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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
up vote 7 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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I agree with you that the lagrangian can't be an operator because the second derivative terms that spoil down the notion of $L(x(t),\dot{x}(t),...)$. But if the action is promoted to $S[\hat{x}(t)]$, with $\hat{x}(t)$ in the Heisenberg's formalism. The condition $\delta S =0$ now can be give us the Heisenberg equation of motion. Something like $0=\int_{t_a}^{t_b}(\dot{q}-[H,p])\delta q + (\dot{p}-[H,q])\delta p + ...$. – Nogueira Dec 5 '15 at 20:59
Dear Nogueira, I may be wrong but it seems to me that you are simply sloppy about the difference between c-numbers and operators. For example, is your action $S$ in your formalism a c-number, or an operator? I think that both answers lead to problems with your claims. If it is an operator, then it can't really be extremized because an operator generally has infinitely many eigenvalues etc. What does it mean that the whole operator is minimized? If it is a c-number, you need to add a trace or something like that. But you will still have problems to define delta p etc. – Luboš Motl Dec 6 '15 at 9:15
I'm aware of this problems. What I'm thinking is in the Schwinger principle. Here the increment of an action is promoted to an operator that generate all possible variations and the expectation value of this operator in time ordering give us the usual action inside the path integral. The variations: on operators now is something like $\delta p = [G,p] \delta \epsilon $. – Nogueira Dec 6 '15 at 14:31
Dear Lubos, Now I'm trying to define this "variation of action operator" by a more correct way. I'm taking the matrix elements $\langle 2|T{\delta_{\epsilon} \hat{S}}|1\rangle = \int [d\phi] \delta_{\epsilon} S[\phi] \exp (\frac{i}{\hbar}S[\phi]) $. I'm facing problems only when I make time variation on the space-like boundaries of the action. – Nogueira Dec 7 '15 at 5:22
Hi, before you fall into the formalism, I don't understand why you expect something like that to exist or be helpful. Quantization is either about turning quantities into operators, or integrating over all histories with actions defining the weights. In some sense, you are trying to do both at the same moment. In Feynman's approach, it is shown that it is not strictly true that only the extemum histories contribute. All contribute. You are trying to contradict it and only allow "one" history - but that's really against the interference prescribed by quantum mechanics. – Luboš Motl Dec 18 '15 at 15:02

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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