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The Jordan-Wigner transformation allows one to map a spin theory to a fermionic theory and, according to wikipedia, it is an example of an S-duality. In turn, according to the wiki page for the S-duality; “S-duality is useful because it relates a theory with coupling constant $g$ to an equivalent theory with coupling constant $1/g$”.

Take now for example the 1D XY Heisenberg model $$ H = J \sum_i (1+g)S_i^x S^x_{i+1} + (1-g) S_i^yS^y_{i+1} $$ and perform a Jordan-Wigner transform: $$ H= \frac{J}{2} \sum_i \left( f^{\dagger}_{i+1} f_i+ f_i^{\dagger} f_{i+1}+ g(f_{i+1}f_i +f^{\dagger}_i f^{\dagger}_{i+1}) \right) $$ where can I see this $g \rightarrow 1/g$ behaviour Of the S-duality?

Thanks!

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    $\begingroup$ In a less strict sense, the S-duality is a duality between a strongly coupled theory and a weakly coupled one. That is certainly achieved when the Jordan-Wigner transform is applied to certain models, such as the XY and Ising models (which become noninteracting fermionic models). However, the Wikipedia text seems to imply it's the transformation itself that is an example of an S-duality. I'm not sure how to interpret that. $\endgroup$
    – Anyon
    Jan 27, 2021 at 15:25
  • $\begingroup$ @Anyon47 Indeed one can turn the second Hamiltonian in my answer to a non interacting fermionic Hamiltonian via a Bogoliubov transformation. This implies "$g\rightarrow 0$" in some sense. Do you mean this is an S-duality because it maps a (strongly) interacting system to a non-interacting system and that's it or does an an S-duality need to have exactly the $g\rightarrow 1/g$ behaviour it says on Wikipedia? $\endgroup$
    – FriendlyLagrangian
    Jan 27, 2021 at 16:11
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    $\begingroup$ Yes, in some sense the second Hamiltonian in your example has $g\rightarrow 0$, where $g$ is a coupling strength in a perturbative series (i.e. it could be different from the anisotropy parameter). Hopefully someone with more expertise in this area will weigh in on whether that constitutes an S-duality or not. I will say that the BBS book (Becker, Becker & Schwarz) defines S-duality as requiring $g\rightarrow 1/g$, but based on what I've read Polchinski seems to allow for a more general duality mapping. $\endgroup$
    – Anyon
    Jan 27, 2021 at 17:21

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