Undergraduate classical mechanics introduces both Lagrangians and Hamiltonians, while undergrad quantum mechanics seems to only use the Hamiltonian. But particle physics, and more generally quantum field theory seem to only use the Lagrangian, e.g. you hear about the Klein-Gordan Lagrangian, Dirac Lagrangian, Standard Model Lagrangian and so on.

Why is there a mismatch here? Why does it seem like only Hamiltonians are used in undergraduate quantum mechanics, but only Lagrangains are used in quantum field theory?

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    $\begingroup$ Both methods are equivalent and are used, to tell the truth. Momenta and coordinates had been used before QM in the old (Bohr) quantization, remember quantization of the phase space $\int dpdq$. $\endgroup$ – Vladimir Kalitvianski Mar 4 '12 at 17:57
  • $\begingroup$ Perhaps it's worth noting that the Lagrangian/path integral approach is very poorly suited to the study of bound state problems. Just try the hydrogen atom with the Lagrangian approach, even Feynman couldn't do it! $\endgroup$ – KF Gauss Jun 1 '18 at 16:57

In order to use Lagrangians in QM, one has to use the path integral formalism. This is usually not covered in a undergrad QM course and therefore only Hamiltonians are used. In current research, Lagrangians are used a lot in non-relativistic QM.

In relativistic QM, one uses both Hamiltonians and Lagrangians. The reason Lagrangians are more popular is that it sets time and spacial coordinates on the same footing, which makes it possible to write down relativistic theories in a covariant way. Using Hamiltonians, relativistic invariance is not explicit and it can complicate many things.

So both formalism are used in both relativistic and non-relativistic quantum physics. This is the very short answer.

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    $\begingroup$ There is also an covariant Hamiltonian formalism of field theory, in which the phase space is infinite dimensional or one uses the language of multisymplectic. Either way the mathematics is too sophisticated to be covered in (under)graduate courses. $\endgroup$ – Tobias Diez Mar 5 '12 at 23:10
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    $\begingroup$ Why must path integrals be used in order to use the Lagrangian formalism for QM? $\endgroup$ – Stan Shunpike Feb 6 '15 at 4:32
  • $\begingroup$ @StanShunpike If you apply the Lagrangian formalism to quantum mechanics you end up with the path integral formalism. That's how Feynman discovered it. If you use the Hamiltonian formalism, you end up with the usual canonical formulation. If you use the Newtonian formalism, well, you end up with Bohmian mechanics. C.f. physicstravelguide.com/frameworks $\endgroup$ – Tim Dec 13 '17 at 10:00

As Weinberg points in his QFT book, in the Hamiltonian formalism it is easier to check the unitarity of the theory because unitarity is directly related to evolution, while in the Lagrangian formalism the symmetries that mix space with time are more explicit. Therefore the Hamiltonian formalism is usually more convenient in non-relativistic and galilean quantum theories.

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    $\begingroup$ What do you mean by more explicit? $\endgroup$ – Stan Shunpike Feb 6 '15 at 4:31
  • $\begingroup$ @StanShunpike In order for a theory to be Poincare invariant, the Lagrangian needs to be a Poincare scalar, what it is easy to see. The equivalent condition in the Hamiltonian formalism is that there is a Poincare algebra with the Hamiltonian as the zero component of the 4-momentum. This condition needs to be checked, as it is not elemental to see. $\endgroup$ – Diego Mazón Feb 8 '15 at 2:00

I would say because of the way you efficiently solve problems as well as pedagogy. Both are used in both cases though.

The Hamiltonian operator approach emphasises the spectrum aspects of quantum mechanics, which the student is introduced to at this point $-$ but here is a Lagrangian

$$\mathcal{L}\left(\psi, \mathbf{\nabla}\psi, \dot{\psi}\right) = \mathrm i\hbar\, \frac{1}{2} (\psi^{*}\dot{\psi}-\dot{\psi^{*}}\psi) - \frac{\hbar^2}{2m} \mathbf{\nabla}\psi^{*} \mathbf{\nabla}\psi - V( \mathbf{r},t)\,\psi^{*}\psi$$

for the Schrödinger equation $$\frac{\partial \mathcal{L}}{\partial \psi^{*}} - \frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial\frac{\partial \psi^{*}}{\partial t}} - \sum_{j=1}^3 \frac{\partial}{\partial x_j} \frac{\partial \mathcal{L}}{\partial\frac{\partial \psi^{*}}{\partial x_j}} = 0.$$

The Lagrangian (density) is especially relevant for the path integral formulation, and in some way closer to bring out symmetries of a field theory. Noether theorem and so on. $-$ but I remember Peskin & Schröders book on quantum field theory starts out with the Hamiltonian approach and introduces path integral methods only 300 pages in.


I think the Hamiltonian approach is emphasized in undergraduate due more to habit and the influence of Dirac, rather than due to any profound mathematical reason. The Hamiltonian is also easier to teach because it is compatible with classical intuitions of time.

