# How do boundary conditions change during a spin transformation?

I am currently reading the following review paper:

(1) Two Dimensional Model as a Soluble Problem for Many Fermions by Schultz et. al.

Equation (3.2), which is reproduced below, introduces the Jordan-Wigner transformations. $$n$$ corresponds to the site number. Suppose there are $$N$$ sites.

$$c_n = \prod_{j = 1}^{n - 1} \exp(i\pi \sigma^+_j \sigma^-_j) \sigma^-_n \hspace{0.95in} c^\dagger_n = \sigma^+_n \prod_{j = 1}^{n - 1}\exp(-i\pi \sigma^+_j \sigma^-_j)$$

Equations (3.12a) and (3.12b) provide some boundary conditions for these new operators $$c$$ and $$c^\dagger$$. These are produced below.

$$c_{N+1} = c_1 (\text{periodic}) \hspace{0.95in} c_{N+1} = -c_1 (\text{anti-periodic})$$

I am trying to explore what these boundary conditions on $$c$$ imply about the boundary conditions for spins. To explore this, I used the following

\begin{align} \sigma^+_n \sigma^-_n &= \frac{1}{4}\big[ \sigma_n^x + i \sigma_n^y \big]\big[ \sigma_n^x - i \sigma_n^y \big] = \frac{1}{2}[1 + \sigma^z_n] \\ \exp(i\pi\sigma^z_n/2) &= i\sigma^z_n \end{align}

to conclude

\begin{align} c_n &= \prod_{j = 1}^{n - 1} \big[- \sigma_j^z\big]\sigma^-_n\\ c^\dagger_n &= \sigma^+_n \prod_{j = 1}^{n - 1}\big[ - \sigma^z_j\big] \end{align}

Thus we see that $$c_{N+1} = (-1)^N\prod_{j = 1}^{N} \big[\sigma_j^z\big]\sigma^-_{N+1}$$. Now it seems that $$c_1 = \sigma_1^-$$ First Question: Is my expression for $$c_1$$ correct?

Now suppose $$N$$ is even. Hence $$c_{N+1} = \prod_{j = 1}^{N} \big[\sigma_j^z\big]\sigma^-_{N+1}$$ Second Question: Can I say that $$\prod_{j = 1}^{N} \big[\sigma_j^z\big] = 1$$? If so, why? If not, why not?

In general: What Does Periodic Boundary Conditions for Fermions Say about Boundary Conditions for Spins?

First, indeed $$c_1=\sigma_1^-$$ is correct.

Now to the general question. Note that $$\sigma_n^z=(-1)^{c_n^\dagger c_n}=1-2c_n^\dagger c_n$$. By inverting the transformation, we can write

$$\sigma_n^-=\prod_{j=1}^{n-1}(-1)^{c_n^\dagger c_n} c_n.$$

In particular, take $$n=N+1$$, it becomes $$\sigma_{N+1}^-=\prod_{j=1}^N (-1)^{c_j^\dagger c_j} c_{N+1}$$.

Now observe that $$(-1)^{\sum_{j=1}^N c_j^\dagger c_j}=(-1)^{N_f}$$, where $$N_f=\sum_{j=1}^N c_j^\dagger c_j$$ is a conserved quantity, called the fermion parity, for any fermionic Hamiltonian with local interactions, so we can treat it as a c-number. In the spin representation, we have $$(-1)^{N_f}=\prod_{j=1}^N \sigma_j^z$$ Again this must be a global symmetry of the spin model, otherwise the Jordan-Wigner transformation does not work.

Suppose we impose the following boundary condition on the fermion: $$c_{N+1}=s c_1$$ where $$s=\pm 1$$ for periodic/anti-periodic BCs. Then the boundary condition on the $$\sigma_n^\pm$$ operator becomes

$$\sigma_{N+1}^\pm = (-1)^{N_f} s \sigma_1^\pm$$

This relation defines the map between boundary conditions of the fermion and spin operators. If you impose PBC on $$c$$'s, the spins can still have P/AP BC depending on the total fermion parity/$$Z_2$$ charge.