# Jordan-Wigner Transformations on fermionic system

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

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

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

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?

• What boundary conditions do you have? Nov 17, 2021 at 20:39
• J > 0 and periodic boundary conditions Nov 17, 2021 at 22:45
• The periodic boundary condition should give a boundary term, like described e.g. here, shouldn't it? Aug 11, 2023 at 14:44

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.