hamiltonian formalism - Eigenvalues of the Lagrangian?

It is often stated that the Lagrangian formalism and the Hamiltonian formalism are equivalent.

We often hear people talk about eigenvalues of Hamiltonians but I have never ever heard a word about eigenvalues of Lagrangians.

Why is this so? Is it not useful? is it not possible to do it?

Answer

I agree with Qmechanic but just to put a different perspective. While one may write down formulae for the Lagrangian, like $$ L = \frac{p^2}{2m} - U(x) $$ which only differs from the Hamiltonian by the minus sign, and while it's possible to simply put hats above all the operators, unlike the Hamiltonian, the Lagrangian isn't a natural operator in any sense.

The reason is simple. In classical physics, the Lagrangian is meant to be nothing else than the integrand that defines the action $$ S = \int dt\, L(t)$$ and the meaning of the action – its defining property – is that it is extremized among all possible trajectories at the trajectories obeying the classical equations of motion: $$ \delta S = 0 $$ To promote the Lagrangian to an operator would mean to promote the action to an operator as well. But if it were so, we would have to compute an operator-valued function of a classical trajectory, $S[x(t)]$. But this is a contradiction because any classical trajectory $x(t)$ is, by assumption, classical, so it is a $c$-number, so any functional calculated out of these $c$-numbers are $c$-numbers as well. They're not operators.

That's why the Lagrangian and the action don't really enter the "operator formalism" of quantum mechanics at all. Instead, the right promotion of the Lagrangian and the action into the world of quantum mechanics is Feynman's description of quantum mechanics in terms of path integrals. In that picture, one directly calculates the probability (transition) amplitudes by summing over all classical trajectories while $\exp(iS/\hbar)$ is the integrand in the sum (integral) over trajectories (histories). This Feynman's picture immediately explains why the action was extremized in classical physics. Near the trajectories that extremize $S$, the value of $S$ is nearly constant to the leading approximation, so these trajectories "constructively interfere", while all other trajectories nearly cancel because their contributions are random phases $\exp(i\phi_{\rm random})$.