diff --git a/tex/energy_transfer/src/index.tex b/tex/energy_transfer/src/index.tex index 9e6a602..e3b2f09 100644 --- a/tex/energy_transfer/src/index.tex +++ b/tex/energy_transfer/src/index.tex @@ -4,18 +4,18 @@ As in~\cite{Hartmann2017Dec} we choose \begin{equation} \label{eq:totalH} - H = H_S + \underbrace{LB^\dagger + L^\dagger B}_{H_I} + H_B + H = H_S + \underbrace{LB^†ger + L^†ger B}_{H_I} + H_B \end{equation} with the system hamiltonian \(H_S\), the bath hamiltonian \begin{equation} \label{eq:bathh} - H_B = \sum_\lambda \omega_\lambda a^\dag a, + H_B = ∑_λ ω_λ a^† a, \end{equation} the bath coupling system operator \(L\) and the bath coupling bath operator \begin{equation} \label{eq:bop} - B=\sum_{\lambda} g_{\lambda} a_{\lambda} + B=∑_{λ} g_{λ}^\ast a_{λ} \end{equation} which define the interaction hamiltonian \(H_I\). @@ -35,25 +35,25 @@ Thus, we need to calculate \begin{eqnarray} \label{eq:calccomm} \begin{aligned} - [H_B,H_I] &= [H_B, LB^\dag + L^\dag B] \\ - &= L[H_B, B^\dag ] + L^\dag [H_B, B] \\ - &= L[H_B, B^\dag ] - \hc. + [H_B,H_I] &= [H_B, LB^† + L^† B] \\ + &= L[H_B, B^† ] + L^† [H_B, B] \\ + &= L[H_B, B^† ] - \hc. \end{aligned} \end{eqnarray} This checks out as the commutator has to be anti-hermitian due to \cref{eq:ehrenfest}. -Using \([H_B, B^\dag ]=\sum_\lambda \omega_\lambda g^\ast_\lambda -a^\dag_\lambda\) it follows that +Using \([H_B, B^† ]=∑_λ ω_λ g_λ +a^†_λ\) it follows that \begin{equation} \label{eq:expcomm} \begin{aligned} - \ev{[H_B,H_I]} &= \sum_\lambda \omega_\lambda g^\ast_\lambda - \ev{La^\dag_\lambda} - \cc - = \sum_\lambda \omega_\lambda g^\ast_\lambda - \ev{La^\dag_\lambda \eu^{\i \omega t}}_I - \cc\\ - &= \frac{1}{\i}\ev{L\partial_t{\sum_\lambda - g^\ast_\lambda a^\dag_\lambda \eu^{\i \omega t}}}_I - \cc - =\frac{1}{\i}\qty(\ev{L\dot{B}^\dag}_I + \cc) + \ev{[H_B,H_I]} &= ∑_λ ω_λ g_λ + \ev{La^†_λ} - \cc + = ∑_λ ω_λ g^\ast_λ + \ev{La^†_λ \eu^{\i ω t}}_I - \cc\\ + &= \frac{1}{\i}\ev{L\partial_t{∑_λ + g^\ast_λ a^†_λ \eu^{\i ω t}}}_I - \cc + =\frac{1}{\i}\qty(\ev{L\dot{B}^†}_I + \cc) \end{aligned} \end{equation} where we switched to the interaction picture with respect to \(H_B\) @@ -62,7 +62,7 @@ In essence this is just the Heisenberg equation for \(H_I\). The expression for \(J\) follows \begin{equation} \label{eq:final_flow} - J(t) = \ev{L^\dag\partial_t B(t) + L\partial_t B^\dag(t)}_I. + J(t) = \ev{L^†\partial_t B(t) + L\partial_t B^†(t)}_I. \end{equation} From this point on, we will assume the interaction picture and drop the @@ -70,55 +70,55 @@ From this point on, we will assume the interaction picture and drop the The two summands yield different expressions in terms of the NMQSD. For use with HOPS with the final goal of utilizing the auxiliary -states the expression \(\ev{L^\dag\partial_t B(t)}\) should be +states the expression \(\ev{L^†\partial_t B(t)}\) should be evaluated. We calculate \begin{equation} \label{eq:interactev} - \ev{L^\dag\partial_t B(t)}=\ev{L^\dag\partial_t B(t)}{\psi(t)} = - \int \braket{\psi(t)}{z}\mel{z}{L^\dag\partial_tB(t)}{\psi(t)}\frac{\dd[2]{z}}{\pi^N}, + \ev{L^†\partial_t