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?
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 $\exp(i\phi_{\rm random})$.
The Hamiltonian is so useful because it is actually the operator providing translation in time (in autonomous systems). We know that any physical quantity on the phase space in the Hamiltonian formalism is evolved like $$\frac{df}{dt} = \{f,H \}$$ Where $\{\}$ is the Poisson bracket. It is thus natural to say $$\frac{d}{dt} = \{\cdot,H\}$$ and a small evolution of a quantity in time is thus $$f(t + \delta t) = f(t) + \frac{d f}{dt} \delta t = (1 + \delta t \frac{d}{dt})f$$ an exact translation by a time $\Delta t$ can be expressed as $${\rm lim}_{N \to \infty} (1+ \frac{\Delta t}{N} \frac{d}{dt})^N = exp(\Delta t \frac{d}{dt}) = exp(\Delta t \{\cdot,H\})$$
When we pass to quantum mechanics, the canonical quantization procedure tells us to substitute $\{,\} \to [\,,]/(i\hbar)$, where $[\,,]$ is the commutator of two operators. Any physical quantity represented by an operator then is evolved as (again keep in mind the system being autonomous) $$A(t+\Delta t) = exp(- i\Delta t [\cdot,H]/\hbar)A(t) $$ This would be the Heisenberg picture of quantum mechanics. A similar argument leads to the fact that any state is evolved as $$|\psi\rangle (t + \Delta t) = exp(-i \Delta t H/\hbar)|\psi\rangle (t)$$ If $\psi$ is by chance an eigenvector of $H$ with eigenvalue $E$, the evolution is trivial $$|\psi\rangle (t + \Delta t) = exp(-i \Delta t E/\hbar) |\psi\rangle (t)$$ That is, only the phase of the state is changing, but any measurable value of the state stays constant. We call this evolution stationary and this is the reason the eigenvalues and eigenvectors of the Hamiltonian are so important.
For a Lagrangian, this is not true - it's eigenvalues do not point to any directly measurable quantity, they do not have any importance in the theory, and in general they would just be extra work to compute.
Comment to the question (v1): Unlike the Hamiltonian $H$ (which is a constant of motion if there is no explicit time dependence), the Lagrangian $L$, as an observable, is typically not conserved in time. Think e.g. of a harmonic oscillator.
Indeed, Hamiltonian density is an operator that reveals the time evolution of the system as stated before.
The Lagrangian density, on the other hand, gives the so-called space-time evolution of the system in the sense that it is invariant under the transformations of space-time (i.e. Lorentz invariance), not time itself. But also it gives how the action principle work in a covariant sense.
As I will point out below, the eigenvalues of the Lagrangian are telling how much the system satisfies the classical equations of motion. Because this is what Lagrangian does, it is the function to describe how the equations of motion would minimize the variation of the action functional. So, its eigenvalues would give, in a rough sense, the classicalness of the motion, or the minimumness of the action.
However, one should note that this sense is only applicable in quantum mechanics because in classical mechanics, everything should satisfy the equations of motion while in quantum mechanics this not a strict case. In quantum mechanics, especially in quantum field theory, there are particles that can be individually off from being in the minimum action principle. But the system as a whole is again on with respect to the minimum action principle.
First of all let's remember that the Lagrangian formalism is covariant, that is, does not change its form when we switch observers or reference frames.
Therefore, one should not consider the time derivative of an observable, $\frac{d \hat{F}}{dt}$, in order to compute the eigenvalues, but should use the derivative with respect to space-time interval $s$ which is $s=\sqrt{c^2t^2-x^2}$ for a generic observer of a frame S(x,t). Sometimes it is called proper time, when considered in time units (dividing it by the speed of light).
So, in terms of relativistic quantum mechanics: $$ \frac{d \hat{F}}{ds}= -i\hbar [ \hat{F}, \hat{L} ] $$ where $\hbar$ is the reduced Planck constant, and $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ is the commutator of operators A and B. Of course the space-time evolution operator would be $\mathbb{U}(\varsigma)=\exp(i \varsigma \, \hat{L} \, / \hbar)$ where $\varsigma$ is a very small space-time interval. Just similar to the canonical formulation. The only difference is that now we are not talking about time evolution but the space-time evolution. The difference is very tricky in relativistic picture.
So, if one would like to write it in classical mechanics and make sense out of it, s/he need to deal with the relativistic Lagrangian equations, namely one of them reads as $$\frac{dx^\mu}{ds}=\frac{\partial L}{\partial p_\mu}$$ where $x^\mu$ is the 4-position and $p_\mu$ is the 4-momentum. Therefore, we could obtain the classical space-time interval derivative of a function as follows: $$ \frac{d F}{ds}= \{ F, L \}_{rel} $$ where $$\{A,B\}_{rel} = \frac{\partial A}{\partial x^\mu} \frac{\partial B}{\partial p_\mu} - \frac{\partial B}{\partial x^\mu} \frac{\partial A}{\partial p_\mu}$$ is the relativistic version of the Poisson brackets. Here, I used a sum over $\mu=0,1,2,3$ which corresponds to space-time indices (see Einstein's summation convension).
Now, let's take one of the simplest relativistic Lagrangian density, one that is linear in momentum and has no potential energy or mass, namely, $$ L= c \sigma^\mu p_\mu $$ where c is the speed of light, and $\sigma^\mu$ are the Pauli matrices together with the unit matrix as the "fourth" component. This Lagrangian describes a massless particle in space-time which can only travel in the speed of light. The reason that there are Pauli matrices here is not related to the spin, but only because it is simpler to compute and think about it. If I started with $L=p^\mu p_\mu$, it would be inconsistent in units, or if I started with $L=mc^2$ it would be too simple for the sense I would like to discuss to answer the question.
If we compute the eigenvalue of this Lagrangian in the good ol' one-particle quantum mechanics, we obtain the eigenvalues of the 4-momentum for different directions in space (Pauli matrices reveal the directionality of the particle by giving -1 when the 3-momentum is negative so that the energy can be positive again). $$ \hat{L} \, \Psi (s) = \ell \, \Psi (s) $$ where it turns out the eigenvalues would be $\ell = E-pc$ if the momentum is positive or $\ell = E + pc$ if the momentum is negative (where I have written them in 1+1 dimension just for simplicity).
Now, one may think that for massless particles it is already $E=\pm pc$, so $\ell=0$. However, this is not the case, because energy would be equal to the momentum only on-shell which means when you have the classical equations of motion satisfied. Otherwise it is called off-shell.
Please note that in a general case the equations of motions need not to hold. For example the particle could be virtual like in quantum field theory. In that case E-pc isn't necessarily zero.
So, the eigenvalues of the Lagrangian is like the off-shellness of a particle or a field, whether it satisfies the equations of motions or not.
In classical mechanics, it is not useful at all because all the objects satisfies the classical equations of motions. However in quantum mechanics, this is not necessarily true. You have virtual particles and so forth. For example a photon is strictly massless on-shell but if it is a virtual photon, as in scattering of two electrons or Casimir effect etc, then the photon can be like it has a mass, contrary to the equations of motion.
