A tiny symmetry transformation may change the Lagrangian $L$ by a total time derivative of some function $f$. This is a basic fact used in the proof of Noether's theorem.
How can we see the effect of this total derivative term in the Hamiltonian framework? Is there a good example to work out? I can't think of one off the top of my head. It just seems strange to me that all this fuss about total derivatives seem to disappear in the Hamiltonian framework.
Answer
I suppose I figured out the "answer" to my very vague question, although the other answers here are also helpful. The "Hamiltonian Lagrangian" is
$$ L = p_i \dot q_i - H. $$ Say we have a conserved charge $Q$, that is $$ \{Q, H\} = 0. $$ If we make the tiny symmetry variation $$ \delta q_i = \frac{\partial Q}{\partial p_i} \hspace{1cm} \delta p_i = - \frac{\partial Q}{\partial q_i} $$ then \begin{align*} \delta L &= - \frac{\partial Q}{\partial q_i} \dot q_i - p_i \frac{d}{dt} \Big( \frac{\partial Q}{\partial p_i} \Big) + \{ H, Q\} \\ &= - \frac{\partial Q}{\partial p_i} \dot q_i - \dot p_i \frac{\partial Q}{\partial p_i} + \frac{d}{dt} \Big( p_i \frac{\partial Q}{\partial p_i} \Big) \\ &= \frac{d}{dt} \Big( p_i \frac{\partial Q}{\partial p_i} - Q\Big) \end{align*}
So we can see that $L$ necessarily changes by a total derivative. When the quantity $p_i \frac{\partial Q}{\partial p_i} - Q = 0$, the total derivative is $0$. This happens when the conserved quantity is of the form $$ Q = p_i f_i(q). $$ Note that in the above case, $$ \delta q_i = f_i(q) $$ That is, symmetry transformations which do not "mix up" the $p$'s with the $q$'s have no total derivative term in $\delta L$.
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