# Qubit system coupled to a bath of quantum harmonic oscillators

It is well known that when we consider a probe harmonic oscillators (called system) that is coupled to a reservoir of N harmonic oscillators, i.e. the Hamiltonian is written as the following, the system is actually exactlyl solvable $$H=H_S+H_R+H_{SR}= \hbar \omega_0(a^{\dagger}a+1/2)+\sum_{j=1}^{N}\hbar \omega_j(b^{\dagger}_j b_j+1/2) +\hbar(a\sum_{j=1}^{N}k^{*}_j b^{\dagger}_j+a^{\dagger}\sum_{j=1}^{N}k_j b_j)$$ Where $$H_S$$ describes the probe (system), and $$H_R$$ describes the reservoir, while $$H_{SR}$$ describes the coupling.

By following the procedure in this papaer, we know that this Hamiltonian is equivalent to a set of uncoupled harmonic oscillators $$H=\sum_{\mu=0}^{N}\hbar\alpha_\mu c^{\dagger}_{\mu}c_{\mu}+ C \quad C= \hbar(\frac{\omega_0}{2}+\sum_{j=1}^{N}\frac{\omega_j}{2})$$ Where $$c_{\mu}=\phi_{\mu}a+\sum_{n=1}^{N}\psi_{\mu n}b_{n} \quad (\mu=0,1,2...N)$$

I have two questions concerning this:

1. How to transform between the number states of $$a/b$$ and number states of $$c$$ ? That is, for the ladder operators $$a/b/c$$, we can define their number states respectively $$a|n,a>=\sqrt{n}|n-1,a> \quad b_i|n,b_i>=\sqrt{n}|n-1,b_i> \quad c_\mu|n,c_\mu>=\sqrt{n}|n-1,c_\mu>$$ How should I express $$|n,a>$$ in terms of $$|n,c>$$?

2 Could this formalism be extended to, for example, a qubit coupled to a harmonic oscillator bath? i.e.

$$H=H_S+H_R+H_{SR}= -\frac{1}{2}\hbar \omega_0\sigma_z+\sum_{j=1}^{N}\hbar \omega_j(b^{\dagger}_j b_j+1/2) +\hbar(\sigma_{-}\sum_{j=1}^{N}k^{*}_j b^{\dagger}_j+\sigma_{+}\sum_{j=1}^{N}k_j b_j)$$

Or perhaps it is impossible since qubit represents fermionic degrees of freedom, would it possible if a qubit is replaced by a spin-1 particle?

1. You can construct them by successive application of ladder operators from the ground state. That is $$|n, c_\mu\rangle \propto (c_\mu^\dagger)^n |0\rangle = (\phi^*_{\mu}a^\dagger+\sum_{n=1}^{N}\psi^*_{\mu n}b^\dagger_{n})^n|0\rangle \,.$$ You can then compute the desired overlap matrix elements in terms of the expansion coefficients.
• @TanTixuan unfortunately the spin-spin model doesn't significantly relax the problem when one qubit is coupled with $N$ other qubits. The one-to-one problems are solvable (eg Jaynes Cummings model) and you can do $N$ qubits equally coupled to one bosonic mode (Tavis Cummings), but the general problems are all hard. Of course in finite dimensions you can diagonalize your Hamiltonian but remember that its size increases exponentially with the number of qubits Mar 16, 2022 at 17:27