Proof of equivalence YouTube

One coordinate and one momentum

For simplicity, we consider a one dimensional system, that is, no indices. Generalising to many coordinate-momenta pairs is just an exercise in keeping track of indices in polynomial expressions.

We want to show, from the axioms, that $$\left\{A, B\right\} = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}$$

The coordinate rules are now $$\begin{align*}\left\{ q, q \right\} &= 0\\\left\{ p, p \right\} &= 0\\\left\{ q, p \right\} &= 1 \quad \left( \Rightarrow \left\{ p, q \right\} = -1 \right)\end{align*}$$

Functions of one coordinate, the same for both

For simple functions - either $F(q)$ and $G(q)$ or $F(p)$ and $G(p)$, we show that the Hamiltonian definition follows from the axioms. That is, $$\left\{ F, G \right\} = \frac{\partial F}{\partial q}\frac{\partial G}{\partial p} - \frac{\partial F}{\partial p}\frac{\partial G}{\partial q}$$

We first show that, if $F = F(q)$, that is, no dependence on momentum, then $$\left\{ q, F(q) \right\} = 0$$

We use induction to show it's true for a general polynomial term, $q^n$, and then generalise to $F$. For induction we need to establish the result for at least one index, so consider $q^0 = 1$.

$$\begin{align*}&\left\{ q, q \right\} = 1 \left\{ q, q \right\} + \left\{ q, 1 \right\} q &\quad (\text{product rule})\\&\:0 = 1 \cdot 0 + \left\{ q, 1 \right\}q &\quad (\text{coordinate rule})\\\Rightarrow & \left\{ q, 1 \right\} = 0 &\end{align*}$$

We could have also noted that one of the coordinate rules gives $$\left\{ q, q^1 \right\} = \left\{ q, q \right\} = 0$$

Next, we suppose that $\left\{ q, q^n \right\} = 0$. Then, $$\begin{align*}\left\{ q, q^{n + 1} \right\} &= q\left\{ q, q^n \right\} + \left\{ q, q \right\}q^n\\&= q \cdot 0 + 0 \cdot q^n\\&= 0\end{align*}$$

Thus, for functions $F = F(q)$, $\left\{ q, F(q) \right\} = 0$ and a similar argument shows that, for $G = G(p)$, $\left\{ p, G(p) \right\} = 0$. This is easily generalised to $$\left\{ F(q), G(q) \right\} = \left\{ F(p), G(p) \right\} = 0$$

This corrresponds to the Hamiltonian definition since $$\frac{\partial F(q)}{\partial q}\frac{\partial G(q)}{\partial p} - \frac{\partial F(q)}{\partial p}\frac{\partial G(q)}{\partial q}= \frac{\partial F}{\partial q}\cdot 0 - 0 \cdot \frac{\partial G}{\partial q} = 0$$

Functions of one coordinate, different for each

Suppose now we have $F(q)$ and $G(p)$. We would like to show that $$\begin{align*}\left\{ F(q), G(p) \right\}&= \frac{\partial F(q)}{\partial q}\frac{\partial G(p)}{\partial p} – \frac{\partial F(q)}{\partial p}\frac{\partial G(p)}{\partial q}\\&= \frac{\partial F}{\partial q}\frac{\partial G}{\partial p}\end{align*}$$ and similarly (except for the minus sign) $$\begin{align*}\left\{ F(p), G(q) \right\}&= \frac{\partial F(p)}{\partial q}\frac{\partial G(q)}{\partial p} – \frac{\partial F(p)}{\partial p}\frac{\partial G(q)}{\partial q}\\&= -\frac{\partial F}{\partial p}\frac{\partial G}{\partial q}\end{align*}$$

It's simplest to start with $\left\{ q, G(p) \right\} \to \left\{ q, p^n \right\}$ (in polynomial terms). We want to show that $$\frac{\partial q}{\partial q}\frac{\partial G}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial G}{\partial q} = 1 \cdot \frac{\partial G}{\partial p} - 0 \cdot 0 = \frac{\partial G}{\partial p}$$

We use induction again. We have already seen that $$\left\{ q, p^0 \right\} = \left\{ q, 1 \right\} = 0 = \frac{\partial}{\partial p}(p^0)$$ and from the coordinates rules, $$\left\{ q, p^1 \right\} = \left\{ q, p \right\} = 1 = \frac{\partial}{\partial p}(p^1)$$

Now suppose that $$\left\{ q, p^n \right\} = \frac{\partial}{\partial p}(p^n)$$ then $$\begin{align*}\left\{ q, p^{n + 1} \right\}&= p \left\{ q, p^n \right\} + \left\{ q, p \right\} p^n\\&= p \frac{\partial}{\partial p}(p^n) + \frac{\partial p}{\partial p} p^n\\&= p \left(n p^{n - 1} \right) + 1 \cdot p^n\\&= (n + 1)p^n\\&= \frac{\partial}{\partial p}(p^{n + 1})\end{align*}$$

Generalising to any function, $$\left\{ q, G(p) \right\} = \frac{\partial G}{\partial p}$$

Similarly, except for a minus sign, we have $$\left\{ p, G(q) \right\} = -\frac{\partial G}{\partial q}$$

