text:
\section{Introduction}

Nonadiabatic coupling elements account for electron-nucleus interactions that are neglected in the Born-Oppenheimer\cite{Born1927} (BO) approximation. These elements couple different electronic states through the nuclear kinetic energy operator. While mostly negligible in ground state chemistry, coupling elements are required when considering molecular dynamics in excited electronic states. Excited state dynamics often involves regions of nuclear space where  electronic states are nearly or exactly degenerate, causing a  breakdown of the BO separation.\cite{Zhu2016,Curchod2018} Accurately describing nonadiabatic coupling elements is therefore important for reliable predictions in photochemistry.

The coupled cluster method is one of the most accurate electronic structure methods, both for ground and excited state properties,\cite{bartlett2007coupled,Krylov2008,helgaker2014molecular,Loos2020a} but it has not found widespread use for predicting excited state dynamics. This is primarily because standard coupled cluster methods give a nonphysical description of regions close to electronic degeneracies, or conical intersections\cite{hattig2005structure,kohn2007can,kjoenstad2017crossing}. This issue can be traced to the method's non-Hermiticity, which seems to imply that coupled cluster methods cannot be used for nonadiabatic dynamics. However, this is not the case. As we have shown in recent work, the method can be constrained to give a correct physical description of excited state conical intersections while retaining the standard non-Hermitian formalism and presumably its accuracy\cite{kjoenstad2017resolving,kjoenstad2019invariant}. These developments may lead to renewed interest in coupled cluster dynamics.

Nonadiabaticity, as described by coupled cluster methods, has been considered by several authors. The formula for the vector coupling was first derived by Christiansen,\cite{Christiansen1999nonadiabatic} who applied the $Z$-vector substitution method\cite{handy1984evaluation} on a biorthonormal expression for the vector coupling,
\begin{align}
    \b{F}_{mn}^I = \Dbraket{\tilde{\psi}_m}{\nabla_I \psi_n}, \quad \Dbraket{\tilde{\psi}_m}{\psi_n} = \delta_{mn}, \label{eq:biorthonormal_F}
\end{align}
where $(\tilde{\psi}_k,\psi_k)$ refers to the left and right $k$th electronic states, and $I$ to a nucleus.
However, Christiansen's paper\cite{Christiansen1999nonadiabatic} did not include an implementation of the coupling. The vector coupling was later rederived by Tajti and Szalay\cite{Tajti2009} by differentiating the corresponding  $m$-to-$n$ transition element of the electronic Hamiltonian. Their derivation is closely related to that given by Ichino \emph{et al.}\cite{Ichino2009} for the quasidiabatic interstate coupling. Tajti and Szalay\cite{Tajti2009} also gave an implementation of the vector coupling at the singles and doubles level (CCSD\cite{Purvis1982}). These papers on the vector coupling\cite{Christiansen1999nonadiabatic,Tajti2009} did not include a discussion of the nuclear Schrödinger equations in coupled cluster theory, where the coupling elements enter.

The correct formula for the vector coupling has been a subject of some controversy.
Tajti and Szalay\cite{Tajti2009} argued that the biorthonormal formula in Eq.~\eqref{eq:biorthonormal_F} is incorrect. As they correctly noted, the vector coupling changes with the norm of the left and right states. A similar observation had been made in an earlier paper on the diagonal BO correction.\cite{Gauss2006}
Since the vector coupling varies with the norm of the states, the full-CC vector coupling is different from the full configuration interaction (CI) limit, where left and right states are identical and usually  normalized. They therefore suggested that normalizing the states was necessary. Furthermore, since the derivative can either act on the left or on the right state, they suggested using an average of the two\cite{Tajti2009}. If true, these observations are troubling: they imply that computing the vector coupling has a computational cost that scales as full-CI due to the normalization factors for the right states. In practice, the normalization factors are therefore approximated. However, it is unfortunate if one must resort to approximations other than the truncation level of the coupled cluster method (e.g., singles and doubles). The need for normalization factors was also assumed in the recent CCSD implementation by Faraji \emph{et al}.\cite{Faraji2018}

One of the main objectives of the present paper is to establish that normalization is not necessary. The reason is that normalization is a special case of an invertible transformation of the electronic basis. Such transformations do not change the expansion space in the Born-Huang expansion\cite{Born1954}
and therefore do not change the molecular wave function. In particular, the coefficients in the Born-Huang expansion---that is, the nuclear wave functions---absorb the transformation of the electronic states. In a recent paper, Shamasundar\cite{Shamasundar2018} also noted that the predicted dynamics must not depend on the normalization of the underlying electronic wave functions. The vector coupling does depend on normalization, but this should not be considered a problem because this quantity is not an observable. Since normalization of wave functions is not necessary, the biorthonormal formula in Eq.~\eqref{eq:biorthonormal_F} is a valid option. In this work, we derive the biorthonormal coupled cluster vector and scalar couplings using the Lagrangian approach developed by Hohenstein.\cite{Hohenstein2016}

The second main objective of the paper is to give a framework for nonadiabatic dynamics using coupled cluster methods. In particular, we argue that the biorthonormal formalism for electronic wave functions implies a biorthonormal formalism for the molecular wave function. Hence, we must determine left and right nuclear wave functions and the nuclear motion is described by two sets of nuclear Schrödinger equations. The result is a molecular wave function pair ($\tilde{\Psi}, \Psi$), where observable quantities are given by the usual biorthonormal formulas. In this contribution, we describe theoretical aspects relevant for nonadiabatic dynamics. Implementation of the various quantities is postponed to a future publication.

\section{Theory}
The total wave function of a molecular system can be expressed as an expansion over the electronic wave functions. The coefficients of this Born-Huang expansion defines the nuclear wave functions. These are determined by inserting the expansion in the Schrödinger equation and projecting out the electronic components. To formulate the corresponding procedure for coupled cluster theory, we first review the description of the electronic states.

\subsection{Electronic wave functions in coupled cluster theory}
In the equation of motion coupled cluster formalism,
a set of left and right electronic states are considered. These are defined as\citep{Stanton1993}
\begin{align}
    \bra{\tilde{\psi}_n} &= \sum_{\mu \geq 0} \st{L}_{\mu}^n \bra{\mu} \exp(-T), \label{eq:EOM_left} \\
    \ket{\psi_n} &= \sum_{\mu \geq 0} \exp(T) \ket{\mu} \st{R}_\mu^n, \quad n = 0, 1, 2, \ldots \label{eq:EOM_right}
\end{align}
The states are not identical, in general, but they satisfy the biorthonormality condition
\begin{align}
    \Dbraket{\tilde{\psi}_m}{\psi_n} = \delta_{mn}. \label{eq:state_biorthonormality}
\end{align}
The scalars $\st{L}_\mu^n$ and $\st{R}_\mu^n$ are state amplitudes, giving weights to the configurations
\begin{align}
    \ket{\mu} &= \tau_\mu \ket{\mathrm{HF}} \\
    \bra{\mu} &= \bra{\mathrm{HF}} \tilde{\tau}_{\mu}^\dagger, \quad \mu \geq 0,
\end{align}
where $\tau_\mu$ and $\tilde{\tau}_{\mu}$ with $\mu > 0$ are excitation operators relative to the Hartree-Fock state $\ket{\mathrm{HF}}$, while $\tau_0 = \tilde{\tau}_0 = \mathbb{I}$ is the identity operator. The ket and bra bases, $\{ \ket{\mu} \}$ and $\{ \bra{\mu}\}$, span the same subspace and are normally required to satisfy the biorthonormality relation
\begin{align}
\Dbraket{\mu}{\nu} = \delta_{\mu\nu}, \quad \mu,\nu \geq 0.
\end{align}
One special case is that the left and right bases are identical and hence orthonormal. However, using different left and right basis is sometimes convenient (e.g., in spin-adapted formulations\citep{helgaker2014molecular}).
Finally, we have the exponential part of the parametrization, defined by the cluster operator
\begin{align}
    T = \sum_{\mu > 0} t_\mu \tau_\mu.
\end{align}
The scalars $t_\mu$ are called cluster amplitudes.

Given the parametrization in Eqs.~\eqref{eq:EOM_left} and \eqref{eq:EOM_right}, how are the parameters determined? First one assumes that the right ground state can be written
\begin{align}
    \ket{\psi_0} = \exp(T) \ket{\mathrm{HF}}.
\end{align}
Then the time-independent Scrödinger equation, expressed as
\begin{align}
    \exp(-T) H \exp(T)\ket{\mathrm{HF}} = E_0 \ket{\mathrm{HF}},
\end{align}
is projected onto the bra basis. The operator
\begin{align}
    \bar{H} = \exp(-T) H \exp(T)
\end{align}
is known as the similarity transformed Hamiltonian. This projection procedure gives an expression for the ground state energy and equations for determining the amplitudes,
\begin{align}
    E_0 &= \Tbraket{\mathrm{HF}}{\bar{H}}{\mathrm{HF}} \label{eq:E_0} \\
    \Omega_\mu  &= \Tbraket{\mu}{\bar{H}}{\mathrm{HF}} = 0, \quad \mu > 0.
\end{align}
The state amplitudes are determined by making the pseudo expectation values
\begin{align}
    E_n(\bst{L}_n, \bst{R}_n) = \Tbraket{\tilde{\psi_n}}{\bar{H}}{\psi_n}, \quad n = 0,1,2,\ldots \label{eq:pseudo_exp}
\end{align}
stationary under the binormality condition given in Eq.~\eqref{eq:state_biorthonormality}. This constrained optimization problem is conveniently formulated in terms of the state Lagrangians
\begin{align}
\begin{split}
    \mathscr{L}_n(\bst{L}_n, \bst{R}_n, \bar{E}_n) &= \Tbraket{\tilde{\psi_n}}{\bar{H}}{\psi_n} + \bar{E}_n (1- \Dbraket{\tilde{\psi}_n}{\psi_n}) \\
    &= \bst{L}_n^T \bar{\b{\mathcal{H}}} \bst{R}_n + \bar{E}_n (1 - \bst{L}_n^T \bst{R}_n),
\end{split}
\end{align}
where we have defined the Hamiltonian matrix
\begin{align}
    \bar{\mathcal{H}}_{\mu\nu} = \Tbraket{\mu}{\bar{H}}{\nu}, \quad \mu,\nu \geq 0.
\end{align}
Stationarity of the Lagrangians imply
\begin{align}
    1 &= \bst{L}_n^T \bst{R}_n \\
    \b{0} &= \bst{L}_n^T \bar{\b{\mathcal{H}}} - \bar{E}_n \bst{L}_n^T \\
    \b{0} &= \bar{\b{\mathcal{H}}} \bst{R}_n - \bar{E}_n \bst{R}_n.
\end{align}
Writing $\bar{E}_n = E_n = E_0 + \omega_n$, we see that the latter two equations read
\begin{align}
    \bst{L}_n^T \bst{A} &= \omega_n \bst{L}_n^T \\
    \bst{A} \bst{R}_n   &= \omega_n \bst{R}_n, \quad n = 0,1,2,\ldots,
\end{align}
where $\bst{A} = \bar{\bst{H}} - E_0 \b{I}$. Identifying the pseudo expectation value in Eq.~\eqref{eq:pseudo_exp} with the energy, we see that $\omega_n$ is the excitation energy of the $n$th state, where it is understood that $\omega_0 = 0$ for the ground state. The matrix $\bst{A}$ can be expressed as
\begin{align}
    \bst{A} = \begin{pmatrix}
                0     & \b{\eta}^T \\
                \b{0} & \b{A}
            \end{pmatrix}
\end{align}
where
\begin{align}
    \eta_\nu   &= \Tbraket{\mathrm{HF}}{[\bar{H}, \tau_\nu]}{\mathrm{HF}} \\
    A_{\mu\nu} &= \Tbraket{\tilde{\mu}}{[\bar{H}, \tau_\nu] }{\mathrm{HF}}.
\end{align}
where $\b{A}$ is called the coupled cluster Jacobian matrix.\citep{Koch1990} The eigenvalues of $\b{A}$ are the non-zero excitation energies, i.e.~$\omega_n$ with $n = 1,2,3,\ldots$

\subsection{The Born-Huang expansion of the total wave function and the nuclear Schrödinger equations}
With the electronic states described, we now turn to the expansion of the total wave function.
The Born-Huang expansion expresses the total wave function in terms of the left and right electronic bases given in Eqs.~\eqref{eq:EOM_left} and \eqref{eq:EOM_right}. Notice that this implies a biorthonormal description of the total wave function, since we can expand in both the left and right states. Hence we have a left and a right total wave function
\begin{align}
    \Psi(\b{r}, \b{R}, t) &= \sum_n \chi_n(\b{R},t) \psi_n(\b{r}; \b{R}) \label{eq:state_right} \\
    \tilde{\Psi}(\b{r}, \b{R}, t) &= \sum_n \tilde{\chi}_n(\b{R},t) \tilde{\psi}_n(\b{r}; \b{R}), \label{eq:state_left}
\end{align}
with associated left and right nuclear wave functions $\tilde{\chi}_n$ and $\chi_n$, and
\begin{align}
    1 = \Dbraket{\tilde{\Psi}}{\Psi} = \sum_{mn} \Tbraket{\tilde{\chi}_m}{\Dbraket{\tilde{\psi_m}}{\psi_n}}{\chi_n} = \sum_{mn} \delta_{mn} \Dbraket{\tilde{\chi}_m}{\chi_n} = \sum_n \Dbraket{\tilde{\chi}_n}{\chi_n},
\end{align}
where we have assumed biorthonormal electronic states in the third equality.
Expectation values are defined through the standard expression\citep{Koch1990,Stanton1993}
\begin{align}
    \braket{\Omega} = \Tbraket{\tilde{\Psi}}{\Omega}{\Psi}, \quad \Omega = \Omega^\dagger. \label{eq:expectation_value}
\end{align}

To derive the equations for the nuclear wave functions, one normally projects the total Schrödinger equation on the electronic basis. In this respect, a biorthonormal description is advantageous; for practical coupled cluster models, where the excitation space is truncated to some excitation order, projection of the \emph{right} Schrödinger equation is done onto the \emph{left} electronic basis, leading to computationally tractable expressions that scale as expected for the given model (e.g.~$\mathcal{O}(N^6)$ for CCSD).

