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.


I) Disclaimer. As a purist, I disapprove of the common praxis to call the implication $$ \tag{1} \{Q,H\}+\frac{\partial Q}{\partial t}~=~0 \qquad\Rightarrow\qquad \frac{dQ}{dt}~\approx~0.$$ for a 'Hamiltonian version of Noether's theorem', cf. my Phys.SE answers here & here. My reason is that the implication (1) is merely a trivial consequence of Hamilton's equations, nothing more.

II) Instead, a 'Hamiltonian version of Noether's theorem' should refer to quasi-symmetries of a Hamiltonian action $$ S_H[q,p] ~:=~ \int \! dt ~ L_H(q,\dot{q},p,t), \tag{2}$$ and their corresponding conservation laws. Here $L_H$ is the so-called Hamiltonian Lagrangian $$ L_H(q,\dot{q},p,t) ~:=~\sum_{i=1}^n p_i \dot{q}^i - H(q,p,t). \tag{3}$$

III) It is a misunderstanding that all that fuss about total derivatives [...] disappears in the Hamiltonian framework. The Hamiltonian version allows for the Hamiltonian action to only be invariant up to boundary terms (i.e. a so-called quasi-symmetry) just like in the standard Lagrangian formulation of Noether's theorem. See also this related Phys.SE post.

  • $\begingroup$ I suppose my question could then be refined: in the Hamiltonian framework, is there a fundamental difference between a symmetry for which $\delta L = 0$ and one for which $\delta L = \dot f$? $\endgroup$ – user1379857 Jan 22 at 22:56
  • $\begingroup$ You only need the latter condition for Noether's theorem. $\endgroup$ – Qmechanic Jan 22 at 23:00

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 $\delta L$.


The reason we don't talk about "changing the Hamiltonian by a total derivative" is because symmetries and conservation laws are usually handled differently in the Hamiltonian picture.

In Hamiltonian mechanics, any function $f$ on phase space generates a flow on phase space, i.e. a one-parameter family of canonical transformations $(q, p) \to (\tilde{q}(\alpha), \tilde{p}(\alpha))$. The induced rate of change of any other phase space function $g$ is $$\frac{dg}{d\alpha} = \{g, f\}.$$ In particular, the Hamiltonian itself generates time translation, $$\frac{dg}{dt} = \{g, H\}.$$ The statement that $Q(q, p)$ is a conserved quantity is simply $$\{Q, H\} = 0.$$ That is, the time evolution generated by $H$ doesn't change the value of $Q$. The key is that this is equivalent, by the antisymmetry of the Poisson bracket, to $\{H, Q\} = 0$, which states that the canonical transformations generated by $Q$ don't change the values of $H$.

Thus, given an infinitesimal canonical transformation that keeps $H$ the same, its generator is a conserved quantity. This is the closest thing to Noether's theorem you'll usually see in Hamiltonian mechanics. Since it refers to only $H$, not the integral of $H$, there's no need to talk about keeping $H$ invariant up to a total derivative -- it just has to be invariant, period. (But also see Qmechanic's answer, about an action-like formulation where it does appear.)


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