Comment on 'Energy transfer, entanglement and decoherence in a molecular dimer interacting with a phonon bath'

We investigate single-exciton transfer with the Hamiltonian of the total system in the single-exciton manifold

$$\begin{equation} \hat{H}=\hat{H}_{\rm e}+\hat{H}_{\rm ph}+g(\left|1\right\rangle\left\langle 1\right|-\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{b}^{\dagger}+\hat{b}), \end{equation} \tag{ 1 }$$

where $\left |j\right \rangle$ represents a single-exciton state with only molecule j excited and the other molecule is in its ground state. Here, $\hat {H}_{\mathrm { e}}$ is the Hamiltonian of the molecules and $\hat {H}_{\mathrm { ph}}=\hbar \omega \hat {b}^{\dagger }\hat {b}$ is the Hamiltonian of a shared phonon mode with $\hat {b}^{\dagger }$ and $\hat {b}$ denoting its creation and annihilation operators. The state of the total system at initial time t = 0 is taken to be in a product form of $\hat {\rho }(0)=\hat {\rho }_{\mathrm { e}}(0)\otimes {\mathrm { e}}^{-\beta \hat {H}_{\mathrm { ph}}}/{\mathrm { Tr}}[{\mathrm { e}}^{-\beta \hat {H}_{\mathrm { ph}}}]$ where $\hat {\rho }_{\mathrm { e}}(0)$ is the initial single-exciton state and β is the inverse temperature of the shared phonon mode. The exciton transfer dynamics are described by the reduced electronic state of the molecules $\hat {\rho }_{\mathrm { e}}(t)={\mathrm { Tr_{ph}}}[\hat {\rho }(t)]$ where $\hat {\rho }(t)$ is the state of the total system at time t and Tr_ph is the partial trace over the phonon degrees of freedom.

To demonstrate that the influence of the shared phonon mode modeled by equation (1) on the reduced electronic state of molecules $\hat {\rho }_{\mathrm { e}}(t)$ is equivalent to that of independent phonon modes, we consider a two-molecule system where molecule j is coupled to an independent phonon mode associated with $\hat {c}_j$

$$\begin{equation} \hat{\cal H}=\hat{H}_{\rm e}+\hat{\cal H}_{\rm ph}+\hbar\sqrt{2}g\left|1\right\rangle\left\langle 1\right|\otimes(\hat{c}_{1}^{\dagger}+\hat{c}_{1})+\hbar\sqrt{2}g\left|2\right\rangle\left\langle 2\right|\otimes(\hat{c}_{2}^{\dagger}+\hat{c}_{2}), \end{equation} \tag{ 2 }$$

where $\hat {\cal H}_{\mathrm { ph}}=\hbar \omega \hat {c}_{1}^{\dagger }\hat {c}_{1}+\hbar \omega \hat {c}_{2}^{\dagger }\hat {c}_{2}$ is the Hamiltonian of the phonon modes, and the initial state of the total system is given by $\hat {\rho }(0)=\hat {\rho }_{\mathrm { e}}(0)\otimes {\mathrm { e}}^{-\beta \hat {\cal H}_{\mathrm { ph}}}/{\mathrm { Tr}}[{\mathrm { e}}^{-\beta \hat {\cal H}_{\mathrm { ph}}}]$. We consider a unitary transformation (basis change) of the annihilation operators of the phonon modes

$$\begin{equation} \left(\begin{array}{@{}c@{}} \hat{c}_{1}\\ \hat{c}_{2} \end{array}\right) =\frac{1}{\sqrt{2}} \left(\begin{array}{@{}cc@{}} 1 & 1\\ -1 & 1 \end{array}\right) \left(\begin{array}{@{}c@{}} \hat{b}\\ \hat{B} \end{array}\right) =U \left(\begin{array}{@{}c@{}} \hat{b}\\ \hat{B} \end{array}\right). \end{equation} \tag{ 3 }$$

Due to unitarity, UU^† = U^†U = I, and the bosonic commutation relations $[\hat {c}_{1},\hat {c}_{1}^{\dagger }]=1$, $[\hat {c}_{2},\hat {c}_{2}^{\dagger }]=1$, $[\hat {c}_{1},\hat {c}_{2}]=[\hat {c}_{1},\hat {c}_{2}^{\dagger }]=0$, the unitary-transformed operators $\hat {b}$ and $\hat {B}$ satisfy the bosonic commutation relations $[\hat {b},\hat {b}^{\dagger }]=1$, $[\hat {B},\hat {B}^{\dagger }]=1$ and $[\hat {b},\hat {B}]=[\hat {b},\hat {B}^{\dagger }]=0$, with $\hat {b}^{\dagger }\hat {b}+\hat {B}^{\dagger }\hat {B}=\hat {c}_{1}^{\dagger }\hat {c}_{1}+\hat {c}_{2}^{\dagger }\hat {c}_{2}$. Then the Hamiltonian of the total system $\hat {\cal H}$ can be rewritten as

