one simply evaluates the old formula. The continuation below follows the positive crossing.

For the A2A_2 chain, put

Δ=ϕ2ϕ1,η=Δπ2.\Delta=\phi_2-\phi_1, \qquad \eta=\Delta-\frac{\pi}{2}.

Continue along the IMS path from Δ<π/2\Delta<\pi/2 to η>0\eta>0. With the published lateral prescription, restoring both contours to the real line gives

ε~1(θ)=m1eθK1,2L~2(θ)L~2(θ+iη+i0),ε~2(θ)=m2eθK2,1L~1(θ)L~1(θiηi0).\begin{aligned} \widetilde\varepsilon_1(\theta) ={}& |m_1|\ee^\theta -K_{1,2}*\widetilde L_2(\theta) -\widetilde L_2(\theta+\ii\eta+\ii0), \\ \widetilde\varepsilon_2(\theta) ={}& |m_2|\ee^\theta -K_{2,1}*\widetilde L_1(\theta) -\widetilde L_1(\theta-\ii\eta-\ii0). \end{aligned}

These signs are for increasing Δ\Delta in the IMS pseudoenergy and bank convention. Reversing the path reverses the crossed residues. Translating to Page 3’s plus-left inverse-jump convention additionally applies the bank involution fixed on Page 4; one must not mix the two sign ledgers.

The equations are no longer closed on real-axis values: their sources sample L1,L2L_1,L_2 on displaced lines. One can keep those off-axis values as auxiliary unknowns, or change coordinates so the new active charge is explicit.

A composite charge closes the wall-crossed A₂ system

Section titled “A composite charge closes the wall-crossed A₂ system”

The geometric A2A_2 continuation contains the composite cycle

γ12=γ1+γ2.\gamma_{12}=\gamma_1+\gamma_2.

In the alternating IMS mass convention,

m12=m1im2=m12eiϕ12.m_{12}=m_1-\ii m_2 = |m_{12}|\ee^{\ii\phi_{12}}.

This is simply central-charge addition expressed after the parity-dependent phase choices; it is not a universal formula for arbitrary mass conventions. Introduce the wall-crossed coordinates

Y1n(θ)=Y1(θ)1+Y2(θπi2),Y2n(θ)=Y2(θ)1+Y1(θ+πi2),Y12n(θ)=Y1(θ)Y2(θπi2)1+Y1(θ)+Y2(θπi2).\begin{aligned} Y_1^{\mathrm n}(\theta) &= \frac{ Y_1(\theta) }{ 1+Y_2(\theta-\frac{\pi\ii}{2}) }, \\ Y_2^{\mathrm n}(\theta) &= \frac{ Y_2(\theta) }{ 1+Y_1(\theta+\frac{\pi\ii}{2}) }, \\ Y_{12}^{\mathrm n}(\theta) &= \frac{ Y_1(\theta)Y_2(\theta-\frac{\pi\ii}{2}) }{ 1+Y_1(\theta)+Y_2(\theta-\frac{\pi\ii}{2}) }. \end{aligned}

Set εan=logYan\varepsilon_a^{\mathrm n}=-\log Y_a^{\mathrm n}, shift each new function by its mass phase, and then suppress the superscript n\mathrm n. With

Ka,b±(u)=Ka,b ⁣(u±πi2),K_{a,b}^{\pm}(u) = K_{a,b}\!\left(u\pm\frac{\pi\ii}{2}\right),

the closed system is

ε~1=m1eθK1,2L~2K1,12+L~12,ε~2=m2eθK2,1L~1K2,12L~12,ε~12=m12eθK12,1L~1K12,2L~2.\begin{aligned} \widetilde\varepsilon_1 ={}& |m_1|\ee^\theta -K_{1,2}*\widetilde L_2 -K_{1,12}^{+}*\widetilde L_{12}, \\ \widetilde\varepsilon_2 ={}& |m_2|\ee^\theta -K_{2,1}*\widetilde L_1 -K_{2,12}*\widetilde L_{12}, \\ \widetilde\varepsilon_{12} ={}& |m_{12}|\ee^\theta -K_{12,1}^{-}*\widetilde L_1 -K_{12,2}*\widetilde L_2. \end{aligned}

The third pseudoenergy is not an arbitrary numerical stabilizer. It is the coordinate of a definite composite charge with a definite classical drive. In this model, the kernel-pole residue, the new WKB discontinuity, and the enlarged BPS spectrum are three descriptions of one geometric continuation.

