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We know a space-translation generator can be written as: \begin{equation} T(\textbf{r}_{0})|\alpha\rangle=e^{-i\frac{\textbf{p}\cdot\textbf{r} _{0}}{\hbar}}|\alpha\rangle=|\alpha'\rangle. \end{equation} If the system is space-translation invariant, then \begin{align} H|\alpha'\rangle&=i\hbar\partial_{t}|\alpha'\rangle=i\hbar\partial_{t}T|\alpha\rangle=Ti\hbar\partial_{t}|\alpha\rangle=TH|\alpha\rangle=THT^{+} |\alpha'\rangle\\ &\rightarrow THT^{+}=H \rightarrow HT=TH. \end{align} If $T^{+}=T^{-1}$ as in this case.
This is the reason why a symmetry needs to have a generator that commutes with the Hamiltonian.

However this makes sense only if the generator itself is indipendente in time, otherwise there should be another term in here: \begin{equation} \partial_{t}T|\alpha\rangle=(\partial_{t}T)|\alpha\rangle+T(\partial_{t}|\alpha\rangle) \end{equation} And the commutation relation changes.

Does it ever happen or is it proven a generator of a Conservative quantity needs to be time independent?

ZeroTheHero
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Matteo
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  • Neither operators nor states are physical objects, and it is merely a matter of convention which will be time-dependent (check the three "pictures" of quantum mechanics). – user3257624 Aug 10 '22 at 09:09

1 Answers1

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  1. A conserved quantity/constant of motion does by definition not change in time, cf. e.g. this Phys.SE post.

  2. However, a conserved quantity could depend explicit on time.

    Example: A free particle with Hamiltonian $H=\frac{p^2}{2m}$. It is straightforward to check that $$q-\frac{p}{m}t$$ is a conserved quantity with explicit time dependence.

Qmechanic
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  • p is the translation operator, however when you apply a translation with T=exp(stuff) you need that stuff not to be time dependant, or else the commutative relation changes as shown before. So my question is whether that translation generator you need to apply on a state (or operator, don't know the right terminology. That T I applied, to sum up) could be or not time dependant. Not the conserved quantity itself. – Matteo Aug 02 '22 at 10:55
  • Oh, thanks. Just read in the first link you sent that that T needs not to be explicitly time dependant. – Matteo Aug 02 '22 at 11:04