# The Bethe Ansatz

#### Equations for the thermal equilibrium statee.h.eq

Let us now apply the formalism of the thermodynamic Bethe Ansatz to the isotropic antiferromagnet. Recall the BGT equations h.bgtl:

\begin{equation*} \phi_j (\lambda^j_\alpha) - \frac{1}{N} \sum_{k=1}^{N_s} \sum_{\beta = 1}^{M_k} \Phi_{jk} (\lambda^j_\alpha - \lambda^k_\beta) = \frac{2\pi}{N} I^j_\alpha \end{equation*}

with fundamental kernels h.phin

\begin{equation*} \phi_n(\lambda) = 2\mbox{atan} \frac{2\lambda}{n} \end{equation*}

and the string-string scattering phase shift h.phijk

\begin{equation*} \Phi_{jk} (\lambda) = (1 - \delta_{jk}) \phi_{|j-k|} (\lambda) + 2\phi_{|j-k|+2} (\lambda) + ... + 2\phi_{j+k-2} (\lambda) + \phi_{j+k} (\lambda). \end{equation*}

We will make use of the derivative kernels h.an

\begin{equation*} a_n (\lambda) = \frac{1}{2\pi} \frac{d}{d\lambda}\phi_n(\lambda) = \frac{1}{2\pi} \frac{n}{\lambda^2 + n^2/4}, \hspace{10mm} a_0 (\lambda) \equiv \delta (\lambda) \end{equation*}

and

$$A_{n m} (\lambda) = \frac{1}{2\pi} \frac{d}{d\lambda}\Phi_{n m}(\lambda). \tag{h.anm}\label{h.anm}$$

In terms of these, the Bethe equations take the form

\begin{equation*} a_n (\lambda) = \rho_n (\lambda) + \rho_n^h (\lambda) + \sum_{m=1}^\infty A_{nm} * \rho_m (\lambda). \end{equation*}

The energy of a state with distributions $$\rho_n (\lambda)$$ of $$n$$-strings, $$n = 1, 2, 3, \ldots$$ is given by

\begin{equation*} \frac{E}{N} = -\frac{h}{2} + \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda ~d_n (\lambda) \rho_n (\lambda) \end{equation*}

where we have defined the driving term

\begin{equation*} d_n (\lambda) = hn - \pi J a_n (\lambda). \end{equation*}

This comes from coupling the external field $$h$$ to the total $$z$$ projection of spin,

\begin{equation*} \frac{S^z_{tot}}{N} = \frac{1}{2} - \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda ~n \rho_n (\lambda). \end{equation*}

The Yang-Yang entropy of the state is given by

\begin{equation*} \frac{S}{N} = \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda \left[ (\rho_n + \rho_n^h) \ln (\rho_n + \rho_n^h) - \rho_n \ln \rho_n - \rho_n^h \ln \rho_n^h \right]. \end{equation*}

The free energy density per site can thus be written

\begin{equation*} f = \frac{E - TS}{N} = -\frac{h}{2} + \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda \left[d_n \rho_n - T (\rho_n + \rho_n^h) \ln (\rho_n + \rho_n^h) + T \rho_n \ln \rho_n + T \rho_n^h \ln \rho_n^h \right]. \end{equation*}

The variation of this free energy is written as

\begin{equation*} \delta f = \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda \left\{ \left[ d_n - T \ln (1 + \rho_n^h/\rho_n) \right] \delta \rho_n - T \ln (1 + \rho_n/\rho_n^h) \delta \rho_n^h \right\}. \end{equation*}

The variations $$\delta \rho_n$$ and $$\delta \rho_n^h$$ are however not independent, but linked by the following constraint (from the Bethe equations):

\begin{equation*} \delta \rho_n^h (\lambda) = -\delta \rho_n (\lambda) - \sum_{m=1}^\infty A_{nm} * \delta \rho_m (\lambda). \end{equation*}

Substituting this into $$\delta f$$ and using the fact that $$A_{nm} = A_{mn}$$, we obtain

\begin{equation*} \delta f \!=\! \sum_{n=1}^\infty \int_{-\infty}^\infty \!\!d\lambda ~\delta \rho_n (\lambda) \left[ d_n (\lambda) - T \ln (\rho_n^h (\lambda)/\rho_n(\lambda)) + \sum_{m=1}^\infty A_{nm} \!*\! T \ln \left[ 1 + \rho_m(\lambda)/\rho_m^h(\lambda) \right] \right]. \end{equation*}

The condition of thermodynamic stability leads to the requirement that the expression in square brackets vanish for all $$\lambda$$. Using the notation

\begin{equation*} \eta_n (\lambda) \equiv \rho_n^h (\lambda)/\rho_n (\lambda), \end{equation*}

thermodynamic equilibrium thus imposes the conditions

$$\ln \eta_n (\lambda) = \frac{d_n(\lambda)}{T} + \sum_{m=1}^\infty A_{nm} * \ln (1 + \eta_m^{-1} (\lambda)). \tag{h.tba1}\label{h.tba1}$$

The equilibrium value of the free energy itself can now be calculated, explicitly using the Bethe equations to write $$\rho^h$$ in terms of $$\rho's$$. The result is

\begin{align*} f = -\frac{h}{2} + &\sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda ~\times \nonumber \\ &\left\{-T \ln (1 + \eta_n^{-1} (\lambda)) a_n (\lambda) + \left[ d_n - T \ln \eta_n + \sum_m A_{nm} * T \ln (1 + \eta_m^{-1} (\lambda)) \right] \right\}. \end{align*}

Noticing that the term in square brackets vanishes due to the equilibrium condition, we finally get the substantially more concise form

$$f = -\frac{h}{2} - T \sum_{n=1}^\infty \int_{-\infty}^\infty d\lambda a_n (\lambda) \ln (1 + \eta_n^{-1} (\lambda)). \tag{h.fn}\label{h.fn}$$

This can be further rewritten in the following way. Starting from the $$n=1$$ case of the Bethe equations,

\begin{equation*} \ln (1 + \eta_1 (\lambda)) = \frac{h - \pi J a_1 (\lambda)}{T} + \sum_{m=1}^\infty (a_{m-1} + a_{m+1}) * \ln (1 + \eta_m^{-1} (\lambda)), \end{equation*}

and operating $$\int_{-\infty}^\infty d\lambda s(\lambda)$$ on this (with $$s(\lambda) = \frac{1}{2\cosh \pi \lambda}$$), we obtain

\begin{equation*} \int_{-\infty}^\infty d\lambda s(\lambda) \ln (1 + \eta_1 (\lambda)) = \frac{h}{2T} - \frac{\pi J}{T} \int_{-\infty}^\infty d\lambda s(\lambda) a_1 (\lambda) + \sum_{m=1}^\infty a_m (\lambda) \ln (1 + \eta_m^{-1} (\lambda)). \end{equation*}

Substituting this in the free energy (recognizing the ground-state energy in the second term on the right-hand side) gives the yet more economical form

$$f = -J \ln 2 - T \int_{-\infty}^\infty d\lambda s(\lambda) \ln (1 + \eta_1 (\lambda)). \tag{h.f1}\label{h.f1}$$