## The variational principle in quantum mechanics, lectures 9 and 10

In lecture 9 we continued reading the papers on matrix product states that we began in lectures 7 and 8.

1. Lecture 10: the time-dependent variational principle

In this lecture we will discuss the time-dependent variation principle (TDVP), which is a powerful method to simulate the nonequilibrium dynamics of a general quantum system while remaining within a given variational class. The TDVP is due, as far as I’m aware, to Dirac. Additionally, describing the TDVP isn’t especially difficult, so it is rather surprising that this elegant method it hasn’t made into standard textbooks. The general framework of the time-dependent variational principle can be found in [P. Kramer and M. Saraceno, Geometry of the Time-Dependent Variational Principle in Quantum Mechanics (Springer-Verlag, Berlin) (1981)].

These notes can be found in pdf format here.

Recall that the time-dependent Schrödinger equation (TDSE) can be derived by extremizing an action functional ${S\{\overline{\psi}(t),\psi(t)\}=\int_{t_{1}}^{t_{2}} dt\, L(\overline{\psi}(t),\psi(t),t)}$ with Lagrangian

$\displaystyle L(\overline{\psi}(t),\psi(t),t)=\frac{i}{2}\langle\psi(t)|\dot{\psi}(t)\rangle-\frac{i}{2}\langle\dot{\psi}(t)|\psi(t)\rangle-\langle\psi(t)|H(t)|\psi(t)\rangle. \ \ \ \ \ (1)$

For the sake of brevity, we henceforth omit the time dependence of ${H}$ and thus of ${L}$. Stationarity of the action under independent variations of ${|\psi\rangle}$ and ${\langle\psi|}$ in the full Hilbert space ${\mathcal{H}}$ yields the TDSE and its complex conjugate. But when we only have access to a subspace or manifold ${\mathcal{M}\subset \mathcal{H}}$, we can still use the calculus of variations with respect to this action to define a time evolution of states ${|\psi\rangle\in \mathcal{M}}$; this is the essence of the TDVP. To this end, we restrict to variations in the tangent plane of ${\mathcal{M}}$ at the point ${|\psi\rangle}$. Assume that the manifold ${\mathcal{M}}$ can be parametrized as

$\displaystyle \mathcal{M}=\{|\psi(\mathbf{z})\rangle,\mathbf{z}\in \mathbb{C}^{n}\}, \ \ \ \ \ (2)$

where we assume the dependence on the ${n}$ complex parameters ${z^{i}}$ to be analytic and explicitly denote the anti-analytic dependence ${\langle \psi(\overline{\mathbf{z}})|}$. Furthermore, we introduce the notation ${\partial_{i}}$ for ${\partial/\partial z^{i}}$ and ${\partial_{\overline{\jmath}}=\partial/\partial \overline{\mathbf{z}}^{\overline{\jmath}}}$, where we always use barred indices for the complex conjugate variables ${\overline{\mathbf{z}}}$. Requiring stationarity of ${S\{\overline{\mathbf{z}}(t),\mathbf{z}(t)\}}$ with respect to a variation ${\overline{\mathbf{z}}(t)\rightarrow \overline{\mathbf{z}}(t)+\overline{\mathbf{\delta z}}(t)}$ results in the following Euler-Lagrange equations

$\displaystyle i G_{\overline{\imath}j}(\overline{\mathbf{z}}(t),\mathbf{z}(t)) \dot{z}^{j}(t) = \langle\partial_{\overline{\imath}} \psi(\overline{\mathbf{z}}(t))|H|\psi(\mathbf{z}(t))\rangle, \ \ \ \ \ (3)$

where ${G_{\overline{\imath}j}}$ is the Gram matrix or overlap matrix of the tangent vectors of ${\mathcal{M}}$:

$\displaystyle G_{\overline{\imath}j}(\overline{\mathbf{z}},\mathbf{z})=\langle\partial_{\overline{\imath}} \psi(\overline{\mathbf{z}})|\partial_j \psi(\mathbf{z})\rangle. \ \ \ \ \ (4)$

We can also interpret this as the metric, indeed it is the pullback of the flat metric to ${\mathcal{M}}$, and — assuming linear independence of the tangent vectors — define the inverse metric as ${G^{i\overline{\jmath}}(\overline{\mathbf{z}},\mathbf{z})}$, such that ${G^{i\overline{\jmath}}(\overline{\mathbf{z}},\mathbf{z})G_{\overline{\jmath}k}(\overline{\mathbf{z}},\mathbf{z})=\delta^{i}_{\ k}}$. The TDVP thus results in

$\displaystyle i\dot{z}^{i}(t) = G^{i\overline{\jmath}}(\overline{\mathbf{z}}(t),\mathbf{z}(t))\langle\partial_{\overline{\jmath}} \psi(\overline{\mathbf{z}}(t))|H|\psi(\mathbf{z}(t))\rangle \ \ \ \ \ (5)$

and its complex conjugate.

