While exploring the Hamiltonian of the x-y model within the framework of quantum mechanics, I encountered the following transformation using the Jordan-Wigner transformation (JWT):

\begin{align} H_{xy} &= -J \sum_{j} S_j ^+ S_{j+1} ^- + S_j ^+ S_{j-1} ^- \xrightarrow{JWT}. \\ &-J \sum_{j} \exp(i\phi_j) d^{\dagger}_j \exp(-i\phi_{j+1}) d_{j+1} + \exp(i\phi_j) d^{\dagger}_j \exp(-i\phi_{j-1}) d_{j-1}. \end{align}

Here's my query: Given that the phases $\phi_j$ commute with the spin operators $S_j^-$ and $S_j^+$, can we express the transformed Hamiltonian in a simplified form as:

\begin{align} H_{xy} &= -J \sum_{j} d^{\dagger}_j \exp(i\phi_j) \exp(-i\phi_{j+1}) d_{j+1} + d^{\dagger}_j \exp(i\phi_j) \exp(-i\phi_{j-1}) d_{j-1} \xrightarrow{?}\\ &-J \sum_{j} d^{\dagger}_j d_{j+1} + d^{\dagger}_j d_{j-1}. \end{align}

My reasoning for this potential simplification stems from the commuting nature of $\phi_j$ with the spin operators. However, I'd like to verify if the product $\exp(i\phi_j) \exp(-i\phi_{j+1})$ necessarily evaluates to unity under these circumstances.

  • $\begingroup$ You would need $\phi_j=\phi_{j+1}$ (different subscripts!) which isn't a given. $\endgroup$
    – DaftWullie
    Commented May 20 at 4:51


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