# What is a symmetry in the path-integral formulation of non-relativistic quantum mechanics?

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.

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