One can also obtain the Euler-Lagrange equations [Eq. (3)] geometrically, by looking for the coefficients ${\dot{z}^{i}(t)}$ which minimize

$\displaystyle \left\lVert \dot{z}^{i}(t)|\partial_i \psi(\mathbf{z}(t))\rangle-H|\psi(\mathbf{z}(t))\rangle\right\rVert.$

This minimization is obtained by the orthogonal projection of ${H|\psi(\mathbf{z}(t))\rangle}$ onto the tangential plane ${T_{\mathbf{z}(t)}\mathcal{M}}$, defined by

$\displaystyle T_{\mathbf{z}}\mathcal{M}=\mathrm{span}\left\{|\partial_i \psi(\mathbf{z})\rangle,i=1,\ldots,n\right\}. \ \ \ \ \ (6)$

The orthogonal projector ${\widehat{P}_{T_{\mathbf{z}}\mathcal{M}}}$ is indeed given by

$\displaystyle \widehat{P}_{T_{\mathbf{z}}\mathcal{M}}(\overline{\mathbf{z}},\mathbf{z})=|\partial_i \psi(\mathbf{z})\rangle G^{i\overline{\jmath}}(\overline{\mathbf{z}},\mathbf{z}) \langle \partial_{\overline{\jmath}} \psi(\overline{\mathbf{z}})|, \ \ \ \ \ (7)$

where the inverse of the Gram matrix appears in order to obtain ${\widehat{P}_{T_{\mathbf{z}}\mathcal{M}}^{2}= \widehat{P}_{T_{\mathbf{z}}\mathcal{M}}}$.

Whereas Hamiltonian evolution in ${\mathcal{H}}$ is unitary and thus norm-preserving, this is no longer guaranteed for the projected evolution. In order to ensure norm preservation, we can define a modified Lagrangian ${\widetilde{L}(\overline{\psi}(t),\psi(t))=L(\overline{\psi}(t),\psi(t))/\langle{\psi(t)|\psi(t)}}$, which results in

$\displaystyle \widetilde{L}(\overline{\mathbf{z}}(t),\mathbf{z}(t))=\frac{i}{2}\left(\dot{z}^{j}(t)\partial_{j}-\dot{\overline{\mathbf{z}}}^{\overline{\jmath}}(t)\partial_{\overline{\jmath}}\right)\ln N(\overline{\mathbf{z}}(t),\mathbf{z}(t))-H(\overline{\mathbf{z}}(t),\mathbf{z}(t)) \ \ \ \ \ (8)$

where

$\displaystyle N(\overline{\mathbf{z}},\mathbf{z})=\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle, \ \ \ \ \ (9)$

and

$\displaystyle H(\overline{\mathbf{z}},\mathbf{z})=\frac{\langle\psi(\overline{\mathbf{z}})|H|\psi(\mathbf{z})\rangle}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}. \ \ \ \ \ (10)$

Stationarity under variations ${\overline{\mathbf{z}}(t)+\overline{\delta \mathbf{z}}(t)}$ now results in the Euler-Lagrange equations

$\displaystyle i \widetilde{G}_{\overline{\imath}j}(\overline{\mathbf{z}}(t),\mathbf{z}(t))\dot{z}^{j}(t)=H_{\overline{\imath}}(\overline{\mathbf{z}}(t),\mathbf{z}(t)), \ \ \ \ \ (11)$

and complex conjugates, where we have introduced the modified Gram matrix

$\displaystyle \widetilde{G}_{\overline{\imath}j}(\overline{\mathbf{z}},\mathbf{z})=\partial_{\overline{\imath}}\partial_{j}\ln N(\overline{\mathbf{z}},\mathbf{z})=\frac{G_{\overline{\imath}j}(\overline{\mathbf{z}},\mathbf{z})}{N(\overline{\mathbf{z}},\mathbf{z})}-\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle\langle\psi(\overline{\mathbf{z}})|\partial_{j}\psi(\mathbf{z})\rangle}{N(\overline{\mathbf{z}},\mathbf{z})^{2}} \ \ \ \ \ (12)$

