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?


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.

  • $\begingroup$ Thank you for the pointers, this is very appreciated. Is there an undergraduate level text I can learn this from? $\endgroup$ Sep 15 '20 at 13:36
  • 1
    $\begingroup$ I don't recall seeing a textbook dedicated to this problem. $\endgroup$
    – Qmechanic
    Sep 15 '20 at 14:24
  • $\begingroup$ What's a good introductory source, in that case? $\endgroup$ Sep 15 '20 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.