Tags: Dynamical Spin Structure Factor

Example: One up-spin

We compute the dynamical spin structure factor. See this post for an intro and the derivation of the formulas in the general \(N\) up-spin case. The dynamical spin structure factor is given by the following formula

\[S^{zz}(q, \omega) = \frac{1}{L} \sum_{j, j' =1}^L e^{- i q (j-j')} \int_{- \infty}^{\infty}dt e^{i \omega t} \langle S_j^z(t) S_{j'}^z(0) \rangle\]

We focus on computing the observable inside the integrand. In this post, we gave a formula for the observable:

\[\begin{align} & \langle S_j^z(t) S_{j'}^z \rangle = \langle \Psi \mid e^{i t H /\hbar } S_j^z e^{-i t H /\hbar } S_{j'}^z \mid \Psi \rangle\\ &= \frac{(-1)^{j' \notin y}}{2} \sum_{x \in X} \frac{(-1)^{j \notin x}}{2} \sum_{[\zeta], [\xi] \in \Xi} \overline{\ell(y, \zeta) u(\zeta, x) } \ell(y, \xi) u(\xi, x) e^{-i t (E(\xi)- E(\zeta) )}. \end{align}\]

for the general \(N\) up-spin case. We treat the \(N=1\) case here, where many of the formulas simplify.

Functions

The Bethe Equations become

\[\xi^L = 1.\]

Thus, the solutions of the Bethe equations are the \(L^{th}\) roots of unity. We set \(\eta = \exp(2 \pi i /L)\) be a primitive \(L^{th}\) roots of unity so that

\[\eta^{k} = e^{2 \pi i k /L},\]

for \(k=1, 2, \dots, L\), are all the solutions of the Bethe equations. Additionally, we have

\[\begin{align} u(\xi, x) = \xi^x, \quad \ell(y, \xi) = \xi^{-y}/L, \quad E(\xi) = \xi + \xi^{-1} - 2 \Delta \end{align}\]

Computation

We have

\[\begin{align} & \langle S_j^z(t) S_{j'}^z \rangle = \langle \Psi \mid e^{i t H /\hbar } S_j^z e^{-i t H /\hbar } S_{j'}^z \mid \Psi \rangle\\ &= \frac{(-1)^{j' \notin y}}{2} \sum_{x \in X} \frac{(-1)^{j \notin x}}{2} \sum_{[\zeta], [\xi] \in \Xi} \overline{\ell(y, \zeta) u(\zeta, x) } \ell(y, \xi) u(\xi, x) e^{-i t (E(\xi)- E(\zeta) )}\\ &= \frac{(-1)^{j' \neq y}}{2 } \sum_{x = 1}^L \frac{(-1)^{j \neq x}}{2 } \sum_{k1, k2 = 1}^L L^{-2}\eta^{(k1-k2)(x-y)} e^{\tfrac{- i t}{\hbar}(\cos (2\pi k1/L) - \cos(2 \pi k2/L))}. \end{align}\]

Next, we integrate over time

\[\begin{align} &\int_{- \infty}^{\infty}dt e^{i \omega t} \langle S_j^z(t) S_{j'}^z(0) \rangle\\ &= \lim_{T \rightarrow \infty} \int_{- T}^{T}dt e^{i \omega t} \langle S_j^z(t) S_{j'}^z(0) \rangle\\ &= \lim_{T \rightarrow \infty} \int_{- T}^{T}dt \frac{(-1)^{j' \neq y}}{2 } \sum_{x = 1}^L \frac{(-1)^{j \neq x}}{2 } \sum_{k1, k2 = 1}^L L^{-2}\eta^{(k1-k2)(x-y)} e^{\tfrac{- i t}{\hbar}(\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar)}\\ &= \lim_{T \rightarrow \infty} \frac{(-1)^{j' \neq y}}{2 } \sum_{x = 1}^L \frac{(-1)^{j \neq x}}{2 } \sum_{k1, k2 = 1}^L L^{-2}\eta^{(k1-k2)(x-y)} \int_{- T}^{T}dt e^{\tfrac{- i t}{\hbar}(\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar)}\\ &= \lim_{T \rightarrow \infty} \frac{(-1)^{j' \neq y}}{2 } \sum_{x = 1}^L \frac{(-1)^{j \neq x}}{2 } \sum_{k1, k2 = 1}^L \frac{- 2 \hbar L^{-2}\eta^{(k1-k2)(x-y)}}{\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar} \mathrm{Im} \left( e^{\tfrac{- i T}{\hbar}(\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar)} \right) \\ &= \lim_{T \rightarrow \infty} \frac{(-1)^{j' \neq y}}{2 } \sum_{x = 1}^L \frac{(-1)^{j \neq x}}{2 } \sum_{k1, k2 = 1}^L \frac{ 2 \hbar L^{-2}\eta^{(k1-k2)(x-y)} }{\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar} \sin \left( \tfrac{T}{\hbar}(\cos (2\pi k1/L) - \cos(2 \pi k2/L) - \omega \hbar) \right). \end{align}\]

At this point, it is hard to tell what the \(T \rightarrow \infty\) will give.

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