The linear response equations for the variational Lagrangian

In order to find the first-order parameter and multiplier response in the variational Lagrangian theory, we start by taking the total derivative of equation variational Lagrangian conditions and then apply the chain rule, from which we find an equation for the first-order wave function response: \[\begin{equation} \require{physics} \sum_j^x \qty( \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}, \boldsymbol{\lambda} ) }{\lambda_a}{p_j} }_{ \vb{p}^\star(\boldsymbol{\eta}_0), \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) } ) \qty( \eval{ \pdv{ p_j^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) = - \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0), \boldsymbol{\lambda} ) }{\lambda_a}{\eta_m} }_{ \boldsymbol{\eta_0}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) } \label{eq:Lagrangian_eq_for_wf} \thinspace , \end{equation}\] which can also be cast into the form of Hooke’s law: \[\begin{equation} \vb{k}_{\vb{p}} \thinspace \vb{x} = - \vb{F}_{\vb{p}} \thinspace . \end{equation}\]

The parameter response force can in this case be calculated as \[\begin{equation} \qty( \vb{F}_{\vb{p}} )_{am} = \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0), \boldsymbol{\lambda} ) }{\lambda_a}{\eta_m} }_{ \boldsymbol{\eta_0}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) } = \eval{ \pdv{ f_a ( \boldsymbol{\eta}, \vb{p}^\star(\boldsymbol{\eta}_0) ) }{\eta_m} }_{\boldsymbol{\eta}_0} \end{equation}\] and the parameter response force constant as: \[\begin{equation} \qty( \vb{k}_{\vb{p}} )_{ai} = \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}, \boldsymbol{\lambda} ) }{\lambda_a}{p_j} }_{ \vb{p}^\star(\boldsymbol{\eta}_0), \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) } = \eval{ \pdv{ f_a(\boldsymbol{\eta}_0, \vb{p}) }{p_i} }_{\vb{p}^\star(\boldsymbol{\eta})} \thinspace . \end{equation}\]

For the first-order multiplier response, we immediately find an equation for the first-order response of the Lagrangian multipliers after taking the appropriate total derivative: \[\begin{equation} \begin{split} & \sum_a^y \qty( \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}, \boldsymbol{\lambda}) }{\lambda_a}{p_i} }_{ \vb{p}^\star(\boldsymbol{\eta}_0), \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) } ) \qty( \eval{ \pdv{ \lambda_a^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \\ & \hspace{12pt} =- \eval{ \pdv{ \mathscr{L}( \boldsymbol{\eta}, \vb{p}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) ) }{p_i}{\eta_m} }_{ \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0) } - \sum_j^x \qty( \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0)) }{p_i}{p_j} }_{ \vb{p}^\star(\boldsymbol{\eta}_0) } ) \qty( \eval{ \pdv{ p_j^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \thinspace , \end{split} \end{equation}\] which can be recast as \[\begin{align} \vb{k}_{\boldsymbol{\lambda}} \thinspace \vb{y} &= - \vb{F}_{\boldsymbol{\lambda}} \\ &= - \vb{G}_{\boldsymbol{\lambda}} - \vb{A}_{\boldsymbol{\lambda}} \thinspace \vb{x} \thinspace . \end{align}\]

The multiplier response force constant is the transpose of the parameter response force constant: \[\begin{equation} \qty( \vb{k}_{\boldsymbol{\lambda}} )_{ia} = \qty( \vb{k}_{\vb{p}} )_{ai} \thinspace , \end{equation}\] and in order to calculate the multiplier response force, we need the following expressions: \[\begin{align} \qty( \vb{G}_{\boldsymbol{\lambda}} )_{im} &= \eval{ \pdv{ \mathscr{L}( \boldsymbol{\eta}, \vb{p}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) ) }{p_i}{\eta_m} }_{ \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0) } \\ &= \eval{ \pdv{ E(\boldsymbol{\eta}, \vb{p}) }{p_i}{\eta_m} }_{ \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0) } + \sum_a^y \lambda_a^\star(\boldsymbol{\eta}_0) \qty( \eval{ \pdv{ f_a(\boldsymbol{\eta}, \vb{p}) }{p_i}{\eta_m} }_{ \boldsymbol{\eta}_0, \vb{p}^\star(\boldsymbol{\eta}_0) } ) \end{align}\] and \[\begin{align} \qty( \vb{A}_{\boldsymbol{\lambda}} )_{ij} &= \eval{ \pdv{ \mathscr{L} ( \boldsymbol{\eta}_0, \vb{p}, \boldsymbol{\lambda}^\star(\boldsymbol{\eta}_0) ) }{p_i}{p_j} }_{ \vb{p}^\star(\boldsymbol{\eta}_0) } \\ & = \eval{ \pdv{ E(\boldsymbol{\eta}_0, \vb{p}) }{p_i}{p_j} }_{ \vb{p}^\star(\boldsymbol{\eta}_0) } + \sum_a^y \lambda_a^\star(\boldsymbol{\eta}_0) \qty( \eval{ \pdv{ f_a(\boldsymbol{\eta}_0, \vb{p}) }{p_i}{p_j} }_{\vb{p}^\star(\boldsymbol{\eta}_0)} ) \thinspace . \end{align}\]