Historically, Dirac argued strongly for the primacy of the Hamiltonian, literally until shortly before his death. My own interpretation of an oblique reprimand of the Lagrangian that Dirac made in his Lectures on Quantum Mechanics (1966) (a great read!) is that Dirac was unhappy with the fame that Feynman was acquiring, although Dirac was always so reserved in expressing discontent with other physicists that it's very hard to say for sure. Dirac's downplaying the value of Lagrangian approach is of course highly ironic, since it was Dirac who first showed that the classical Lagrangian can be applied to QM, in an early paper [1]. It was that same paper that many years later inspired and unleashed Feynman's remarkable QED work.

[1] P. A. M. Dirac, The Lagrangian in Quantum Mechanics, Phys. Zs. Sowjetunion 3 (1933) No. 1; reprinted in: J. Schwinger (Ed.), Selected Papers on Quantum Electrodynamics, 1958, No. 26

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    $\begingroup$ Dirac didn't give a hoot about fame. He famously wanted to reject the Nobel prize, but was told that would just make him more famous. He was annoyed for the same reason that he gave up the path integral--- he couldn't figure out what to do in the case that the Hamiltonian wasn't quadratic in the momenta. This was ignored by Feynman as well, and it is only resolved by a more general view of path integration than that available in Feynman's work. The quadratic momentum case is enough for field theory, unfortunately, so people don't notice that the formalism as usually presented is incomplete. $\endgroup$ – Ron Maimon Mar 24 '12 at 7:54
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    $\begingroup$ @RonMaimon interesting, thanks! I was not aware of that specific concern by Dirac. You wouldn't happen to have a quick reference on that, would you?... And overall, the more original work I've read by Dirac, the more my jaw drops. He was an amazing and (I think) under-appreciated thinker, even given his substantial fame. $\endgroup$ – Terry Bollinger Mar 24 '12 at 8:12
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    $\begingroup$ Bolinger: Dirac is a great physicist, he is the founder of high energy physics, but I think people already recognize that. The story I read regarded the infamous Pocono conference (or shelter island, I forget which is which) where Feynman presented path-integrals and diagrams. Dirac commented that this formalism is not apparently unitary. In his lectures on field theory from the 1960s, he makes the case that quadratic momenta are the only thing the path integral handles. I might be getting the cites wrong, I read it a long, long time ago. $\endgroup$ – Ron Maimon Mar 24 '12 at 8:33
  • $\begingroup$ @RonMaimon There is a path integral formulation of the Hamiltonian formalism. It is equivalent to the usual formulation for theories quadratic in the momenta. $\endgroup$ – orbifold May 16 '12 at 20:55
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    $\begingroup$ @orbifold: And many people said (stupidly) that the p-q path integral is not well defined because p and q don't commute (for example, Sidney Coleman used to say this, it's totally wrong). Further, Dirac would have considered the p-q path integral to be equivalent to the canonical formalism (which it is), since it picks out a p-q decomposition and a time-decomposition. It's only Feynman's form (after doing the p integral) that is covariant under relativity. $\endgroup$ – Ron Maimon May 17 '12 at 0:16

In few words

  1. Unitarity of evolution operator U(t) is easy to see with Hamiltonian formalism.
  2. Lorentz invariance of S-matrix (scattering matrix) is easy to see with Lagrangian formalism.

It is not true "that QFT and particle physics rely instead on Lagrangian"

The generator of time translations in quantum theory is the Hamiltonian, not the Lagrangian; therefore, we need a Hamiltonian to study evolution of the quantum system.

As mentioned in the Volume 1 of Weinberg's textbook on QFT, chapter 7:

It is the Hamiltonian formalism that is needed to calculate the S-matrix (whether by operator or path-integral methods) but it is not always easy to choose Hamiltonians that yield a Lorentz-invariant S-matrix.

The point of the Lagrangian formalism is that it makes it easy to satisfy Lorentz invariance and other symmetries: a classical theory with a Lorentz-invariant Lagrangian density will when canonically quantized lead to a Lorentz-invariant quantum theory. That is, we shall see here that such a theory allows the construction of suitable quantum mechanical operators that satisfy the commutation relations of the Poincaré algebra, and therefore leads to a Lorentz-invariant S-matrix.

Therefore the usual recipe consists on postulating some Lagrangian, checking it satisfies certain basic properties, then deriving a Hamiltonian from that Lagrangian, and finally using this Hamiltonian to compute the elements of the S-matrix.


In my opinion, the existing choice between the canonical (Hamiltonian) and path-integral (Lagrangian) formalisms is a far-reaching consequence of particle-wave dualism in QM. The first emphasizes the spectral aspects, the second can be viewed as a deep generalization of Fermat's principle for rays propagation in optics. Since most of experiments in particle physics represent some sort of scattering, the wave aspects are usually more important, hence, the Lagrangian formalism is much more adequate for the practical use.


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