# Poisson Brackets of Harmonic Oscillator

## Theorem

Let $P$ be a classical harmonic oscillator.

Let the real-valued function $\map x t$ be the position of $P$, where $t$ is time.

Then $P$ has the following Poisson brackets:

$\sqbrk {x, p} = 1$
$\sqbrk {x, H} = \dfrac p m$
$\sqbrk {p, H} = - k x$

## Proof

The standard Lagrangian of $P$ is:

$L = \dfrac 1 2 \paren {m {\dot x}^2 - k x^2}$.

The canonical momentum is:

$p = \dfrac {\partial L} {\partial \dot x} = m \dot x$

The Hamiltonian associated to $L$ in canonical coordinates reads:

$H = \dfrac {p^2} {2 m} + \dfrac k 2 x^2$

Then:

$\sqbrk {x, p} = \dfrac {\partial x} {\partial x} \dfrac {\partial p} {\partial p} - \dfrac {\partial p} {\partial x} \dfrac {\partial x} {\partial p} = 1$
$\sqbrk {x, H} = \dfrac {\partial x} {\partial x} \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial p} - \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial x} \dfrac {\partial x} {\partial p} = \dfrac p m$
$\sqbrk {p, H} = \dfrac {\partial p} {\partial x} \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial p} - \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial x} \dfrac {\partial p} {\partial p} = - k x$

$\blacksquare$