Last time we saw the Hamiltonian formulation of mechanics

where $\H = \H(q,p,t)$ is the Hamiltonian function on phase space, which may also depend on time.

In this post we discuss the Poisson bracket, which is a convenient tool for rewriting Hamiltonian dynamics, and which will make the connection between symmetry and conservation law more clear. The Poisson bracket is formally equivalent to the commutators in quantum mechanics, as well as to the Lie bracket of the Hamiltonian vector fields.

As before, we follow Susskind’s presentation in the Classical Mechanics course, Lectures 7 and 8.


Poisson bracket

Given two smooth functions on phase space

we define their Poisson bracket as

which is again a smooth function on phase space. (Here the domain is really the cotangent bundle $\T^\ast Q = \bigcup_{q \in Q} \T_q^\ast Q \cong Q \times \R^d$ if $Q$ is a $d$-dimensional manifold.)

Notice that we can write the Poisson bracket as a bilinear form using the symplectic matrix:

So we see that the Poisson bracket is antisymmetric:

and in particular it is zero on the diagonal:

It is also a derivation, which means it satisfies Leibniz’ product rule:

and furthermore, it satisfies Jacobi’s identity:

These properties endow the space of smooth functions on the phase space with a Poisson algebra structure (e.g., see this).

For example, the Poisson brackets of the “canonical coordinates” $(q_i, p_i)$ in phase space are:

for all $1 \le i,j \le d$, and


Hamiltonian dynamics

Now let $(q_t, p_t)$ evolve according to Hamilton’s equations above. Then for any function $F = F(q,p)$, its time derivative along the curve is

This is the Poisson bracket formulation of mechanics, that for any (time-independent) function $F$:

and we can compute everything we want to know about the dynamics by choosing $F$ appropriately. For example, by choosing $F(q,p) = q$ we get

and by choosing $F(q,p) = p$ we get

so we recover Hamilton’s equations. Notice that if the Hamiltonian $\H$ itself is time-independent, then:

which says that the Hamiltonian, or the energy, is conserved.

More generally, if $F = F(q,p,t)$ is time-dependent, its total time derivative also involves the partial derivative:

so for example, if the Hamiltonian $\H$ depends on time, then total and partial time derivatives coincide:


Symmetry transformation

The relation (for time-independent $F$)

says that the effect of taking the Poisson bracket of $F$ with the Hamiltonian $\H$ is to give the change in $F$ when we change time $t$, which is the conjugate variable to the Hamiltonian. Thus, we say that the Hamiltonian is the generator of the time translation.

More generally, the Poisson bracket of $F$ with an arbitrary generator (function) $G \colon Q \times \R^d \to \R$ gives the change in $F$ when we make a change to the variable conjugate to $G$. I’m not sure how to formulate this precisely, but Susskind gave the following examples:

  • Momentum: If $G(q,p) = p$ is momentum, then

    which is the change in $F$ when we change the position $q$, which is the variable conjugate to momentum $p$. So momentum generates the translation transformation on $q$.

  • Angular momentum: Now suppose we are in three dimension, $Q = \R^3$, and we label $q = (x,y,z)$ as usual, and $p = (p_x,p_y,p_z)$. Recall that a rotation around the $z$-axis with angle $\epsilon$ gives the transformation:

    which gives rise to the conserved quantity:

    which is the $z$-component of the angular momentum vector $L = (L_x, L_y, L_z)$, where the other components are $L_x = y p_z - z p_y$ and $L_y = z p_x - x p_z$, corresponding to rotation around the $x$- and $y$-axes.

    Now observe what happens when we take the Poisson bracket of $F = x$ with $G = L_z$:

    by the product rule and since most of the terms vanish. Similarly, we can use $F = y$ and $F = z$ to get:

    which is the same (up to the factor $\epsilon$) as the rotation transformation above.

    Furthermore, we can also compute:

    and even:

    Thus, the angular momentum $L_z$ generates rotation around the $z$-axis, and similarly for $L_x$ and $L_y$.


Symmetry and conservation law

Suppose we have a generator $G$ that is conserved under the Hamiltonian dynamics, which means:

Now we can turn the equation around and write (since $-0 = 0$):

which means when we do a small transformation associated to $G$, the Hamiltonian $\H$ doesn’t change. That means $G$ is a symmetry of the Hamiltonian.

This is the connection between symmetry and conservation law. The first equation above says $G$ doesn’t change when we perform the transformation associated to $\H$, which is time translation, whereas the second equation says $\H$ is invariant under the transformation associated to $G$. The first says that $G$ is a conserved quantity, while the second says $G$ is a symmetry of the Hamiltonian, and they are equivalent.

(Notice that the above is symmetry of the Hamiltonian, whereas previously we talked about symmetry of the Lagrangian. I suppose they are equivalent, though there is the additional term $p \dot q = p \part{\H}{p}$, which should always be preserved(?).)