For a degree-(r+1)(r+1) polynomial, the minimal chamber has rr adjacent cycles. The maximal chamber can contain

r(r+1)2\frac{r(r+1)}{2}

cycles joining all ordered pairs of turning points. Intermediate chambers have intermediate charge sets. The count is a result for this polynomial geometry; it is not the node count of an arbitrary GMN problem.

The ordered product survives while its factors mutate

Section titled “The ordered product survives while its factors mutate”

Let e1e2=1e_1\mathbin{\cdot}e_2=1. In the book’s inverse-jump convention, the elementary automorphisms act on x=Ve1x=\mathcal V_{e_1} and y=Ve2y=\mathcal V_{e_2} by

Se1I:(x,y)(x,y1+x),Se2I:(x,y)(x(1+y),y).\begin{aligned} \mathfrak S_{e_1}^{\mathrm I} &: (x,y) \longmapsto \left(x,\frac{y}{1+x}\right), \\ \mathfrak S_{e_2}^{\mathrm I} &: (x,y) \longmapsto \left(x(1+y),y\right). \end{aligned}

With the rightmost map acting first, the pentagon identity is

Se2ISe1I=Se1ISe1+e2ISe2I.\boxed{ \mathfrak S_{e_2}^{\mathrm I} \circ \mathfrak S_{e_1}^{\mathrm I} = \mathfrak S_{e_1}^{\mathrm I} \circ \mathfrak S_{e_1+e_2}^{\mathrm I} \circ \mathfrak S_{e_2}^{\mathrm I} }.

Both sides send the generators to

(x(1+y),y1+x+xy).\left( x(1+y), \frac{y}{1+x+xy} \right).

The identity says that the ordered automorphism of a convex sector is unchanged when the two active rays swap order, provided the composite factor is inserted. It does not say that individual coordinates have no Stokes jumps, nor does the algebra by itself prove that a proposed ODE has the required active charges. That geometric input comes from its WKB spectral network or BPS structure.

The relation to the integral equation can now be stated precisely:

  • a contour residue is an analytic fact about a chosen representation;
  • a charge mutation is discrete geometric data;
  • the KS identity is the compatibility condition ensuring that the sectorial RH problem reconstructs the same continued coordinate;
  • the three are identified in the polynomial example only after the Page 4 node–charge, phase, refinement, and bank dictionaries are passed.

An excited-state divisor crossing generally changes none of the BPS indices in this identity. It chooses a different solution or adds state roots to the same integral operator.

For a generic quartic polynomial, r=3r=3. The minimal chamber begins with the adjacent cycles

γ1,γ2,γ3.\gamma_1, \qquad \gamma_2, \qquad \gamma_3.

Continuing to the maximal chamber activates the three additional cycles

γ12=γ1+γ2,γ23=γ2+γ3,γ123=γ1+γ2+γ3.\gamma_{12}=\gamma_1+\gamma_2, \qquad \gamma_{23}=\gamma_2+\gamma_3, \qquad \gamma_{123}=\gamma_1+\gamma_2+\gamma_3.

The generic maximal-chamber TBA therefore has six functions, in agreement with r(r+1)/2=6r(r+1)/2=6. In the alternating IMS convention their new drives are fixed by

m12=m1im2,m23=m3im2,m123=m1+m3im2.\begin{aligned} m_{12}&=m_1-\ii m_2, \qquad m_{23}=m_3-\ii m_2, \\ m_{123}&=m_1+m_3-\ii m_2. \end{aligned}

These are the period sums written after the same parity-dependent factors of i\ii used in the A2A_2 example. They are not guessed effective masses. The full six-equation system is needed at generic quartic moduli.

