Jordan-Wigner Transformations on fermionic system

Question

I've been trying to use Jordan-Wigner Transformations on a given fermionic Hamiltonian. The given Hamiltonian is: $$ \hat{H}= -\sum_{m=1}^{N}(J_z \hat{S}_{m}^{z} \hat{S}_{m+1}^{z} + \frac{J_{\perp}}{2}(\hat{S}_{m}^{+}\hat{S}_{m+1}^{-}+\hat{S}_{m}^{-}\hat{S}_{m+1}^{+}))$$

and the form of the Jordan-Wigner Transformations given are: $$\hat{S}_{m}^{+}=\hat{c}_{m}^{\dagger} e^{i \pi \sum_{j<m}\hat{n}_{j}} $$

$$\hat{S}_{m}^{-}= e^{-i \pi \sum_{j<m}\hat{n}_{j}} \hat{c}_{m}$$

$$\hat{S}_{m}^{z}=\hat{c}_{m}^{\dagger}\hat{c}_{m} - \frac{1}{2}$$

The answer that I manage to get/how far I get in my calculation is:

$$\hat{H}= -\sum_{m=1}^{N}(J_z (\frac{1}{4} - \frac{1}{2}\hat{c}_{m}^{\dagger}\hat{c}_{m} - \frac{1}{2}\hat{c}_{m+1}^{\dagger}\hat{c}_{m+1} +\hat{c}_{m}^{\dagger}\hat{c}_{m}\hat{c}_{m+1}^{\dagger}\hat{c}_{m+1}) + \frac{J_{\perp}}{2}(\hat{c}_{m}^{\dagger}\hat{c}_{m+1} + \hat{c}_{m}\hat{c}_{m+1}^{\dagger})).$$

However, the answer given in the textbook is:

$$\hat{H}= -\sum_{m=1}^{N}(J_z (\frac{1}{4} - \hat{c}_{m}^{\dagger}\hat{c}_{m} +\hat{c}_{m}^{\dagger}\hat{c}_{m}\hat{c}_{m+1}^{\dagger}\hat{c}_{m+1}) + \frac{J_{\perp}}{2}(\hat{c}_{m}^{\dagger}\hat{c}_{m+1} + \hat{c}_{m+1}^{\dagger}\hat{c}_{m})).$$

Is anyone able to help me make the last step from my answer to the given answer?

Answer 1

For the $J_z$ part you're basically there already. Just note that with periodic boundary conditions $c^\dagger_{N+1}=c^\dagger_1$ and you're free to change the summation index, so $$ \sum_{m=1}^{N} \hat{c}_{m+1}^{\dagger}\hat{c}_{m+1} = \sum_{m=1}^{N} \hat{c}_{m}^{\dagger}\hat{c}_{m}. $$ For the $J_\perp$ part you've made a sign error for one of the terms. A good way for you to locate the error might be to Jordan-Wigner-transform both $S_{m+1}^+S_m^-$ and $S_m^-S_{m+1}^+$. Obviously these should give the same result, but you may find one more straightforward than the other.