Relationship between symmetries and quantum operators of classical quantities?

Question

I noticed this the other day. I don't really know "what" this means, I'd love to understand.

• The energy operator is $\hat E = -i \hbar \frac{\partial}{\partial t}$. Conservation of energy is a consequence of time symmetry.
• The momentum operator is $i \hbar \frac{\partial}{\partial x}$. Conservation of momentum is a consequence of space symmetry.
• The angular momentum operator is $-i \hbar (r \times \nabla)$. Conservation of angular momentum is a consequence of rotational symmetry, which 'feels related' to curl: $r \times \nabla$.

Is the "general form" of any quantum mechanical operator of a given classical quantity $Q$, whose conservation law is given by a symmetry in some 'direction '$d$ going to be proportial to $\hat Q \equiv i \hbar \frac{\partial}{\partial d}$?

If not, why do the energy and momentum operator have their symmetries in the derivative? is there a reason?

Answer 1

There is a lot of truth behind OP's observations, which are backed up by the following facts:

• An infinitesimal symmetry $\delta$ with symmetry parameter $\epsilon$ is generated by a Noether charge $\hat{Q}$ in the sense that $\delta=\epsilon [\hat{Q},\cdot]$, cf. e.g. this Phys.SE post.

• The symmetry parameter $\epsilon$ can often be associated with a variable/coordinate $q$ of theory.

• If $[\hat{Q},q]\propto {\bf 1}$ is proportional to the identity operator we can go to the corresponding Schroedinger representation $\hat{Q}\propto\frac{\partial}{\partial q}$ in $q$-space.