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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?

Qmechanic
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Siddharth Bhat
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1 Answers1

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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.

Qmechanic
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