By inserting the $\Psi$ in Eq.~\eqref{eq:state_right} into the time dependent Schrödinger equation,
\begin{align}
    H \Psi = i \frac{\mathrm{d} \Psi}{\mathrm{d} t},
\end{align}
and projecting it onto the left electronic basis, we get a coupled set of equations for the right nuclear wave functions $\chi_n$.
These nuclear Schrödinger equations can be expressed as
\begin{align}
    (i \frac{\mathrm{d}}{\mathrm{d} t} - E_m) \chi_m = \sum_{I,n} \frac{1}{2 M_I} (\delta_{mn}\nabla_I^2 + G_{mn}^I + 2 \b{F}_{mn}^I \cdot \nabla_I) \chi_{n}, \label{eq:nuclear_SE_right}
\end{align}
where we have suppressed the $\b{R}$ and $t$ dependence for readability. The nonadiabatic coupling vectors in Eq.~\eqref{eq:nuclear_SE_right} are given in the biorthonormal basis:
\begin{align}
    G_{mn}^I &= \Dbraket{\tilde{\psi}_m}{\nabla_I^2 \psi_n} \label{eq:scalar_coupling} \\
    \b{F}_{mn}^I &= \Dbraket{\tilde{\psi}_m}{\nabla_I \psi_n}. \label{eq:vector_coupling}
\end{align}
These are called the scalar and vector couplings, respectively.

In analogous fashion, we derive the nuclear Schrödinger equations for the left nuclear wave functions from the complex conjugated Schrödinger equation
\begin{align}
    H \tilde{\Psi} = -i \frac{\mathrm{d} \tilde{\Psi}}{\mathrm{d} t}. \label{eq:conjugated_TDSE}
\end{align}
Inserting Eq.~\eqref{eq:state_left} into Eq.~\eqref{eq:conjugated_TDSE}, and projecting onto the left electronic basis, leads to
\begin{align}
    (-i \frac{\mathrm{d}}{\mathrm{d} t} - E_m) \tilde{\chi}_m = \sum_{I,n} \frac{1}{2 M_I} (\delta_{nm}\nabla_I^2 + \tilde{G}_{nm}^I + 2 \tilde{\b{F}}_{nm}^I \cdot \nabla_I) \tilde{\chi}_{n}, \label{eq:nuclear_SE_left}
\end{align}
where
\begin{align}
    \tilde{G}_{nm}^I &= \Dbraket{\nabla_I^2 \tilde{\psi}_n}{\psi_m} \\
    \tilde{\b{F}}_{nm}^I &= \Dbraket{\nabla_I \tilde{\psi}_n}{\psi_m}.
\end{align}

The nuclear Schrödinger equations may be expressed in the more compact matrix notation
\begin{align}
    (i \frac{\mathrm{d} }{\mathrm{d} t} - \b{E}) \b{\chi} &= \sum_I \frac{1}{2 M_I} (\b{I}\nabla_I^2 + \b{G}_I + 2 \b{F}_I \cdot \nabla_I)\b{\chi} \label{eq:right_nuclear_SE_matrix} \\
    (-i \frac{\mathrm{d} }{\mathrm{d} t} - \b{E}) \tilde{\b{\chi}}  &= \sum_I \frac{1}{2 M_I} (\b{I}\nabla_I^2 + \tilde{\b{G}}_I + 2 \tilde{\b{F}}_I \cdot \nabla_I)\tilde{\b{\chi}},
\end{align}
where $\b{E}$ is a diagonal matrix with the electronic energies on the diagonal, $\b{I}$ is the identity matrix, $\b{\chi}$ is a vector containing the right nuclear wave functions, and $\b{G}_I$ and $\b{F}_I$ are matrices consisting of the scalar and vector couplings of the $I$th nucleus, respectively. The quantities with a tilde are similarly defined.

This matrix notation has been used to illuminate some relations to gauge theories in the nuclear Schrödinger equations; Pacher \emph{et al}.\citep{Pacher1989} found that the vector coupling can be seen to serve a role analogous to the vector potential in electromagnetism. In the present work, it serves as a useful notation for dealing with basis transformations and the vector algebra needed to demonstrate invariance under such transformations.

\subsection{Basis invariance and the special case of norm invariance}
In the literature on nonadiabatic coupling vectors in coupled cluster theory, normalization is often considered problematic. The reason is that the left and right states are binormal in the coupled cluster formalism. Compared to the nonadiabatic couplings in full-CI theory, where the states are normalized, the full coupled cluster limit is ``incorrect'' because the value of the couplings depend on the geometry-dependent normalization constants. While this suggests that one should normalize the states, doing so is not straightforward. The computational cost of the normalization factor scales as full-CI for the right electronic states:\citep{Ichino2009}
\begin{align}
    N_R^n &= \Dbraket{\psi_n}{\psi_n} \\
    N_L^n &= \Dbraket{\tilde{\psi}_n}{\tilde{\psi}_n}.
\end{align}
Since one cannot evaluate $N_R^n$ in general, some have suggested $N_R^n = (N_L^n)^{-1}$ or $N_R^n = N_L^n = 1$ as alternatives. The former gives the full-CI limit while the latter simply assumes the standard binormality.\citep{Ichino2009,Faraji2018}

Binormality is not an issue from the point of view of dynamics. Changing the norm of the electronic states is a special case of a basis transformation of the electronic basis. As such, the Born-Huang expansion and the projection equations are equivalent in the transformed and untransformed bases. Changes in the electronic basis are absorbed in the expansion coefficients, i.e., the nuclear wave functions. In the special case of normalization, the right electronic wave functions are divided by $N_R^n$ while the right nuclear wave functions are multiplied by $N_R^n$. The total wave function is invariant under such transformations.

More precisely, consider invertible transformations of the left and right electronic bases. In vector notation, these transformations can be expressed as
\begin{align}
    \tilde{\b{\psi}}' &= \tilde{\b{\psi}} \b{N} \\
    \b{\psi}' &= \b{\psi} \b{M},
\end{align}
where the matrices $\b{M}$ and $\b{N}$ are assumed to be smooth invertible matrix functions of the nuclear coordinates. For notational simplicity, we have let the left and right wave function vectors be row vectors. Transformed quantities are denoted by a prime.
In the transformed basis, the total left and right wave function have the Born-Huang expansions
\begin{align}
        \Psi'(\b{r}, \b{R}, t) &= \sum_n \chi_n'(\b{R},t) \psi_n'(\b{r}; \b{R}) \label{eq:state_right_transformed} \\
        \tilde{\Psi}'(\b{r}, \b{R}, t) &= \sum_n \tilde{\chi}_n'(\b{R},t) \tilde{\psi}_n'(\b{r}; \b{R}). \label{eq:state_left_transformed}
\end{align}
We wish to show that the wave function in the transformed basis is identical to that obtained in the untransformed basis; that is, $\Psi' = \Psi$ and $\tilde{\Psi}' = \tilde{\Psi}$. The conclusion that follows is that the choice of electronic basis does not change the predictions of the theory. In other words, it is perfectly appropriate to use the biorthonormal description that is standard in coupled cluster theory.\citep{Koch1990}

Before proceeding, we define some notation. In the transformed basis,
we have to account for the non-unit overlap of the electronic wave functions. Hence, when projecting the time-dependent Schrödinger equation onto the electronic basis, we get electronic overlap matrix elements. We define these elements as
\begin{align}
    S_{mn} = \Dbraket{\tilde{\psi}_m'}{\psi_n'} = \sum_{kl} \Dbraket{\tilde{\psi}_k N_{km}}{\psi_l M_{ln}} = \sum_{kl} N_{km}^\ast \delta_{kl} M_{ln} = (\b{N}^\dagger \b{M})_{mn}.
\end{align}
Similarly, the electronic Hamiltonian matrix is not necessarily diagonal:
\begin{align}
    (\b{H}_e)_{mn} = \Dbraket{\tilde{\psi}_m'}{H_e \psi_n'} = \sum_{kl} \Dbraket{\tilde{\psi}_k N_{km}}{H_e \psi_l M_{ln}} = \sum_{kl} N_{km}^\ast E_{kl} M_{ln} = (\b{N}^\dagger \b{E} \b{M})_{mn}.
\end{align}

We show the equivalence for the right wave functions. The proof for the left wave function is identical. Following the standard procedure, we now insert the transformed wave function in Eq.~\eqref{eq:state_right_transformed} into the Schrödinger equation and project onto the transformed left electronic wave functions.
The result is the right nuclear Schrödinger equation
\begin{align}
    \bigl( \b{N}^\dagger \b{E} \b{M} - i \frac{\mathrm{d}}{\mathrm{d} t} \b{S}  \bigr)\b{\chi}' = \sum_I \frac{1}{2 M_I} (\b{S} \nabla_I^2  + \b{G}_I' + 2 \b{F}_I' \cdot \nabla_I) \b{\chi}'. \label{eq:right_nuclear_SE_transformed}
\end{align}
If the total wave function is invariant, and $\b{\psi}' = \b{\psi} \b{M}$, then we must have nuclear wave functions that cancel the transformation of the electronic wave functions:
\begin{align}
    \b{\chi}' = \b{M}^{-1} \b{\chi}. \label{eq:transformed_nuclear_wave_functions}
\end{align}
Indeed, with $\b{\chi}'$ as given in Eq.~\eqref{eq:transformed_nuclear_wave_functions}, we have
\begin{align}
    \Psi' = \sum_k \psi_k' \chi_k' = \sum_{klm} \psi_l M_{lk} M_{km}^{-1} \chi_m = \sum_l \psi_l \chi_l = \Psi.
\end{align}

Let us confirm that Eq.~\eqref{eq:transformed_nuclear_wave_functions} is in fact a solution to the transformed nuclear Schrödinger equation in Eq.~\eqref{eq:right_nuclear_SE_transformed}. We begin by relating the old and new nonadiabatic coupling terms. The gradient of the electronic wave functions transform as
\begin{align}
    \nabla_I \psi_l' = \sum_m \nabla_I (\psi_m M_{ml}) = \sum_m \Bigl( (\nabla_I \psi_m) M_{ml} + \psi_m (\nabla_I M_{ml}) \Bigr).
\end{align}
Hence, the vector couplings can be written as
\begin{align}
    (\b{F}_I')_{kl} = \Dbraket{\tilde{\psi}_k'}{\nabla_I \psi_l'} = \sum_n N_{nk}^\ast \Dbraket{\tilde{\psi}_n}{\nabla_I \psi_l'} = \sum_{nm} \Bigl( N_{nk}^\ast F_{nm}^I M_{ml} + N_{nk}^\ast \delta_{nm} \nabla_I M_{ml} \Bigr).
\end{align}
In more compact matrix notation, we have
\begin{align}
    \b{F}_I' = \b{N}^\dagger \b{F}_I \b{M} + \b{N}^\dagger (\nabla_I \b{M}).
\end{align}
Similarly, the Laplacian of the electronic wave functions transform as
\begin{align}
    \nabla_I^2 \psi_l' = \nabla_I \cdot \nabla_I \psi_l' = \sum_m \Bigl( (\nabla_I^2 \psi_m) M_{ml} + 2 (\nabla_I \psi_m)\cdot (\nabla_I M_{ml}) + \psi_m (\nabla_I^2 M_{ml}) \Bigr),
\end{align}
implying that the scalar couplings transform as
\begin{align}
    \b{G}_I' = \b{N}^\dagger \b{G}_I \b{M} + 2 \b{N}^\dagger \b{F}_I \cdot (\nabla_I \b{M}) + \b{N}^\dagger (\nabla_I^2 \b{M}).
\end{align}

