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If we have that Lagrangian is equal to L=T-V and Hamiltonian is just H=T+V my question is if for every lagrangian we could consider is just the energy of a particle with a potential -V(x) instead of V(x) what would happen if V(-x)=-V(x)?

In fact if we substitute velocities by momenta in the Lagrangian teh Euler-Lagrange equation for a particle is just (putting -V(x) instead of V(x)) the Hamilton equation [tex] \dot p =-\frac{dH}{dp} [/tex].

then it occurred to me if we could use this fact to quantizy the dynamics of a particle using Lagrangians as if they were the pseudo-energy of the system to get energy levels.

In fact if we substitute velocities by momenta in the Lagrangian teh Euler-Lagrange equation for a particle is just (putting -V(x) instead of V(x)) the Hamilton equation [tex] \dot p =-\frac{dH}{dp} [/tex].

then it occurred to me if we could use this fact to quantizy the dynamics of a particle using Lagrangians as if they were the pseudo-energy of the system to get energy levels.

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