B(t)}=\ev{L^†\partial_t B(t)}{\psi(t)} = + ∫ \braket{\psi(t)}{z}\mel{z}{L^†\partial_tB(t)}{\psi(t)}\frac{\dd[2]{z}}{\pi^N}, \end{equation} where \(N\) is the total number of environment oscillators and -\(z=\qty(z_{\lambda_1}, z_{\lambda_2}, \ldots)\). +\(z=\qty(z_{λ_1}, z_{λ_2}, \ldots)\). To that end, \begin{equation} \label{eq:nmqsdficate} \begin{aligned} - \mel{z}{\partial_tB(t)}{\psi(t)} &= \sum_\lambda g_\lambda - \qty(\partial_t \eu^{-\i\omega_\lambda - t})\partial_{z^\ast_\lambda}\ket{\psi(z^\ast,t)} \\ - &= \int_0^t \sum_\lambda g_\lambda - \qty(\partial_t \eu^{-\i\omega_\lambda - t})\pdv{\eta_s^\ast}{z^\ast_\lambda}\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}\\ - &= -\i\int_0^t\dot{\alpha}(t-s)\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}, + \mel{z}{\partial_tB(t)}{\psi(t)} &= ∑_λ g_λ + \qty(\partial_t \eu^{-\iω_λ + t})\partial_{z^\ast_λ}\ket{\psi(z^\ast,t)} \\ + &= ∫_0^t ∑_λ g_λ + \qty(\partial_t \eu^{-\iω_λ + t})\pdv{\eta_s^\ast}{z^\ast_λ}\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}\\ + &= -\i∫_0^t\dot{\alpha}(t-s)\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}, \end{aligned} \end{equation} -where \(\eta^\ast_t\equiv -\i \sum_\lambda g^\ast_\lambda -z^\ast_\lambda \eu^{\i\omega_\lambda t}\). +where \(\eta^\ast_t\equiv -\i ∑_λ g^\ast_λ +z^\ast_λ \eu^{\iω_λ t}\). With this we can write \begin{equation} \label{eq:steptoproc} - \ev{L^\dag\partial_t B(t)} = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta, - t)}L^\dag\int_0^t\dd{s} \dot{\alpha}(t-s)\fdv{\eta^\ast_s} \ket{\psi(\eta^\ast,t)}. + \ev{L^†\partial_t B(t)} = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta, + t)}L^†∫_0^t\dd{s} \dot{\alpha}(t-s)\fdv{\eta^\ast_s} \ket{\psi(\eta^\ast,t)}. \end{equation} Defining \begin{equation} \label{eq:defdop} -D_t = \int_0^t\dd{s} \alpha(t-s)\fdv{\eta^\ast_s} +D_t = ∫_0^t\dd{s} \alpha(t-s)\fdv{\eta^\ast_s} \end{equation} as in~\cite{Suess2014Oct} we can write \begin{equation} \label{eq:final_flow_nmqsd} J(t) = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta, - t)}L^\dag\dot{D}_t\ket{\psi(\eta^\ast,t)} + \cc, + t)}L^†\dot{D}_t\ket{\psi(\eta^\ast,t)} + \cc, \end{equation} where we've used that the integral in \(D_t\) can be expanded over the whole real axis. If we assume \(\alpha = \exp(-w t)\) then \begin{equation} \label{eq:hopsj} J(t) = \i \mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta, - t)}wL^\dag\ket{\psi^{(1)}(\eta^\ast,t)} + \cc., + t)}wL^†\ket{\psi^{(1)}(\eta^\ast,t)} + \cc., \end{equation} where \(\ket{\psi^{(1)}(\eta^\ast,t)}\) is the first HOPS hierarchy state. This can be generalized to any BCF that is a sum of exponentials. @@ -126,7 +126,7 @@ state. This can be generalized to any BCF that is a sum of exponentials. Interestingly one finds that \begin{equation} \label{eq:alternative} - \ev{L\partial_t B^\dag(t)} = \i\int\frac{\dd[2]{z}}{\pi^N} + \ev{L\partial_t B^†(t)} = \i∫\frac{\dd[2]{z}}{\pi^N} \dot{\eta}_t^\ast \mel{\psi(\eta,t)}{L}{\psi(\eta^\ast,t)}. \end{equation} However, this