The question is about:

(1) whether giving a Lagrangian is sufficient enough to (uniquely) well-define a Hamiltonianian quantum theory with a Hilbert space?

The answer should be Yes, or No.

If yes, then I suppose one can write any of its state function $|\Psi \rangle$ in the Hilbert space, with a Hamiltonian $H$, satisfies:

$ i \partial_t |\Psi \rangle=H |\Psi \rangle $

Then, whether the situation changes if one replacing Lagrangian $L$ to an action $S$ or a path integral or partition function $Z$, i.e.

(2) whether this $L$, $S$ or $Z$ to which level (uniquely) well-defines a Hamiltonianian quantum theory with a Hilbert space? Or, they cannot well-define it yet?

(3) The even further question is whether a Lagrangian is enough to well-define Hamiltonianian quantum theory with a Hilbert space on the discrete lattice? From this arXiv paper 1305.1045, it is obvious that even if we have a Lagrangian for the standard model, it is NOT enough to well-define a Hamiltonianian quantum theory with a Hilbert space on the lattice NON-PERTURBATIVELY.


It will be important to explain why this is so. I may be wrong, but I had impression reading some post comment about questions by an expert at Physics Stackexchange(Dr. Luboš Motl?) stating that knowing Lagrangian/action is everything about knowing the physical system. Sorry if I made a mistake here.

But my interactions with other experts in the field of condensed matter, often stated that Lagrangian is not yet enough. The quantization, Hamiltonian and Hilbert space etc is needed.

It will be nice to touch the issues of gauge theory, whether giving a gauge theory Lagrangian can define a Hamiltonianian quantum theory with a Hilbert space? As I know the 2+1D $Z_N$ gauge theory written as 2+1D U(1)xU(1) Chern-Simons theory is discussed in this post. In that sense givging a U(1)xU(1) Chern-Simons theory can mean $Z_N$ gauge theory (with discrete $Z_N$ symmetry) or something else(two copies of U(1) symmetry).


  • $\begingroup$ ps. I will be happy just hearing comments/answers about (1)(2), where the (3) defining on the lattice or defining non-perturbatively can be saved for the later. Thanks. :) $\endgroup$
    – wonderich
    Commented Oct 23, 2013 at 4:57

1 Answer 1


I can answer (1) and (2). The answer is: NO. Passing form classical mechanics to quantum one requires, in general, to add more information. There is no rigorous machinery allowing one to write the quantum corresponding of a classical object. Physically speaking, this is because quantum structures are more fundamental in Nature than classical ones.

Mathematically speaking the problems arise when you have to define precise self-adjoint operators in order to exploit the spectral machinery. I mean the following. Every formal rule from classical to quantum picture, concerning observables - i.e., a machinery associating classical observables to corresponding operators, describing the "same" observables at quantum level - gives rise to, at most symmetric operators. Instead proper self-adjoint operators are necessary to describe quantum observables.

As a matter of fact it happens that a symmetric operator admits one, many or none self-adjoint extension. The second case is the most common. Here, some further physical information must enter the formalism to chose the "right" self-adjoint extension. This further information is not included in the Lagrangian or Hamiltonian classical description of the considered system.

  • $\begingroup$ I presume that OP includes identification of statistics of the fields (bosonic/fermionic) as part of the quantization / regularization procedure given together with the Lagrngian? $\endgroup$
    – Slaviks
    Commented Dec 16, 2013 at 18:23
  • $\begingroup$ Yes, you are right, that is a further problem. However it also involves the notion of spin (in view of spin statistics theorem) that is a completely quantum one. Is there a classical corresponding, for instance, of Dirac field? $\endgroup$ Commented Dec 16, 2013 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.