$$\begin{eqnarray} &&\fl \hat{\cal H}=\hat{H}_{\rm e}+\hat{\cal H}_{\rm ph}+\hbar g(\left|1\right\rangle\left\langle 1\right|-\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{b}^{\dagger}+\hat{b})+\hbar g(\left|1\right\rangle\left\langle 1\right|+\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{B}^{\dagger}+\hat{B}), \end{eqnarray} \tag{ 4 }$$

where $\hat {\cal H}_{\mathrm { ph}}=\hbar \omega \hat {b}^{\dagger }\hat {b}+\hbar \omega \hat {B}^{\dagger }\hat {B}$. Note that the phonon mode associated with $\hat {B}$ does not influence the exciton transfer. This is because: (i) the reduced electronic state $\hat {\rho }_{\mathrm { e}}(t)={\mathrm { Tr_{ph}}}[\hat {\rho }(t)]$ is in the single-exciton subspace for all time t, i.e. $\hat {\rho }_{\mathrm { e}}(t)=\rho _{11}(t)\left |1\right \rangle \left \langle 1\right |+\rho _{22}(t)\left |2\right \rangle \left \langle 2\right |+\rho _{12}(t)\left |1\right \rangle \left \langle 2\right |+\rho _{21}(t)\left |2\right \rangle \left \langle 1\right |$; and (ii) the initial state of the phonon mode associated with $\hat {B}$ is decoupled from the rest of the total system at time t = 0. Then the interaction Hamiltonian $\hbar g(\left |1\right \rangle \left \langle 1\right |+\left |2\right \rangle \left \langle 2\right |)\otimes (\hat {B}^{\dagger }+\hat {B})$ leads to the decoupled dynamics of the phonon associated with $\hat {B}$ and the other degrees of freedom. In this case, the dynamics of the reduced electronic state $\hat {\rho }_{\mathrm { e}}(t)$, i.e. the exciton transfer dynamics, is not influenced even if the phonon mode associated with $\hat {B}$ is removed from $\hat {\cal H}$ such that

$$\begin{eqnarray} &&\hat{\cal H}'=\hat{H}_{\rm e}+\hat{\cal H}'_{\rm ph}+\hbar g(\left|1\right\rangle\left\langle 1\right|-\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{b}^{\dagger}+\hat{b}), \end{eqnarray} \tag{ 5 }$$

where $\hat {\cal H}'_{\mathrm { ph}}=\hbar \omega \hat {b}^{\dagger }\hat {b}$, and the initial state of the total system is given by $\hat {\rho }'(0)=\hat {\rho }_{\mathrm { e}}(0)\otimes {\mathrm { e}}^{-\beta \hat {\cal H}'_{\mathrm { ph}}}/{\mathrm { Tr}}[{\mathrm { e}}^{-\beta \hat {\cal H}'_{\mathrm { ph}}}]$, leading to $\hat {\rho }_{\mathrm { e}}(t)={\mathrm { Tr_{ph}}}[\hat {\rho }(t)]={\mathrm { Tr_{ph}}}[\hat {\rho }'(t)]$. This implies that the influence of the shared phonon mode modeled by equation (1) on exciton transfer is equivalent to that of the independent phonon modes associated with $\hat {c}_{1}$ and $\hat {c}_{2}$ modeled by equation (2).

We now generalize the proof of the equivalence to the case of many phonon modes and show that the shared bath model considered in [1] is equivalent to an independent bath model. In [1], single-exciton transfer was investigated with the Hamiltonian of the total system in the single-exciton manifold

$$\begin{equation} \hat{H}=\hat{H}_{\rm e}+\hat{H}_{\rm ph}+\sum_{\xi}\hbar g_{\xi}(\left|1\right\rangle\left\langle 1\right|-\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{b}^{\dagger}_{\xi}+\hat{b}_{\xi}), \end{equation} \tag{ 6 }$$

where $\hat {H}_{\mathrm { ph}}=\sum _{\xi }\hbar \omega _{\xi }\hat {b}_{\xi }^{\dagger }\hat {b}_{\xi }$ is the Hamiltonian of the phonon bath with $\hat {b}_{\xi }^{\dagger }$ and $\hat {b}_{\xi }$ denoting creation and annihilation operators of a shared phonon mode ξ. The state of the total system at initial time t = 0 was taken to be in a product form of $\hat {\rho }(0)=\hat {\rho }_{\mathrm { e}}(0)\otimes {\mathrm { e}}^{-\beta \hat {H}_{\mathrm { ph}}}/{\mathrm { Tr}}[{\mathrm { e}}^{-\beta \hat {H}_{\mathrm { ph}}}]$ where $\hat {\rho }_{\mathrm { e}}(0)$ is the initial single-exciton state and β is the inverse temperature of the phonon bath. The exciton transfer dynamics were described by the reduced electronic state of molecules $\hat {\rho }_{\mathrm { e}}(t)={\mathrm { Tr_{ph}}}[\hat {\rho }(t)]$.