$\displaystyle =\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\partial_{j}\psi(\mathbf{z})\rangle}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}-\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle\langle\psi(\overline{\mathbf{z}})|\partial_{j}\psi(\mathbf{z})\rangle}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle ^{2}}, \ \ \ \ \ (13)$

and the gradient of the normalized expectation value

$\displaystyle H_{\overline{\imath}}(\overline{\mathbf{z}},\mathbf{z})=\partial_{\overline{\imath}}H(\overline{\mathbf{z}},\mathbf{z})=\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|H|\psi(\mathbf{z})\rangle}{N(\overline{\mathbf{z}},\mathbf{z})}-\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}{N(\overline{\mathbf{z}},\mathbf{z})}H(\overline{\mathbf{z}},\mathbf{z}) \ \ \ \ \ (14)$

$\displaystyle =\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|H|\psi(\mathbf{z})\rangle}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}-\frac{\langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle\langle\psi(\overline{\mathbf{z}})|H|\psi(\mathbf{z})\rangle}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle^{2}}. \ \ \ \ \ (15)$

These expressions can be easily interpreted. Under an infinitesimal variation, the norm or phase of a state ${|\psi\rangle}$ changes if we move in the direction of ${|\psi\rangle}$. Norm conservation is thus obtained when we subtract from every tangent vector ${|\partial_{i}\psi(\mathbf{z})\rangle}$ its component along ${|\psi(\mathbf{z})\rangle}$ by replacing it with ${\widehat{P}_{0}(\overline{\mathbf{z}},\mathbf{z})|\partial_{i}\psi(\mathbf{z})\rangle}$, where the projector ${\widehat{P}_{0}}$ is given by

$\displaystyle \widehat{P}_{0}(\overline{\mathbf{z}},\mathbf{z})=\widehat{1}-\frac{|\psi(\mathbf{z})\rangle\langle \psi(\overline{\mathbf{z}})|}{\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}, \ \ \ \ \ (16)$

and thus ${\widehat{P}_{0}(\overline{\mathbf{z}},\mathbf{z})|\partial_{i}\psi(\mathbf{z})\rangle=|\partial_{i}\psi(\mathbf{z})\rangle-|\psi(\mathbf{z})\rangle N(\overline{\mathbf{z}},\mathbf{z})^{-1} \langle\psi(\overline{\mathbf{z}})|\partial_{i}\psi(\mathbf{z})\rangle}$. We indeed find

$\displaystyle \tilde{G}_{\overline{\imath}j}(\overline{\mathbf{z}},\mathbf{z})=N(\overline{\mathbf{z}},\mathbf{z})^{-1} \langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\widehat{P}_{0}(\overline{\mathbf{z}},\mathbf{z}) |\partial_{j}\psi(\mathbf{z})\rangle \ \ \ \ \ (17)$

and

$\displaystyle H_{\overline{\imath}}(\overline{\mathbf{z}},\mathbf{z})=N(\overline{\mathbf{z}},\mathbf{z})^{-1} \langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|\widehat{P}_{0}(\overline{\mathbf{z}},\mathbf{z}) H|\psi(\mathbf{z})\rangle. \ \ \ \ \ (18)$

