Are symmetries necessarily canonical transformations?

A canonical transformation is defined as a transformation such that afterwards Hamilton's equations still hold.

It can then be shown that this requirement implies that canonical transformations are necessarily generated by some function $$Q$$ which acts on the coordinates via \begin{align} q \to q' &= q + \epsilon \frac{\partial Q}{\partial p} \\ p \to p' &= p - \epsilon \frac{\partial Q}{\partial q} \tag{1} \end{align} (See for example, page 102 in Tong's notes.)

A symmetry in the Hamiltonian formalism is defined as a transformation which is generated by a function $$Q$$ for which $$\{Q,H\}=0 \tag{2}$$ holds, where $$H$$ denotes the Hamiltonian.

Are symmetries solely defined by Eq. (2) or do they need to fulfill additionally Eq. (1)? In other words, are symmetries necessarily canonical transformations?

• Briefly, eq. (2) has nothing to do with the definition of canonical transformations. – Qmechanic Feb 20 '19 at 12:43
• @Qmechanic cool thanks. But is there any error in my reasoning above? Because clearly I have a problem finding a conceptual difference between the two. Both concepts seem to map solutions of the EOM to solutions of the EOM and are therefore describe what I would call symmetries. – jak Feb 20 '19 at 12:55
• ...a symmetry generator must satisfy (2) and defines a canonical tfmation via (1). Not all canonical tfmations, (1), need be symmetries. What is your point? – Cosmas Zachos Feb 20 '19 at 16:03
• Yes, symmetries are especially transparent canonical transformations, as detailed in this one. – Cosmas Zachos Feb 20 '19 at 22:53

Consider canonical transformations $$\Phi_\lambda:p_i,q^j \to p'_i,q'^j$$ depending smoothly on a parameter $$\lambda$$ such that $$\lambda=0$$ reduces to an identity. Such a transformation is at least locally expressible (for $$\lambda \approx 0$$) as $$\Phi_\lambda(z) = \exp(\lambda \{z,G\})$$ where $$G$$ is some generating function and $$z=(p_i,q^j)$$. To linear order in $$\lambda$$ you can express this relation as $$p'_i = p_i + \lambda \{p_i,G\}+\mathcal{O}(\lambda^2) = p_i - \lambda \frac{\partial G}{\partial q^i}+\mathcal{O}(\lambda^2)$$ $$q'^j = q^j + \lambda \{q^j,G\}+\mathcal{O}(\lambda^2) = q^j + \lambda \frac{\partial G}{\partial p_j}+\mathcal{O}(\lambda^2)$$ You can also view $$G$$ as a sort of "Hamiltonian" and $$q'^j(\lambda),p'_i(\lambda)$$ as the trajectories corresponding to its Hamilton's equations.
Now one can ask what sort of infinitesimal transform of this type leaves the Hamiltonian invariant. It is easy to show that under this transform the Hamiltonian gets push-forwarded as $$H'(p_i,q^j)\equiv H\left(p'_k(p_i,q^j),q'^l(p_i,q^j)\right) = H(p_i,q^j) + \lambda \{H,G\} + \mathcal{O}(\lambda^2)$$ (I would recommend to verify the last equation to understand what is happening here.)
One last note: a phase-space symmetry is not quite the same thing as a spatial symmetry (a symmetry on the $$q^j$$ space). However, it is easy to show that when a physical problem has a spatial configuration symmetry characterized by an infinitesimal transform $$q'^j = q^j + \epsilon V^j(q^j)$$, then this is lifted to a phase-space symmetry by using the generating function $$G = \sum_j V^jp_j$$.