The symmetric quartic first uses parity to identify ε~3=ε~1\widetilde\varepsilon_3=\widetilde\varepsilon_1 and ε~23=ε~12\widetilde\varepsilon_{23}=\widetilde\varepsilon_{12}, leaving four functions labelled 1,2,12,1231,2,12,123. The monic pure-quartic point is more special: its enhanced Z4\mathbb Z_4 symmetry further gives

ε~1=ε~12,ε~2=ε~123.\widetilde\varepsilon_1 = \widetilde\varepsilon_{12}, \qquad \widetilde\varepsilon_2 = \widetilde\varepsilon_{123}.

Only after the wall crossings and these orbit reductions does one obtain the two-function equation matched explicitly on Page 4. There are consequently two different reductions in play:

minimal A3 functional systemreflection fold2 functions,maximal six-charge WKB systemZ4 charge orbits2 functions.\begin{array}{c} \text{minimal }A_3\text{ functional system} \xrightarrow{\text{reflection fold}} 2\text{ functions}, \\ \text{maximal six-charge WKB system} \xrightarrow{\mathbb Z_4\text{ charge orbits}} 2\text{ functions}. \end{array}

Page 4 proved that the two endpoints agree after the rapidity, mass, kernel, and nonlinear-logarithm passports are transported. The present page supplies the missing chamber route to the WKB endpoint. Folding the minimal three-cycle equation directly would skip the composite charges and would not describe a generic continuation through the intervening walls.

On the wall, a prescription is part of the answer

Section titled “On the wall, a prescription is part of the answer”

At the instant a logarithmic branch point or kernel pole lies on the reference contour, the ordinary real integral is undefined. Three choices must not be conflated:

  • an upper or lower indentation gives a lateral value and a signed full residue relative to the opposite lateral contour;
  • a principal value can be appropriate when a real or conjugation symmetry makes the two lateral values comparable;
  • a median prescription may add a nonlinear half-jump and is not in general the arithmetic mean of two complex solutions.

For a simple pole, a contour drawn exactly through the pole can be represented by a principal value plus a half-residue, but the sign is set by the indentation. For a logarithmic zero, the branch cut and the continued sheet of LL must also be recorded. A statement such as “take half the source at the wall” is incomplete without those choices.

A continuation algorithm that preserves the passports

Section titled “A continuation algorithm that preserves the passports”

For either an excited state or a changing WKB chamber:

  1. begin at a point where the equation and analytic class are verified;
  2. continue parameters in small steps and use the previous solution as the next seed;
  3. reconstruct the unknown off the contour and track zeros of every nonlinear factor as well as poles of every shifted kernel;
  4. record the direction, multiplicity, bank, and winding of each crossing;
  5. add the residue primitive and solve its root condition, or enlarge the charge basis if the geometric wall-crossing dictionary requires it;
  6. verify the functional relation or ordered sector product on both sides of the wall;
  7. test continuity of a normalization-independent observable before assigning physical state labels;
  8. retain both lateral solutions at a wall until a physical prescription selects one.

The continuation path matters. Returning to the same numerical parameter after circling a branch point can land on a different solution sheet. A state label should therefore include the seed, path homotopy, root integers, and final boundary condition—not only the final masses.

Calling every source an excited particle. A kernel-pole residue can be forced by changing a WKB mass phase even when no state divisor crosses. Inspect which singularity moved before interpreting the source.

Changing the BPS spectrum because a kernel looks singular. A genuine mutation needs a charge lattice, active indices, and an ordered-product identity. Contour algebra alone supplies none of those discrete facts.

Solving the sourced equation without the root condition. The source positions are zeros of the solution being sought. Treating them as fitted constants generally produces a function with the wrong divisor.

Dropping the lateral sign at the wall. The IMS formulas and Page 3 use opposite bank conventions. Transfer the Page 4 involution before comparing residue signs.

Assuming the final parameter fixes the state. Analytic continuation can have monodromy, and different paths can reach different solution sheets at the same parameter value.

