quantum field theory - Connection between conserved charge and the generator of a symmetry

I'm trying to understand the connection between Noether charges and symmetry generators a little better. In Schwartz QFT book, chapter 28.2, he states that the Noether charge $Q$ generates the symmetry, i.e. is identical with the generator of the corresponding symmetry group. His derivation of this goes as follows: Consider the Noether charge

\begin{equation} Q= \int d^3x J_0(x) = \int d^3 x \sum_m \frac{\delta L}{\delta \dot \phi_m} \frac{\delta \phi_m}{\delta \alpha} \end{equation}

which is in QFT an operator and using the canonical commutation relation $$[ \phi_m(x) ,\pi_n(y)]=i \delta(x-y)\delta_{mn},$$ with $\pi_m=\frac{\delta L}{\delta \dot \phi_m}$ we can derive

\begin{equation} [Q, \phi_n(y)] = - i \frac{\delta\phi_n(y)}{\delta \alpha}. \end{equation}

From this he concludes that we can now see that "$Q$ generates the symmetry transformation".

Can anyone help me understand this point, or knows any other explanation for why we are able to write for a symmetry transformation $e^{iQ}$, with $Q$ the Noether charge (Which is of course equivalent to the statement, that Q is the generator of the symmetry group)?

To elaborate a bit on what I'm trying to understand: Given a symmetry of the Lagrangian, say translation invariance, which is generated, in the infinite dimensional representation (field representation) by differential operators $\partial_\mu$. Using Noethers theorem we can derive a conserved current and a quantity conserved in time, the Noether charge. This quantity is given in terms of fields/ the field. Why are we allowed to identitfy the generator of the symmetry with this Noether charge?

Any ideas would be much appreciated

Answer

Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$ T_g = \exp(tX) $$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$ (g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other hand the variation of $\phi$ is defined as the first order contribution under the group action, e.g $$ g\cdot\phi = \phi + \frac{\delta \phi}{\delta t}t+\mathcal{O}(t^2) $$ Since in physics we like generators to be hermitian, rather than anti-hermitian one sends $X\mapsto iX$ and establishes $$ [X,\phi] = -i\frac{\delta \phi}{\delta t} $$

Also, this answer and links therein ought to help you further.