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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?

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You might be interested in this question. – ACuriousMind Jul 19 '14 at 14:38

3 Answers 3

up vote 6 down vote accepted

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 $\exp(i\phi_{\rm random})$.

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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.

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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.

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