To demonstrate that the shared phonon bath modeled by equation (6) is equivalent to an independent bath model, we start with the Hamiltonian of the total system in the presence of independent phonon modes

$$\begin{equation} \hat{\cal H}=\hat{H}_{\rm e}+\hat{\cal H}_{\rm ph}+\sum_{j=1}^{2}\sum_{\xi}\hbar \sqrt{2}g_{\xi}\left|j\right\rangle\left\langle j\right|\otimes(\hat{c}_{j\xi}^{\dagger}+\hat{c}_{j\xi}), \end{equation} \tag{ 7 }$$

where $\hat {\cal H}_{\mathrm { ph}}=\sum _{j=1}^{2}\sum _{\xi }\hbar \omega _{\xi }\hat {c}_{j\xi }^{\dagger }\hat {c}_{j\xi }$. We now consider a unitary transformation of the annihilation operators of the phonon modes

$$\begin{equation} \left(\begin{array}{@{}c@{}} \hat{c}_{1\xi}\\ \hat{c}_{2\xi} \end{array}\right) =\frac{1}{\sqrt{2}} \left(\begin{array}{@{}cc@{}} 1 & 1\\ -1 & 1 \end{array}\right) \left(\begin{array}{@{}c@{}} \hat{b}_{\xi}\\ \hat{B}_{\xi} \end{array}\right) =U \left(\begin{array}{@{}c@{}} \hat{b}_{\xi}\\ \hat{B}_{\xi} \end{array}\right). \end{equation} \tag{ 8 }$$

Due to unitarity, UU^† = U^†U = I, and the bosonic commutation relations $[\hat {c}_{1\xi },\hat {c}_{1\xi }^{\dagger }]=1$, $[\hat {c}_{2\xi },\hat {c}_{2\xi }^{\dagger }]=1$, $[\hat {c}_{1\xi },\hat {c}_{2\xi }]=[\hat {c}_{1\xi },\hat {c}_{2\xi }^{\dagger }]=0$, the unitary-transformed operators $\hat {b}_{\xi }$ and $\hat {B}_{\xi }$ satisfy the bosonic commutation relations $[\hat {b}_{\xi },\hat {b}_{\xi }^{\dagger }]=1$, $[\hat {B}_{\xi },\hat {B}_{\xi }^{\dagger }]=1$, $[\hat {b}_{\xi },\hat {B}_{\xi }]=[\hat {b}_{\xi },\hat {B}_{\xi }^{\dagger }]=0$, with $\hat {b}_{\xi }^{\dagger }\hat {b}_{\xi }+\hat {B}_{\xi }^{\dagger }\hat {B}_{\xi }=\hat {c}_{1\xi }^{\dagger }\hat {c}_{1\xi }+\hat {c}_{2\xi }^{\dagger }\hat {c}_{2\xi }$. Then the Hamiltonian of the total system $\hat {\cal H}$ can be rewritten as

$$\begin{eqnarray} &&\fl \hat{\cal H}=\hat{H}_{\rm e}+\hat{\cal H}_{\rm ph}+\sum_{\xi}\hbar g_{\xi}(\left|1\right\rangle\left\langle 1\right|-\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{b}_{\xi}^{\dagger}+\hat{b}_{\xi})\nonumber\\&&+\sum_{\xi}\hbar g_{\xi}(\left|1\right\rangle\left\langle 1\right|+\left|2\right\rangle\left\langle 2\right|)\otimes(\hat{B}_{\xi}^{\dagger}+\hat{B}_{\xi}), \end{eqnarray} \tag{ 9 }$$

where $\hat {\cal H}_{\mathrm { ph}}=\sum _{\xi }\hbar \omega _{\xi }\hat {b}_{\xi }^{\dagger }\hat {b}_{\xi }+\sum _{\xi }\hbar \omega _{\xi }\hat {B}_{\xi }^{\dagger }\hat {B}_{\xi }$. Similar to the case of a few phonon modes, the phonon modes associated with $\hat {B}_{\xi }$ do not influence the exciton transfer dynamics even if they are removed from $\hat {\cal H}$. This implies that the influence of the shared phonon modes $\hat {b}_{\xi }$ modeled by equation (6) on exciton transfer is equivalent to that of the independent phonon modes associated with $\hat {c}_{1\xi }$ and $\hat {c}_{2\xi }$ modeled by equation (7). This shows that the shared bath model considered in [1] is equivalent to an independent bath model.