The gradient and Laplacian of $\b{\chi}'$ is derived in the same way as for the electronic states, giving
\begin{align}
    \nabla_I \chi_l' &= \sum_m \bigl( M_{lm}^{-1} (\nabla_I \chi_m) +  (\nabla_I M_{lm}^{-1})\chi_m \bigr) \\
    \nabla_I^2 \chi_l' &= \sum_m \bigl( M_{lm}^{-1} (\nabla_I^2 \chi_m) + 2  (\nabla_I M_{lm}^{-1}) \cdot (\nabla_I \chi_m) + (\nabla_I^2 M_{lm}^{-1})\chi_m \bigr).
\end{align}
Thus, we have the following contributions on the right hand side of the nuclear Schrödinger equation:
\begin{align}
    \b{S} \nabla_I^2 \b{\chi}' &= \b{N}^\dagger (\nabla_I^2 \b{\chi} + 2 \b{M} (\nabla_I \b{M}^{-1}) \cdot \nabla_I \b{\chi} + \b{M} (\nabla_I^2 \b{M}^{-1}) \b{\chi} ) \\
    \b{G}_I' \b{\chi}' &= \b{N}^\dagger (\b{G}_I \b{\chi} + 2 \b{F}_I \cdot (\nabla_I \b{M})\b{M}^{-1}\b{\chi} + (\nabla_I^2 \b{M})\b{M}^{-1}\b{\chi}) \\
\begin{split}
    2 \b{F}_I'\cdot\nabla_I\b{\chi}' &= \b{N}^\dagger ( 2 \b{F}_I \cdot \nabla_I \b{\chi} + 2 \b{F}_I \cdot \b{M} (\nabla_I \b{M}^{-1}) \b{\chi} \\
    &\quad\quad 2 (\nabla_I \b{M})\b{M}^{-1} \cdot \nabla_I \b{\chi} + 2 (\nabla_I\b{M})\cdot (\nabla_I \b{M}^{-1})\b{\chi}).
\end{split}
\end{align}
Though somewhat involved, most of the terms cancel when added together. In fact, since
\begin{align}
    0 &= \nabla_I (\b{M} \b{M}^{-1}) = (\nabla_I \b{M}) \b{M}^{-1} + \b{M} (\nabla_I \b{M}^{-1}) \\
    0 &= \nabla_I^2 (\b{M} \b{M}^{-1}) = (\nabla_I^2 \b{M}) \b{M}^{-1} + \b{M} (\nabla_I^2 \b{M}^{-1}) + 2 (\nabla_I \b{M}) \cdot (\nabla_I \b{M}^{-1}),
\end{align}
we can write
\begin{align}
\begin{split}
    \b{S} \nabla_I^2 \b{\chi}' + \b{G}_I' \b{\chi}' + 2 \b{F}_I'\cdot\nabla_I\b{\chi}' &= \b{N}^\dagger \bigl(
                                     \nabla_I^2 \b{\chi} + \b{G}_I \b{\chi} + 2 \b{F}_I \cdot \nabla_I \b{\chi} \\
                                &+ 2 \nabla_I ( \b{M} \b{M}^{-1} ) \cdot \nabla_I \b{\chi} + \nabla_I^2 ( \b{M} \b{M}^{-1} )\b{\chi} \\
                                   &+  2 \b{F}_I \cdot \nabla_I ( \b{M} \b{M}^{-1} ) \b{\chi} \bigr) \\
                                   &= \b{N}^\dagger (
                                     \nabla_I^2 \b{\chi} + \b{G}_I \b{\chi} + 2 \b{F}_I \cdot \nabla_I \b{\chi}).
\end{split}
\end{align}
In other words, with $\b{\chi}' = \b{M}^{-1} \b{\chi}$, the right nuclear Schrödinger equation simplifies to
\begin{align}
    \bigl( \b{N}^\dagger \b{E} - i \frac{\mathrm{d}}{\mathrm{d} t} \b{N}^\dagger  \bigr)\b{\chi} = \b{N}^\dagger \sum_I \frac{1}{2 M_I} ( \nabla_I^2  + \b{G}_I + 2 \b{F}_I \cdot \nabla_I) \b{\chi}, \label{eq:right_nuclear_SE_transformed_2}
\end{align}
which, upon premultiplication by $\b{N}^{-\dagger}$, is seen to be equivalent to the original right nuclear Schrödinger equation in Eq.~\eqref{eq:right_nuclear_SE_matrix}.

Since all the derivation steps we have made are reversible, we have shown that $\b{\chi}$ is a solution to the untransformed nuclear Schrödinger equation if and only if $\b{\chi}'$ is a solution to the transformed Schrödinger equation. The total right wave function is therefore invariant with respect to transformations of the electronic basis, $\Psi' = \Psi$.

One consequence of basis invariance is that the nonadiabatic couplings can be derived in the standard biorthonormal formalism. To  derive expressions for these elements, we must first consider the geometry dependence of the many-body  operators.

\subsection{Geometry dependence of the many-body operators}
The scalar and vectors couplings, see Eqs.~\eqref{eq:scalar_coupling} and \eqref{eq:vector_coupling}, involve differentiation of the electronic wave functions with respect to the nuclear coordinates $\b{x}$. To evaluate these, we need to consider the dependence of both the wave function parameters and the many-body operators. The operator's dependence is handled through orbital connections which relates orbitals at neighbouring geometries. Note that there is no unique orbital connection; many-body operators are expressed with respect to a specific orthonormal orbital basis, but at each geometry there are an infinite number of such bases related by unitary transformations. For reasons that will become clear, we will use the so-called natural connection. Our presentation will follow closely that given by of Olsen \emph{et al}.\cite{olsen1995orbital}

When evaluating derivatives at $\b{x}_0$, we need to relate the basis at $\b{x}_0$ to some basis at $\b{x} = \b{x}_0 + \Delta \b{x}$. Suppose the molecular orbitals (MOs) at $\b{x}_0$ are
\begin{align}
    \phi_m(\b{x}_0) = \sum_\alpha  \chi_\alpha(\b{x}_0) C_{\alpha m}(\b{x}_0),
\end{align}
where $C_{\alpha m}$ are orbital coefficients and  $\chi_\alpha$ are atomic orbitals. The unmodified MOs (UMOs) are defined by freezing the orbital coefficients,
\begin{align}
    \phi_m^u(\b{x}) = \sum_\alpha C_{\alpha m}(\b{x}_0) \chi_\alpha(\b{x}).
\end{align}
The UMOs are not orthonormal, however:
\begin{align}
    S_{mn}(\b{x}) = \Dbraket{\phi_m^u(\b{x})}{\phi_n^u(\b{x})}, \quad S_{mn}(\b{x}_0) = \delta_{pq}.
\end{align}
Hence, UMOs are related to orthonormalized MOs (or OMOs) through
\begin{align}
    \phi_m(\b{x}) = \sum_n \phi_n^u(\b{x}) T_{nm}(\b{x}),
\end{align}
where the connection matrix $\b{T}(\b{x})$ satisfies $\b{T}(\b{x}_0) = \b{I}$ and
\begin{align}
    \b{T}(\b{x})^\dagger \b{S}(\b{x}) \b{T}(\b{x}) = \b{I}. \label{eq:orthogonality_connection_mat}
\end{align}
In the natural connection, $\b{T}$ is chosen to be
\begin{align}
    \b{T}(\b{x}) = \b{W}(\b{x})^{-1} (\b{W}(\b{x}) \b{S}(\b{x}) \b{W}(\b{x})^\dagger)^{1/2} = \b{W}(\b{x})^{-1} \b{\Delta}(\b{x}),
\end{align}
where
\begin{align}
    W_{mn}(\b{x}) = \Dbraket{\phi_m^u(\b{x}_0)}{\phi_n^u(\b{x})}.
\end{align}
The natural connection minimizes the change in the orthonormalized orbitals at $\b{x}$ relative to the orbitals at $\b{x}_0$.

Let us now relate the orbital space at $\b{x}$ to the orbital space at $\b{x}_0$. In order to do so, we need to consider a \emph{complete} orbital basis (denoted by indices $pq\ldots$), which we  partition into the OMO basis ($mn\ldots$) and the orthogonal complement orbitals, or OCOs ($uv\ldots$). For complete bases, we can write
\begin{align}
    \phi_p(\b{x}) = \sum_q \phi_q(\b{x}_0) U_{qp}(\b{x}), \quad U_{qp}(\b{x}) = \Dbraket{\phi_q(\b{x}_0)}{\phi_p(\b{x})}.
\end{align}
Occupation number states at $\b{x}$ can thus be expressed as
\begin{align}
    \ket{\Phi(\b{x})} = U(\b{x}) \ket{\Phi(\b{x}_0)}, \label{eq:unitary_ON_state}
\end{align}
with
\begin{align}
    U(\b{x}) = \exp(-b(\b{x})), \quad b(\b{x}) = \sum_{pq} b_{pq}(\b{x}) a_p^\dagger(\b{x}_0) a_q(\b{x}_0),
\end{align}
where $b(\b{x})$ is the anti-Hermitian operator with $b_{pq}(\b{x})$ defined such that $\b{U}(\b{x}) = \exp(-\b{b}(\b{x}))$. The many-body operators can be expanded as
\begin{align}
    a_p^\dagger(\b{x}) = \sum_q a_q^\dagger(\b{x}_0) U_{qp}(\b{x}).
\end{align}
To evaluate derivatives with respect to some specific $x$, we expand operators about $\b{x}_0$,
\begin{align}
    a_p^\dagger(\b{x}) &= a_p^\dagger + a_p^{(1)\dagger} \Delta x + \frac{1}{2} a_p^{(2)\dagger} (\Delta x)^2 + \ldots \\
    b(\b{x}) &= b^{(1)} \Delta x + \frac{1}{2} b^{(2)} (\Delta x)^2 + \ldots,
\end{align}
where
\begin{align}
    b^{(n)} = \sum_{pq} b^{(n)}_{pq} a_p^\dagger a_q.
\end{align}
Here we have let $a_p^\dagger \equiv a_p^\dagger(\b{x}_0)$ and suppressed the $\b{x}$-dependence of the derivatives. It will be useful to split operator contributions in the OMO ($\parallel$) and OCO blocks ($\perp$):
\begin{align}
    a_p^\dagger(\b{x}) = \sum_m a_m^\dagger(\b{x}_0) U_{mp}(\b{x}) + \sum_u a_u^\dagger(\b{x}_0) U_{up}(\b{x}) = a_{p\parallel}^\dagger(\b{x}) + a_{p\perp}^\dagger(\b{x}).
\end{align}

Let us evaluate
\begin{align}
    f_{IJ} &= \Dbraket{\Phi_I(\b{x}_0)}{\frac{\partial}{\partial x} \Phi_J(\b{x})}\Big\vert_0.
\end{align}
Using Eq.~\eqref{eq:unitary_ON_state}, we get
\begin{align}
    f_{IJ} &= \Tbraket{\Phi_I(\b{x}_0)}{\frac{\partial U}{\partial x}\Big\vert_0}{ \Phi_J(\b{x}_0)} = - \Tbraket{\Phi_I(\b{x}_0)}{b^{(1)}}{\Phi_J(\b{x}_0)}.
\end{align}
To simplify further, we note that $U_{mn} = \Delta_{mn}$ is Hermitian in the natural connection. Since the $U_{uv}$ block can similarly be chosen to be Hermitian, we have\cite{olsen1995orbital}
\begin{align}
b_{mn} = b_{uv} = 0 \implies b_{mn}^{(k)} = b_{uv}^{(k)} = 0 \label{eq:b_mn_uv}
\end{align}
and so
\begin{align}
    f_{IJ} &= - \sum_{mn} b_{mn}^{(1)} \Tbraket{\Phi_I(\b{x}_0)}{a_m^\dagger a_n}{\Phi_J(\b{x}_0)} = 0.
\end{align}
In general, $f_{IJ}$ is non-zero with connections other than the natural connection.

Next, we consider the second derivative
\begin{align}
    g_{IJ} &= \Dbraket{\Phi_I(\b{x}_0)}{\frac{\partial^2}{\partial x^2} \Phi_J(\b{x})}\Big\vert_0 = \Tbraket{\Phi_I(\b{x}_0)}{\frac{\partial^2 U}{\partial x^2}\Big\vert_0}{\Phi_J(\b{x}_0)},
\end{align}
which can be written
\begin{align}
\begin{split}
    g_{IJ} &= \Tbraket{\Phi_I(\b{x}_0)}{-b^{(2)} + b^{(1)}b^{(1)}}{\Phi_J(\b{x}_0)} = \Tbraket{\Phi_I(\b{x}_0)}{b^{(1)}b^{(1)}}{\Phi_J(\b{x}_0)}.
\end{split}
\end{align}
In the final equality, we have used Eq.~\eqref{eq:b_mn_uv}. Now, notice that since
\begin{align}
    b^{(1)} = \sum_{um} b_{um}^{(1)} a_u^\dagger a_m + \sum_{mu} b_{mu}^{(1)} a_m^\dagger a_u = - \sum_m a_{m\perp}^{(1)\dagger} a_m + \sum_m a_m^\dagger a_{m\perp}^{(1)},
\end{align}
the only non-zero $b^{(1)}b^{(1)}$ contribution is the one that first creates an electron in the complementary space and then destroys it. Thus,
\begin{align}
\begin{split}
    g_{IJ} &= -\sum_{mn} \Tbraket{\Phi_I(\b{x}_0)}{a_m^\dagger a_{m\perp}^{(1)} a_{n\perp}^{(1)\dagger} a_n}{\Phi_J(\b{x}_0)} \\
    &= -\sum_{mn} \Tbraket{\Phi_I(\b{x}_0)}{a_m^\dagger [a_{m\perp}^{(1)}, a_{n\perp}^{(1)\dagger}]_+ a_n}{\Phi_J(\b{x}_0)}.
\end{split}
\end{align}
The commutator can be expressed as
\begin{align}
    [a_{m\perp}^{(1)}, a_{n\perp}^{(1)\dagger}]_+ =  \sum_{uv} U^{(1)\ast}_{um} U^{(1)}_{vn} [a_u, a_v^\dagger]_+ = \sum_u U_{um}^{(1)\ast} U_{un}^{(1)} = \sum_u \Dbraket{\phi_m^{(1)}}{\phi_u(\b{x}_0)}\Dbraket{\phi_u(\b{x}_0)}{\phi_n^{(1)}}. \label{eq:commutator_w_inner_proj}
\end{align}
Moreover, since
\begin{align}
    \Dbraket{\phi_m(\b{x}_0)}{\phi_n^{(1)}} = U_{mn}^{(1)} = -b_{mn}^{(1)} = 0,
\end{align}
the inner projection in Eq.~\eqref{eq:commutator_w_inner_proj} is equivalent to the identity and so
\begin{align}
    [a_{m\perp}^{(1)}, a_{n\perp}^{(1)\dagger}]_+ = \Dbraket{\phi_m^{(1)}}{\phi_n^{(1)}}.
\end{align}
Hence, we get the final result
\begin{align}
    g_{IJ} = -\sum_{mn} \Tbraket{\Phi_I(\b{x}_0)}{a_m^\dagger a_n}{\Phi_J(\b{x}_0)} \Dbraket{\phi_m^{(1)}}{\phi_n^{(1)}}.
\end{align}
The formulas for $f_{IJ}$ and $g_{IJ}$ are valid for occupation number states but allow for generalization to general wave functions. We will be concerned with evaluating partial derivatives with respect to $x$ for wave functions of the form
\begin{align}
    \ket{\psi_k(\b{x})} = \sum_I c_{Ik}(\b{x}) \ket{\Phi_I(\b{x})}.
\end{align}
Since the $c_{Ik}$ depend implicitly on $\b{x}$, we have $\partial c_{Ik} / \partial x = 0$. Thus,
\begin{align}
\begin{split}
    f_{kl} &= \Tbraket{\psi_k(\b{x}_0)}{\frac{\partial}{\partial x}}{\psi_l(\b{x})} \Big\vert_0 \\
    &= \sum_{IJ} c_{Ik}^{\ast}(\b{x}_0)  \Tbraket{\Phi_I(\b{x}_0)}{\frac{\partial}{\partial x}}{\Phi_J(\b{x})} \Big\vert_0 c_{Jl}(\b{x}_0) \\ &= \sum_{IJ} c_{Ik}^{\ast}(\b{x}_0)  f_{IJ} c_{Jl}(\b{x}_0) \\
    &= 0
\end{split}
\end{align}
and
\begin{align}
\begin{split}
    g_{kl} &= \Tbraket{\psi_k(\b{x}_0)}{\frac{\partial^2}{\partial x^2}}{\psi_l(\b{x})} \Big\vert_0 \\
    &= \sum_{IJ} c_{Ik}^{\ast}(\b{x}_0)  g_{IJ} c_{Jl}(\b{x}_0) \\
    &= -\sum_{mn} \Tbraket{\psi_k(\b{x}_0)}{a_m^\dagger a_n}{\psi_l(\b{x}_0)} \Dbraket{\phi_m^{(1)}}{\phi_n^{(1)}}.
\end{split}
\end{align}

