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Why does Quantum Field Theory use usually Lagrangians rather than Hamiltonains?

I heard many reasons, but I'm not sure which is true.

Some say it's just a matter of beauty, so Lagrangians are more beautiful because they don't break/separate the space-time variables (so space-time is a single variable, like in the Klein-Gordon Lagrangian and Hamiltonian).

Some say that Hamiltonians are not always Lorentz invariant.

Could someone explain in a little bit more details?

Thanks.

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marked as duplicate by Emilio Pisanty, Qmechanic Sep 25 '13 at 17:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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First of all, in condensed-matter physics one often uses the Hamiltonian instead, so when you say that QFT "usually" uses Lagrangians, you really mean relativistic QFT. In any case, Weinberg emphasizes in his book that each of the two formalisms has its own virtues: the Hamiltonian formalism makes unitarity manifest, while the Lagrangian formalism highlights the symmetries of the system. –  Tomáš Brauner Sep 25 '13 at 11:09
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This question has been asked before. –  NikolajK Sep 25 '13 at 11:18
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Essentially a duplicate of physics.stackexchange.com/q/21866/2451 –  Qmechanic Sep 25 '13 at 11:48

1 Answer 1

up vote 7 down vote accepted

You have already mentioned the correct reason---the Lagrangian is manifestly Lorentz-invariant whereas the Hamiltonian is not. Since a relativistic field theory must be build of Lorentz-invariant quantities only the Lagrangian approach is good.

Compare for example the expressions for a free real scalar field $\phi$ $$ \mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi $$ This is Lorentz-invariant, because the Lorentz index $\mu$ is contracted in this way. The Hamiltonian for this theory is $$ H=\frac{1}{2}\dot\phi^2+\frac{1}{2}(\vec\nabla\phi)^2 $$ which is not manifestly Lorentz-invariant.

Is there any other reason?

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