# About the rigour of replacing spins by hardcore Bosons

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 $$\lim_{t\to\infty}\lim_{U\to\infty} e^{-iHt} = \lim_{U\to\infty}\lim_{t\to\infty} e^{-iHt}$$.

• 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? Apr 25, 2019 at 10:25
• 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. Apr 26, 2019 at 0:10
• @RubenVerresen I guess this is probably one way to phrase it Apr 26, 2019 at 7:01
• @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 $U$? Apr 28, 2019 at 16:40
• @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 Apr 29, 2019 at 7:48