For partial derivatives of the energy, we also have to account for the explicit $\b{x}$-dependence of the Hamiltonian. We express the OMO Hamiltonian as
\begin{align}
    H = \sum_{pq} h_{pq}(\b{x}) E_{pq}(\b{x}) + \frac{1}{2} \sum_{pqrs} g_{pqrs}(\b{x}) e_{pqrs}(\b{x}),
\end{align}
where both the integrals and the operators depend on $\b{x}$. However, the dependence of the operators can be ignored in energy derivatives because matrix elements of occupation number states are constant:
\begin{align}
    \Dbraket{\Phi_I(\b{x})}{\Phi_J(\b{x})} = \Tbraket{\Phi_I(\b{x}_0)}{U(\b{x})^\dagger U(\b{x})}{\Phi_J(\b{x}_0)} = \Dbraket{\Phi_I(\b{x}_0)}{\Phi_J(\b{x}_0)}
\end{align}
In particular, elements involving $E_{pq}(\b{x})$ and $e_{pqrs}(\b{x})$ are linear combinations of such  overlaps and therefore give no contributions in energy derivatives\cite{Helgaker1992}. The integrals are related to the UMO basis as
\begin{align}
    h_{pq}(\b{x}) &= \sum_{mn} T_{mp}(\b{x})^\ast h_{mn}^{u}(\b{x}) T_{nq}(\b{x}) \\
    g_{pqrs}(\b{x}) &= \sum_{mnkl} T_{mp}(\b{x})^\ast T_{nq}(\b{x})^\ast g_{mnkl}^{u}(\b{x}) T_{kr}(\b{x}) T_{ls}(\b{x}).
\end{align}
By differentiating $\b{T} \b{W} = \b{W}^\dagger \b{T}^\dagger$ and $\b{I} = \b{T}^\dagger \b{S} \b{T}$, we find that
\begin{align}
    \b{T}^{(1)} = - \b{W}^{(1)}.
\end{align}
Consequently, the partial derivative of the Hamiltonian can be written
\begin{align}
    H^{(1)} = H^{(1)}_u - \{ W^{(1)}, H \},
\end{align}
where $H^{(1)}_u$ is the derivative of the UMO Hamiltonian and
\begin{align}
\{ W^{(1)}, H \} = \sum_{pq} j_{pq} E_{pq} + \frac{1}{2} \sum_{pqrs} j_{pqrs} e_{pqrs},
\end{align}
where
\begin{align}
    j_{pq} &= \sum_{m} (W^{(1)}_{pm} h_{mq} + h_{pm} W^{(1)}_{mq}) \\
    j_{pqrs} &= \sum_{m} (W^{(1)}_{pm} g_{mqrs} + W^{(1)}_{qm} g_{pmrs} + g_{pqms} W^{(1)}_{mr} + g_{pqrm} W^{(1)}_{ms}).
\end{align}
The $\b{W}^{(1)}$ matrix, given by
\begin{align}
    W^{(1)}_{pq} = \frac{\partial W_{pq}}{\partial x} \Big\vert_0 = \sum_{\alpha\beta} C_{\alpha p}(\b{x}_0) C_{\beta q}(\b{x}_0) \int \chi_{\alpha}(\b{x}_0) \frac{\partial \chi_\beta}{\partial x} \Big\vert_0 \mathrm{d} \b{r},
\end{align}
is analogous to $S^{(1)}_{pq} = \partial S_{pq} / \partial x \vert_0$ in the symmetric connection $\b{T} = \b{S}^{-1/2}$.

This concludes our discussion of how the geometry dependence of the many-body operators affects energy derivatives and nonadiabatic coupling elements. We refer the reader to Helgaker and Jørgensen\cite{Helgaker1992} for more details regarding connections and energy derivatives and to Olsen \emph{et al}.\cite{olsen1995orbital} for more on the natural connection. In the next section, we derive expressions for the nonadiabatic elements in coupled cluster theory.

\subsection{Nonadiabatic coupled cluster couplings in a Lagrangian formalism}
To obtain a Lagrangian for the vector coupling, Hohenstein\citep{Hohenstein2016} defined a quantity whose first derivatives are identical to components of the vector coupling. Although Hohenstein used it for  configuration interaction theory, the observation generalizes straightforwardly to coupled cluster theory. The quantity is the partially frozen overlap
\begin{align}
    \mathscr{O}_{mn}(\b{x}) = \Dbraket{\tilde{\psi}_m(\b{x}_0)}{\psi_n(\b{x})},
\end{align}
in terms of which we have
\begin{align}
    (\b{F}_{mn}^I)_i = \DBraket{\tilde{\psi}_m(\b{x})}{\frac{\mathrm{d}}{\mathrm{d} x_i} \psi_n(\b{x})}\Big\vert_0 = \frac{\mathrm{d}}{\mathrm{d} x_i} \mathscr{O}_{mn}(\b{x})\Big\vert_0, \quad i \in I, \label{eq:hohenstein_vector}
\end{align}
and
\begin{align}
    G_{mn}^I = \DBraket{\tilde{\psi}_m(\b{x})}{\sum_{i\in I} \frac{\mathrm{d}^2}{\mathrm{d} x_i^2} \psi_n(\b{x})}\Big\vert_0 = \sum_{i\in I} \frac{\mathrm{d}^2}{\mathrm{d} x_i^2} \mathscr{O}_{mn}(\b{x}) \Big\vert_0. \label{eq:hohenstein_scalar}
\end{align}
For convenience, we write $i \in I$ to signify that $x_i$ is one of the three coordinates at nucleus $I$ ($x, y,$ or $z$). Clearly, the vector and scalar couplings are derivatives of the partially frozen overlap and may therefore be evaluated using a Lagrangian. Note that the overlap $\mathscr{O}_{mn}(\b{x})$ depends on $\b{x}_0$. We suppress this dependency for notational simplicity.

The overlap is expressed in terms of coupled cluster wave functions, which depend on $\b{x}$ but also on a set of wave function parameters $\b{\lambda}$ (which themselves depend on $\b{x}$). Written out in terms of wave function parameters, the overlap is given by
\begin{align}
    \mathscr{O}_{mn}(\b{x}, \b{\lambda}) = \Tbraket{\tilde{\psi}_m(\b{x}_0)}{\exp(-\kappa)\exp(T)}{\st{R}_n},
\end{align}
where
\begin{align}
    \bra{\tilde{\psi}_m(\b{x}_0)} = \bra{\st{L}_m} \exp(-T) \Big\vert_0
\end{align}
and
\begin{align}
    \kappa = \sum_{p > q} \kappa_{pq} (E_{pq} - E_{qp}) = \sum_{p > q} \kappa_{pq} E_{pq}^-.
\end{align}
The $\kappa$ operator accounts for orbital rotations, meaning changes in the Hartree-Fock orbitals, where, by definition, we have $\kappa(\b{x}_0) = 0$. Following the standard recipe, we add the equations (denoted by $\mathscr{E}_{mn}$) that determine the parameters as constraints with associated Lagrangian multipliers (denoted by $\b{\gamma}$),
\begin{align}
    \mathscr{L}_{mn}(\b{\lambda}, \b{x}, \b{\gamma}) = \mathscr{O}_{mn}(\b{\lambda}, \b{x}) + \b{\gamma}^T \mathscr{E}_{mn}(\b{\lambda}, \b{x}), \quad m \neq n,
\end{align}
where $\b{\lambda}$ and $\b{\gamma}$ are determined for every $\b{x}$ by stationarity:
\begin{align}
    \frac{\partial \mathscr{L}_{mn}}{ \partial \gamma_k}  &= (\mathscr{E}_{mn})_k = 0 \label{eq:stationarity_multipliers} \\
    \frac{\partial \mathscr{L}_{mn}}{ \partial \lambda_k} &= 0. \label{eq:stationarity_parameters}
\end{align}
The derivatives of this Lagrangian are identical to the derivatives of the frozen overlap (since $\mathscr{E}_{mn} = 0$). One advantage of the Lagrangian formalism is that it automatically incorporates the $2n + 1$ and $2n + 2$ rules for $\b{\lambda}$ and $\b{\gamma}$, respectively.
In particular,
\begin{align}
    (\b{F}_{mn}^I)_i = \frac{\mathrm{d} \mathscr{L}_{mn}}{\mathrm{d} x_i}\Big\vert_0 = \frac{\partial \mathscr{L}_{mn}}{\partial x_i}\Big\vert_0, \quad i \in I,
\end{align}
where the final equality follows from stationarity, see Eqs.~\eqref{eq:stationarity_multipliers} and \eqref{eq:stationarity_parameters}. Denoting partial derivatives with respect to geometrical coordinates as
\begin{align}
    a^{(i)} = \frac{\partial a}{\partial x_i} \Big\vert_0, \quad a^{(i,j)} = \frac{\partial^2 a}{\partial x_i \partial x_j} \Big\vert_0,
\end{align}
we can write
\begin{align}
    (\b{F}_{mn}^I)_i = \mathscr{L}_{mn}^{(i)}, \quad i \in I.
\end{align}
Furthermore, if we let
\begin{align}
    f_\alpha = \frac{\partial \mathscr{L}_{mn}}{\partial \lambda_\alpha} \Big\vert_{\b{\lambda} = \b{\lambda}_0}, \quad H_{\alpha\beta} =  \frac{\partial^2 \mathscr{L}_{mn}}{\partial \lambda_\alpha \partial \lambda_\beta} \Big\vert_0, \label{eq:gradient_hessian_parameter}
\end{align}
then the scalar coupling can be expressed as (see Appendix A)
\begin{align}
    G_{mn}^I = \sum_{i\in I} \frac{\mathrm{d}^2 \mathscr{L}_{mn}}{\mathrm{d} x_i^2}\Big\vert_0 = \sum_{i \in I} \Bigl( \mathscr{L}_{mn}^{(i,i)} + \sum_{\alpha\beta} \lambda_\alpha^{(i)} H_{\alpha\beta} \lambda_\beta^{(i)} + 2 \sum_\alpha \lambda_\alpha^{(i)} f_\alpha^{(i)} \Bigr). \label{eq:scalar_coupling_Lagrangian}
\end{align}
Clearly, $\b{F}_{mn}^I$ and $G_{mn}^I$ are similar in complexity to the energy gradient and Hessian. However, $G_{mn}^I$ is somewhat simpler than the energy Hessian because the first derivatives of the parameters ($\b{\lambda}^{(i)}$) can be considered one at a time.

