# Euler-Lagrange Equations for Molecular Dynamics

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?

• Is it because we usually denote $x=(x_1,x_2,...,x_n)$ as a row vector and bra vectors are equivalent? Jul 27, 2019 at 10:50

## 1 Answer

Short answer: It doesn't matter if you start with bra- or ket-vectors..


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.