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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.

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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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  • $\begingroup$ @MarkMitchison I'm hoping I didn't miss anything. Did you have stronger reasons to think the bosonic case fails? $\endgroup$ – Emilio Pisanty May 24 '15 at 20:39
  • $\begingroup$ Excellent answer, thanks Emilio. I guess the unboundedness of the Hamiltonian could be remedied by a sufficiently strong non-linearity in the impurity Hamiltonian. Regarding my reasons for believing the bosonic case fails, they can be summarised succinctly as "laziness". $\endgroup$ – Mark Mitchison May 24 '15 at 20:45
  • $\begingroup$ Fair enough ;). I guess adding a term $C(\hat n)^2$ to the hamiltonian, with $C>g^2/\omega_b$ would probably fix it, but you're going to have an interesting time finding physical models which implement that. It's a bizarre hamiltonian in the first place - I can see why you'd want it for finite systems, but the bosonic case is just weird. A coupling of the form $\hat x\,\hat n$? No wonder it's unbounded. $\endgroup$ – Emilio Pisanty May 24 '15 at 20:52
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    $\begingroup$ Actually the Hamiltonian $\hat{n}\hat{x}$ describes, for example, the radiation pressure force in cavity optomechanics. This was precisely the system in which I was originally interested, before being distracted by this multi-mode problem. In the single-mode case, I expect that the unboundedness is probably remedied by non-linearities of the opto-mechanical oscillator (i.e. the $b$ mode in your notation) when the displacements are large. $\endgroup$ – Mark Mitchison Jun 3 '15 at 17:36

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