# The Bethe Ansatz

#### Properties of solutions to the Bethe equationsc.l.p

Theorem 1: all solutions of the Bethe equations for the interacting Bose gas are real for $$c > 0$$.

Proof: consider $$e^{i \lambda L}$$. Observe that $$|e^{i\lambda L}| \leq 1$$ for $$\mbox{Im}~ \lambda \geq 0$$ and $$|e^{i\lambda L}| \geq 1$$ for $$\mbox{Im}~ \lambda \leq 0$$. Moreover, $$|(\lambda + ic)/(\lambda-ic)| \geq 1$$ for $$\mbox{Im}~ \lambda \geq 0$$ and $$c > 0$$, and then also $$|(\lambda + ic)/(\lambda-ic)| \leq 1$$ for $$\mbox{Im}~ \lambda \leq 0$$.

Consider a set $$\{ \lambda_j \}$$, $$j = 1, ..., N$$, solution to the Bethe equations for a given proper set of quantum numbers $$\{ I_j \}$$. Define the momentum with maximal imaginary part $$\lambda_{max} \in \{ \lambda_j \}$$, such that $$\mbox{Im}~ \lambda_{max} \geq \mbox{Im}~ \lambda_j$$, $$j = 1, ..., N$$. The Bethe equation for $$\lambda_{max}$$ gives

\begin{equation*} |e^{i\lambda_{max} L} | = \left| \prod_j \frac{\lambda_{max} - \lambda_j + ic}{\lambda_{max} - \lambda_j - ic} \right| \geq 1 \end{equation*}

meaning that we must have $$\mbox{Im}~ \lambda_{max} \leq 0$$, so $$\mbox{Im}~ \lambda_j \leq 0 ~\forall ~j$$. Defining $$\lambda_{min}$$ in a similar way, we can show that $$\mbox{Im}~ \lambda_j \geq 0 ~\forall ~j$$, so $$\mbox{Im}~ \lambda_j = 0 ~\forall ~j$$ $$\Box$$.

Theorem 2: for $$c > 0$$ and for a given proper set of quantum numbers $$\{ I_j\}$$, the solution for the set $$\{ \lambda_j \}$$ exists and is unique (Yang and Yang '69 1969.Yang.JMP.10).

Proof: The Bethe equations are equivalent to the extremum conditions for the so-called Yang-Yang action

$$S (\{ \lambda \}) = \frac{L}{2} \sum_j \lambda_j^2 + \sum_{j < l} \Phi (\lambda_j - \lambda_l) - 2\pi I_j \lambda_j, \tag{l.yya}\label{l.yya}$$

where

\begin{equation*} \Phi (\lambda) = \int_0^\lambda d\lambda' \phi (\lambda') = 2\lambda ~\mbox{atan}~ \frac{\lambda}{c} - c \ln (1 + \lambda^2/c^2). \end{equation*}

That is, equating

\begin{equation*} \partial_{\lambda_j} S (\{ \lambda \}) = \lambda_j L + \sum_l \phi (\lambda_j - \lambda_l) - 2\pi I_j \end{equation*}

to zero for all directions yields the set of Bethe equations.

Consider now the Hessian of the Yang-Yang action, namely the matrix of second-order derivatives

\begin{align*} S_{jl} \equiv \frac{\partial^2 S}{\partial_{\lambda_j} \partial_{\lambda_l}} &= \partial_{\lambda_j} \left\{\lambda_l L + \sum_{m \neq l} 2~\mbox{atan}~ \frac{\lambda_l - \lambda_m}{c} - 2\pi I_l \right\} \nonumber \\ &= \delta_{jl} \left\{ L + \sum_{m=1}^N \frac{2c}{(\lambda_j - \lambda_m)^2 + c^2} \right\} - \frac{2c}{(\lambda_j - \lambda_l)^2 + c^2} \end{align*}

For any real vector $${\bf v}$$ with nonzero norm, we can build the quadratic form

\begin{align*} \sum_{j,l} v_j S_{jl} v_l &= \sum_{j=1}^N v_j^2 L + \sum_{j,m=1}^N v_j^2 \frac{2c}{(\lambda_j - \lambda_m)^2 + c^2} - \sum_{j,l=1}^N v_j v_l \frac{2c}{(\lambda_j - \lambda_l)^2 + c^2} \nonumber \\ &= \left.\sum_{j=1}^N v_j^2 L + \sum_{j>l=1}^N (v_j - v_l)^2 \frac{2c}{(\lambda_j - \lambda_l)^2 + c^2}\right|_{c > 0} > 0. ~~~~~ \end{align*}

Thus, $$S$$ is a strictly concave function in $$N$$-dimensional space, and the solution to the extremum condition (the Bethe equations) for a given proper set of quantum numbers is unique $$\Box$$.

Theorem 3: for $$c > 0$$, if $$I_j > I_l$$ then $$\lambda_j > \lambda_l$$. If $$I_j = I_l$$ then $$\lambda_j = \lambda_l$$.

Proof: subtracting the Bethe equation for $$\lambda_l$$ from that for $$\lambda_j$$ gives

$$\lambda_j - \lambda_l + \frac{1}{L} \sum_{m=1}^N \left[ \phi (\lambda_j - \lambda_m) - \phi(\lambda_l - \lambda_m)\right] = \frac{2\pi}{L} (I_j - I_l). \tag{l.ld}\label{l.ld}$$

Since the $$\phi$$ kernel is monotonic, the first and second terms on the left-hand side of this equation have the same sign, directly proving the theorem.

Theorem 4: for $$c > 0$$, the rapidity differences are bounded by

$$\frac{2\pi}{L} \frac{|I_j - I_l|}{1 + \frac{2n}{c}} \leq |\lambda_j - \lambda_l| \leq \frac{2\pi}{L} |I_j - I_l|. \tag{l.bd}\label{l.bd}$$

Proof: defining the Cauchy kernel

$$\frac{1}{2\pi} \frac{d \phi (\lambda)}{d\lambda} = \frac{1}{\pi} \frac{c}{c^2 + \lambda^2} \equiv {\cal C} (\lambda), \tag{l.ck}\label{l.ck}$$

we note that

\begin{equation*} 0 \leq {\cal C} (\lambda) \leq \frac{1}{\pi c}, \hspace{10mm} \lambda \in \mathbb{R} \end{equation*}

and thus that

\begin{equation*} \phi (\lambda_1) - \phi (\lambda_2) = 2\pi \int_{\lambda_2}^{\lambda_1} d\lambda ~{\cal C} (\lambda) \leq \frac{2}{c} (\lambda_1 - \lambda_2). \end{equation*}

Substituting this in equation l.ld then immediately completes the proof.