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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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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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