# Boundary value condition used during Jordan-Wigner transformation for a $1 D$ Ising chain

In the section 4.1 of Quantum Computation by Adiabatic Evolution, Farhi et al proposes a quantum adiabatic algorithm to solve the $2$-SAT problem on a ring. To compute the complexity of the algorithm the authors computed the energy gap between the ground and first excited states of the adiabatic Hamiltonian.

The adiabatic Hamiltonian is defined as $$\tilde{H} (s) = (1-s) \sum^n_{j=1}(1-\sigma^{(j)}_x) + s \sum^n_{j=1}\frac{1}{2} (1-\sigma^{(j)}_z \sigma^{(j+1)}_z )$$

To prove the correctness of the algorithm, the authors consider an operator which negates the value of the bits in the $z$ axis. $$G = \prod^n_{j=1}\sigma^{(j)}_x$$

Then the authors start the steps of Jordan-Wigner transformation. The fermionic operators are defined as follows.

$$b_j = \sigma_x^1 \sigma_x^2 \ldots \sigma_x^{j-1} \sigma_-^{j} \mathbf{ 1}^{j+1} \ldots \mathbf{ 1}^n \\ b^\dagger_j = \sigma_x^1 \sigma_x^2 \ldots \sigma_x^{j-1} \sigma_+^{j} \mathbf{ 1}^{j+1} \ldots \mathbf{ 1}^n$$ where $$\sigma_{\pm} = \sigma_x \pm i \sigma_y$$.

Then the authors keep mentioning few more identities. Then at the beginning of page 14, they say

Since we will restrict ourselves to the $G = 1$ sector, (4.10) and (4.11) are only consistent if $b_{n+1} = −b_1$ , so we take this as the definition of $b_{n+1}$.

My questions: I understand that $G$ is an operator which commutes with the adiabatic Hamiltonian and it negates the values of the qubits in the $z$ axis. But, I don't understand what the definition of the 'sector' of $G$ is, when the sector is $G = 1$, what other sectors are possible, and finally why $b_{n+1} = b_1$ cannot be true.

My attempt: In the computational basis i.e. along $z$ axis, $\sigma^x |1\rangle = 1\cdot|0\rangle$ and $\sigma^x |0\rangle = 1\cdot|1\rangle$. So, the eigenvalue is always $1$. Is this the sector, the authors are talking about?

• Note that $G^2=1$ so there are two eigenvalues that are +-1. To see the -1 sector, consider a singlet state. – Mikael Fremling Jun 9 '16 at 6:57

To see that there are two sector, corresponding to the eigenvalues of $G$ note that $G^2=1$ since

$$G^2 = (\prod^n_{j=1}\sigma^{(j)}_x)^2 = \prod^n_{j=1}(\sigma^{(j)}_x)^2 = 1$$

Thus there are two eigenvalues to $G$ that are $\pm1$. These sectors need not be of the same size. Consider just 2 spins in theire singlet and triplet configurations.

The singlet is $$|s\rangle = |\uparrow,\downarrow\rangle - |\downarrow,\uparrow\rangle$$ while the triplets are $$|t,\pm\rangle = |\downarrow\downarrow\rangle \pm |\uparrow,\uparrow\rangle$$ $$|t,0\rangle = |\uparrow,\downarrow\rangle + |\downarrow,\uparrow\rangle$$

The effect of $G$ on these are $$G|s\rangle = |\downarrow,\uparrow\rangle-|\uparrow,\downarrow\rangle = -|s\rangle$$ and

$$G|t,\pm\rangle = |\uparrow,\uparrow\rangle\pm|\downarrow\downarrow\rangle = \mp |t,\pm\rangle$$ $$G|t,0\rangle = |\downarrow,\uparrow\rangle+|\uparrow,\downarrow\rangle = +|t,0\rangle$$

Which sows that it's easy to construct both positive and negative eigenvectors of $G$.