2 fixed notation

As already mentioned, the conserved quantity is A itself.

I'll elaborate a little to see the full analogy between the classical and quantum mechanical case.

Let's assume that the lagrangian depends on the coordinates $q_i\in R^n$ and their first derivatives. The classical hamiltonian $H(p,q)$ which is a function on the phase space $T^{*}R^n$ can be obtained through the Legendre transform.

Now if a transformation preserves the Lagrangian, you will see that there exists a function $A(p,q)$ on $T^{*}R^n$ whose canonical Poisson bracket with the Hamiltonian vanishes: ${A(p,q), H(p,q)\{A(p,q), H(p,q)\} = 0$. When you implement this transformation on the coordinates, you get the transformation law you started with: ${A(p,q),q_i} \{A(p,q),q_i\} = K_i(q)$,

Thus, what Noether theorem really does is to allow you compute a function $A(p,q)$ which canonically generates the transformation you started with. This function is conserved under the classical evolution since its Poisson brackets with the hamiltonian vanish.

1

As already mentioned, the conserved quantity is A itself.

I'll elaborate a little to see the full analogy between the classical and quantum mechanical case.

Let's assume that the lagrangian depends on the coordinates $q_i\in R^n$ and their first derivatives. The classical hamiltonian $H(p,q)$ which is a function on the phase space $T^{*}R^n$ can be obtained through the Legendre transform.

Now if a transformation preserves the Lagrangian, you will see that there exists a function $A(p,q)$ on $T^{*}R^n$ whose canonical Poisson bracket with the Hamiltonian vanishes: ${A(p,q), H(p,q)} = 0$. When you implement this transformation on the coordinates, you get the transformation law you started with: ${A(p,q),q_i} = K_i(q)$,

Thus, what Noether theorem really does is to allow you compute a function $A(p,q)$ which canonically generates the transformation you started with. This function is conserved under the classical evolution since its Poisson brackets with the hamiltonian vanish.