Why not use the Lagrangian, instead of the Hamiltonian, in nonrelativistic QM?

Question

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-Gordon 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 Lagrangians are used in quantum field theory?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 of the value of the Lagrangian approach is deeply ironic since Dirac was the one who first showed that the classical Lagrangian can be applied to QM [1]. It was that same obscure 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

Answer 5

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.

Answer 6

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.

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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.

Answer 7

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.

Answer 8

Hamiltonians are popular in non-relativistic formulations, where the fact the they are not relativistic invariants objects is not important. Lagrangians are popular in particle physics because they are Relativistic Invariants. This only facilitates everything because you do not have to be concerned about what is the new form of the Lagrangian when you are analyzing systems where the relativistic effects occur. Evidently, in High-Energy Physics, everything is mostly in a relativistic regime. If we use Hamiltonians for analyzing relativistic regimes, we would have to be concerned about the changes of the Hamiltonian at such regimes. I hope that this helps.