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Suppose $\mathbb{U}$ is a unitary operator acting on the Hilbert space of states representing a symmetry transformation such as rotation, translation, etc. $\mathbb{U}$ is said to be a symmetry of non-relativistic quantum mechanics (NRQM) if leaves the Hamiltonian $H$ invariant i.e., $$\mathbb{U}^\dagger H\mathbb{U}=H.$$ This is the statement of symmetry in NRQM in its Hamiltonian formulation.

What is the statement of symmetry in NRQM in its Lagrangian or path-integral formulation? In this case, one uses Lagrangian instead of Hamiltonians and the operators don't make an explicit appearance.

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There is nothing special about Lorentz invariance in QFT. Condensed matter theorists work with non-relativistic QFTs all the time. The answer to your question is the same whether or not you have Lorentz invariance. A symmetry is a transformation which leaves invariant the measure $[D\phi] e^{iS[\phi]}$ of the path integral which defines a theory.

If you'd like an example, consider the translation symmetry of a free non-relativistic particle with path integral

$$\int [D x(t)] \exp(i \int dt \frac12 m \dot x^2)$$ which is invariant under $x(t) \to x(t) + \delta x$.

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In the path integral formulation of NRQM a symmetry of the theory is one which leaves the measure of the path integral invariant (i.e. the jacobian of the transformation is 1). If the measure remains invariant, then a simple change of variables inside the path integral ensures any transition amplitudes will be the same.

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  • $\begingroup$ I know this true in quantum field theory. It would be great if you can give an example from NRQM where the path-integral measure is really not invariant under some transformation. @CsarAlgebra $\endgroup$ – SRS Jul 1 '17 at 15:31

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