Poisson brackets YouTube

Phase space

Phase space has a rich structure relating to quantities defined on the phase space that remain invariant under certain types of transformations.

These transformations can be thought of as flows in phase space. For example, we can consider a trajectory of a system as a flow - an evolution over time. We found that the Hamiltonian was invariant under this time-translation.

There were other symmetries, which were all transformations of the system's coordinates, but we now want to consider symmetries involving the phase space coordinates. To do this, we define Poisson brackets as a set of axioms.

Axioms

Let $A(q_i, p_i), B(q_i, p_i)$ be functions of the phase space coordinates. We define the Poisson bracket of $A$ with $B$, written $\left\{A, B\right\}$, to be the function that satifies the following axioms:

Anti-symmetry: $$\left\{ A, B \right\} = -\left\{ B, A \right\}$$

Linearity: For numbers $\alpha, \beta$, $$\left\{ \alpha A + \beta B, C \right\} = \alpha \left\{ A, C \right\} + \beta \left\{ B, C \right\}$$

Product rule: $$\left\{ AB, C \right\} = A \left\{ B, C \right\} + \left\{ A, C \right\} B$$

From these three rules we can deduce that $$\begin{align*}\left\{ A, A \right\} &= 0\\\left\{ A, \alpha B + \beta C \right\} &= \alpha \left\{ A, B \right\} + \beta \left\{ A, C \right\}\\\left\{ A, BC \right\} &= B \left\{ A, C \right\} + \left\{ A, B \right\} C\end{align*}$$

Coordinate rules: $$\begin{align*}\left\{ q_i, q_j \right\} &= 0\\\left\{ p_i, p_j \right\} &= 0\\\left\{ q_i, p_j \right\} &= \delta_{ij} \quad \left( \Rightarrow \left\{ p_i, q_j \right\} = - \delta_{ij} \right)\end{align*}$$

Time postulate: For a given Hamiltonian, $$\dot{A} = \left\{ A, H \right\}$$

Equivalence with the Hamiltonian definition

In a previous lecture, we defined a Poisson bracket of two functions of the phase space coordinates, $A(q_i, p_i), B(q_i, p_i)$, to be $$\left\{A, B\right\} = \sum_i \left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i} - \frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)$$ and that if we were given the Hamiltonian, $H$, of a system, then Hamilton's equations are equivalent to $$\dot{A} = \left\{A, H\right\}$$

The above axioms are equivalent to the Hamiltonian definition and the time postulate can derived from Hamilton's equations.

Hamiltonian definition $\Rightarrow$ axioms

It's a straightforward exercise to show the axioms follow directly from the Hamiltonian definition.

Axioms $\Rightarrow$ Hamiltonian definition

It's a bit more involved going the other way. Since the functions we are considering are all smooth (by which we mean continuous and infinitely differentiable) functions, they can be approximated to any order by polynomials.

Functions in physics can all be represented by Fourier series, which are infinite sums of sines and cosines. These, in turn, are defined by the exponential function - itself an infinite sum of polynomial terms.

Hence, if we can prove the equivalence of the axioms and the definition for general polynomial terms, then we can immediately generalise to any function.

I have gone a bit further in trying to show why the axioms imply the Hamiltonian definition than Professor Susskind did in the lecture - probably a bit more than necessary. See the next section.

A simple example

Given the equivalence to Hamilton's form, the axioms hold for all systems for which the principle of least actions hold, which, for Professor Susskind, means all known classical systems.

For a simple example, given the Hamiltonian (for a particle in free space), $$H = \sum_i \frac{p_i^2}{2m}$$ then $$\begin{align*}\dot{p_j} &= \left\{ p_j, H \right\}\\&= \left\{ p_j, \sum_i \frac{p_i^2}{2m} \right\}\\&= 0\end{align*}$$ meaning conservation of momentum, and $$\begin{align*}\dot{x_j} &= \left\{ x_j, H \right\}\\&= \left\{ x_j, \sum_i \frac{p_i^2}{2m} \right\}\\&= \left\{ x_j, \frac{p_j^2}{2m} \right\}\\&= \frac{\partial}{\partial p_j} \left(\frac{p_j^2}{2m} \right)\\&= \frac{p_j}{m}\end{align*}$$ which are results we have already established.