Polaron transform of the Frenkel exciton model

1. Frenkel exciton model

Consider a Frenkel exciton model with $n$ exciton sites, each coupled to a local harmonic bath. The full Hamiltonian is

\[ \hat{H} = \sum_n \left( \varepsilon_n + \sum_\nu \frac{g_{n,\nu}^2}{2 \omega_{n,\nu}^2} \right) \lvert n\rangle\langle n\rvert + \sum_n \sum_{m\neq n} \Delta_{n,m} \left( \lvert n\rangle\langle m\rvert + \lvert m\rangle\langle n\rvert \right) \]

\[ \quad + \sum_{n,\nu} \frac{1}{2} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big) - \sum_{n,\nu} g_{n,\nu} \hat{R}_{n,\nu} \lvert n\rangle\langle n\rvert. \tag{1} \]

We identify:

• $\varepsilon_n$ – the (renormalized) energy of site $n$;
• $\Delta_{n,m}$ – electronic coupling between sites $n$ and $m$;
• $\hat{R}_{n,\nu}$, $\hat{P}_{n,\nu}$ – position and momentum of the $\nu$th bath mode on site $n$;
• $\omega_{n,\nu}$ – bath mode frequency;
• $g_{n,\nu}$ – linear system–bath coupling constant.

The bath and system–bath parts are

\[ \hat{H}_b = \sum_{n,\nu} \frac{1}{2} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big), \qquad \hat{H}_{sb} = - \sum_{n,\nu} g_{n,\nu} \hat{R}_{n,\nu} \lvert n\rangle\langle n\rvert, \]

and the diagonal system part (including the counterterm) is

\[ \hat{H}_d = \sum_n \left( \varepsilon_n + \sum_\nu \frac{g_{n,\nu}^2}{2 \omega_{n,\nu}^2} \right) \lvert n\rangle\langle n\rvert. \]

The off-diagonal system couplings are

\[ \hat{H}_\text{off} = \sum_n \sum_{m\neq n} \Delta_{n,m} \left( \lvert n\rangle\langle m\rvert + \lvert m\rangle\langle n\rvert \right). \]

3. Transform of $\lvert n\rangle\langle n\rvert$

The diagonal projector transforms as

\[ e^{\hat{S}} \lvert n\rangle\langle n\rvert e^{-\hat{S}} = \lvert n\rangle\langle n\rvert + [\hat{S},\lvert n\rangle\langle n\rvert] + \frac{1}{2!} [\hat{S},[\hat{S},\lvert n\rangle\langle n\rvert]] + \cdots. \tag{3} \]

The first commutator is

\[ [\hat{S},\lvert n\rangle\langle n\rvert] = \sum_{k,\mu} M_{k,\mu}\hat{P}_{k,\mu} [\lvert k\rangle\langle k\rvert,\lvert n\rangle\langle n\rvert]. \]

\[ [\lvert k\rangle\langle k\rvert,\lvert n\rangle\langle n\rvert] = \lvert k\rangle\langle k\rvert n\rangle\langle n\rvert - \lvert n\rangle\langle n\rvert k\rangle\langle k\rvert = \lvert k\rangle\langle n\rvert \delta_{k,n} - \lvert n\rangle\langle k\rvert \delta_{k,n}. \]

Hence

\[ [\hat{S},\lvert n\rangle\langle n\rvert] = \sum_\mu \big( M_{n,\mu}\hat{P}_{n,\mu}\lvert n\rangle\langle n\rvert - M_{n,\mu}\hat{P}_{n,\mu}\lvert n\rangle\langle n\rvert \big) = 0. \tag{4} \]

All higher commutators vanish, so

\[ e^{\hat{S}} \lvert n\rangle\langle n\rvert e^{-\hat{S}} = \lvert n\rangle\langle n\rvert. \tag{5} \]

5. Transform of $\hat{H}_b + \hat{H}_{sb}$ and choice of $M_{n,\nu}$

5.1 Bath Hamiltonian

Using the transformed operators in $\hat{H}_b$, we write

\[ e^{\hat{S}} \hat{H}_b e^{-\hat{S}} = \frac{1}{2} \sum_{n,\nu} e^{\hat{S}} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big) e^{-\hat{S}}. \]

