The independent boson model consists of the following Hamiltonian: $$ H_s = E \sigma^z $$ $$ H_b = \sum_k \omega_k b^{\dagger}_kb_k $$ $$H_{sb} = \sigma^z \sum_k (g_k b_k + g_k^{\ast}b^{\dagger}_k).$$ The model describes a single spin-1/2 impurity with Pauli operators $\sigma^{x,y,z}$ linearly coupled to an infinity of bosonic modes $b_k$. Importantly, the interaction $H_{sb}$ commutes with $H_s$.
The model is exactly solvable by introducing a state-dependent displacement:
$$ U = \exp \left[ \sigma^z \sum_k (g_k^{\ast}b^{\dagger}_k - g_k b_k)\right],$$ leading to the transformed Hamiltonian $$U H U^{\dagger} = E^{\prime} \sigma^z + \sum_k \omega_k b^{\dagger}_kb_k + \mathrm{const.}$$ where $E^{\prime}$ is the renormalised impurity energy. Similar tricks allow one to compute time evolution etc. The solutions can be found in detail in Mahan's book Many-Particle Physics.
Note that there exists an equivalence between a spin-1/2 particle and a single fermionic mode, i.e. we can rewrite the above Hamiltonian by replacing $\sigma^z \to c^{\dagger} c$, where $\{c,c^{\dagger}\} = 1$ are fermionic ladder operators. The resulting model is equivalent up to a shift of the equilibrium position of the oscillators.
However, when $c$ is instead taken to be bosonic, the solution fails. The fermionic/spin solution relies on the fact that $(c^{\dagger}c)^2 = c^{\dagger}c$, which ultimately stems from the fact that the fermionic Hilbert space has 2 states. In contrast, the Hilbert space of a bosonic mode is infinite-dimensional.
Is the independent boson model always exactly solvable so long as the Hilbert space of the impurity is finite-dimensional?
I mean precisely the following: imagine replacing $\sigma^z$ with $S^z$, the $z$ projection of a spin with total angular momentum $S > 1/2$. Is the model exactly solvable? Constructive answers which describe the form of the solution would be great, or any references to where this problem has already been solved in the literature.
Answer
I'm not completely sure why you think that the bosonic mode fails, but it seems to me that the answer is definitely yes. The system is solvable in both the finite-dimensional and the bosonic case; the problem with the bosonic case is that the solution is ugly, because the hamiltonian is ugly.
Take a hamiltonian of the form $$ H=E_0S+S\sum_k (g_k b_k+g^*b_k^\dagger)+\sum_k\omega_k b_k^\dagger b_k, $$ where $S$ is so far unspecified. Let $|s⟩$ be an eigenstate of $S$, with $S|s⟩=s|s⟩$, and look for an eigenstate of the form $|\psi⟩=|s⟩|\phi⟩$. This eigenstate ought to exist because of the structure of the hamiltonian, but for now you can simply look at the action of $H$ on states of that form: \begin{align} H|\psi⟩ &= \left[\left(E_0+\sum_k (g_k b_k+g^*b_k^\dagger)\right)s+\sum_k\omega_k b_k^\dagger b_k\right]|s⟩|\phi⟩ \\ &= \left[E_0s-\sum_k\omega_k \delta_k\delta_k^* +\sum_k\omega_k (b_k+\delta_k)^\dagger(b_k+\delta_k)\right]|s⟩|\phi⟩, \end{align} where $\delta_k=sg_k/\omega_k$.
It is now easy to find eigenstates of this form - all you need is to take $|\phi⟩$ as a product of eigenstates of each displaced number operator $(b_k+\delta_k )^\dagger (b_k+\delta_k)$. This is easy to write as \begin{align} |\phi⟩ &= \bigotimes_k D(-\delta_k)|n_k⟩ = \exp\left(\sum_k(\delta_k^*b_k-\delta_k b_k)\right)|\{n_k\}⟩ \\&= \exp\left(s\sum_k\frac{g_k^*b_k-g_k b_k}{\omega_k}\right)|\{n_k\}⟩. \end{align} This eigenstate then has energy $$ E=sE_0-s^2\sum_k |g_k|^2/\omega_k +\sum_k\omega_k n_k. $$
In some ways, you're sort of done. This is enough to provide a basis of eigenstates of $H$, and there is definitely no problem if $S$ is finite-dimensional. On the other hand, I imagine you still want a canonical-transformation formulation of this. To do that, you can write $$ |\psi⟩=\exp\left(S\sum_k\frac{g_k^*b_k-g_k b_k}{\omega_k}\right)|s⟩|\{n_k\}⟩=U|s⟩|\{n_k\}⟩, $$ via a unitary transformation $$ U=\exp\left(S\sum_k\frac{g_k^*b_k-g_k b_k}{\omega_k}\right). $$ This means that the transformed hamiltonian acts as \begin{align} U^\dagger HU|s⟩|\{n_k\}⟩ &= U^\dagger H|\psi⟩ = \left[sE_0-s^2\sum_k |g_k|^2/\omega_k +\sum_k\omega_k n_k\right]U^\dagger |\psi⟩ \\ & = \left[sE_0-s^2\sum_k |g_k|^2/\omega_k +\sum_k\omega_k n_k\right]|s⟩|\{n_k\}⟩ \\ & = \left[SE_0-S^2\sum_k |g_k|^2/\omega_k +\sum_k\omega_k b_k^\dagger b_k \right]|s⟩|\{n_k\}⟩ \end{align} on a basis, and therefore has that same action everywhere: \begin{align} U^\dagger HU = SE_0-S^2\sum_k |g_k|^2/\omega_k +\sum_k\omega_k b_k^\dagger b_k. \tag1 \end{align} This works regardless of the dimensionality of $S$, as far as I can tell. I can't completely rule out funky business with the unitarity of $U$ in the infinite-dimensional case but I honestly can't see where the problem would come in from.
That said, the bosonic case $S=a^\dagger a$ is obviously problematic, because the transformed hamiltonian in $(1)$ is unbounded from below. as soon as you have one nonzero $g_k$. However, I think this is a problem with $H$ itself rather than the transformation.
To see this, consider the case where a single $g_k$ is nonzero, so $$ H=\omega_s a^\dagger a+(gb+g^*b^\dagger)a^\dagger a +\omega_b b^\dagger b. $$ Consider further the state $|\chi⟩=|n⟩|-n g/\omega_b⟩$, i.e. the number state $|n⟩$ tensor-times a coherent state at $-ng/\omega_b$ for the bath, and calculate its energy expectation value: \begin{align} ⟨\chi|H|\chi⟩ &= \omega_s n +n(g(-ng/\omega_b)+g(-ng/\omega_b)) +\omega_b|ng/\omega_b|^2 = \omega_s n-n^2g^2/\omega_b. \end{align} This is arbitrarily negative for sufficiently large $n$, which proves that there is no ground state with finitely negative energy.
So: the problem with a boson-boson interaction of this form is not that the hamiltonian is not diagonalizable (which it patently is), but that the interaction is unphysical.
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