Are symmetries necessarily canonical transformations?

Question

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?

Answer 1

First of all, it is false that transformations that preserve the form of Hamilton equations are canonical. Canonical transformations are defined as (passive) transformations in the space of phases that preserve the symplectic form. That is a smaller class of transformations. Secondly, Symmetries are (active) canonical transformations whose generating function commutes with the Hamiltonian (actually the definition is a bit more general).

To answer your question consider a symmetry: it is canonical and its generating function commutes with the Hamiltonian. Next add a function to the Hamiltonian that does not commute with the generating function.

The transformation remains canonical (it continues to preserve the symplectic form) but it ceases to be a symmetry.

In summary, the symmetry requirement is not necessary in the definition of canonical transformation, whereas a symmetry is by definition an (active) canonical transformation.

Answer 2

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.)

In other words, when a function is a generator of an infinitesimal symmetry of the phase space, it is also an integral of motion and vice versa. (Do note that there are also discrete symmetries which cannot be characterized in this way.)

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 $G = \sum_j V^jp_j$.

Answer 3

The point is that for a transformation to have a corresponding charge/generator, it should be a symplectic symmetry (canonical transformation).

Suppose that $z^a$ is a coordinate system on the phase space. One classical choice is to take $z=(q,p)$, but it is not necessary. The phase space is endowed with a symplectic form $\Omega_{ab}$ (an antisymmetric 2 form which is closed $\partial_{[a}\Omega_{bc]}=0$ and non nondegenerate). Then an infinitesimal symmetry transformation is given by $$z\to z'^a=z^a+\xi^a$$ To define the associated generator (charge) $Q_\xi$, we consider the equation $$\partial_a Q_\xi=\Omega_{ab}\xi^b$$ For $Q_\xi$ to exist, we should have $\partial_a\partial_b Q_\xi-\partial_b\partial_aQ_\xi=0$. It is straightforward to check that this condition is equivalent to $$\cal L_\xi \Omega=0$$ That is, a symmetry of the Hamiltonian satisfying ${\cal L}_\xi H=0$, is associated to a well defined generator (charge) $Q_\xi$ only if it is also a symmetry of the symplectic form. And symplectic symmetry is just another name for canonical transformation.

Answer 4

Any reversible variable change maps solutions of the EOM to solutions of the EOM.

EOM themselves have N integrals of motion: N different combinations of the variables that equal to constants, for example, to the N initial data, whatever whether you manage to integrate them explicitly or not.