Using $e^{\hat{S}}\hat{P}_{n,\nu}e^{-\hat{S}} = \hat{P}_{n,\nu}$ and $e^{\hat{S}}\hat{R}_{n,\nu}e^{-\hat{S}} = \hat{R}_{n,\nu} - i M_{n,\nu}\lvert n\rangle\langle n\rvert$:

\[ e^{\hat{S}} \hat{H}_b e^{-\hat{S}} = \frac{1}{2} \sum_{n,\nu} \hat{P}_{n,\nu}^2 + \frac{1}{2} \sum_{n,\nu} \omega_{n,\nu}^2 \big( \hat{R}_{n,\nu} - i M_{n,\nu}\lvert n\rangle\langle n\rvert \big)^2. \]

\[ = \frac{1}{2} \sum_{n,\nu} \hat{P}_{n,\nu}^2 + \frac{1}{2} \sum_{n,\nu} \omega_{n,\nu}^2 \big( \hat{R}_{n,\nu}^2 - 2 i M_{n,\nu} \hat{R}_{n,\nu}\lvert n\rangle\langle n\rvert - M_{n,\nu}^2 \lvert n\rangle\langle n\rvert \big). \tag{12} \]

So

\[ e^{\hat{S}} \hat{H}_b e^{-\hat{S}} = \frac{1}{2} \sum_{n,\nu} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big) - i \sum_{n,\nu} \omega_{n,\nu}^2 M_{n,\nu} \hat{R}_{n,\nu} \lvert n\rangle\langle n\rvert \]

\[ \quad - \frac{1}{2} \sum_{n,\nu} \omega_{n,\nu}^2 M_{n,\nu}^2 \lvert n\rangle\langle n\rvert. \]

5.2 System–bath interaction

The system–bath Hamiltonian transforms as

\[ e^{\hat{S}} \hat{H}_{sb} e^{-\hat{S}} = e^{\hat{S}} \left( -\sum_{n,\nu} g_{n,\nu} \hat{R}_{n,\nu} \lvert n\rangle\langle n\rvert \right) e^{-\hat{S}}. \]

Using the previous results,

\[ e^{\hat{S}} \hat{H}_{sb} e^{-\hat{S}} = -\sum_{n,\nu} g_{n,\nu} \big( e^{\hat{S}} \hat{R}_{n,\nu} e^{-\hat{S}} \big) \big( e^{\hat{S}} \lvert n\rangle\langle n\rvert e^{-\hat{S}} \big) \]

\[ = -\sum_{n,\nu} g_{n,\nu} \big( \hat{R}_{n,\nu} - i M_{n,\nu}\lvert n\rangle\langle n\rvert \big) \lvert n\rangle\langle n\rvert \]

\[ = -\sum_{n,\nu} g_{n,\nu}\hat{R}_{n,\nu} \lvert n\rangle\langle n\rvert + i \sum_{n,\nu} g_{n,\nu} M_{n,\nu} \lvert n\rangle\langle n\rvert. \tag{13} \]

5.3 Choosing $M_{n,\nu}$

Combining the transformed $\hat{H}_b$ and $\hat{H}_{sb}$, the coefficient in front of $\hat{R}_{n,\nu}\lvert n\rangle\langle n\rvert$ is

\[ - i \omega_{n,\nu}^2 M_{n,\nu} - g_{n,\nu}. \]

To eliminate the linear system–bath coupling, we choose $M_{n,\nu}$ such that this term vanishes:

\[ - i \omega_{n,\nu}^2 M_{n,\nu} - g_{n,\nu} = 0 \quad\Rightarrow\quad M_{n,\nu} = i\,\frac{g_{n,\nu}}{\omega_{n,\nu}^2}. \tag{15} \]

Using this, and collecting the diagonal terms, we find

\[ e^{\hat{S}}(\hat{H}_b + \hat{H}_{sb}) e^{-\hat{S}} = \frac{1}{2} \sum_{n,\nu} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big) - \sum_{n,\nu} \frac{g_{n,\nu}^2}{2 \omega_{n,\nu}^2} \lvert n\rangle\langle n\rvert. \tag{16} \]

Together with the diagonal system part, the transformed diagonal + bath sector is

\[ e^{\hat{S}}(\hat{H}_d + \hat{H}_b + \hat{H}_{sb}) e^{-\hat{S}} = \sum_n \varepsilon_n \lvert n\rangle\langle n\rvert + \frac{1}{2} \sum_{n,\nu} \big( \hat{P}_{n,\nu}^2 + \omega_{n,\nu}^2 \hat{R}_{n,\nu}^2 \big). \tag{18} \]

Thus, after the polaron shift, the diagonal system energies are bare ($\varepsilon_n$) and the bath is a collection of unshifted harmonic oscillators.