Why not using Lagrangian, instead of Hamiltonian, in non relativistic QM?

Question

When we studied classical mechanics on the undergraduate level, on the level of Taylor, we covered Hamiltonian as well as Lagrangian mechanics.

Now when we studied QM, on the level of Griffiths, we always dealt with Hamiltonians not Lagrangians. Why is that? I know that QFT and particle physics rely instead on Lagrangian!!

1. Could someone provide some insights on why, it seems that, "only" Hamiltonians are used in undergrad QM.

2. Also it would be great if providing some insights on why "only" Lagrangians appear in high energy physics (KG Lagrangian, Dirac Lagrangian, Standard Model Lagrangian...etc)

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

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 3

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 he published in an obscure Russian journal. It was that same paper that many years later inspired and unleashed Feynman's remarkable QED work.

Answer 4

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 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.