To proceed, we must define the Lagrangian $\mathscr{L}_{mn}$ in detail. The conditions $\mathscr{E}_{mn}$ include all equations that must be solved to evaluate the overlap $\mathscr{O}_{mn}$. These are (a) the Hartree-Fock equations, (b) the amplitude equations, and (c) the eigenvalue equations for the right state amplitudes. Written out in full, we have
\begin{align}
\begin{split}
    \mathscr{L}_{mn} &= \mathscr{O}_{mn} + \b{\gamma}^T \mathscr{E}_{mn} \\
    &= \mathscr{O}_{mn} + \bar{\b{\kappa}}^T \b{F}_c + \b{\zeta}^T \b{\Omega} +\b{\beta}_n^T(\bar{\bst{H}} - E_n )\bst{R}_n + \bar{E}_n (1 - \Dbraket{\st{L}_n^{0}}{\st{R}_n}), \label{eq:Lagrangian_conditions}
\end{split}
\end{align}
where we have introduced multipliers associated with the different sets of equations, $\bar{\b{\kappa}}$, $\b{\zeta}$, as well as $\b{\beta}_n$ and $\bar{E}_n$. We have also introduced the Brillouin condition
\begin{align}
    (\b{F}_c)_{pq} = \Tbraket{\mathrm{HF}}{[E_{pq}^-, H(\kappa)]}{\mathrm{HF}}, \quad p > q,
\end{align}
where
\begin{align}
    H(\kappa) = \exp(\kappa) H \exp(-\kappa).
\end{align}
Furthermore, the similarity transformed Hamiltonian in $\b{\Omega}$ and $\bar{\bst{H}}$ is given by
\begin{align}
    \bar{H} = \bar{H}(\kappa) = \exp(-T) H(\kappa) \exp(T)
\end{align}
and the $n$th electronic energy defined as
\begin{align}
    E_n = \Tbraket{\st{L}_n^{(0)}}{\bar{H}}{\st{R}_n}.
\end{align}

With $\mathscr{L}_{mn}$ defined, we can now consider the equations for the zeroth order multipliers. These are determined from the zeroth order terms of the $\b{\lambda}$ stationarity, Eq.~\eqref{eq:stationarity_parameters}. To keep our notation simple, we will denote the zeroth order terms as $\b{\gamma}^{(0)} \equiv \b{\gamma}$ and $\b{\lambda}^{(0)} \equiv \b{\lambda}$, where it should be understood from context when these are $\b{\gamma}$ and $\b{\lambda}$ evaluated at $\b{x}_0$. Differentiation with respect to the state parameters gives
\begin{align}
    \frac{\partial \mathscr{L}_{mn}}{\partial \bst{R}_n}\Big\vert_0 &= 0 = \bst{L}_m^T + \b{\beta}_n^T (\bar{\bst{H}} - E_n \b{I}) - \bigl(\bar{E}_n + E_n \, \b{\beta}_n^T \b{R}_n \bigr)\bst{L}_n^T = 0. \label{eq:stationarity_right_amplitudes}
\end{align}
To solve this equation, we note that if we let
\begin{align}
    \bar{E}_n = - E_n \, \b{\beta}_n^T \b{R}_n,
\end{align}
the equation for $\b{\beta}_n$ becomes
\begin{align}
    \b{\beta}_n^T (\bar{\bst{H}} - E_n \b{I}) &= -\bst{L}_m^T.
\end{align}
Thus, we have
\begin{align}
    \b{\beta}_n &= (E_n - E_m)^{-1} \bst{L}_m.
\end{align}

Next we consider stationarity with respect to $\b{t}$. This can be expressed as
\begin{align}
    \frac{\partial \mathscr{L}_{mn}}{\partial \b{t}}\Big\vert_0 = 0 = {^t}\b{\eta} + \b{\zeta}^T \b{A}, \label{eq:stationarity_cluster}
\end{align}
where
\begin{align}
    {^{t}}\eta_\mu = \Tbraket{\bar{\st{L}}_m}{\tau_\mu}{\bar{\st{R}}_n} +
    \Tbraket{\bar{\beta}_n}{[H,\tau_\mu]}{\bar{\st{R}}_n}
\end{align}
and where we have introduced the notation
\begin{align}
    \ket{\bar{X}} &= \exp(T)\ket{X} \\
    \bra{\bar{X}} &= \bra{X} \exp(-T).
\end{align}

Finally, we have stationarity with respect to $\b{\kappa}$, which can be written
\begin{align}
    \frac{\partial \mathscr{L}_{mn}}{\partial \b{\kappa}}\Big\vert_0 = 0 = {^\kappa}\b{\eta} +  \bar{\b{\kappa}}^T \b{A}^\mathrm{HF},\label{eq:stationarity_kappa}
\end{align}
where
\begin{align}
\begin{split}
    {^{\kappa}}\eta_{rs} = &-\Tbraket{\bar{\st{L}}_m}{ E_{rs}^- }{\bar{\st{R}}_n} + \Tbraket{\bar{\zeta}}{[E_{rs}^-,H]}{\mathrm{CC}} + \Tbraket{\bar{\beta}_n}{ [E_{rs}^-, H] }{\bar{\st{R}}_n}
\end{split}
\end{align}
and
\begin{align}
    A_{pq,rs}^\mathrm{HF} &= \Tbraket{\mathrm{HF}}{[E_{pq}^-,[E_{rs}^-,H]]}{\mathrm{HF}}.
\end{align}

With the zeroth order multipliers determined, we can derive the expression for the vector coupling. By partially differentiating $\mathscr{L}_{mn}$, we find that
\begin{align}
\begin{split}
    (\b{F}_{mn}^I)_i = \mathscr{L}_{mn}^{(i)} &= (E_n - E_m)^{-1} \Tbraket{\st{L}_m}{\bar{H}^{(i)}}{\st{R}_n} \\
    &+ \Tbraket{\zeta}{\bar{H}^{(i)}}{\mathrm{HF}} + \Tbraket{\mathrm{HF}}{[\bar{\kappa},H^{(i)}]}{\mathrm{HF}}, \quad i \in I,
\end{split} \label{eq:derivative_coupling_cc}
\end{align}
where
\begin{align}
    \bar{H}^{(i)} = \exp(-T) H^{(i)} \exp(T)
\end{align}
and where quantities at $\b{x}_0$ are denoted as $y^{(0)} \equiv y$ (e.g., we denote $T^{(0)}$ as $T$).

The  vector coupling given in Eq.~\eqref{eq:derivative_coupling_cc} has also been identified by other authors. It was derived by Christiansen,\cite{Christiansen1999nonadiabatic} who assumed biorthonormality and used $Z$-vector substitution\cite{handy1984evaluation} on the expression for the vector coupling. Tajti and Szalay\cite{Tajti2009} identified the same expression indirectly via $Z$-vector substitution on derivatives of Hamiltonian transition elements. However, they also argued\cite{Tajti2009} that the coupling should not be given by Eq.~\eqref{eq:derivative_coupling_cc} but rather be averaged and expressed with normalized states. As we have shown, Eq.~\eqref{eq:derivative_coupling_cc} is a valid choice due to norm invariance and represents the vector coupling in the right nuclear Schrödinger equations.
For the left Schrödinger equations, we can make use of the identity
\begin{align}
    \Dbraket{\tilde{\psi}_m}{\psi_n} = \delta_{mn} \implies \b{F}_{mn}^I = - \tilde{\b{F}}_{mn}^I. \label{eq:vector_to_vector_tilde}
\end{align}
Before moving on to the scalar coupling, we note that although the
$Z$-vector substitution method is equivalent to the Lagrangian technique, the latter method gives, in our opinion, an especially elegant way of deriving the  coupling elements.

For the scalar coupling, we must determine the first derivatives of the parameters. Equations for these are obtained as the first order terms of the multiplier stationarity conditions. In the case of $\b{t}$, we have
\begin{align}
    \frac{\mathrm{d} \b{\Omega}}{\mathrm{d} x_i} \Big\vert_0 = 0 = {^t}\b{\xi}^{(i)} + \b{A} \b{t}^{(i)},
\end{align}
where
\begin{align}
    {^t}\xi_\mu^{(i)} = \Tbraket{\mu}{\bar{H}^{(i)}}{\mathrm{HF}} + \Tbraket{\mu}{\exp(-T)[\kappa^{(i)},H]\exp(T)}{\mathrm{HF}}.
\end{align}
In the case of $\b{\kappa}$, we similarly have
\begin{align}
    \frac{\mathrm{d} \b{F}_c}{\mathrm{d} x_i}\Big\vert_0 = 0 = {^\kappa}\b{\xi}^{(i)} + \b{A}^\mathrm{HF} \b{\kappa}^{(i)},
\end{align}
where
\begin{align}
    {^\kappa}\xi^{(i)}_{rs} = \Tbraket{\mathrm{HF}}{[E_{rs}^-, H^{(i)}]}{\mathrm{HF}}.
\end{align}
The binormality condition implies
\begin{align}
    \frac{\mathrm{d}}{\mathrm{d} x_i} ( 1 - \Dbraket{\st{L}_n^{0}}{\st{R}_n} ) \Big\vert_0 = 0 = - \bst{L}_n^T \bst{R}_n^{(i)}, \label{eq:binormality_1st_response}
\end{align}
while the eigenvalue condition implies
\begin{align}
    \frac{\mathrm{d}}{\mathrm{d} x_i} \Bigl( (\bar{\bst{H}} - E_n )\bst{R}_n \Bigr) \Big\vert_0 &= 0 = {^{\st{R}_n}}\b{\xi}^{(i)} + (\bar{\bst{H}} - E_n)\bst{R}_n^{(i)} \label{eq:response_R}
\end{align}
Here we have defined
\begin{align}
    {^{\st{R}_n}}\b{\xi}^{(i)} &= \Bigl( \bst{Y}^{(i)} - \bst{L}_n^T \bst{Y}^{(i)} \bst{R}_n \Bigr) \bst{R}_n
\end{align}
where
\begin{align}
    \st{Y}^{(i)}_{\mu\nu} = \Tbraket{\mu}{\bar{H}^{(i)}}{\nu} + \Tbraket{\mu}{[\bar{H},T^{(i)}]}{\nu} + \Tbraket{\mu}{[\kappa^{(i)},\bar{H}]}{\nu}, \quad \mu,\nu \geq 0.
\end{align}

With the derivatives of the parameters determined, let us next consider $f_\alpha^{(i)}$ and $H_{\alpha\beta}$, see Eq.~\eqref{eq:gradient_hessian_parameter}. Recall that the $\alpha$ and $\beta$ indices refer to the parameters $\lambda_{\alpha}$ and $\lambda_{\beta}$. The gradient $\b{f}$ is given by the zeroth order equations for the multipliers, that is, Eqs.~\eqref{eq:stationarity_right_amplitudes}, \eqref{eq:stationarity_cluster}, and \eqref{eq:stationarity_kappa}, with $\b{\lambda} = \b{\lambda}_0$ but allowing for $\b{x} \neq \b{x}_0$.
Partially differentiating these terms with respect to $x_i$ gives $\b{f}
^{(i)}$. The blocks of the $\sum_{\alpha} f_\alpha^{(i)} \lambda_\alpha^{(i)}$ contributions to $G_{mn}^I$ may be written
\begin{align}
    {^{\st{R}_n}}f^{(i)}_\mu (\bst{R}_n^{(i)})_\mu = \frac{\partial^2 \mathscr{L}_{mn}}{\partial x_i \partial \st{R}_\mu^n} \Big\vert_0  (\bst{R}_n^{(i)})_\mu &=
    \Tbraket{\bar{\beta}_n}{H^{(i)} - E_n^{(i)}}{\bar{\st{R}}_n^{(i)}}
\end{align}
and
\begin{align}
    {^t}f_\mu^{(i)} t_\mu^{(i)} = \frac{\partial^2 \mathscr{L}_{mn}}{\partial x_i \partial t_\mu} \Big\vert_0  t_\mu^{(i)} &= \Tbraket{\bar{\zeta}}{[H^{(i)}, T^{(i)}]}{\mathrm{CC}} + \Tbraket{\bar{\beta}_n}{[H^{(i)}, T^{(i)}]}{\bar{\st{R}}_n}
    \\
\begin{split}
    {^\kappa}f_{pq}^{(i)} \kappa_{pq}^{(i)} = \frac{\partial^2 \mathscr{L}_{mn}}{\partial x_i \partial \kappa_{pq}} \Big\vert_0  \kappa_{pq}^{(i)} &= \Tbraket{\bar{\zeta}}{[\kappa^{(i)},H^{(i)}]}{\mathrm{CC}} +
    \Tbraket{\bar{\beta}_n}{[\kappa^{(i)},H^{(i)}]}{\bar{\st{R}}_n} \\
    &+ \Tbraket{\mathrm{HF}}{[\bar{\kappa},[\kappa^{(i)}, H^{(i)}]]}{\mathrm{HF}}
\end{split}
\end{align}
where repeated indices implies summation. For contributions to $G_{mn}^I$ involving the parameter Hessian $H_{\alpha\beta} = \partial^2 \mathscr{L}_{mn} / \partial \lambda_\alpha \partial \lambda_\beta \vert_0$, we have, for terms involving $\b{t}$ and $\b{\kappa}$,
\begin{align}
\begin{split}
     t_\mu^{(i)} H_{\mu\nu}^{tt} t_\nu^{(i)} &= t_\mu^{(i)} \frac{\partial^2 \mathscr{L}_{mn}}{\partial t_\mu \partial t_\nu} \Big\vert_0 t_\nu^{(i)} \\
     &= \Tbraket{\bar{\st{L}}_m}{(T^{(i)})^2}{\bar{\st{R}}_n} \\
     &+ \Tbraket{\bar{\zeta}}{[[H, T^{(i)}],T^{(i)}]}{\mathrm{CC}} \\
    &+
    \Tbraket{\bar{\beta}_n}{[[H, T^{(i)}], T^{(i)}]}{\bar{\st{R}}_n}
\end{split}
\end{align}
as well as
\begin{align}
\begin{split}
    \kappa_{rs}^{(i)} H_{rs\nu}^{\kappa t} t_\nu^{(i)} &= \kappa_{rs}^{(i)}\frac{\partial^2 \mathscr{L}_{mn}}{\partial \kappa_{rs} \partial t_\nu} \Big\vert_0 t_\nu^{(i)} \\
    &= -\Tbraket{\bar{\st{L}}_m}{\kappa^{(i)} T^{(i)}}{\bar{\st{R}}_n} \\
    &+ \Tbraket{\bar{\zeta}}{[[\kappa^{(i)},H],T^{(i)}]}{\mathrm{CC}} \\
    &+  \Tbraket{\bar{\beta}_n}{[[\kappa^{(i)},H],T^{(i)}]}{\bar{\st{R}}_n}
\end{split}
\end{align}
and
\begin{align}
\begin{split}
    \kappa_{pq}^{(i)} H_{pqrs}^{\kappa \kappa} \kappa_{rs}^{(i)} &= \kappa_{pq}^{(i)}\frac{\partial^2 \mathscr{L}_{mn}}{\partial \kappa_{pq} \partial \kappa_{rs}} \Big\vert_0 \kappa_{rs}^{(i)} \\
    &= \Tbraket{\bar{\st{L}}_m}{(\kappa^{(i)})^2}{\bar{\st{R}}_n} \\
    &+ \Tbraket{\bar{\zeta}}{[\kappa^{(i)},[\kappa^{(i)},H]]}{\mathrm{CC}} \\
    &+ \Tbraket{\bar{\beta}_n}{[\kappa^{(i)},[\kappa^{(i)},H]]}{\bar{\st{R}}_n} \\
    &+ \Tbraket{\mathrm{HF}}{[\bar{\kappa},[\kappa^{(i)},[\kappa^{(i)}, H]]]}{\mathrm{HF}}.
\end{split}
\end{align}

