It is said that since the path-integral formulation of quantum mechanics/or quantum field theory uses the Lagrangian rather than the Hamiltonian, as the fundamental quantity, it preserves all the symmetries of the theory. How can one understand this? I mean, what could be an example in which the Hamiltonian fails to preserve the symmetry of the theory (i.e., action)?
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/37535/2451 , physics.stackexchange.com/q/78508/2451 and links therein. $\endgroup$ – Qmechanic♦ Nov 27 '15 at 13:34
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In relativistic QFT for instance, the Hamiltonian does not respect Lorenz invariance.
That is not to say that the symmetry is not present, in the Hamiltonian. Of course it is, the Lagrangian and Hamiltonian are equivalent ways of analysing the same thing.
But the Hamiltonian does not have Lorenz invariance autocratically built into it, as it singles out the time direction as a special direction.
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$\begingroup$ Usually interaction hamiltonian coincides with interaction lagrangian, so that it preserves Lorentz invariance as well as lagrangian. Also the central object in QFT is S-operator, which is expressed through interaction hamiltonian. $\endgroup$ – Name YYY Jan 2 '16 at 15:30
