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

Question

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.

Answer 1

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 $x(t) \to x(t) + \delta x$.

Answer 2

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.