The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$.

The definition of the exponential for an unbounded self-adjoint operator requires theorems from Borel functional calculus that say that for every measurable function $f$ on the reals the expression $f(O)$ for $O$ a self-adjoint operator defines a unique operator with the property that $f(O)v_\lambda = f(\lambda) v_\lambda$ for every eigenstate $v_\lambda$ with eigenvalue $\lambda$. Naively, you might even take this as the *definition* of $f(O)$.

You can find the proofs of these assertions for example in chapter VIII of *"Methods of Modern Mathematical Physics"* by Reed and Simon.

You are right to be confused. Local gauge invariance has to do with locality, but not in an obviously direct way.

Take the action for EM coupled to a charged scalar.

$$\mathcal{L}_{gauge} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + (D_\mu\phi)^*D^\mu\phi,$$

where $F_{\mu\nu}(t,x) = \partial_\mu A_\nu(t,x) - \partial_\nu A_\mu(t,x)$ and $D_\mu\phi(t,x) = \partial_\mu\phi(t,x) -ieA_\mu(t,x)\phi(t,x)$. I have put the spacetime labels to emphasise the following: *every term in the lagrangian features fields evaluated at a single spacetime event*.

This way we know for sure that the resulting action

$$S_{EM} := \int dtd^3x ~~\mathcal{L}_{gauge}(t,x)$$

is local. Why? Change the fields in one region of spacetime, and the action changes only there.

Ok, but what does this have to do with *local gauge invariance?* Can't we write down the action $S_{EM}$ directly in terms of the Lorentz tensor $F^{\mu\nu}$ (ie the E and B fields)? Yes, but it won't be written explicitly in terms of a local Lagrangian density. *We* know it is local, but it would not be apparent from the way it is formulated.

It turns out that this is a general pattern: the fundamental interactions are local, and can be written in local form only using gauge fields. I don't know a deep explanation for this, I think we just take it as a brute fact.

## Best Answer

If you are dealing with a multipartite state $\lvert\Psi\rangle$, then the distinction between

localandnon-localoperations is an important one for example for the study of entanglement.Consider for simplicity a

bipartitestate $\lvert\Psi\rangle$, that is, a state that can be written as $\lvert\Psi\rangle=\sum_{ij}c_{ij}\lvert i\rangle\otimes\lvert j\rangle$. Alocaloperation $A\equiv (A'\otimes I)$ is one that only acts on one part of the system. For example, an operation of the form $$A\lvert\Psi\rangle=\sum_{ij} c_{ij}(A'\lvert i\rangle)\otimes\lvert j\rangle$$ is local in that it only affects the first part of the system. As another example, an operation of the form $A'\otimes B'$ is also local, and its action on aproductstate $\lvert\psi\rangle\otimes\lvert\phi\rangle$ would read $$\lvert\psi\rangle\otimes\lvert\phi\rangle\to (A'\lvert\psi\rangle)\otimes(B'\lvert\phi\rangle).$$ One notable property of local (unitary) operations is that they do not affect the entanglement (in a way that can be made precise).Global operations are operations that are not local.

For a two-qubit system, an example of a global operation can be a CNOT, while a local operation is any single-qubit gate.