In 1+1D Ising model with a transverse field defined by the Hamiltonian
There is a duality transformation which defines new Pauli operators $\mu^x_i$ and $\mu^z_i$ in a dual lattice
then these $\mu_i^x$ and $\mu_i^z$ satisfy the same commutation and anti-commutation relations of $\sigma^x_i$ and $\sigma^z_i$, and the original Hamiltonian can be written in terms of $\mu_i^x$ and $\mu_i^z$ as
At this stage, many textbooks will tell us since $\sigma$'s and $\mu$'s have the same algebra relations, the right hand side of the last equation is nothing but $H(h,J)$. My confusions are
1) Does that the operators having the same algebra really imply that $H(J,h)$ and $H(h,J)$ have the same spectrum? We know for a given algebra we can have different representations and these different representations may give different results. For example, the angular momentum algebra is always the same, but we can have different eigenvalues of spin operators.
2) This is related to the first confusion. Instead of looking at the algebra of the new operators, we can also look at how the states transform under this duality transformation. In the eigenbasis of $\mu_i^x$, if I really consider it as a simple Pauli matrix, the state $|\rightarrow\rangle$ corresponds to two states in the original picture, i.e. $|\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\rangle$. The same for state $|\leftarrow\rangle$. In the $\mu_i^z$ basis, the correspondence is more complicated. A state corresponds to many states in the original picture, and the number of the corresponding states depend on the position of this state. Therefore, this duality transformation is not unitary, which makes me doubt whether $H(J,h)$ and $H(h,J)$ should have the same spectrum. Further, what other implication may this observation lead to? For example, doing one duality transformation is a many-to-one correspondence, then doing it back should still be a many-to-one correspondence, then can we recover the original spectrum?
3) Another observation is we in the above $\mu_i^z$ involves a string of operators on the left side, we can equally define it in terms of a string of operators on the right side, so it seems there is an unobservable string. What implication can this observation lead to? Is this unobservable string related to the unobservable strings in Levin-Wen model?