If the manifold ${\mathcal{M}}$ allows for norm and phase variations of states, i.e. if ${|\psi\rangle\in T\mathcal{M}}$, then we can define the contravariant vector ${\psi^{i}}$ such that ${\psi^{i}|\partial_{i}\psi\rangle=|\psi\rangle}$. In this paragraph, we henceforth omit all arguments ${\mathbf{z}}$ and ${\overline{\mathbf{z}}}$ for the sake of brevity. By definition we have that ${\widehat{P}_{0}|\partial_{i}\psi\rangle\psi^{i}=0}$ and we can conclude that ${\widetilde{G}_{\overline{\imath}j}}$ has an eigenvalue zero, since ${\widetilde{G}_{\overline{\imath}j}\psi^{j}=0=\overline{\psi}^{\overline{\imath}}G_{\overline{\imath}j}}$, from which we immediately obtain the corresponding eigenvector. (Note that ${G_{\overline{\imath}j}}$ is a Hermitian matrix.) We now also define the covariant vector ${\psi_{\overline{\imath}}=G_{\overline{\imath}j}\psi^{j}=\langle\partial_{\overline{\imath}} \psi|\psi\rangle}$ so that ${\overline{\psi}_{i}\psi^{i}=\langle\psi|\psi\rangle=N}$. With these definitions, we can write ${\widetilde{G}_{\overline{\imath}j}=N^{-1} G_{\overline{\imath}j}-N^{-2}\psi_{\overline{\imath}}\overline{\psi}_{j}}$. Even though ${\widetilde{G}_{\overline{\imath}j}}$ is not invertible, we can still define a pseudo-inverse as ${\widetilde{G}^{i\overline{\jmath}}=N G^{i\overline{\jmath}}-\psi^{i}\overline{\psi}^{\overline{\jmath}}}$, so that ${\widetilde{G}^{i\overline{\jmath}} \widetilde{G}_{\overline{\jmath}k}=\delta^{i}_{\ k}-N^{-1}\psi^{i}\overline{\psi}_{k}}$ and ${\widetilde{G}_{\overline{\imath}j} \widetilde{G}^{j\overline{k}}=\delta^{\ \overline{k}}_{\overline{\imath}}-N^{-1}\psi_{\overline{\imath}}\overline{\psi}^{\overline{k}}}$. Since we can rewrite ${H_{\overline{\imath}}}$ as ${N^{-1} (\delta^{\ \overline{k}}_{\overline{\imath}}-N^{-1}\psi_{\overline{\imath}}\overline{\psi}^{\overline{k}}) \langle\partial_{\overline{k}}\psi|H|\psi\rangle}$, we are allowed to apply this pseudo-inverse to the Euler-Lagrange equations in order to obtain

$\displaystyle i \dot{z}^{i}(t)= \widetilde{G}^{i\overline{\jmath}}(\overline{\mathbf{z}}(t),\mathbf{z}(t)) H_{\overline{\jmath}}(\overline{\mathbf{z}}(t),\mathbf{z}(t)). \ \ \ \ \ (19)$

In principle, the component of ${\dot{\mathbf{z}}(t)}$ along the zero eigenspace of ${\widetilde{G}}$ can be chosen freely but, with the particular solution in the equation above, we satisfy ${\overline{\psi}_{i}\dot{z}^{i}(t)=\langle\psi|\partial_{i}\psi\rangle\dot{z}^{i}(t)=0}$, which is the required condition for norm (and phase) conservation.

If the manifold ${\mathcal{M}}$ does not contain the freedom to change the norm and phase of a state, the modified metric ${\widetilde{G}}$ would have the same rank as the original metric ${G}$, which we assumed to be invertible. In particular, if ${|\psi(\mathbf{z})\rangle\perp T_{\mathbf{z}}\mathcal{M}}$, then ${\widetilde{G}_{\overline{\imath},j}(\overline{\mathbf{z}},\mathbf{z})=N(\overline{\mathbf{z}},\mathbf{z})^{-1} G_{\overline{\imath},j}(\overline{\mathbf{z}},\mathbf{z})}$ and ${H_{\overline{\imath}}(\overline{\mathbf{z}},\mathbf{z})= N(\overline{\mathbf{z}},\mathbf{z})^{-1} \langle\partial_{\overline{\imath}}\psi(\overline{\mathbf{z}})|H|\psi(\mathbf{z})\rangle}$. The Euler-Lagrange equations following from ${S\{\overline{\mathbf{z}}(t),\mathbf{z}(t)\}}$ or from ${\widetilde{S}\{\overline{\mathbf{z}}(t),\mathbf{z}(t)\}}$ are then identical.

Finally, by defining for every pair of functions ${f(\overline{\mathbf{z}},\mathbf{z})}$ and ${g(\overline{\mathbf{z}},\mathbf{z})}$ a Poisson bracket

$\displaystyle \{f,g\}=\partial_{i}f \widetilde{G}^{i\overline{\jmath}}\partial_{\overline{\jmath}}g-\partial_{i}g \widetilde{G}^{i\overline{\jmath}}\partial_{\overline{\jmath}}f \ \ \ \ \ (20)$

we can write down the Euler-Lagrange equations as

$\displaystyle \dot{z}^{i} =i \{H,z^{i}\}, \ \ \ \ \ (21)$

and

$\displaystyle \dot{\overline{\mathbf{z}}}^{\overline{\imath}}=i \{H,\overline{\mathbf{z}}^{\overline{\imath}}\}. \ \ \ \ \ (22)$

For every operator ${\widehat{O}}$ acting on ${\mathcal{H}}$, we can define the expectation value ${O(\overline{\mathbf{z}},\mathbf{z})=\langle\psi(\overline{\mathbf{z}})|\widehat{O}|\psi(\mathbf{z})\rangle/\langle\psi(\overline{\mathbf{z}})|\psi(\mathbf{z})\rangle}$ so that its time evolution is governed by ${\dot{O}=i\{H,O\}}$. The manifold ${\mathcal{M}}$ is thus a symplectic manifold. From the antisymmetry of the Poisson bracket we find ${\{H,H\}=0}$, which implies that the energy of the state ${|\psi\rangle\in\mathcal{M}}$ is conserved under exact integration of the TDVP equations.