Next we have terms involving right state and the cluster amplitudes and orbital rotations:
\begin{align}
    \begin{split}
        t_\mu^{(i)} H_{\mu \nu}^{t \st{R}_n}  (\bst{R}_n^{(i)})_\nu &= t_{\mu}^{(i)} \frac{\partial^2 \mathscr{L}_{mn}}{\partial t_{\mu} \partial (\bst{R}_n)_\nu} \Big\vert_0 (\bst{R}_n^{(i)})_\nu \\
        &=  \Tbraket{\bar{\st{L}}_m}{T^{(i)}}{\bar{\st{R}}_n^{(i)}} \\ &+\Tbraket{\bar{\beta}_n}{[H, T^{(i)}]}{\bar{\st{R}}_n^{(i)}}  \\
        &- \Dbraket{\beta_n}{\st{R}_n^{(i)}}\Tbraket{\bar{\st{L}}_n}{[H, T^{(i)}]}{\bar{\st{R}}_n}
    \end{split} \\
    \begin{split}
        \kappa_{pq}^{(i)} H_{pq \nu}^{\kappa \st{R}_n}  (\bst{R}_n^{(i)})_\nu &= \kappa_{pq}^{(i)} \frac{\partial^2 \mathscr{L}_{mn}}{\partial \kappa_{pq} \partial (\bst{R}_n)_\nu} \Big\vert_0 (\bst{R}_n^{(i)})_\nu \\
        &= -  \Tbraket{\bar{\st{L}}_m}{\kappa^{(i)}}{\bar{\st{R}}_n^{(i)}} \\ &+ \Tbraket{\bar{\beta}_n}{[\kappa^{(i)}, H]}{\bar{\st{R}}_n^{(i)}}  \\
        &- \Dbraket{\beta_n}{\st{R}_n^{(i)}}\Tbraket{\bar{\st{L}}_n}{[\kappa^{(i)},H]}{\bar{\st{R}}_n}.
    \end{split}
\end{align}
Finally, we have the partial derivative of the Lagrangian, which can be written
\begin{align}
\begin{split}
    \mathscr{L}^{(i,i)}_{mn} &= - \sum_{pq} \Tbraket{\bar{\st{L}}_m}{E_{pq}}{\bar{\st{R}}_n} \Dbraket{\phi_p^{(i)}}{\phi_q^{(i)}} +  \Tbraket{\bar{\zeta}}{H^{(i,i)}}{\mathrm{CC}} \\
    &+ \Tbraket{\mathrm{HF}}{[\bar{\kappa}, H^{(i,i)}]}{\mathrm{HF}} +  \Tbraket{\bar{\beta}_n}{H^{(i,i)}}{\bar{\st{R}}_n},
\end{split}
\end{align}

Written in compact notation, the scalar coupling may be evaluated as
\begin{align}
    \begin{split}
        G_{mn}^I &= \Dbraket{\bar{\st{L}}_m}{\bar{R}_n^{(i,i)}} +\Tbraket{\bar{\zeta}}{K^{(i,i)}}{\mathrm{CC}}
        + \Tbraket{\bar{\beta}_n}{K^{(i,i)}}{\bar{\st{R}}_n} \\
        &+ \Tbraket{\mathrm{HF}}{[\bar{\kappa}, J^{(i,i)}]}{\mathrm{HF}}
+ \Tbraket{\bar{\beta}_n}{L^{(i)} - \braket{L^{(i)}}_n}{\bar{\st{R}}_n^{(i)}}
    \end{split} \label{eq:scalar_coupling}
\end{align}
where we have let
\begin{align}
    K^{(i,i)} &= J^{(i,i)} + 2 [J^{(i)},T^{(i)}] + [[H, T^{(i)}],T^{(i)}] \\
    J^{(i)} &= H^{(i)} + [\kappa^{(i)}, H] \\
    L^{(i)} &= 2 (H^{(i)} + [\kappa^{(i)}, H] + [H, T^{(i)}]) \\
    \braket{L^{(i)}}_n &= \Tbraket{\bar{\st{L}}_n}{L^{(i)}}{\bar{\st{R}}_n} \\
    J^{(i,i)} &= H^{(i,i)} + 2 [\kappa^{(i)}, H^{(i)}] + [\kappa^{(i)},[\kappa^{(i)}, H]]
\end{align}
as well as
\begin{align}
\begin{split}
    \Dbraket{\bar{\st{L}}_m}{\bar{R}_n^{(i,i)}} &= \Tbraket{\bar{\st{L}}_m}{(\kappa^{(i)})^2 - 2\kappa^{(i)} T^{(i)} + (T^{(i)})^2}{\bar{\st{R}_n}} -  2\Tbraket{\bar{\st{L}}_m}{\kappa^{(i)}}{\bar{\st{R}}_n^{(i)}} \\
    &+ 2\Tbraket{\bar{\st{L}}_m}{T^{(i)}}{\bar{\st{R}}_n^{(i)}} - \sum_{pq} \Tbraket{\bar{\st{L}}_m}{E_{pq}}{\bar{\st{R}}_n} \Dbraket{\phi_p^{(i)}}{\phi_q^{(i)}}.
\end{split} \label{eq:derivative_terms}
\end{align}

Throughout the derivations above, we have considered the off-diagonal coupling elements ($m \neq n$). The diagonal terms can be derived from the Lagrangian
\begin{align}
\begin{split}
    \mathscr{L}_{nn} &= \mathscr{O}_{nn} + \b{\gamma}^T \mathscr{E}_{nn} \\
    &= \mathscr{O}_{nn} + \bar{\b{\kappa}}^T \b{F}_c + \b{\zeta}^T \b{\Omega} +\b{\beta}_n^T(\bar{\bst{H}} - E_n )\bst{R}_n + \bar{E}_n (1 - \Dbraket{\st{L}_n^{0}}{\st{R}_n}), \label{eq:Lagrangian_conditions_diagonal}
\end{split}
\end{align}
which gives the slightly different $\st{R}_n$ stationarity condition
\begin{align}
    0 = (1 + \bar{E}_n - E_n \b{\beta}_n^T \bst{R}_n) \bst{L}_n^T + \b{\beta}_n^T(\bar{\bst{H}} - E_n ).
\end{align}
Here we again select $\bar{E}_n$ to make the first term vanish, giving
\begin{align}
    \b{\beta}_n^T = \bst{L}_n^T.
\end{align}
Other than this change, the derivation of the scalar coupling is virtually unchanged. Terms involving differentiation of $\mathscr{O}_{nn}$ has the left state $\bra{\st{L}_n}$ in the bra instead of $\bra{\st{L}_m}$ (e.g., in the stationarity conditions for the zeroth order multipliers). In particular, the expression in Eq.~\eqref{eq:scalar_coupling} is valid with $m = n$.

Unlike for the vector coupling, there is no convenient relationship between $G_{mn}^I$ and $\tilde{G}_{mn}^I$. To derive the latter quantity, we may consider the Lagrangian
\begin{align}
\begin{split}
    \mathscr{L}_{mn} &= \mathscr{O}_{mn} + \b{\gamma}^T \mathscr{E}_{mn} \\
    &= \mathscr{O}_{mn} + \bar{\b{\kappa}}^T \b{F}_c + \b{\zeta}^T \b{\Omega} +\bst{L}_m^T(\bar{\bst{H}} - E_m )\b{\beta}_m + \bar{E}_m (1 - \Dbraket{\st{L}_m}{\st{R}_m^{0}}), \label{eq:Lagrangian_conditions_tilde}
\end{split}
\end{align}
where
\begin{align}
    \mathscr{O}_{mn} &= \Tbraket{\st{L}_m}{\exp(-T)\exp(\kappa)}{\psi_n(\b{x}_0)} \\
    E_m &= \bst{L}_m^T \bar{\bst{H}} \bst{R}_m^0.
\end{align}
The $\st{L}_m$ stationarity then gives
\begin{align}
    0 = \bst{R}_n + (\bar{\bst{H}} - E_m )\b{\beta}_m + (\bar{E}_m - E_m \bst{L}_m^T \b{\beta}_m) \bst{R}_m,
\end{align}
from which we again have $\bar{E}_m = E_m \bst{L}_m^T \b{\beta}_m$ and thus
\begin{align}
    \b{\beta}_m = -(E_n - E_m)^{-1} \bst{R}_n.
\end{align}
The equations for the zeroth order multipliers are derived as before, with the result that the multipliers change their sign, thus giving the result in Eq.~\eqref{eq:vector_to_vector_tilde} for the vector coupling. For the derivative of the parameters, we have the same equations for $\b{t}^{(i)}$ and $\b{\kappa}^{(i)}$. For the derivative of $\st{L}_m$, we must solve the equation
\begin{align}
   \frac{\mathrm{d}}{\mathrm{d} x_i} \Bigl( \bst{L}_m^T(\bar{\bst{H}} - E_m ) \Bigr)\Big\vert_0 = 0 = {^{\st{L}_m}}\b{\xi}^{(i)T} + \bst{L}_m^{(i)T}(\bar{\bst{H}} - E_m),
\end{align}
which is analogous to Eq.~\eqref{eq:response_R}. In contributions involving $\Tbraket{\beta_n}{\ldots}{\st{R}_n}$ in $G_{mn}^I$, we obtain similar expressions involving $\Tbraket{\st{L}_m}{\ldots}{\beta_m}$ in the case of $\tilde{G}_{mn}^I$. The end-result is
\begin{align}
    \begin{split}
        \tilde{G}_{mn}^I &= \Dbraket{\bar{\st{L}}_m^{(i,i)}}{\bar{R}_n} +\Tbraket{\bar{\zeta}}{K^{(i,i)}}{\mathrm{CC}}
        + \Tbraket{\bar{\st{L}}_m}{K^{(i,i)}}{\bar{\beta}_m} \\
        &+ \Tbraket{\mathrm{HF}}{[\bar{\kappa}, J^{(i,i)}]}{\mathrm{HF}}
+ \Tbraket{\bar{\st{L}}_m^{(i)}}{L^{(i)} - \braket{L^{(i)}}_m}{\bar{\beta}_m},
    \end{split} \label{eq:scalar_coupling_tilde}
\end{align}
with
\begin{align}
    \begin{split}
        \Dbraket{\bar{\st{L}}_m^{(i,i)}}{\bar{R}_n} &= \Tbraket{\bar{\st{L}}_m}{(\kappa^{(i)})^2 - 2  T^{(i)} \kappa^{(i)} + (T^{(i)})^2}{\bar{\st{R}_n}} +  2\Tbraket{\bar{\st{L}}_m^{(i)}}{\kappa^{(i)}}{\bar{\st{R}}_n} \\
    &- 2\Tbraket{\bar{\st{L}}_m^{(i)}}{T^{(i)}}{\bar{\st{R}}_n} - \sum_{pq} \Tbraket{\bar{\st{L}}_m}{E_{pq}}{\bar{\st{R}}_n} \Dbraket{\phi_p^{(i)}}{\phi_q^{(i)}}.
    \end{split}
\end{align}
Finally, $\tilde{G}_{nn}^I$ is obtained in a similar manner to $G_{nn}^I$, see Eq.~\eqref{eq:Lagrangian_conditions_diagonal} and the surrounding text.

This concludes our derivation of the  coupled cluster scalar coupling. To the best of our knowledge, equations for this coupling have not been presented in the literature before (with $m \neq n$). Diagonal terms were also considered by Gauss \emph{et al}.\cite{Gauss2006} from a different starting point.
The scalar coupling is often omitted in dynamics simulations, but  its potential influence on nonadiabatic dynamics has been highlighted in recent years (see Curchod and Martínez\cite{Curchod2018} and references therein).

\section{Concluding remarks}
The norm of the electronic states changes the value of nonadiabatic coupling elements but does not change the molecular wave function. The biorthonormal formula assumed by Christiansen\cite{Christiansen1999nonadiabatic} is therefore a valid choice for coupled cluster dynamics with the appropriate nuclear Schrödinger equations. More generally, we have shown that the wave function is invariant under invertible transformations of the electronic basis.
Of course, the biorthonormal couplings are not directly comparable to the coupling elements of an Hermitian method with normalized states, such as CI or full-CI. However, this reflects the basis-dependence of the couplings and not the validity of the biorthonormal formalism.

We therefore derive a set of nuclear Schrödinger equations assuming  biorthonormal projection onto the electronic basis. Combined with expressions derived for the vector and scalar couplings, these nuclear Schrödinger equations serve as a starting point for the application of nonadiabatic dynamics methods to coupled cluster theory.

Our derivations have been restricted to standard coupled cluster theory. However, the Lagrangian formalism is easily extended to similarity constrained coupled cluster methods,\cite{kjoenstad2017resolving,kjoenstad2019invariant} which are suited to describe relaxation through a conical intersection between excited states. The application to ground state intersections is less straightforward, but may be accessible with approaches that use a different reference than the closed-shell Hartree-Fock state.\cite{Lefrancois2017}

\begin{acknowledgement}
We thank Todd J.~Martínez for enlightening discussions. We also acknowledge funding from the Marie Sk{\l}odowska-Curie European Training Network ``COSINE - COmputational Spectroscopy In Natural sciences and Engineering'', Grant Agreement No. 765739, and the Research Council of Norway through FRINATEK projects 263110 and 275506.

