7
$\begingroup$

As the title implies, why is it that the most common formalisms we use in quantum mechanics prefer to describe systems in the terms of a Hamiltionian instead of a Lagrangian?

Is there some convenience to defining our systems one way over the other? Are there cases I'm not aware of where Lagrangian formalism is preferred?

$\endgroup$
  • $\begingroup$ Would be interesting for someone to give a pro argument for using Lagrangians. $\endgroup$ – BMS Nov 22 '13 at 3:12
  • 1
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/21866/2451 and links therein. $\endgroup$ – Qmechanic Nov 22 '13 at 11:04
3
$\begingroup$

It's because they're based on the historical approach: Schroedinger's equation.

Schroedinger's equation was discovered on its own before we knew about canonical quantization. Dirac came up with the canonical quantization rules which re-wrote (and generalized) Schroedinger's equation into the familiar one we have today, $\hat{H} \psi = i \dot{\psi}$.

That said, there is an approach which uses the action (and thus the Lagrangian or Lagrangian density) due to Feynman: The path integral approach. This approach has as its biggest advantage the ability to be reconciled with Special Relativity, which proved much too difficult a task for extensions of the Schroedinger equation (the Dirac Equation was the most successful attempt, but wasn't general enough to describe some phenomena).

This is what is used in the most advanced quantum physics such as quantum field theory, quantum electrodynamics being the best example. But unless you're interested in high-energy particle physics or really advanced condensed matter physics, the traditional quantum mechanics is sufficient.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

I don't have an answer for why there is no simple Lagrangian formulation, but I can explain some of why a Hamiltonian one is easy. Part of the way to go from Classical Mechanics to Quantum is by replacing Poisson brackets with commutators, and observables with operators on Hilbert space and their expectation values. So the equation

$\frac{d}{dt} f(q, p, t) = \left\{ f,H \right\} + \frac{\partial f}{\partial t} $

becomes the quantum

$\frac{d}{dt} \langle f \rangle = -i \langle\left[f,H \right]\rangle + \langle \frac{\partial f}{\partial t} \rangle.$

So the Hamiltonian is convenient because it gives the time evolution of operators, states, and expectation values directly. Also, because the Hamiltonian is a conserved quantity, stationary states (i.e. those that do not evolve in time) will be eigenvectors of the Hamiltonian, and eigenvalue problems are easy.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This answer is much closer to what I'm looking for! Isn't the Lagrangian also a conserved quantity whenever the Hamiltonian is conserved? Or am I not remembering the Lagrangian correctly? $\endgroup$ – user28754 Nov 22 '13 at 2:44
  • 1
    $\begingroup$ If the lagrangian does not depend explicitly on time, then the hamiltonian is conserved. You can tell the lagrangian because the potential energy goes down, the kinetic term goes up, both of which increase the lagrangian. $\endgroup$ – ZachMcDargh Nov 22 '13 at 3:13
0
$\begingroup$

I can think of several reasons for why using Hamiltonians is preferred, but the most important, I'ld say, is that you need to use path integral formalism in order to formulate (non relativistic) QM in terms of the Lagrangian, which, for an undergrad course, is a bit of an overkill.

Also, many of the most renowned equations in QM like, say, the Schrödinger Equation, use the Hamiltonian: $\hat H \Psi=\hat E \Psi$ Thus, although it is possible, why change it? It would be quite a pain to do so.

For what I understand, however, modern QM relies heavily on both the Hamiltonian and the Lagrangian formalism.

Hope it helped!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for offering some thoughts on the matter! Overkill for an undergrad course is unfortunately not much of a rationale. As for the the Schrödinger Equation, is there a particular reason why using Hamiltonian formulation results in something 'cleaner' or more usable than Lagrangian? $\endgroup$ – user28754 Nov 22 '13 at 1:54
  • $\begingroup$ Sorry for not being much more rational, but I wasn't sure what type of answer you where looking for :p. Anyways, yes, there are several reasons as to why use the Hamiltonian, first of all it provides an answer in terms of momentum, and allows us to predict further onto the future, also, it is derived from a Lagrangian so it can be reversed back into one. As for the Schrödinger Equation, the Hamiltonian explains the time-evolution of the wave function, in terms of plank's and i, that is, which is clearly advantageous when dealing with the evolution of a system. $\endgroup$ – Demian Licht Nov 22 '13 at 2:00
  • $\begingroup$ Not a problem! I am looking for something more specific, however. Lagrangian formalism provides a perfect framework for examining the evolution of a system in time with classical systems - why would a Hamiltonian allow us to better see into the future than a Lagrangian? Sadly, 'clearly advantageous' isn't much of an rationale. Clearly advantageous as opposed to what? $\endgroup$ – user28754 Nov 22 '13 at 2:04
  • $\begingroup$ Ok, I think I get to where you're going, my best explanation would be that the Hamiltonian is easier to diagonalize for it has less degrees of freed, sadly I am not capable of fully explaining this over the a thread, you could try to access this page: mathpages.com/home/kmath523/kmath523.htm or contact me, I would be more than happy to help you! $\endgroup$ – Demian Licht Nov 22 '13 at 2:09