Physics Asked on November 26, 2020
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., $$mathbb{U}^dagger Hmathbb{U}=H.$$ 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.
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.
Answered by CStarAlgebra on November 26, 2020
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 $[Dphi] e^{iS[phi]}$ 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 $x(t) to x(t) + delta x$.
Answered by Dwagg on November 26, 2020
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