How do I derive the eigenvalues of the 1D Heisenberg model? (Bethe Ansatz) I've been trying to work through Introduction to the Bethe Ansatz I (by Michael Karbach and Gerhard Muller) in  spare time and I am having trouble deriving the eigenvalues given in equation (5) for the r=1 (1 spin down) block of the Hamiltonian.
The Hamiltonian is :
$$H=-J\sum_{n=1}^N \left[\frac{1}{2}\left(S_n^+S_{n+1}^-+S_n^-S_{n+1}^+\right)+S_n^zS_{n+1}^z\right]$$
The eigenvalues I should arrive at are:
$$E-E_o=J(1-\cos(k)) \phantom{XXXX}  (5)$$
With eigenvectors: 
$$| \psi\rangle=\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle \phantom{XXXXX} (4)$$
Where $E_o$ is the energy of the state in which all spin are aligned and E is the energy of a state with one flipped spin at site n. (The lattice spacing here is 1)
All the necessary information to solve the problem appears to be given in the preceding paragraphs of the paper I linked but I have been unable to put it all together. I'm working on the second part of question 1 in the document and I have to "show that the states (4) are eigenvectors 
of H with eigenvalues (5).
 A: Ok I think I got it:
$$\langle \psi' | H| \psi\rangle =\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^+S_{i+1}^-)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle+\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^-S_{i+1}^+)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle +\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-J\sum_{i=1}^NS_i^zS_{i+1}^z)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle $$
Looking at each line of $\langle \psi' | H| \psi\rangle$ seperately:
$$\sum_{n'=1}^N\frac{-J}{2N}\langle n'|e^{-ikn' }(\sum_{i=1}^NS_i^+S_{i+1}^-)\sum_{n=1}^Ne^{ikn}|n\rangle=\frac{-J}{2N}\sum_{n'=1}^{N}\langle n'| (\sum_{i=1}^NS_i^+S_{i+1}^-)(e^{ik(1-n')}| 1\rangle + e^{ik(2-n')}| 2\rangle +...+e^{ik(N-n')}| N\rangle )$$
$$=\frac{-J}{2N}\sum_{n'=1}^N\langle n' |((e^{ik(1-n')}| 2\rangle+e^{ik(2-n')}| 3\rangle +...+e^{ik(N-n')}| 1\rangle) $$
$$=\frac{-J}{2}e^{-ik} $$
Following the same procedure for the second part I obtain:
$$\sum_{n'=1}^N\frac{-J}{2N}\langle n'|e^{-ikn' }(\sum_{i=1}^NS_i^-S_{i+1}^+)\sum_{n=1}^Ne^{ikn}|n\rangle =\frac{-J}{2N}\sum_{n'=1}^N \langle n' | (e^{ik(1-n')}| N \rangle+e^{ik(2-n')}| 1\rangle +...+e^{ik(N-n')}|N-1\rangle ) $$
$$=-\frac{J}{2}e^{ik}$$
Lastly:
$$\sum_{n'=1}^N\frac{1}{N}\langle n'|e^{-ikn' }(-J\sum_{i=1}^NS_i^zS_{i+1}^z)\sum_{n=1}^Ne^{ikn}|n\rangle =\frac{1}{N}\sum_{n'=1}^N\langle n' |(-J(\frac{-1}{4}+\frac{(N-2)}{4}-\frac{1}{4})e^{ik(1-n')}|1\rangle-J(-1+\frac{N}{4})e^{ik(2-n')}|2\rangle-J(-1+\frac{N}{4})e^{ik(N-n')}|N\rangle)$$
$$=J+E_o$$
Putting this all together I obtain:
$$E=-\frac{J}{2}e^{-ik}-\frac{J}{2}e^{ik}+J+E_o$$
$$E-E_o=J(1-cos(k))$$