seems numerically problematic because \(\eta^\ast\) is not @@ -135,11 +135,11 @@ that we utilize the first hierarchy states that are already being calculated as a byproduct. In the language of~\cite{Hartmann2021Aug} we can generalize to -\(\alpha(t) = \sum_i G_i \eu^{-W_i t}\) and thus +\(\alpha(t) = ∑_i G_i \eu^{-W_i t}\) and thus \begin{equation} \label{eq:hopsflowrich} - J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta, - t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)} + \cc, + J(t) = ∑_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta, + t)}L^†\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)} + \cc, \end{equation} where \(\psi^{\vb{e}_\mu}\) is the \(\mu\)-th state of the first hierarchy and \(\bar{g}_\mu\) is an arbitrary scaling introduced in @@ -154,30 +154,30 @@ In the spirit of the usual derivation of the nonlinear NMQSD we write \begin{equation} \label{eq:newb} \begin{aligned} - \ev{L^\dag\dot{B(t)}} &= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} + \ev{L^†\dot{B(t)}} &= ∫ \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} \braket{\psi}{z}\!\braket{z}{\psi} - \frac{\braket{\psi(t)}{z}\!\mel{z}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z}\!\braket{z}{\psi}} + \frac{\braket{\psi(t)}{z}\!\mel{z}{L^†\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z}\!\braket{z}{\psi}} \\ - &= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} - \frac{\mel{z(t)}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z(t)}\!\braket{z(t)}{\psi}}, + &= ∫ \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} + \frac{\mel{z(t)}{L^†\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z(t)}\!\braket{z(t)}{\psi}}, \end{aligned} \end{equation} -where \(z_{\lambda}^{*}(t)=z_{\lambda}^{*}+\i g_{\lambda} \int_{0}^{t} -\dd{s} \eu^{-\i \omega_{\lambda} s}\ev{L^\dagger}_{s}\). +where \(z_{λ}^{*}(t)=z_{λ}^{*}+\i g_{λ} ∫_{0}^{t} +\dd{s} \eu^{-\i ω_{λ} s}\ev{L^†ger}_{s}\). We find that next steps are the same as in \cref{sec:nonlin} by noting \begin{equation} \label{eq:deriv_trick} - \eval{\partial_{z^\ast_\lambda}}_{z^\ast=z_\lambda^\ast(t)} = - \int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=z^\ast_\lambda(t)} + \eval{\partial_{z^\ast_λ}}_{z^\ast=z_λ^\ast(t)} = + ∫_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_λ}}_{z^\ast=z^\ast_λ(t)} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))} = - \int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=0} + ∫_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_λ}}_{z^\ast=0} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))}, \end{equation} which does alter the definition of \(D_t\) but results in the same HOPS equations. The shifted process \(\tilde{\eta}^\ast= \eta^\ast(z^\ast(t),t)=\eta^\ast(t) + -\int_0^t\dd{s}\alpha^\ast(t-s)\ev{L^\dag}_{\psi_s}\) appears directly +∫_0^t\dd{s}\alpha^\ast(t-s)\ev{L^†}_{\psi_s}\) appears directly in the NMQSD equation but results only in a slight change in the functional derivative. Note however that \begin{equation} @@ -196,7 +196,7 @@ Therefore, \label{eq:newbcontin} J(t) = -\i - \mathcal{M}_{\tilde{\eta}^\ast}\frac{\mel{\psi(\tilde{\eta},t)}{L^\dag\dot{\tilde{D}}_t}{\psi(\tilde{\eta}^\ast,t)}}{\braket{\psi(\tilde{\eta},t)}{\psi(\tilde{\eta}^\ast,t)}} + \mathcal{M}_{\tilde{\eta}^\ast}\frac{\mel{\psi(\tilde{\eta},t)}{L^†\dot{\tilde{D}}_t}{\psi(\tilde{\eta}^\ast,t)}}{\braket{\psi(\tilde{\eta},t)}{\psi(\tilde{\eta}^\ast,t)}} + \cc, \end{equation} where the dependence on \(\tilde{\eta}\) is symbolic and to be @@ -206,9 +206,9 @@ Again we express the result in the language of~\cite{Hartmann2021Aug} to obtain \begin{equation} \label{eq:hopsflowrich} - J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} + J(t) = ∑_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\frac{\bra{\psi^{(0)}(\eta, - t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)}}{\bra{\psi^{(0)}(\eta, + t)}L^†\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)}}{\bra{\psi^{(0)}(\eta, t)}\ket{\psi^{0}(\eta^\ast,t)}} + \cc. \end{equation} @@ -225,7 +225,7 @@ method described in~\cite{Hartmann2017Dec}. The shift operator \begin{equation} \label{eq:shiftop} - \vb{D}(y) = \bigotimes_\lambda \eu^{y_\lambda a_\lambda^\dag-y^\ast_\lambda a_\lambda} + \vb{D}(y) = \bigotimes_λ \eu^{y_λ a_λ^†-y^\ast_λ a_λ} \end{equation} the ground state of the environment into an arbitrary coherent state @@ -240,55 +240,55 @@ thermal initial bath as \begin{equation} \label{eq:shiftbath} \rho = - \prod_\lambda\qty(\int\dd[2]{y_\lambda} - \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda}) - U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag. + \prod_λ\qty(∫\dd[2]{y_λ} + \frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ}) + U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^† U(t)^†. \end{equation} -The usual step is now to insert \(\id =\vb{D}(y)\vb{D}^\dag(y)\) to +The usual step is now to insert \(\id =\vb{D}(y)\vb{D}^†(y)\) to arrive at a new time translation operator \begin{equation} \label{eq:utilde} - \tilde{U}(t) = \vb{D}^\dag(y)U(t)\vb{D}(y) + \tilde{U}(t) = \vb{D}^†(y)U(t)\vb{D}(y) \end{equation} and to interpret the integral in \cref{eq:shiftbath} in a monte-carlo sense which leads to a stochastic contribution to the system Hamiltonian \begin{equation} \label{eq:thermalh} - H_{\mathrm{sys}}^{\mathrm{shift}}=L \xi^{*}(t)+L^{\dagger} \xi(t) + H_{\mathrm{sys}}^{\mathrm{shift}}=L \xi^{*}(t)+L^{†ger} \xi(t) \end{equation} with the stochastic process \begin{equation} \label{eq:xiproc} - \xi(t):=\sum_{\lambda} g_{\lambda} y_{\lambda} \eu^{-\mathrm{i} \omega_{\lambda} t} + \xi(t):=∑_{λ} g_{λ} y_{λ} \eu^{-\mathrm{i} ω_{λ} t} \end{equation} with corresponding moments \(\mathcal{M}(\xi(t))=0=\mathcal{M}(\xi(t) \xi(s))\) and \[ -\mathcal{M}\left(\xi(t) \xi^{*}(s)\right)=\frac{1}{\pi} \int_{0}^{\infty} \mathrm{d} \omega \bar{n}(\beta \omega) J(\omega) e^{-\mathrm{i} \omega(t-s)}. +\mathcal{M}\left(\xi(t) \xi^{*}(s)\right)=\frac{1}{\pi} ∫_{0}^{\infty} \mathrm{d} ω \bar{n}(\beta ω) J(ω) e^{-\mathrm{i} ω(t-s)}. \] Remember that we want to calculate \begin{equation} \label{eq:whatreallymatters} \begin{aligned} - \ev{L^\dag\dot{B}(t)} &= \tr[L^\dag\dot{B}(t)\rho(t)] \\ - &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} - \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\tr[L^\dag\dot{B}(t) - U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag] . + \ev{L^†\dot{B}(t)} &= \tr[L^†\dot{B}(t)\rho(t)] \\ + &=\prod_λ\qty(∫\dd[2]{y_λ} + \frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\tr[L^†\dot{B}(t) + U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^† U(t)^†] . \end{aligned} \end{equation} To recover the zero temperature formulation of this expectation value we -again insert a \(\id\), but have to commute \(\vb{D}(y)^\dag\) past +again insert a \(\id\), but have to commute \(\vb{D}(y)^†\) past \(\dot{B}(t)\). This leads to the expression \begin{equation} \label{eq:pureagain} \begin{aligned} - \ev{L^\dag\dot{B}(t)} &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} - \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\\ - &\qquad\times\tr[L^\dag(\dot{B}(t) + \dot{\xi}(t)) - \vb{D}^\dag(y) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}^\dag(y)U(t)^\dag\vb{D}(y)] \\ - &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} - \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\tr[L^\dag\qty{\dot{B}(t) + \dot{\xi}(t)} - \tilde{U}(t)\ketbra{\psi}\otimes\ketbra{0} \tilde{U}(t)^\dag]. + \ev{L^†\dot{B}(t)} &=\prod_λ\qty(∫\dd[2]{y_λ} + \frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\\ + &\qquad\times\tr[L^†(\dot{B}(t) + \dot{\xi}(t)) + \vb{D}^†(y) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}^†(y)U(t)^†\vb{D}(y)] \\ + &=\prod_λ\qty(∫\dd[2]{y_λ} + \frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\tr[L^†\qty{\dot{B}(t) + \dot{\xi}(t)} + \tilde{U}(t)\ketbra{\psi}\otimes\ketbra{0} \tilde{U}(t)^†]. \end{aligned} \end{equation} which returns us to the zero temperature formalism with a transformed @@ -297,7 +297,7 @@ Hamiltonian and the replacement \label{eq:breplacement} B(t) \rightarrow B(t) + \xi(t) \end{eqnarray} -which plausibly corresponds to the \(L^\dag\) part of \(H_I + H_{\mathrm{sys}}^{\mathrm{shift}}\). +which plausibly corresponds to the \(L^†\) part of \(H_I + H_{\mathrm{sys}}^{\mathrm{shift}}\). The appearance of \(\dot{\xi}(t)\) may cause concern. However, for twice differentiable \(\mathcal{M}(\xi(t)\xi^\ast(s))\) the sample @@ -322,13 +322,13 @@ Alternatively we can calculate Now, \begin{equation} \label{eq:hshcomm} - [H_{\mathrm{sys}}^{\mathrm{shift}}, H_I] = \xi(t) [L^\dag, L] - B(t)^\dag + \xi^\ast(t) [L, L^\dag] B + [H_{\mathrm{sys}}^{\mathrm{shift}}, H_I] = \xi(t) [L^†, L] + B(t)^† + \xi^\ast(t) [L, L^†] B \end{equation} and therefore \begin{equation} \label{eq:finalex} - \ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I]} = -i \mathcal{M}_{\eta^\ast}\mel{\psi}{\xi(t)^\ast[L,L^\dag]D_t}{\psi}. + \ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I]} = -i \mathcal{M}_{\eta^\ast}\mel{\psi}{\xi(t)^\ast[L,L^†]D_t}{\psi}. \end{equation} This is an expression that we can easily evaluate with the HOPS method.