In literature one sometimes find that spins are replaced by hardcore bosons. Formally one replaces spin operators $\sigma^- \leftrightarrow a$, $\sigma^+ \leftrightarrow a^\dagger$, $\sigma_z \leftrightarrow a^\dagger a - 1/2$ and appends a term $U a^\dagger a^\dagger aa$ to the spin Hamiltonian. The limit $U \to \infty$ in the end splits the energy of states with more than 2 bosons away from the $\{|0\rangle,|1\rangle\}$ manifold. Identifying $|0\rangle \leftrightarrow |\downarrow\rangle$, $|1\rangle \leftrightarrow |\uparrow\rangle$ then gives a mapping between the bosonic and spin system.

My question is: how rigorous is this mapping? Can there be systems (i.e. Hamiltonians) or states for which the spin treatment yields different results than the bosonic treatment?

A possible example I had in mind: Consider some Hamiltonian $H(\sigma^-, \sigma^+, \sigma_z)$ and do the mapping to $H_B = H(a, a^\dagger, a^\dagger a - 1/2) + U a^\dagger a^\dagger aa$. Maybe time evolution $H_B$ can be easily solved using coherent states, thus I decompose some initial state into the coherent state basis, do the time evolution there, project the results onto the $\{|0\rangle, |1\rangle\}$-manifold and then take the limit $U\to \infty$. However, the coherent states always have a non-zero overlap with the "bad manifold" $\{|n\rangle | n\ge 2\}$. Thus this procedure works only, if no probability flows from the $\{|n\rangle | n\ge 2\}$-manifold to the $\{|0\rangle, |1\rangle\}$-manifold.

No, it is easy to show, that all states in $\{|n\rangle | n\ge 2\}$ are approximative eigenstates of $H_B$ with eigenenergy $U$ and the full eigenstates with eigenenergy $\mathcal{O}(U)$ have an overlap $\mathcal{O}(1/U)$ with the states from $\{|0\rangle, |1\rangle\}$. Thus, in the limit $U\to\infty$ the states $\{|0\rangle, |1\rangle\}$ and $\{|n\rangle | n\ge 2\}$ belong to disjoint subspaces of $H_B$'s eigenspace. Formally, this would suffice for any time evolution for some time $t$.

However, for asymptotic states (i.e., $t\to\infty$) im not so sure. Comming from a pertubation theory perspective I can argue that for each infinitesimal time step states from the $\{|0\rangle, |1\rangle\}$-manifold can be mapped to the $\{|n\rangle | n\ge 2\}$-manifold with some rate $\mathcal{O}(1/U)$ and vice versa. Doing infinitely many of these infinitesimal time steps can allow that these transition add up to something $\mathcal{O}(1)$. This is, we don't necessarily know whether $\lim_{t\to\infty}\lim_{U\to\infty} e^{-iHt} = \lim_{U\to\infty}\lim_{t\to\infty} e^{-iHt}$.

  • $\begingroup$ What is the difference other than using a different symbol for your creation operators? What matters are the CCR of spins and hardcore bosons, and these are---by definition---the same. Maybe your question is instead: starting from the CCR for (softcore) bosons, how does one derive the effective CCR for hardcore bosons in the limit of on-site repulsion? $\endgroup$ Apr 25, 2019 at 10:25
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    $\begingroup$ I'm not sure, but I think your reasoning makes sense. Superexchange processes would go as ~$J^2/U$ (where $J$ is the boson hopping), so if you took limits so that $U,t \rightarrow \infty$ but $J^2 t /\hbar U \neq 0$ these processes might still occur. But maybe all this says is that you have to take your limits carefully. $\endgroup$
    – Rococo
    Apr 26, 2019 at 0:10
  • $\begingroup$ @RubenVerresen I guess this is probably one way to phrase it $\endgroup$
    – manthano
    Apr 26, 2019 at 7:01
  • $\begingroup$ @RubenVerresen 🍻 I interpreted it kind of like what we were talking about with softening constraints... are there any phase separations at infinite $U$ that don't persist to any finite $U$? $\endgroup$ Apr 28, 2019 at 16:40
  • $\begingroup$ @RyanThorngren That is not something I have looked into yet, so I must admit that I don't know. However, my model at hand is essentially non-interacting driven spins, similar to Jaynes-Cummings, so I assume there is no phase separation present $\endgroup$
    – manthano
    Apr 29, 2019 at 7:48


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