# The Bethe Ansatz

#### Excitations at zero field: spinonsg.h.e

Besides the zero-field ground state, it is possible to analytically construct eigenstates with finite numbers of excitations. We consider here the simplest type of excited states, which are obtained by upturning one of the down spins and considering purely real, finite rapidity solutions to the Bethe equations.

Let us begin again with the finite lattice of an even number $$N$$ of sites. We specialize to the subsector with $$M = N/2 - 1$$ downturned spins. According to h.im, the maximal allowable quantum number for real finite rapidities is then $$I^{\mbox{max}}_{N/2-1} = N/4$$. Since we must put $$M = N/2 - 1$$ particles in $$2I^{\mbox{max}}_{N/2-1} + 1 = N/2 + 1$$ available slots, we can construct

\begin{equation*} N_{2sp} = \frac{N(N+2)}{8} \end{equation*}

different states, which will be called 2-spinon states 1981.Faddeev.PLA.85. The set of quantum numbers of these states are completely characterized by the position of the two holes in the quantum number distribution, which we write $$I^h_1, I^h_2$$. The presence of these holes means that we can write the $$x$$-space hole distribution as $$\rho_h(x) = \frac{1}{N} \delta(x - \frac{I^h_1}{N}) + \frac{1}{N} \delta(x - \frac{I^h_2}{N})$$. By defining the hole rapidities $$\lambda^h_i = \lambda(x = I^h_i/N)$$, this distribution becomes $\rho_h (\lambda) = \frac{1}{N} \delta(\lambda - \lambda^h_1) + \frac{1}{N} \delta(\lambda - \lambda^h_2)$ in rapidity space. Equation l.bec for a 2-spinon state labeled by hole rapidities $$\lambda^h_i$$ thus becomes $a_1(\lambda) = \rho(\lambda) + a_2 * \rho(\lambda) + \frac{1}{N} \delta(\lambda - \lambda^h_1) + \frac{1}{N} \delta(\lambda - \lambda^h_2)$ where the convolution notation is $$a * b (\lambda) = \int_{-\infty}^{\infty} d\lambda' a(\lambda - \lambda') b(\lambda')$$. Fourier transforming gives $\rho_{\rm 2sp}(\omega; \lambda^h_1, \lambda^h_2) = \rho_g (\omega) - \frac{\rho_{\rm sp} (\omega)}{N} (e^{i\omega\lambda^h_1} + e^{i\omega\lambda^h_2}),$ where $$\rho_g (\omega)$$ is the ground state rapidity distribution and $$\rho_{\rm sp} (\omega) = \frac{1}{1 + a_2 (\omega)} = \frac{1}{1 + e^{-|\omega|}}$$ is the spinon density function. The inverse Fourier transform finally yields

\begin{equation*} \rho_{\rm 2sp} (\lambda; \lambda^h_1, \lambda^h_2) = \rho_g (\lambda) - \frac{1}{N} \rho_{\rm sp} (\lambda - \lambda^h_1) - \frac{1}{N} \rho_{\rm sp} (\lambda - \lambda^h_2) \end{equation*}

where the spinon density distribution is given in terms of the Dirac delta distribution corrected by reals part of digamma functions:

$$\rho_{\rm sp} (\lambda) = \delta(\lambda) - \frac{1}{2\pi} \Re (\psi_0 (1 + i\lambda/2) - \psi_0 (1/2 + i\lambda/2)). \tag{h.r2sp}\label{h.r2sp}$$

Knowing the rapidity distribution, we can now compute the energy and momentum of a 2-spinon state over that of the ground state:

\begin{equation*} E_{\rm 2sp} (\lambda^h_1, \lambda^h_2) = -NJ\pi \int_{-\infty}^{\infty} d\lambda a_1 (\lambda) [\rho_{\rm 2sp} (\lambda; \lambda^h_1, \lambda^h_2) - \rho_g (\lambda)] \end{equation*}

meaning that

$$E_{\rm 2sp} (\lambda^h_1, \lambda^h_2) = \varepsilon_{\rm sp} (\lambda^h_1) + \varepsilon_{\rm sp}(\lambda^h_2), \tag{h.2spe}\label{h.2spe}$$

where the single spinon energy is

\begin{equation*} \varepsilon_{\rm sp} (\lambda^h) = J \pi \int_{-\infty}^{\infty} d\lambda a_1 (\lambda) \rho_{\rm sp} (\lambda - \lambda^h) = J\pi \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} a_1 (\omega) \rho_{\rm sp}(\omega) e^{i\omega \lambda^h} = \frac{J\pi}{2\cosh \pi \lambda^h}. \end{equation*}

The momentum of the 2-spinon state over that of the ground state is similarly expressed as

$$P_{\rm 2sp} (\lambda^h_1, \lambda^h_2) = p_{\rm sp} (\lambda^h_1) + p_{\rm sp} (\lambda^h_2), \tag{h.2spp}\label{h.2spp}$$

where $p_{\rm sp} (\lambda^h) = -\int_{-\infty}^{\infty} d\lambda \rho_{\rm sp} (\lambda - \lambda^h) [\pi - \phi_1 (\lambda)].$ We have that $$\lim_{\lambda^h \rightarrow \infty} p_{\rm sp} (\lambda^h) = 0$$, and $\frac{dp_{\rm sp}}{d\lambda^h} = 2\pi \int_{-\infty}^{\infty} d\lambda \rho_{\rm sp} (\lambda) a_1 (\lambda) = 2\varepsilon_{\rm sp} (\lambda^h)/J,$ which allows to write $p_{\rm sp} (\lambda^h)= -\int_{\lambda^h}^{\infty} d\lambda \frac{\pi}{\cosh \pi \lambda} = - 2~\mbox{atan} (e^{-\pi \lambda^h}) ~~\in [-\pi, 0].$ In turn, back-substitution into the single spinon energy yields the well-known spinon dispersion relation

$$\varepsilon_{\rm sp} (p) = \frac{J\pi}{2} | \sin p |, \hspace{1cm} p \in [-\pi, 0]. \tag{h.esp}\label{h.esp}$$