We now consider $F(q), G(p)$. We show that $$\left\{ F(q), G(p) \right\} = \frac{\partial F(q)}{\partial q}\frac{\partial G(p)}{\partial p} - \frac{\partial F(q)}{\partial p}\frac{\partial G(p)}{\partial q} = \frac{\partial F}{\partial q} \frac{\partial G}{\partial p} - 0 \cdot 0 = \frac{\partial F}{\partial q} \frac{\partial G}{\partial p}$$

First consider $\left\{ q^0, G(p) \right\} = \left\{ 1, G(p) \right\}$. Note that $$\left\{ q, G(p) \right\} = 1 \cdot \left\{ q, G(p) \right\} + \left\{ 1, G(p) \right\}q$$ which means $$\begin{align*}\left\{ q^0, G(p) \right\} &= \left\{ 1, G(p) \right\}\\&= 0\\&= \frac{\partial }{\partial q}(q^0) \frac{\partial G}{\partial p}\end{align*}$$

We have already shown that $$\left\{ q^1, G(p) \right\} = \left\{ q, G(p) \right\} = \frac{\partial G}{\partial p} =\frac{\partial}{\partial q}(q^1) \frac{\partial G}{\partial p}$$

For the induction step, suppose that $$\left\{ q^n, G(p) \right\} = \frac{\partial}{\partial q}(q^n) \frac{\partial G}{\partial p}$$

Then, $$\begin{align*}\left\{ q^{n + 1}, G(p) \right\}&= q \left\{ q^n, G(p) \right\} + \left\{ q, G(p) \right\} q^n\\&= q \frac{\partial}{\partial q}(q^n) \frac{\partial G}{\partial p} + \frac{\partial G}{\partial p} q^n\\&= q n(q^{n - 1}) \frac{\partial G}{\partial p} + \frac{\partial G}{\partial p} q^n\\&= (n + 1) q^n \frac{\partial G}{\partial p}\\&= \frac{\partial}{\partial q} (q^{n + 1}) \frac{\partial G}{\partial p}\end{align*}$$

In general, we have $$\left\{ F(q), G(p) \right\} = \frac{\partial F}{\partial q} \frac{\partial G}{\partial p}$$ and similarly, $$\left\{ F(p), G(q) \right\} = -\frac{\partial F}{\partial p} \frac{\partial G}{\partial q}$$

Products of functions of one coordinate

Suppose now we have $F(q), F'(p), G'(q)$ and $G(p)$. We would like to show that $$\begin{align*}\left\{ F(q)F'(p), G'(q)G(p) \right\}&= \frac{\partial (FF')}{\partial q}\frac{\partial (G'G)}{\partial p} – \frac{\partial (FF')}{\partial p}\frac{\partial (G'G)}{\partial q}\end{align*}$$

But, using various results above, we see that $$\begin{align*}\left\{ F(q)F'(p), G'(q)G(p) \right\}&= F(q)\left\{ F'(p), G'(q)G(p) \right\} + \left\{ F(q), G'(q)G(p) \right\} F'(p)\\&= F(q)G'(q)\left\{ F'(p), G(p) \right\} + F(q)\left\{ F'(p), G'(q) \right\}G(p)\\&\quad+ G'(q)\left\{ F(q), G(p) \right\} F'(p) + \left\{ F(q), G'(q) \right\} F'(p)G(p)\\&= F(q)\left\{ F'(p), G'(q) \right\}G(p) + G'(q)\left\{ F(q), G(p) \right\} F'(p)\\&= F(q)G(p) \left(-\frac{\partial}{\partial p}\left( F'(p) \right) \frac{\partial}{\partial q}\left( G'(q) \right) \right) + G'(q) F'(p) \left(\frac{\partial }{\partial q} \left(F(q) \right) \frac{\partial }{\partial p} \left(G(p) \right) \right)\\&= \frac{\partial }{\partial q} \left(F(q)F'(p) \right) \frac{\partial }{\partial p} \left(G'(q)G(p) \right) - \frac{\partial}{\partial p}\left( F(q)F'(p) \right) \frac{\partial}{\partial q}\left( G'(q)G(p) \right)\end{align*}$$

General functions of both coordinates

The final act is to notice that, again in terms of polynomials, we can write any function, $A(q, p)$ as a sum of products, $$A(q, p) = \sum_{i}A_i(q)A'_i(p)$$ and apply the above result to each of the products.

We end up with the result, that, given the axioms, we can write $$\left\{ A(q, p), B(q, p) \right\} = \frac{\partial A}{\partial q} \frac{\partial B}{\partial p} -\, \frac{\partial A}{\partial p} \frac{\partial B}{\partial q}$$

The time postulate implies Hamilton's Equations

Suppose that a Hamiltonian, $H$ is known for a given system. The time postulate says that, for any function $A = A(q, p)$, we have $$\dot{A} = \left\{ A, H \right\} = \frac{\partial A}{\partial q} \frac{\partial H}{\partial p} -\, \frac{\partial A}{\partial p} \frac{\partial H}{\partial q}$$

But the time derivative can also be written as $$\dot{A} = \frac{\partial A}{\partial q} \dot{q} + \frac{\partial A}{\partial p} \dot{p}$$

Equating the two expressions, and using different choices for $A$, which we can do since the time postulate applies to arbitrary functions, we get $$\begin{align*}\frac{\partial H}{\partial q} &= -\dot{p}\\\frac{\partial H}{\partial p} &= \dot{q}\end{align*}$$ which are Hamilton's equations.