Updating the pseudoenergy but not the observable. Straightening a contour changes every integral over the crossed logarithm. Recompute the energy, period, or determinant residue from the same oriented contour.

Calling a source-free branch the ground state. The Lee–Yang example already has two physically relevant branches of the unsourced equation. The divisor and continuation-path ledgers, not the visible source count, identify the state.

1. Classify the crossing before changing the equation. For each event, name the primary ledger that changes: (a) a zero of 1+eεb1+\ee^{-\varepsilon_b} passes through the rapidity contour; (b) ϕ2ϕ1\phi_2-\phi_1 increases through π/2\pi/2; (c) two active BPS rays align and a composite factor appears; (d) parameters circle a pinch and return without any final contour crossing.

Solution

(a) changes the divisor ledger and adds a root condition. (b) changes the contour representation because a kernel pole crosses. (c) changes the chamber ledger while preserving the ordered sector product. (d) changes the solution sheet and therefore the continuation entry in the observable ledger; the displayed contour and equation may remain unchanged. If two events occur at once, both entries must be recorded independently.

2. Derive an oriented source. Suppose zbz_b has one simple zero at θ\theta_*, the closed difference CbR\mathcal C_b-\mathbb R winds once counterclockwise around it, and

Kab(u)=12πiuLogSab(u).K_{ab}(u) = \frac{1}{2\pi\ii} \partial_u\Log S_{ab}(u).

Find the source created when Cb\mathcal C_b is restored to R\mathbb R.

Solution

Because θLogSab(θθ)=2πiKab\partial_{\theta'}\Log S_{ab}(\theta-\theta')=-2\pi\ii K_{ab}, integration by parts changes the contour difference into

12πiCbRLogSab(θθ)θLogzb(θ) ⁣dθ.-\frac{1}{2\pi\ii} \oint_{\mathcal C_b-\mathbb R} \Log S_{ab}(\theta-\theta') \partial_{\theta'}\Log z_b(\theta')\,\dd\theta'.

The logarithmic derivative has residue 11 at θ\theta_*. The source is therefore

LogSab(θθ).-\Log S_{ab}(\theta-\theta_*).

A clockwise winding, a pole of zbz_b, or the opposite primitive convention reverses the corresponding sign.

3. Reconstruct the Lee–Yang source pair. A transported contour winds counterclockwise around θ0-\theta_0 and clockwise around +θ0+\theta_0. Use Exercise 2 to find the two-source factor and state the equation that determines θ0\theta_0 on the one-particle branch.

Solution

The winding numbers are +1+1 and 1-1. Hence the two contributions are

LogS(θ+θ0)+LogS(θθ0)=LogS(θθ0)S(θ+θ0).-\Log S(\theta+\theta_0) +\Log S(\theta-\theta_0) = \Log\frac{S(\theta-\theta_0)}{S(\theta+\theta_0)}.

The source location must satisfy ε(θ0)=πi\varepsilon(\theta_0)=\pi\ii on the continuously selected branch. Evaluation of the sourced TBA at θ0\theta_0 gives

0=rcoshθ0LogS(2θ0)ϕL(θ0).0 = r\cosh\theta_0 -\Log S(2\theta_0) -\phi*L(\theta_0).

The branch of LogS\Log S is part of this equation.

4. Distinguish a DDV hole from an algebraic factor. The auxiliary function obeys a TQTQ identity and its counting equation has zeros of 1+a1+a. Explain why a hole is not an additional factor in that identity, and list the extra data needed to turn a zero into a state label.

Solution

The TQTQ identity fixes the algebraic divisor of 1+a1+a in terms of transfer and QQ data. A hole is a zero allowed by the counting function but omitted from the occupied Bethe-root set; it is a label relative to a chosen state and contour, not a new multiplicative term. One must record the lip crossed, its multiplicity, the integer in

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

and the boundary determinant or Q-eigenvalue that identifies the state.