\end{acknowledgement}

\section{Appendix A: Lagrangian derivatives}
Here we derive first and second derivatives of the generic Lagrangian
\begin{align}
    \mathscr{L}(\b{\lambda}, \b{x}, \b{\gamma}) = \mathscr{O}(\b{\lambda},\b{x}) + \b{\gamma}^T \mathscr{E}(\b{\lambda}, \b{x})
\end{align}
with respect to $\b{x}$. The parameters and multipliers both depend on $\b{x}$ since they are determined, for a given $\b{x}$, from the stationarity conditions
\begin{align}
    \frac{\partial \mathscr{L}}{\partial \lambda_k} = 0, \quad \frac{\partial \mathscr{L}}{\partial \gamma_k} = 0.
\end{align}
Using Einstein notation, we can write the Taylor expansion of $\mathscr{L}$ about some $\b{x}_0$ as
\begin{align}
\begin{split}
    \mathscr{L}(\b{\lambda}, \b{x}, \b{\gamma}) &= \mathscr{L}_0
    + \frac{\partial \mathscr{L}}{\partial x_k}\Big\vert_0 \Delta x_k + \frac{1}{2} \Delta x_k \frac{\partial^2 \mathscr{L}}{\partial x_k \partial x_l}\Big\vert_0 \Delta x_l + \frac{1}{2} \Delta \lambda_k \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \lambda_l}\Big\vert_0 \Delta \lambda_l \\
    & + \Delta \lambda_k \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial x_l}\Big\vert_0 \Delta x_l +
    \Delta \gamma_k \frac{\partial^2 \mathscr{L}}{\partial \gamma_k \partial x_l}\Big\vert_0 \Delta x_l + \Delta \lambda_k \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \gamma_l}\Big\vert_0 \Delta \gamma_l + \ldots,
\end{split} \label{eq:Taylor_Lagrangian}
\end{align}
where we have ignored terms of order three or higher in $\Delta \b{x}$. These terms do not contribute to the first and second derivatives and are therefore not relevant to the analysis given here.

In the first derivative, only the partial derivative survives,
\begin{align}
    \frac{\mathrm{d} \mathscr{L}}{\mathrm{d} x_i}\Big\vert_0 = \frac{\partial \mathscr{L}}{\partial x_i}\Big\vert_0.
\end{align}
This is due to the stationarity conditions, since they ensure that there are no linear terms in $\Delta \b{\lambda}$ and $\Delta \b{\gamma}$ in the Taylor expansion in Eq.~\eqref{eq:Taylor_Lagrangian}. In the second derivative, it is convenient to introduce notation for derivatives with respect to particular components of $\b{x}$. We let
\begin{align}
    a^{(i)} &= \frac{\partial a}{\partial x_i} \Big\vert_0 \\
    a^{(i,j)} &= \frac{\partial^2 a}{\partial x_i \partial x_j} \Big\vert_0.
\end{align}
Then we can write
\begin{align}
    \frac{\mathrm{d} \mathscr{L}}{\mathrm{d} x_i}\Big\vert_0 = \mathscr{L}^{(i)}.
\end{align}
and
\begin{align}
\begin{split}
    \frac{\mathrm{d}^2 \mathscr{L}}{\mathrm{d} x_i \mathrm{d} x_j} \Big\vert_0 &= \mathscr{L}^{(i,j)} + \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \lambda_l}\Big\vert_0 \lambda_l^{(j)} + \lambda_k^{(j)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial x_i}\Big\vert_0 + \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial x_j}\Big\vert_0 \\
    &+ \gamma_k^{(j)} \frac{\partial^2 \mathscr{L}}{\partial \gamma_k \partial x_i}\Big\vert_0 + \gamma_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \gamma_k \partial x_j}\Big\vert_0 + 2 \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \gamma_l}\Big\vert_0 \gamma_l^{(j)}.
\end{split}
\end{align}
Now,
\begin{align}
    \gamma_l^{(j)}\Bigl( \frac{\partial^2 \mathscr{L}}{ \partial x_i \partial \gamma_l}\Big\vert_0 + \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \gamma_l}\Big\vert_0 \Bigr) = \gamma_l^{(j)} \frac{\mathrm{d} \mathscr{E}_l}{\mathrm{d} x_i}\Big\vert_0 = 0,
\end{align}
by stationarity, so that
\begin{align}
    \frac{\mathrm{d}^2 \mathscr{L}}{\mathrm{d} x_i \mathrm{d} x_j} \Big\vert_0 &= \mathscr{L}^{(i,j)} + \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \lambda_l}\Big\vert_0 \lambda_l^{(j)} + \lambda_k^{(j)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial x_i}\Big\vert_0 + \lambda_k^{(i)} \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial x_j}\Big\vert_0.
\end{align}
To simplify the notation further, we define derivatives with respect to the parameters:
\begin{align}
    f_k = \frac{\partial \mathscr{L}}{\partial \lambda_k}, \quad H_{kl} = \frac{\partial^2 \mathscr{L}}{\partial \lambda_k \partial \lambda_l}\Big\vert_0.
\end{align}
Thus, we get the final expression for the second derivatives:
\begin{align}
    \frac{\mathrm{d}^2 \mathscr{L}}{\mathrm{d} x_i \mathrm{d} x_j} \Big\vert_0 = \mathscr{L}^{(i,j)} + \lambda_k^{(i)} H_{kl} \lambda_l^{(j)} + \lambda_k^{(j)} f_k^{(i)} + \lambda_k^{(i)} f_k^{(j)}.
\end{align}

redpajama_set_name: RedPajamaArXiv


On paper, Edward P. Jones and Dave Eggers seem to have little in common. The former grew up poor in predominantly African-American Northeast D.C., made his critical reputation with a collection of deceptively understated short stories, and even after a National Book Award nomination, continued to labor in relative penury and obscurity. The latter grew up in an affluent Chicago suburb and found commercial success early, with a memoir that placed the Dave Eggers voice – inventive, flashy, ironic – front and center. And yet this literary season has found the two stars aligning in the literary firmament. First, in August, Eggers penned an appreciative and thoughtful Sunday Times review of Jones' new collection All Aunt Hagar's Children – a book which, at least superficially, could not be more different than Eggers' recent collection How We Are Hungry. Then, two weeks ago, Eggers published a novel embodying the very qualities he praised to in Jones' work: "its sweep, its humanity, the unvarnished perfection of its prose and [a] steady and unerring" narrative force. And though it may surprise critics of McSweeney's to hear it, What is the What is the finest American novel I have read since The Known World.
The novel is a gently fictionalized autobiography of Valentino Achak Deng, a living casualty of the ongoing Sudanese civil war. Having fled from his ruined boyhood village on foot, Deng grew up in U.N.-run camps in Ethiopia and Kenya. He settled in Atlanta in 2001, and after a series of setbacks began looking for a writer who might help him tell his story. As stories go, this one is dramatic and wrenching prima facie, and in a two-part article for The Believer, Eggers gave it respectful, even tentative journalistic treatment. But, sensing that this approach placed barriers of "objectivity" between the audience from the material, he decided, boldly and correctly (with apologies to La Kakutani) to recast Deng's story as first-person fiction.
The moon was high when the movement in the grass began and the moon had begun to fall and dim when the shuffling finally stopped. The lion was a simple black silhouette, broad shoulders, its thick legs outstretched, its mouth open. It jumped from the grass, knocked a boy from his feet. I could not see this part, my vision obscured by the line of boys in front of me. I heard a brief wail. Then I saw the lion clearly again as it trotted to the other side of the path, the boy neatly in its jaws. The animal and its prey disappeared into the high grass and the wailing stopped in a moment. The first boy's name was Ariath.
This paragraph alone would be an extraordinary act of self-effacement for a writer given to flourishes, and an extraordinary act of trust on the part of Deng. That they sustain this voice for 475 pages is something like a miracle. The writer speaks from inside his narrator – from his heart, from his gut, from his intellect. And the distance between audience and subject narrows until we feel that we, too, are Valentino Achak Deng, in all of his complexity and contradiction.
Because imperfect as a human being, he makes a perfect protagonist. He is whip-smart yet perpetually naive, generous and selfish, strong and weak, courageous and timid, full of both faith and doubt. In other words, he is a lot like the Dave Eggers of that other fictionalized autobiography, A Heartbreaking Work of Staggering Genius… not because Eggers has played ventriloquist, but because he has tapped into something universal. In the course of the novel, Achak becomes as real to us as we are to ourselves, and we feel his every loss and triumph as though they were our own.
The first half of the book concerns the destruction of the tranquil Dinka homeland in Southern Sudan by agents of the Islamic government in Khartoum and his harrowing walk across the country in the company of thousands of other "Lost Boys." The novel grounds every historical exigency in the dramatic interactions of rounded characters. If the expectation of a simple story of good vs. evil (and some of the political nuances) gets confounded in the process, we can appreciate more fully the quiet heroism of children who talk each other out of suicide, of young teachers who lead groups of boys through minefields and crocodile-infested rivers, of villagers who risk the disapproval of their elders by sharing their food with these unwanted boys. And though it feels inappropriate to render an aesthetic judgment on Deng's experience, his quest for safety generates a narrative force to rival anything in Lord of the Rings. The difference is that there are no invisibility cloaks or magic breads here.
Things get quieter in the second half, as Deng finds some measure of safety in the refugee camps. But his earlier struggles resonate poignantly in his attempts to contact the father he hasn't heard from in a decade, and especially in a visit to the relatively prosperous and stable capital city of Kenya. Without ever editorializing, What is the What reminds us of the brutality the world's millions of impoverished children face daily; how decadent something as simple as a grocery store can look to those who are living on U.N. rice. And calamity continues to bedevil Deng as he waits to be relocated to the U.S. – which will prove to be no promised land.
In a rare instance of overt artistic license, Eggers uses the invasion and robbery of Deng's apartment in Atlanta as a frame for his novel. We return periodically to scenes of Deng being assaulted in his apartment, or filing a police report, or waiting to be treated for his injuries in the ER. His internal monologues – his memories of Africa – are directed at the various characters he meets along the way. For the most part, this device works just fine. We are deprived of the solace of seeing Deng as exotic, someone "over there"; rather, his struggles are ours… and the injustices he faces in America are the ones we perpetrate every day with our impatience, our pettiness, our indifference. And Deng himself is guilty of these human failings. Occasionally, though, Eggers seems to overreach in his transitions between the fictional present and the fictional past, and to milk the robbery too aggressively for suspense. In almost every other particular, however, What is the What's formal features merge perfectly with its moral authority, until it is impossible to speak of artistic "choices." It is equally difficult to analyze the rich relationship the reader develops with Mr. Deng. Like The Known World, and like Deng's life, the book just is. And that's about the highest praise I can think of.
Eggers has been a fixture on the American literary scene for long enough that it's easy to forget he's in his mid-thirties. Like his near-contemporaries Zadie Smith and David Foster Wallace, he has occasionally suffered in his writing from a kind of IQ overload, an analysis-paralysis. His second book (and first novel), You Shall Know Our Velocity was not an unqualified success, and some readers have been rubbed the wrong way by the antic quality of his fiction. They may be tempted to write off What is the What, rather than read it. But its large-heartedness is an antidote to such small-mindedness. It takes us deep inside a person we will never forget and heralds the arrival of a writer who has found himself by looking beyond himself, and who has learned the difference between intelligence and wisdom.
Very interesting, and the comparison to Jones surprises me as he is undoubtedly the real deal whereas Eggers has always been about style over substance in my mind. This sounds like quite a departure for Eggers, though. It is interesting how taking a writer out of his subject matter "comfort zone" can sometimes elevate the level of the writing since he must try so much harder to make it work.
"I don't want you to think that I don't know that you know that I know that you know that this is not me telling the truth, but me trying to tell the truth about how I can't tell the truth."
As soon as I heard about What is the What, I had the feeling that the subject was going to launch Eggers beyond the n-sided polygon of postmodern prose, and it's exhilarating to see that it can be done. I would love to see Wallace make a similar novelistic leap of faith.


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Entertainment 2021-11-24T11:39:01Z
Jay-Z becomes the most-nominated artist in Grammys history
The rapper and businessman picked up nods for his guest appearances on Ye's album "Donda" and DMX's album "Exodus."
Here are the nominees of the 2022 Grammy Awards
The nominees for the 64th Grammys include Olivia Rodrigo, Lil Nas X, and Billie Eilish. The awards show will air live on CBS on January 31, 2022.
Celebrity 2021-11-23T18:39:48Z
20 popular artists who have surprisingly never won a Grammy
It's shocking that talented musicians like Nicki Minaj and Blake Shelton have never won this prestigious award, even though they've been nominated.
Lil Nas X admits he was jealous of Billie Eilish when she won record of the year at the Grammys over 'Old Town Road'
The "Montero (Call Me by Your Name)" rapper opened up about feeling jealous toward the "Your Power" singer after she beat him at the 2020 awards show.
The Grammys are dropping secretive nomination committees after The Weeknd called the awards show 'corrupt'
After The Weeknd demanded "transparency" from the Grammys, the awards show announced it's changing how artists are nominated.
16 people who have won an EGOT, the most coveted award in Hollywood
Only a handful of stars have won the EGOT: an Emmy, a Grammy, an Oscar, and a Tony. Here's the complete list of winners.
Lifestyle 2021-03-28T12:15:03Z
35 of Lady Gaga's most iconic red-carpet looks of all time
Lady Gaga has been a fashion chameleon ever since she rose to fame back in 2008. Here are some of her most daring, bold, and delightful outfits.
Culture 2021-03-17T19:39:23Z
Cardi B and Candace Owens sparred on Twitter after the 'WAP' Grammys performance, and both women threatened lawsuits
Cardi B and Owens have started yet another public Twitter feud following Owen's comments about "WAP" on Tucker Carlson's Fox News show.
Eddie Van Halen's son is 'hurt' by the Grammy's brief tribute to his father
"I know rock isn't the most popular genre," Wolfgang Van Halen wrote, "but I think it's impossible to ignore the legacy my father left."
10 unbelievable Grammys records, from the youngest winner to the most-awarded artists of all time
Beyoncé became the female artist with the most Grammys — 28! — in 2021. Here are nine other unbelievable records in Grammy history.
13 cool details from the Grammys that you might've missed
From Harry Styles' NSFW banana necklace to Taylor Swift's performance Easter eggs, here's what you missed at this year's Grammy Awards.
Style 2021-03-15T16:57:37Z
The best-dressed Grammy award-winning artist the year you were born
See how celebrity style at the awards show has evolved over the years, from Peggy Lee to Taylor Swift.
Taika Waititi confused after winning Grammy for 'Jojo Rabbit': 'I guess they're just giving Grammys to anyone now'
Taika Waititi is now halfway to becoming an EGOT winner after adding a Grammy to the Oscar he won for "Jojo Rabbit" last year.
Grammys host Trevor Noah mocked royal-family tension over Meghan Markle and Prince Harry's Oprah interview
Noah compared the tension among the Grammy nominees to a family reunion at Buckingham Palace following the bombshell interview.
Here are the winners of the 2021 Grammy Awards
The 63rd Grammy Awards were hosted by the "Daily Show" host Trevor Noah. Beyoncé, Ariana Grande, Lady Gaga, and Megan Thee Stallion won awards.
15 rock stars who never won a Grammy but definitely should have
Despite their success, bands including Queen, Journey, and the Grateful Dead never took home a Grammy Award.
35 of the most iconic red carpet looks in Grammys history
The 63rd Grammy Awards are on Sunday, March 14. Here are the best red carpet looks over the last six decades.
10 best new artist Grammy winners who went on to major fame — and 10 who didn't
The Grammy Award for best new artist has gone to superstars like Mariah Carey and The Beatles. Still, not every winner found mainstream success.
The most memorable red carpet look from the year you were born
Ahead of the Golden Globes on Sunday, we took a look back at the most iconic red carpet looks of the last 60 years.
How the $100 billion Hollywood machine is scrambling to come back after its worst year in decades
Insider's entertainment team took a deep dive into how Hollywood is moving forward after movies, TV shows, awards shows, and music had to shift.

