# The Bethe Ansatz

#### Factorized form of TBA equationse.h.f

The fundamental set of TBA equations h.tba1 fully determine the thermal equilibrium state of the $$XXX$$ chain at given magnetic field and temperature. Their form is however slightly inconvenient if one thinks of an actual calculation (for example, a numerical solution) due to the fact that all functions are coupled to all others. The coupling kernel $$A_{nm}$$ is however relatively simple, being given by a relatively simple sum of fundamental kernels. Inverting this kernel might then yield worthwhile simplifications.

Fourier transforms of kernels

Using the Fourier space expression of our fundamental kernel

\begin{equation*} a_n (\omega) = e^{-\frac{|\omega|}{2} n} \end{equation*}

and the definition of the string-string derivative kernel

\begin{equation*} A_{nm} (\omega) = (1 - \delta_{nm}) a_{|n-m|} (\omega) + 2 a_{|n-m|+2} (\omega) + ... + 2 a_{n+m-2} (\omega) + a_{n+m} (\omega), \end{equation*}

we obtain

\begin{equation*} A_{n m} (\omega) = \coth \frac{|\omega|}{2} \left( e^{-\frac{|\omega|}{2}|n-m|} - e^{-\frac{|\omega|}{2}(n+m)}\right) - \delta_{nm}. \end{equation*}
Derivation \begin{align*} A_{n m} &= e^{-\frac{|\omega|}{2} |n-m|} + 2 \sum_{j=1}^{\mbox{min}(n,m) - 1} e^{-\frac{|\omega|}{2} |n-m|+2j} + e^{-\frac{|\omega|}{2} (n+m)} - \delta_{nm} \nonumber \\ &= e^{-\frac{|\omega|}{2} |n-m|} + 2 e^{-\frac{|\omega|}{2} |n-m|} \frac{e^{-|\omega|} - e^{-|\omega| \mbox{min} (n,m)}}{1 - e^{-|\omega|}} + e^{-\frac{|\omega|}{2} (n+m)} - \delta_{nm} \nonumber \\ &= \frac{1}{1 - e^{-|\omega|}} \left( e^{-\frac{|\omega|}{2} |n-m|} + e^{-\frac{|\omega|}{2} (|n-m|+2)} - e^{-\frac{|\omega|}{2} (n+m)} - e^{-\frac{|\omega|}{2} (n+m+2)} \right) - \delta_{nm} \nonumber \\ &= \frac{1 + e^{-|\omega|}}{1 - e^{-|\omega|}} \left( e^{-\frac{|\omega|}{2} |n-m|} - e^{-\frac{|\omega|}{2} (n+m)} \right) - \delta_{nm} \end{align*}

For future convenience, let us define a kernel inverting $$\delta_{nm} + A_{nm}$$. One can verify by direct substitution that the following inversion identity holds:

$$\sum_{l=1}^\infty B_{nl} (\omega) \left[ \delta_{lm} + A_{lm} (\omega)\right] = \delta_{nm}, \tag{h.b1}\label{h.b1}$$

where

$$B_{nm} (\omega) = \delta_{nm} - s(\omega) (\delta_{n, m+1} + \delta_{n, m-1}) \tag{h.bnm}\label{h.bnm}$$

in which we used

$$s(\lambda) = \frac{1}{2\cosh \pi \lambda}, \hspace{10mm} s(\omega) = \frac{1}{2\cosh \frac{\omega}{2}}. \tag{h.s}\label{h.s}$$
Derivation \begin{equation*} \sum_{l=1}^\infty B_{nl} (\omega) \left[ \delta_{lm} + A_{lm} (\omega)\right] = A_{nm} (\omega) - \frac{1}{2\cosh \frac{\omega}{2}} (A_{n+1 m} (\omega) + A_{n-1 m} (\omega)) - (B_{nm}(\omega) - \delta_{nm}). \end{equation*}

We have

\begin{equation*} A_{n+1 m} (\omega) + A_{n-1 m} (\omega) = \coth \frac{|\omega|}{2} \left( e^{-\frac{|\omega|}{2} |n-m + 1|} + e^{-\frac{|\omega|}{2} |n-m-1|} - e^{-\frac{|\omega|}{2} (n+m)} 2\cosh \frac{\omega}{2} \right) \end{equation*}

The term in parentheses is

\begin{equation*} = \left\{ \begin{array}{ll} 2\cosh \frac{\omega}{2} \left( e^{-\frac{|\omega|}{2}|n-m|} - e^{-\frac{|\omega|}{2}(n+m)}\right), \hspace{10mm} & n \neq m, \\ 2 e^{-\frac{|\omega|}{2}} - e^{-|\omega| n} ~2\cosh \frac{\omega}{2}, & n = m. \end{array} \right. \end{equation*}

Therefore, for $$n \neq m$$, we get

\begin{equation*} \sum_{l=1}^\infty B_{nl} (\omega) \left[ \delta_{lm} + A_{lm} (\omega)\right] = A_{nm} (\omega) - A_{nm}(\omega) - B_{nm} (\omega) = -B_{nm} (\omega) \end{equation*}

whereas for $$n = m$$ we obtain (using $$A_{nn} (\omega) = \coth \frac{|\omega|}{2} \left( 1 - e^{-|\omega| n} \right)$$)

\begin{equation*} \sum_{l=1}^\infty B_{nl} (\omega) \left[ \delta_{ln} + A_{ln} (\omega)\right] =\frac{1}{\sinh \frac{|\omega|}{2}} \left( (1 - e^{-|\omega| n}) \cosh \frac{\omega}{2} - e^{-\frac{|\omega|}{2}} + e^{-|\omega| n} ~\cosh \frac{\omega}{2} \right) = 1. \end{equation*}

