The Bethe Ansatz

Let us first consider the one-body function

\begin{equation*} S^{\psi^\dagger \psi} (-k, -\omega) = \frac{1}{L} \int_0^L dx dx' e^{ik (x - x')} \int_{-\infty}^\infty dt e^{-i\omega t} \langle \psi^\dagger(x,t) \psi(x',0) \rangle = \frac{1}{L} \int_{-\infty}^\infty dt e^{-i \omega t} \langle \psi^\dagger_{k} (t) \psi_{k} (0) \rangle \end{equation*}

Its integrated intensity obeys

\[ \frac{1}{L} \sum_k \int_{-\infty}^\infty \frac{d\omega}{2\pi} S^{\psi^\dagger \psi} (k, \omega) = \langle \psi^\dagger (0, 0) \psi(0, 0) \rangle = \rho \]

which, when written in terms of matrix elements and considering averaging over a single state \(\alpha\), gives the sum rule

\[ \sum_{\alpha'} | \langle \alpha' | \psi (0) | \alpha \rangle |^2 = \rho \]

Turning to the spectral function for \(\psi\),

\begin{equation*} A^{\psi \psi^\dagger} (k, \omega) = \frac{1}{L} \int_0^L dx dx' e^{-ik (x - x')} \int_{-\infty}^\infty dt e^{i\omega t} \langle \left[ \psi(x,t), \psi^\dagger(x',0) \right] \rangle \end{equation*}

we can apply the f-sumrule fsr, here taking the form

\[ \int_{-\infty}^\infty \frac{d\omega}{2\pi} ~\omega ~A^{\psi \psi^\dagger} (k, \omega) = \frac{-1}{L} \langle \left[ \left[ H, \psi_k \right], \psi^\dagger_k \right] \rangle. \]

Using HLLk to compute the concatenated commutator gives

\begin{align*} \left[ \left[ H, \psi_k \right], \psi^\dagger_k \right] &= -(k^2 - \mu) \left[ \psi_k, \psi^\dagger_k \right] - \frac{2c}{L^2} \sum_{k_1 q} \left[ \psi^\dagger_{k_1 + q} \psi_{k_1} \psi_{k + q}, \psi^\dagger_k \right] \\ &= -L (k^2 - \mu) - \frac{4c}{L} \sum_q \psi^\dagger_q \psi_q = -L \left( k^2 - \mu + 4c \rho \right). \end{align*}

In terms of matrix elements, using specfun.Lr in the above thus gives (again for averaging over a single state \(\alpha\))

\[ \frac{1}{L} \sum_{\alpha'} \omega_{\alpha' \alpha} \left( | \langle \alpha | \psi_k | \alpha' \rangle |^2 + | \langle \alpha' | \psi_{k} | \alpha \rangle |^2 \right) = k^2 -\mu + 4c\rho \]

Note that one can drop the chemical potential from this equation, by omitting it on both sides (left: in the definition of the energies \(E_{\alpha}\) and \(E_{\alpha'}\)), in view of the commutation relation \(\psi_k \psi^\dagger_k - \psi^\dagger_{k} \psi_{k} = L\).