redpajama_set_name: RedPajamaCommonCrawl


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East of Eli: An Interview – Part 2 of 2
0 February 2, 2019 14:14 by Shanai Momi
Part 1 of this article can be found here. The following interview has been condensed for this article.
You started out in acting, one of the things you might be recognized for is Bring It On, did you always know you wanted to focus on music or was there something that changed?
I've always been a music lover, as much as I love to play, I love to listen, I love to find new bands, I love to support independent music, even mainstream, everything. Acting was great, but it was kind of like a rebound. I grew up playing hockey, it kind of fell apart and I didn't know what to do. I kind of fell into the acting world and it ended up being my mainstream income very fast. I wasn't as passionate about it as I was about music, over time I started to pull back and focus on my music, make that transition.
What was it that scared you about music?
What scared me was the honesty, the authenticity behind it. The most amazing performances we see in film are letting you go behind the front that people put on. They're connecting you to something they feel, they're pulling from an emotion they know, something they've been through. Acting; I have nothing but respect for the art form, but you could kind of play it off that it was the character, I could always hide behind the character. Whereas doing music, it was like this is me.
What message would you want people who are unfamiliar with your work to take away from your music, if you could pick one thing?
It's okay to be you, no matter where you're at. Whether it's good or… There are peaks and valleys. You wouldn't know where it was to stand victorious on a peak if you didn't know the lowest of the lows of a valley. Life tends to have polar opposites; the highest peak in the world, the deepest trench, but as you go through life it's not as drastic. You can be in a good place but be in a low place too. That's part of the journey. My biggest message; I'm not perfect. I've had this long journey to realize I'm okay with my faults, I'm learning, I'm growing, and I'm not giving up on myself.
What are your thoughts on the intimacy of your gigs? At times it feels like you're the only one in the room.
If this ever gets to the stadium size, I will never stop playing these shows. They're too important to me. The big shows are, obviously for business great, and to reach that many people would be incredible. But I honestly love these shows, and I will forever do them. An EOE warning is you might be more part of the show than you think because you're close to the front.
You implied [during the show] that the closest song to your heart right now is Trenches, why is that?
I wanted people to understand I'm not some kind of guru, I'm not perfect. That's why I write these songs, because it can't just be this bad, it's gotta get better. I write from a depressed place. The Siege came from a place where I was ready to quit, to give up on everything. My dad, when he was here, was always on fire about believing in yourself, about self-help stuff. He encouraged me, he was like you can't give up, this is what you're meant to do. That's why the chorus is "Carry on". Then it became a song for my father. It came from a broken place. And so, Trenches, was kind of looking myself in the mirror.
You also said The Silent Kind is your favourite, so why is that?
That song is very dear to my heart because I was in a very broken place. The song was written about a young woman in New York. I had gone out, I'd had a really rough day. It was like two, three in the morning. I saw this woman uncontrollably sobbing. She'd been through a break-up, been cheated on. We walked and talked, and she shared everything. I got her an Uber to her place, gave her a hug, and we parted ways. The next day I grabbed my guitar and within ten minutes sound kinda came out and I wrote about that experience and things I was feeling. I like the silent life, there are no problems. When you're by yourself, your problems kinda go away. It's an honest statement; it's the truth but not the best truth. You can't avoid people. We need connection. But sometimes, dammit, you just want to disappear.
Before Nothing Ordinary, you said that was one you used to hate. What is it that makes artists hate their work, and why does maybe distance or time help that?
I think a lot of the time, artists speak from a place that is beyond their own selves. Nothing Ordinary was beyond my comprehension at the time. I definitely felt it, I definitely wrote it. I was in a place where there was nothing really happy about where I was in life. I started writing the music and the words started coming together, and it felt special at the time, but then because I couldn't believe it for myself. I didn't really like myself so the idea of 'life is extraordinary', felt cheesy or not true. But no, actually it is [true]. It might be so simple it's hard to grasp. People started requesting it and I had to come to grips with that there was something special about this song.
To find out more or purchase tickets to see East of Eli: https://eastofeli.com/
Pandora Ogg
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East of Eli: Part 1 of 2
by Shanai Momi - Feb 2, 2019
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Virgilio Figueroa (Santiago, 1870- Santiago de Chile, 3 de julio de 1939) fue un profesor de historia e historiador chileno.

Biografía
Estudió Historia en la Universidad de Chile, de donde egresó en 1896, dando inicio a su investigación genealógica de personajes chilenos, tema en el cual basó su tesis de grado.

Famoso por su obra Diccionario Histórico, Biográfico y Bibliográfico, que, según las palabras del mismo autor, es «una enciclopedia histórica en la cual se puede encontrar una síntesis de la vida, labor, obras de hombres y mujeres, y su influencia sobre sus ambientes y las épocas posteriores».

La obra de Figueroa tiene cinco tomos, si se incluye el índice.

Obras
 Diccionario Histórico, Biográfico y Bibliográfico: 1800-1930 Tomo I (1925)
 Diccionario Histórico, Biográfico y Bibliográfico: 1800-1930 Tomo II (1928)
 Diccionario Histórico, Biográfico y Bibliográfico: 1800-1930 Tomo III (1930)
 Diccionario Histórico, Biográfico y Bibliográfico: 1800-1930 Tomo IV (1931)

Referencias

Enlaces externos
 en memoriachilena.cl Virgilio Figueroa

Nacidos en Santiago de Chile
Historiadores de Chile
Escritores de Chile del siglo XIX
Escritores de Chile del siglo XX
Fallecidos en Santiago de Chile

redpajama_set_name: RedPajamaWikipedia


carry out and assess credited work placements?
This website includes tools for students, placement hosts and placement tutors in higher education to use when embarking on a work placement. They have been co-developed to facilitate critical engagement with the process of carrying out a work placement from these three perspectives. The 'overarching considerations' and three flow diagrams have been developed in consultation with postgraduate students, representatives of arts organisations and academic teaching staff on MAs in Applied Theatre, Curating, Arts Management and Museums, Galleries and Contemporary Culture. These tools aim to help support ethical, tripartite agreements between the three parties involved – the student, the academy and the host organisation.
The university tutor and host organisation will support the student if any issues arise before and during the placement.
Ensure students with any self-declared disabilities are supported.
Clarify and agree the learning aims and learning outcomes of the placement.
Identify and agree clear roles, expectations and timescales (hours and working patterns) for each of you.
Acknowledge the needs and resources to put together a work placement that is adaptable and flexible to the student.
Agree on outcomes but also acknowledge that these may change depending on the nature of the course, student and the specialised interest.
Consider remuneration for the student, even if legally this is not required, as this can address issues of equality (consider the time and resources it takes to do a placement). Students should not be expected to incur expenses as part of their work placement.
Set up a meeting at the end of the placement between university tutor, host and student focusing on feedback of the process and reflecting on learning aims and outcomes. Individual meetings between student and host, student and university tutor, tutor and host may also be necessary.
Confirm that the university will be solely responsible for assessment. Host organisations are welcome to provide feedback.
Acknowledge the relationship between work and learning and between the practice and academic study.
Clarify intellectual property rights and confidentiality clauses prior to the work placements.
Ensure students and host organisations are aware of the need to make sure that there is a good match between them and that appropriate screening processes are in place.
Ensure that both the university and host organisation has public liability insurance. In the case when the host organisation does not have insurance that covers the student, ensure the student is made aware of this and the need for a health and safety risk assessment.
Clarify who in the university is a point of contact for the host organisation and students and manage the expectations of the students and host organisations.
Develop and maintain a network of hosts (for example set up an annual committee from representatives from arts organisations).
Ensure equality of placement opportunities. Consider the time and resources needed, the networks that students have or do not have and how this may prevent them from carrying out placements, acknowledging that some students will need more support than others.
Ensure host organisations know about the course (e.g. copy of handbook) and ensure that students provide host organisations with information about themselves or give tutors permission to release this information.
If suitable, identify shared collaborations or long term relationship with hosts, in terms of research and outreach.
Ensure the student is aware of any other legal requirements, such as DBS checks if working with children.
Be clear about your learning aims and outcomes for the placement and how it relates to your overall MA studies.
Consider declaring any health issues or disabilities so as to receive the support you need.
Research the organisations you want to do a work placement in and liaise with your university tutor / placement administrator to clarify how to approach the organisation.
Be aware of the time and resources needed to undertake a work placement and identify what kind of support you need.
Provide host organisations with details of any requirements that are associated with hosting you in a work placement for academic credit.
Familiarise yourself with the host organisation's placement policy and code of conduct in relation to guidelines for best practice set by the university and other agencies.
Be mindful that learning means being open-minded and an awareness of the context you are in.
If you broker your own placement, make sure you inform your placement tutor to ensure that the university regulations are adhered to.
Be mindful of the expectations of professionalism that are required of you during your work placement.
Identity a staff member who is the contact and supervisor in the organisation for the work placement student and a timetable of regular meetings.
Provide an induction for the student at the beginning of the work placement including introduction to the organisation, staff, health and safety and any issues of confidentiality.
Acknowledge the student through the work/research they are doing in the organisation rather than as the 'intern' or the 'work placement person'; acknowledging that they are there to learn as well as work (e.g. a label could be Arts Education Placement Student).
Provide an exit interview for the student at the end of the placement that outlines any exit procedures and responsibilities.
If suitable, identify shared collaborations or long term relationship with universities in terms of research and outreach.


Sen. John McCain at the opening of the 53rd Munich Security Conference in Munich, Germany, on February 17.
Republican Sen. John McCain issued a blistering statement Tuesday morning condemning President Donald Trump's reported disclosure of classified information to Russian diplomats during a meeting in the Oval Office last week.
"The reports that the president shared sensitive intelligence with Russian officials are deeply disturbing," said the Arizona senator, who is notably hawkish on Russia.
"Reports that this information was provided by a US ally and shared without its knowledge sends a troubling signal to America's allies and partners around the world and may impair their willingness to share intelligence with us in the future."
The Washington Post reported Monday night that Trump "jeopardized a critical source of intelligence on the Islamic State" in his conversation with Russian Foreign Minister Sergey Lavrov and Ambassador Sergey Kislyak last week. The Post said Trump shared information that "had been provided by a US partner through an intelligence-sharing arrangement considered so sensitive that details have been withheld from allies and tightly restricted even within the US government."
The report drew bipartisan criticism from members of Congress. A spokesman for House Speaker Paul Ryan, who is normally tight-lipped about Trump's behavior, said Ryan "hopes for a full explanation of the facts from the administration."
Republican Sen. Bob Corker also weighed in, telling USA Today that "to compromise a source is something that you just don't do." He continued: "That's why we keep the information that we get from intelligence sources so close ... to prevent that from happening."
McCain criticized Trump's decision to spend time "sharing sensitive information with the Russians" instead of "focusing on Russia's aggressive behavior, including its interference in American and European elections, its illegal invasion of Ukraine and annexation of Crimea, its other destabilizing activities across Europe, and the slaughter of innocent civilians and targeting of hospitals in Syria."
Jim Jeffrey, a former US ambassador who served as deputy national security adviser in the George W. Bush administration, made a similar characterization: "Whatever the president shared with the Russians beyond the absolute minimum raises the specter that he doesn't get how justifiably outraged Americans are at the hacking of the 2016 election. That's the most troubling."
The incident happened last Wednesday, one day after Trump fired the FBI director, James Comey, amid an FBI investigation into whether the Trump campaign colluded with Russians to undermine Hillary Clinton, then the Democratic presidential nominee, in the 2016 election.
Many have noted that Trump, as president, is legally allowed to essentially disclose classified information to whomever he wants. But the fact that the information he shared was not a US secret, but that of an American ally, may complicate his authority to declassify information at will. Matthew Rosenberg of The New York Times told CNN on Monday that the intel came from a close "Middle Eastern ally."
National security experts reacted with alarm to the report, calling it "appalling," "nauseating," and "mind-boggling."
"This is not a garden variety breach," national security experts, including Susan Hennessey and Benjamin Wittes, wrote for Lawfare on Monday. "And outrage over it is not partisan hypocrisy about protecting classified information."