# What does a symmetry that changes the Lagrangian by a total derivative do to the Hamiltonian $H$?

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.

## 3 Answers

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.

• 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 $\delta L = \dot f$? – user1379857 Jan 22 '19 at 22:56
• You only need the latter condition for Noether's theorem. – Qmechanic Jan 22 '19 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.)