5. Locate every phase-shifted kernel pole. For

KΔ(u)=12πcosh(u+iΔ),K_\Delta(u) = \frac{1}{2\pi\cosh(u+\ii\Delta)},

find its poles and residues as functions of Δ\Delta. When does the first one meet the real uu-axis?

Solution

The zeros of coshz\cosh z occur at z=πi(n+12)z=\pi\ii(n+\tfrac12). Therefore

un=πi(n+12)iΔ,nZ.u_n = \pi\ii\left(n+\frac12\right)-\ii\Delta, \qquad n\in\mathbb Z.

Since sinh[πi(n+12)]=i(1)n\sinh[\pi\ii(n+\tfrac12)]=\ii(-1)^n, the residues in the uu plane are

Resu=unKΔ(u)=(1)n2πi.\operatorname*{Res}_{u=u_n}K_\Delta(u) = \frac{(-1)^n}{2\pi\ii}.

For continuation from Δ<π/2|\Delta|<\pi/2, the nearest poles meet the real axis at Δ=±π/2\Delta=\pm\pi/2. Which residue enters the equation still depends on the direction and contour orientation.

6. Recover the composite drive. Suppose

Y1(θ)exp(m1eθ),Y2 ⁣(θπi2)exp(im2eθ).Y_1(\theta) \sim \exp(-m_1\ee^\theta), \qquad Y_2\!\left(\theta-\frac{\pi\ii}{2}\right) \sim \exp(\ii m_2\ee^\theta).

Find the classical mass of their product.

Solution

Multiplication adds the exponents:

Y1(θ)Y2 ⁣(θπi2)exp ⁣[(m1im2)eθ].Y_1(\theta) Y_2\!\left(\theta-\frac{\pi\ii}{2}\right) \sim \exp\!\left[-(m_1-\ii m_2)\ee^\theta\right].

Thus m12=m1im2m_{12}=m_1-\ii m_2. This is the IMS expression of Z12=Z1+Z2Z_{12}=Z_1+Z_2 after its alternating phase convention has been applied.

7. Verify the inverse-jump pentagon. Let x=Ve1x=\mathcal V_{e_1}, y=Ve2y=\mathcal V_{e_2}, and e1e2=1e_1\mathbin{\cdot}e_2=1. Apply both sides of

Se2ISe1I=Se1ISe1+e2ISe2I\mathfrak S_{e_2}^{\mathrm I} \circ\mathfrak S_{e_1}^{\mathrm I} = \mathfrak S_{e_1}^{\mathrm I} \circ\mathfrak S_{e_1+e_2}^{\mathrm I} \circ\mathfrak S_{e_2}^{\mathrm I}

to (x,y)(x,y), with the rightmost factor acting first.

Solution

On the left, the e1e_1 jump first sends yy/(1+x)y\mapsto y/(1+x); the e2e_2 jump must then act on the transformed coordinates. Direct substitution gives

(x,y)(x(1+y),y1+x+xy).(x,y) \longmapsto \left( x(1+y), \frac{y}{1+x+xy} \right).

Using multiplicativity, Ve1+e2=xy\mathcal V_{e_1+e_2}=xy, and applying the three right-hand factors in order gives the same pair. The middle factor is indispensable: omitting it makes the two noncommuting orders different.

8. Audit the quartic chamber route. List the charge sets before and after maximal wall crossing for r=3r=3. Why is the final two-function pure-quartic equation not obtained merely by deleting one of the six charges?

Solution

The minimal set is {γ1,γ2,γ3}\{\gamma_1,\gamma_2,\gamma_3\}. The maximal set is

{γ1,γ2,γ3,γ12,γ23,γ123}.\{\gamma_1,\gamma_2,\gamma_3, \gamma_{12},\gamma_{23},\gamma_{123}\}.

The three composite charges carry their own drives and couplings, so none can simply be discarded at generic moduli. Parity first reduces the six functions to four; the enhanced pure-quartic symmetry then identifies 11 with 1212 and 22 with 123123, leaving two. This maximal-chamber reduction and the reflection fold of the minimal A3A_3 functional system are distinct constructions whose endpoints agree only after the Page 4 passport checks.