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I often read that the Lorentz symmetry is manifest in the path integral formulation but is not in the canonical quantization - what does this really mean?

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up vote 4 down vote accepted

Manifest Lorentz symmetry means that one can see Lorentz invariance directly from the way the theory is formulated; typically when space and time are treated on the same footing as components of a 4-vector. In these cases, the Lorentz group generators are represented in a simple way (hence the ''manifest'' symmetry), but it is far less trivial to find a corresponding Hilbert space of state vectors on which the interacting energy-momentum 4-vector acts.

However, a theory can be Lorentz invariant in a more indirect way, such as in the canonical formalism, where a Hilbert space and an associated Hamiltonian is specified directly. Then Lorentz invariance is established by proving the (then far less trivial) existence of 6 generators satisfying the commutation relations for the Lorentz generators, such that the interacting Hamiltonian and the free momentum generators transform jointly as a 4-vector.

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The canonical formulation is based on a Hamiltonian framework which requires the definition of a time coordinate. So, all of the quantities you calculate depend on this choice of time, and so it is not obvious that everything is Lorentz invariant.

Path integrals respect lorentz invariance from start to finish, as they are based on a Lagrangian framework.

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The action requires the definition of a time coordinate as well. I guess you mean the Lagrangian is chosen such that the action is Lorentz invariant which is easy to show. – Physiks lover Sep 16 '12 at 13:35
The Hamiltonian formalism requires choosing a time coordinate, and singling out time derivatives and treating them specially as you create canonical momenta and perform a Legendre transform. A $\partial^{\mu}\phi\partial_{\mu}\phi$ term will lose its explicit lorentz invariance and become $\pm \left(\dot \phi^{2} + \partial_{i}\phi\partial^{i}\phi\right)$, with the choice of $\pm$ depending on whether you choose the metric to be $+---$ or $-+++$. – Jerry Schirmer Sep 16 '12 at 14:09
By specifying "$x^{0}$ is the time component", you've just chosen a time coordinate. You do this always when you choose coordinates that are non-null (i.e., not light-like)! The trick is the Hamiltonian formalism preserves Lorentz invariance when we keep the choice arbitrary, i.e., just specify the topology of spacetime as space+time... – Alex Nelson Sep 30 '12 at 6:12
@AlexNelson: Except you can take the variation with respect to $\phi$ and $\partial_{\mu}\phi$, which <b>is</b> manifestly Lorentz invariant. Doing this in the Hamiltonian formalism is impossible, since space and time variables are treated inherently differently. – Jerry Schirmer Sep 30 '12 at 6:45

In Lorentz invariant theories:

  • The Lagrangian density is a Lorentz scalar.

  • The Hamiltonian is the generator of time translations (as Jerry Schirmer correctly says, one needs a specific time variable to begin with) and thus transforms as the zero component of a four vector. And it is less obvious to say if a quantity is a zero component of a four vector than to say if it is a scalar. The best way to say if a field theory with a given Hamiltonian is Lorentz invariant is to work out the commutators (or Poisson brackets if the theory is classical) of the Hamiltonian with the boost and angular momentum generators to check if they close the Lorentz algebra.

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