The Bethe Ansatz

We can also consider the limit \(c \rightarrow 0^+\). Here,

\begin{equation*} \lim_{c \rightarrow 0+} {\cal C} (\lambda) = \delta (\lambda) \end{equation*}

so

\begin{equation*} \epsilon(\lambda) = \lambda^2 - \mu - T \ln [1 + e^{-\epsilon(\lambda)/T}], \hspace{10mm} \\ \epsilon(\lambda) = T \ln [e^{(\lambda^2 - \mu)/T} - 1] \end{equation*}

and therefore

\begin{equation*} \rho(\lambda) = \frac{1}{2\pi} \frac{1}{e^{(\lambda^2 - \mu)/T} - 1}, \hspace{10mm} \rho_h(\lambda) = \frac{1}{2\pi} \end{equation*}

and the free energy

\begin{equation*} g(T, \mu) = T \int_{-\infty}^{\infty} \frac{d\lambda}{2\pi} \ln \left[ 1 - e^{-(\lambda^2 - \mu)/T}\right] \end{equation*}

coincides (as expected) with that of ideal free bosons.