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In the Car-Parrinello (CP) method for molecular dynamics simulation, the Euler-Lagrange equations are given as

$$ \begin{aligned} \frac { d } { d t } \frac { \partial \mathcal { L } _ { \mathrm { CP } } } { \partial \left\langle \dot { \psi } _ { i } \right| } & = \frac { \partial \mathcal { L } _ { \mathrm { CP } } } { \partial \left\langle \psi _ { i } \right| }.\end{aligned} $$

Why are we using the bra-vectors of the basis functions $\left\langle \psi _ { i } \right|$ here, instead of the ket vectors?

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  • $\begingroup$ Is it because we usually denote $x=(x_1,x_2,...,x_n)$ as a row vector and bra vectors are equivalent? $\endgroup$ – exp ikx Jul 27 at 10:50
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Short answer: It doesn't matter if you start with bra- or ket-vectors..

$\newcommand{\bra}[1]{\langle #1 \rvert} \newcommand{\ket}[1]{\lvert #1 \rangle} \newcommand{\LCP}[0]{\mathcal{L}_{\mathrm{CP}}}$ Longer answer:

When you set up the Euler-Langrange equation for the bra-vectors $\bra{\psi_i}$ $$ \frac{d}{dt} \frac{\partial\LCP}{\partial\bra{\dot{\psi_i}}} = \frac{\partial\LCP}{\partial\bra{\psi_{i}}},$$ then you get differential equations for the time-evolution of the ket-vectors $\ket{\psi_i(t)}$.

On the other hand, when you chose to set up the Euler-Langrange equations for the ket-vectors $\ket{\psi_i}$ $$ \frac{d}{dt} \frac{\partial\LCP}{\partial\ket{\dot{\psi_i}}} = \frac{\partial\LCP}{\partial\ket{\psi_{i}}},$$ then you get differential equations for the time-evolution of the bra-vectors $\bra{\psi_i(t)}$.

You can work out the details and see: In the end the differential equations for $\ket{\psi_i(t)}$ and $\bra{\psi_i(t)}$ look the same. Hence you can freely choose which one you prefer (by convention most people prefer the differential equations for the ket-vectors). The physical results from both are the same.

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