The symplectic properties of the time-dependent variational principle also conserve other symmetries. Assume that the Hamiltonian is invariant under the action of a symmetry operator ${\widehat{U}}$ (which should be a unitary operator), such that ${[H,\widehat{U}]=0}$. In order to be able to transfer this symmetry to the manifold ${\mathcal{M}}$, we need to assume that for any state ${|\psi(\mathbf{z})\rangle\in\mathcal{M}}$, the action of ${\widehat{U}}$ is mapped to a new state ${|\psi(\mathbf{u}(\mathbf{z}))\rangle=\widehat{U}|\psi(\mathbf{z})\rangle\in\mathcal{M}}$. Because of the unitarity of ${\widehat{U}}$, we have ${N(\overline{\mathbf{u}}(\overline{\mathbf{z}}),\mathbf{u}(\mathbf{z}))= N(\overline{\mathbf{z}},\mathbf{z})}$, from which we obtain

$\displaystyle \partial_{\overline{\imath}}\overline{u}^{\overline{\jmath}}(\overline{\mathbf{z}}) \widetilde{G}_{\overline{\jmath},k}(\overline{\mathbf{u}}(\overline{\mathbf{z}}),\mathbf{u}(\mathbf{z})) \partial_l u^k(\mathbf{z})= G_{\overline{\imath},l}(\overline{\mathbf{z}},\mathbf{z}), \ \ \ \ \ (23)$

The condition ${[H,\widehat{U}]=0}$ also allows to conclude ${H(\overline{\mathbf{u}}(\overline{\mathbf{z}}),\mathbf{u}(\mathbf{z}))=H(\overline{\mathbf{z}},\mathbf{z})}$ and thus

$\displaystyle \partial_{\overline{\imath}}\overline{u}^{\overline{\jmath}}(\overline{\mathbf{z}}) H_{\overline{\jmath}}(\overline{\mathbf{u}}(\overline{\mathbf{z}}),\mathbf{u}(\mathbf{z}))=H_{\overline{\imath}}(\overline{\mathbf{z}},\mathbf{z}) \ \ \ \ \ (24)$

and

$\displaystyle H_{j}(\overline{\mathbf{u}}(\overline{\mathbf{z}}),\mathbf{u}(\mathbf{z}))\partial_{i}u^{j}(\mathbf{z}) =H_{i}(\overline{\mathbf{z}},\mathbf{z}). \ \ \ \ \ (25)$

The metric and the gradient thus transform covariantly under the symmetry transformation and can be used to transform Eq. (11) into

$\displaystyle +i\partial_{\overline{\imath}}\overline{u}^{\overline{\jmath}}(\overline{\mathbf{z}}(t)) \widetilde{G}_{\overline{\jmath},k}(\overline{\mathbf{u}}(\overline{\mathbf{z}}(t)),\mathbf{u}(\mathbf{z}(t))) \frac{d}{d t}u^k(\mathbf{z}(t))=\partial_{\overline{\imath}}\overline{u}^{\overline{\jmath}}(\overline{\mathbf{z}}(t)) H_{\overline{\jmath}}(\overline{\mathbf{u}}(\overline{\mathbf{z}}(t)),\mathbf{u}(\mathbf{z}(t))), \ \ \ \ \ (26)$

