- #1
Cedric Chia
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- TL;DR Summary
- Help me understand the "polaron transformation" frequently encountered in quantum optics
I'm trying to understand the so-called polaron transformation as frequently encountered in quantum optics. Take the following paper as example: "Quantum dot cavity-QED in the presence of strong electron-phonon interactions" by I. Wilson-Rae and A. Imamoğlu. We have the spin-phonon model with cavity desribed by the following Hamiltonian
$$
H=\hbar\omega_{eg}\sigma_{ee}+\hbar\omega_ca^{\dagger}a+\hbar g(\sigma_{eg}a+a^{\dagger}\sigma_{ge})+\sum_k\hbar\omega_kb_k^{\dagger}b_k+\sigma_{ee}\sum_k\hbar\lambda_k(b_k+b_k^{\dagger})+\hbar\Omega_p(\sigma_{eg}e^{-i\omega t}+\sigma_{ge}e^{i\omega t})
$$
The polaron transformation is defined as
$$
H'=e^sHe^{-s}
$$
where ##s=\sigma_{ee}\sum_k\frac{\lambda_k}{\omega_k}(b_k^{\dagger}-b_k)##. I have no difficulty transforming ##H## into ##H'## using the results I found
$$
e^s\sigma_{ge}e^{-s}=\sigma_{ge}\exp{\sum_k}\frac{\lambda_k}{\nu_k}(b_k^{\dagger}-b_k)
$$
and
$$
e^sb_k e^{-s}=b_k-\frac{\lambda_k}{\omega_k}\sigma_{ee}
$$
But I just don't know how to get the following transformed Hamiltonian in terms of ##\left<B\right>##, ##B_+##, ##B_-## and also ##X_g## and ##X_u##:
$$
H^{\prime}=H_{s y s}^{\prime}+H_{i n t}^{\prime}+H_{b a t h}^{\prime}
$$
with
$$
\begin{align}
H_{b a t h}^{\prime}&=\sum_{k} \omega_{k} b_{k}^{\dagger} b_{k}\\
H_{s y s}^{\prime}&=\hbar \omega \sigma_{00}+\hbar \omega_{c} \sigma_{11}+\hbar\left(\omega_{e g}-\Delta\right) \sigma_{22}+\langle B\rangle X_{g}\\
H_{i n t}^{\prime}&=X_{g} \xi_{g}+X_{u} \xi_{u}
\end{align}
$$
with the definition of ##\left<B\right>##, ##B_+##, ##B_-##, ##X_g## and ##X_u## defined as follows:
\begin{align}
X_{g}&=\hbar\left[g\left(\sigma_{21}+\sigma_{12}\right)+\Omega_{p}\left(\sigma_{20}+\sigma_{02}\right)\right]\\
X_{u}&=i \hbar\left[g\left(\sigma_{12}-\sigma_{21}\right)+\Omega_{p}\left(\sigma_{02}-\sigma_{20}\right)\right]\\
B_{\pm}&=\exp \left(\pm \sum_{k} \frac{\lambda_{k}}{\omega_{k}}\left(b_{k}-b_{k}^{\dagger}\right)\right)\\
\xi_{g}&=\frac{1}{2}\left(B_{+}+B_{-}-2\langle B\rangle\right)\\
\xi_{u}&=\frac{1}{2 i}\left(B_{+}-B_{-}\right)
\end{align}
What is the physical intuition of introducing these operators? In particular, why do we need to introduce ##X_u## and ##\xi_u## with an imaginary number ##i## at the front where in the original Hamiltonian ##H## there wasn't even any ##i##?
This polaron transformation approach has been adopted by many recent studies so I really want to understand what's happening clearly but I couldn't find any lecture notes or textbooks on this. I would appreciate any help, explanation or book/paper recommendation greatly. Thank you.
$$
H=\hbar\omega_{eg}\sigma_{ee}+\hbar\omega_ca^{\dagger}a+\hbar g(\sigma_{eg}a+a^{\dagger}\sigma_{ge})+\sum_k\hbar\omega_kb_k^{\dagger}b_k+\sigma_{ee}\sum_k\hbar\lambda_k(b_k+b_k^{\dagger})+\hbar\Omega_p(\sigma_{eg}e^{-i\omega t}+\sigma_{ge}e^{i\omega t})
$$
The polaron transformation is defined as
$$
H'=e^sHe^{-s}
$$
where ##s=\sigma_{ee}\sum_k\frac{\lambda_k}{\omega_k}(b_k^{\dagger}-b_k)##. I have no difficulty transforming ##H## into ##H'## using the results I found
$$
e^s\sigma_{ge}e^{-s}=\sigma_{ge}\exp{\sum_k}\frac{\lambda_k}{\nu_k}(b_k^{\dagger}-b_k)
$$
and
$$
e^sb_k e^{-s}=b_k-\frac{\lambda_k}{\omega_k}\sigma_{ee}
$$
But I just don't know how to get the following transformed Hamiltonian in terms of ##\left<B\right>##, ##B_+##, ##B_-## and also ##X_g## and ##X_u##:
$$
H^{\prime}=H_{s y s}^{\prime}+H_{i n t}^{\prime}+H_{b a t h}^{\prime}
$$
with
$$
\begin{align}
H_{b a t h}^{\prime}&=\sum_{k} \omega_{k} b_{k}^{\dagger} b_{k}\\
H_{s y s}^{\prime}&=\hbar \omega \sigma_{00}+\hbar \omega_{c} \sigma_{11}+\hbar\left(\omega_{e g}-\Delta\right) \sigma_{22}+\langle B\rangle X_{g}\\
H_{i n t}^{\prime}&=X_{g} \xi_{g}+X_{u} \xi_{u}
\end{align}
$$
with the definition of ##\left<B\right>##, ##B_+##, ##B_-##, ##X_g## and ##X_u## defined as follows:
\begin{align}
X_{g}&=\hbar\left[g\left(\sigma_{21}+\sigma_{12}\right)+\Omega_{p}\left(\sigma_{20}+\sigma_{02}\right)\right]\\
X_{u}&=i \hbar\left[g\left(\sigma_{12}-\sigma_{21}\right)+\Omega_{p}\left(\sigma_{02}-\sigma_{20}\right)\right]\\
B_{\pm}&=\exp \left(\pm \sum_{k} \frac{\lambda_{k}}{\omega_{k}}\left(b_{k}-b_{k}^{\dagger}\right)\right)\\
\xi_{g}&=\frac{1}{2}\left(B_{+}+B_{-}-2\langle B\rangle\right)\\
\xi_{u}&=\frac{1}{2 i}\left(B_{+}-B_{-}\right)
\end{align}
What is the physical intuition of introducing these operators? In particular, why do we need to introduce ##X_u## and ##\xi_u## with an imaginary number ##i## at the front where in the original Hamiltonian ##H## there wasn't even any ##i##?
This polaron transformation approach has been adopted by many recent studies so I really want to understand what's happening clearly but I couldn't find any lecture notes or textbooks on this. I would appreciate any help, explanation or book/paper recommendation greatly. Thank you.