I am considering spin-Boson model with the total Hamiltonian given by \begin{align} \hat{H}_{SE} &= \hat{H}_S + \hat{H}_E + \hat{H}_I. \end{align} where \begin{align} \hat{H}_S &= \frac{\Omega}{2}\hat{\sigma}_3, \\ \hat{H}_E &= \sum_k \left[\frac{\hat{p}_k^2}{2m_k} + \frac{1}{2}m_k\omega_k^2\hat{q}_k^2\right], \\ \hat{H}_I &= \lambda \hat{A} \otimes \hat{B}, \end{align} with \begin{align} \hat{A} &= a_3\hat{\sigma}_3 - a_1\hat{\sigma}_1, \\ \hat{B} &= \sum_k c_k \hat{q}_k. \end{align} where $\hat{\sigma}_i$ is the Pauli matrix. Also, $\hat{p}_k$ and $\hat{q}_k$ are the momentum and position operators of the $k$th Bosonic oscillator with mass $m_k$, frequency $\omega_k$ and SE coupling strength $c_k$. The spectral density is defined as \begin{align} J(\omega) = \sum_k \frac{c_k^2}{m_k\omega_k}\delta(\omega-\omega_k). \end{align}
My question is:
- Is the long time expectation value of the system energy ($\lim_{t\to \infty} \langle\sigma_3(t)\rangle$) known to be independent of the variable $a_3$ at least in the perturbative limit in the coupling parameter $\lambda$? (Please read What I tried section to see why I think this should be true.)
- If yes, how can one approach proving this result rigorously? Are you aware of any literature in this direction?
Any insights or references on methods to demonstrate this would be appreciated.
What I tried:
This result holds at least for second order correction to the steady state population (see eqn 46 of our paper, which is essentially independent of $a_3$). Moreover, this result should also be true because of the intuition that dephasing should not influence the asymptotic energies of the qubit. In the case of pure dephasing (i.e., $a_1 = 0$), the population of the spin does not even evolve in time because we have $[\sigma_3, \hat{H}_{SE}] = 0$. But, when system is not undergoing pure dephasing (i.e., $a_1 \neq 0$), at intermediate time, the evolution of $\langle \sigma_3(t) \rangle$ does depend upon the value of $a_3$, since now, we have $[\sigma_3, \hat{H}_{SE}] \neq 0$. In fact, this numerical plot confirms the same. We note that in the numerical plot, the asymptotic values of the $\langle \sigma_3(t) \rangle$ for the two plots is interestingly close, where the difference could be because of numerical error.
Some specific things I tried (Following part can be skipped): eqn 21.46 and 21.54 of Weiss's book (3rd edition) gives the $\langle \sigma_i (t) \rangle$ exactly in terms of path integrals. But I was not able to obtain this result from this equation.
I also tried using Polaron transform to obtain this result using the unitary \begin{align} U(\alpha) = \exp(\sigma_3 \otimes (\alpha a^\dagger - \alpha^* a)) \end{align} This unitary is able to prove this result for pure dephasing case (for which, the result is already trivial), but the argument gets more complicated for the general case. Here is the detailed calculations:
Let the full Hamiltonian of a qubit coupled to a harmonic oscillator be given as \begin{align} H = \sigma_i + a^\dagger a + \sigma_i \otimes (a + a^\dagger) \end{align} My goal is to transform it so that it becomes easier to study. \begin{align} U(\alpha) = \exp(\sigma_i \otimes (\alpha a^\dagger - \alpha^* a)) \end{align} I choose $A = \sigma_i \otimes \alpha a^\dagger$ and $B = -\sigma_i \otimes \alpha^* a$. Then we have \begin{align} [A,B] &= -\alpha \alpha^* [a^\dagger,a]\\ &= \alpha \alpha^* \end{align} Then we have \begin{align} U(\alpha) &= e^{-|\alpha|^2/2} e^{A}e^{B}\\ &= e^{|\alpha|^2/2} e^{B}e^{A} \end{align}
\begin{align} -\frac{\partial U(\alpha)}{\partial \alpha^*} &= U(\alpha) (\alpha/2 + \sigma_i \otimes a)\\ &= (\sigma_i \otimes a - \alpha/2) U(\alpha) \end{align} Then \begin{align} \implies U(-\alpha) (\sigma_i \otimes a) U(\alpha) &= \sigma_i \otimes a + \alpha\\ \implies U(-\alpha) (\sigma_i \otimes a^\dagger) U(\alpha) &= \sigma_i \otimes a^\dagger + \alpha^* \end{align}
This means \begin{align} \implies U(-\alpha) (\sigma_i \otimes (a + a^\dagger )) U(\alpha) &= \sigma_i \otimes (a + a^\dagger) + \alpha + \alpha^*\\ \end{align}
But what will be the action on $a^\dagger a$? We have \begin{align} a^\dagger a &= (\sigma_i \otimes a^\dagger) (\sigma_i \otimes a)\\ \implies U(-\alpha) a^\dagger a U(\alpha) &= U(-\alpha) (\sigma_i \otimes a^\dagger)U(\alpha) U(-\alpha) (\sigma_i \otimes a) U(\alpha)\\ &= a^\dagger a + \sigma_i \otimes (\alpha^* a + \alpha a^\dagger) + \alpha \alpha^* \end{align}
Then, the total Hamiltonian transforms as \begin{align} \sigma_i + a^\dagger a + \sigma_i \otimes (a + a^\dagger) &\to \sigma_i + a^\dagger a + \sigma_i \otimes ((\alpha^* + 1) a + (\alpha + 1)a + \alpha + \alpha^* + \alpha \alpha^* \end{align}
If we choose $\alpha = -1$, then the extra term is just -1. This just produces an energy shift.
What is the meaning of this transformation?
We went to a new basis in which the system and the bath are uncoupled. Let us evolve the system in this basis. Then, after a long time, we go back to the usual frame? $\sigma_i$ will remain the same because this operator is unaffected by this Unitary as well as the transformed Hamiltonian.
To solve the full problem, we need to put in the $a_1$ term as well. Also, given that \begin{align} U &= \exp(\sigma_3 \otimes (a - a^\dagger)) \end{align} we need to calculate what is \begin{align} U^\dagger (\sigma_1 \otimes (a^\dagger + a)) U \end{align}