and its complex conjugate. By using the injectivity of the map ${\mathbf{u}(\mathbf{z})}$, we can eliminate the Jacobians ${\partial_{\overline{\imath}} \overline{u}^{\overline{\jmath}}}$ and ${\partial_k u^j}$ in order to obtain the correct flow equations in terms of the new coordinates ${(\mathbf{u}(t),\overline{\mathbf{u}}(t))}$. One case that is not covered by this general derivation is when ${\widehat{U}}$ is an anti-linear operator, since ${\mathbf{u}}$ will then depend on ${\mathbf{z}}$ anti-holomorphically. Anti-linear transformations appear in quantum mechanics exclusively for time-reversal transformations. Let us denote ${\widehat{R}|\psi(\mathbf{z})\rangle=|\psi(\mathbf{r}(\overline{\mathbf{z}}))\rangle}$ with ${\widehat{R}}$ the operator of an elementary time-reversal transformation. Because of the anti-unitarity of ${\widehat{R}}$ and its commutation relation with the Hamiltonian (i.e. ${[H,\widehat{R}]=0}$ since we assume ${H}$ to be time-reversal invariant), we obtain ${N(\overline{\mathbf{r}}(\mathbf{z})),\mathbf{r}(\overline{\mathbf{z}}))=\overline{N(\overline{\mathbf{z}},\mathbf{z})}=N(\overline{\mathbf{z}},\mathbf{z})}$ and ${H(\overline{\mathbf{r}}(\mathbf{z}),\mathbf{r}(\overline{\mathbf{z}}))=\overline{H(\overline{\mathbf{z}},\mathbf{z})}=H(\overline{\mathbf{z}},\mathbf{z})}$, which yields

$\displaystyle \partial_{i}\overline{r}^{\overline{\jmath}}(\mathbf{z}) \widetilde{G}_{\overline{\jmath},k}(\overline{\mathbf{r}}(\mathbf{z}),\mathbf{r}(\overline{\mathbf{z}})) \partial_{\overline{l}} r^k(\overline{\mathbf{z}})= \widetilde{G}_{\overline{l},i}(\overline{\mathbf{z}},\mathbf{z}), \ \ \ \ \ (27)$

and

$\displaystyle \partial_{i}\overline{r}^{\overline{\jmath}}(\mathbf{z}) H_{\overline{\jmath}}(\overline{\mathbf{r}}(\mathbf{z}),\mathbf{r}(\overline{\mathbf{z}}))=H_{i}(\overline{\mathbf{z}},\mathbf{z}), \ \ \ \ \ (28)$

$\displaystyle H_{j}(\overline{\mathbf{r}}(\mathbf{z}),\mathbf{r}(\overline{\mathbf{z}}))\partial_{\overline{\imath}}r^{j}(\overline{\mathbf{z}})=H_{\overline{\imath}}(\overline{\mathbf{z}},\mathbf{z}). \ \ \ \ \ (29)$

These relations convert Eq. (11) into

$\displaystyle \begin{cases} +i\frac{d}{dt}\overline{r}^{\overline{\jmath}}(\mathbf{z}(t)) \widetilde{G}_{\overline{\jmath},k}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t))) \partial_{\overline{\imath}} r^k(\overline{\mathbf{z}}(t)) = H_{k}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t)))\partial_{\overline{\imath}}r^{k}(\overline{\mathbf{z}}(t)),\\ -i \partial_i\overline{r}^{\overline{k}}(\mathbf{z}(t)) \widetilde{G}_{\overline{k},j}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t)))\frac{d}{dt} r^{j}(\overline{\mathbf{z}}(t)) = \partial_{i}\overline{r}^{\overline{k}}(\mathbf{z}(t)) H_{\overline{k}}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t))), \end{cases} \ \ \ \ \ (30)$

or, by eliminating the Jacobian of the transformation,

$\displaystyle \begin{cases} -i \widetilde{G}_{\overline{\imath},j}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t)))\frac{d}{d} r^{j}(\overline{\mathbf{z}}(t)) = H_{\overline{\imath}}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t))),\\ +i\frac{d}{dt}\overline{r}^{\overline{\jmath}}(\mathbf{z}(t)) \widetilde{G}_{\overline{\jmath},i}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t))) = H_{i}(\overline{\mathbf{r}}(\mathbf{z}(t)),\mathbf{r}(\overline{\mathbf{z}}(t))) \end{cases} \ \ \ \ \ (31)$

Note that the signs of the two equations have been switched, which is necessary to revert the time evolution of the new coordinates ${(\overline{\mathbf{r}}(t),\mathbf{r}(t))}$. For a time-reversal invariant Hamiltonian ${H}$ and a variational manifold ${\mathcal{M}}$ that contains the time-reversed state ${\widehat{R}|\psi(\mathbf{z})\rangle\in\mathcal{M}}$ for each of its elements ${|\psi(\mathbf{z})\rangle\in\mathcal{M}}$, the flow equations of the time-dependent variational principle are also time-reversal invariant.