In classical mechanics:
If a Lagrangian $\mathcal{L}$ is preserved by an infinitesimal change in the state space variables $q_i \to q_i + \varepsilon K_i(q)$, this leads to only second order change in the Lagrangian:
$$ 0 = \frac{d\mathcal{L}}{d\varepsilon} = \sum_i \left( \frac{\partial \mathcal{L}}{\partial q_i}K_i + \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \dot{K}_i \right) = \frac{d}{dt}\left(\sum_i \frac{\partial \mathcal{L}}{\partial \dot{q}_i} K_i \right). $$
Then we get our conserved momentum because the rate of change on the right side is $0$.
In quantum mechanics, an observable $A$ commuting with the Hamiltonian, i.e. with $[\hat{H},A] = 0$, corresponds to a symmetry of the time-independent Schrödinger equation $\hat{H}\Psi = E \Psi$. How do we compute the conserved quantity related to $A$? In particular, what is the conserved quantity associated with the identity operator?
Best Answer
In hindsight, Noether's theorem is a dramatic hint of quantum mechanics. Mariano is completely correct in his comment that the conserved quantity is $A$ itself, but it deserves a bit of explanation.
A classical probabilistic system is characterized by an algebra of random variables. You could consider the Boolean random variables, in which case the algebra is a $\sigma$-algebra $\Omega$. Or you could consider real or complex random variables; if you take the bounded ones then the algebra is $L^\infty(\Omega)$. In quantum probability, you have the same sort of thing, except that the algebra of bounded complex random variables is a non-commutative von Neumann algebra. One choice with special properties is the algebra $\mathcal{B}(\mathcal{H})$ of all bound operators on a Hilbert space $\mathcal{H}$.
The special property of $\mathcal{B}(\mathcal{H})$ is that all automorphisms are inner, so that any symmetry $A$ of a quantum dynamical system is necessarily also a random variable that you can measure. This does not happen classically, nor even for other non-commutative von Neumann algebras. Even without writing down a time-independent Schrodinger equation, it makes Noether's theorem trivial, because the symmetry $A$ must be conserved if you interpret it as a quantum random variable. Unlike in the classical case, $A$ doesn't even need to generate or come from a continuous group action.
For example, the parity operator (which negates all three coordinates of space) is a conserved quantity of electromagnetism, so it leads to a (two-valued) conserved quantity in quantum electrodynamics which is also called parity. The discrete symmetry also exists classically as a symmetry of Maxwell's equations (if you are careful to negate the magnetic field vectors twice), but the classical Noether's theorem doesn't apply.
Anyway, the identity operator is the trivial random variable that is always 1, as Aaron says.