We moreover have the identity

$$\sum_{m=1}^\infty B_{nm} (\omega) a_m (\omega) = \delta_{n,1} s(\omega). \tag{h.ba}\label{h.ba}$$
Derivation \begin{equation*} \sum_{m=1}^\infty B_{nm} (\omega) a_m (\omega) = a_n (\omega) - \frac{1}{2\cosh \frac{\omega}{2}} (a_{n+1} (\omega) + (1 - \delta_{n,1}) a_{n-1} (\omega)) = 0 + \delta_{n,1} s (\omega). \end{equation*}

To obtain a simpler form of the TBA equations, let us start by rewriting the equilibrium conditions h.tba1 as

\begin{equation*} \ln (1 + \eta_n (\lambda)) - \sum_{m=1}^\infty \left[ \delta_{nm} + A_{nm} \right] * \ln (1 + \eta_m^{-1} (\lambda)) = \frac{h n - \pi J a_n (\lambda)}{T}. \end{equation*}

Convolving this with $$\sum B$$, using

\begin{equation*} \sum_{m=1}^\infty B_{nm} * m = \sum_m m \int_{-\infty}^\infty d\lambda B_{nm} (\lambda) = \sum_m m B_{nm} (\omega = 0) = 0 \end{equation*}

and the identities h.b1 and h.ba, we obtain the simple system

\begin{align} & \ln \eta_1 (\lambda) = -\frac{\pi J}{T} s(\lambda) + s * \ln (1 + \eta_2 (\lambda)), \nonumber \\ & \ln \eta_n (\lambda) = s * \left[\ln (1 + \eta_{n-1} (\lambda)) + \ln (1 + \eta_{n+1} (\lambda)) \right], \hspace{10mm} n > 1. \tag{h.tbaf}\label{h.tbaf} \end{align}

Amazingly, the magnetic field does not appear in this infinite set of coupled equations. In fact, we must at this stage recognize that the system we have obtained in h.tbaf is incomplete as it stands. Namely, the functions $$\eta_n$$ for large $$n$$ can take larger and larger values, and as such our sequence of equations is divergent, but the nature of this divergent is set by information not contained in h.tbaf. The asymptotics at large $$n$$ are however easy to read from our original form of the TBA equations h.tba1: as $$n \rightarrow \infty$$, $$\eta_n \rightarrow \infty$$ in such a way that

$$\lim_{n\rightarrow \infty} \frac{\ln \eta_n (\lambda)}{n} = \frac{h}{T}. \tag{h.tbaa}\label{h.tbaa}$$

Setting this asymptotic condition renders the solution of h.tbaf unique.

An equivalent rewriting of these equations makes use of the definitions

\begin{equation*} \eta_n (\lambda) \equiv e^{\varepsilon_n(\lambda)/T} \end{equation*}

to obtain

\begin{align} \varepsilon_1 (\lambda) &= -\pi J s(\lambda) + s * T \ln \left[1 + e^{\varepsilon_2(\lambda)/T}\right], \nonumber \\ \varepsilon_n (\lambda) &= s * T \ln \left[1 + e^{\varepsilon_{n-1}(\lambda)/T} \right] + s * T \ln \left[1 + e^{\varepsilon_{n+1}(\lambda)/T} \right], \hspace{5mm} n > 1, \nonumber \\ \lim_{n \rightarrow \infty} &\frac{\varepsilon_n(\lambda)}{n} = h. \tag{h.tbae}\label{h.tbae} \end{align}

Determining the densities

To determine the densities $$\rho_n(\lambda)$$ and $$\rho_n^h(\lambda)$$, one can for example proceed as follows. The $$\eta_n$$ are functions of $$\lambda$$ and implicitly of $$J, h$$ and $$T$$. From h.tba1, one gets for example the derivative system

$$\frac{\pi}{T} a_n (\lambda) = \frac{-1}{\eta_n} \frac{\partial \eta_n (\lambda)}{\partial J} + \sum_{m=1}^\infty A_{nm} * \left(\frac{-1}{\eta_m (1 + \eta_m)} \frac{\partial \eta_m (\lambda)}{\partial J} \right). \tag{h.tbafd}\label{h.tbafd}$$

By inspection, at the hand of l.bec, on finds that

$$\rho_n (\lambda) = \frac{-T}{\pi} \frac{1}{\eta_n (1 + \eta_n)} \frac{\partial \eta_n (\lambda)}{\partial J}, \hspace{10mm} \rho_n (\lambda) + \rho_n^h (\lambda) = \frac{-T}{\pi} \frac{\partial \ln \eta_n (\lambda)}{\partial J}. \tag{h.tbafr}\label{h.tbafr}$$

To find the densities, one can thus start by solving the factorized system h.tbaf (using the associated asymptotic conditions h.tbaa for the $$\eta_n$$. Thereafter a solution can be found to h.tbafd, and substituted in h.tbafr.