# Poisson Brackets

Note this is a Lie algebra.

Definition: the poisson bracket of $$U,V$$ with respect to the phase space coordinates $$q,p$$ is defined as $$[U,V]_{q,p} = \sum_i\left(\frac{\partial U}{\partial q_i}\frac{\partial V}{\partial p_i} - \frac{\partial U}{\partial p_i}\frac{\partial V}{\partial q_i}\right)$$. So, $$[U,V]_{q,p} = U_qV_p - U_pV_q = \begin{pmatrix} \frac{\partial U}{\partial q} & \frac{\partial U}{\partial p} \end{pmatrix}\cdot\begin{pmatrix} \frac{\partial V}{\partial p} \\ -\frac{\partial V}{\partial q} \end{pmatrix} = \begin{pmatrix} \frac{\partial U}{\partial q} & \frac{\partial U}{\partial p} \end{pmatrix}J\begin{pmatrix} \frac{\partial V}{\partial q} \\ \frac{\partial V}{\partial p} \end{pmatrix} = \frac{\partial U}{\partial \eta}^T J \frac{\partial V}{\partial \eta} = [U,V]_\eta$$.

## Properties

Is this quantity conserved under canonical transformations? It is! As long as you have $$\eta$$ a canonical coordinate, then you don’t need to specify $$\eta$$ or $$q,p$$ on the brackets.

Suppose we have $$\eta\to \zeta$$. So, $$MJM^T = J$$.

Then, $$[U,V]_\eta = \frac{\partial U}{\partial\eta}^TJ\frac{\partial V}{\partial\eta} = \frac{\partial U}{\partial\eta}^TM^TJM\frac{\partial V}{\partial\eta}$$. Note, $$\frac{\partial V}{\partial\eta_i} = \sum_j \frac{\partial V}{\partial \partial\zeta_j}\frac{\partial \zeta_j}{\partial\eta_i} \Rightarrow \frac{\partial V}{\partial\eta} = M^T\frac{\partial V}{\partial\zeta}$$. Then, $$[U,V]_\eta = \frac{\partial U}{\partial\zeta}^TMJM^T\frac{\partial V}{\partial\zeta} = \frac{\partial U}{\partial\zeta}J\frac{\partial V}{\partial\zeta} = [U,V]_\zeta$$.

Suppose, $$U$$ and $$V$$ are canonical coordinates, $$[q_i,q_j] = 0 = [p_i,p_j]$$. For, $$[q_i,p_j] = \delta_{ij} = -[p_i,q_j]$$. So, $$[\eta,\eta] = J$$. Thus, you can use this to determine if a set of coordinates are canonical or not.

1. Anticommutivity: $$[U,V] = -[V,U]$$.
2. $$[U,U] = 0$$.
3. Bilinearity: $$[aU + bV,W] = a[U,W] + b[V,W]$$.
4. Libenitz Rule: $$[UV,W] = U[V,W] + [U,W]V$$.
5. Jacobi Identity: $$[U,[V,W]] + [V,[W,U]] + [W, [U, V]] = \varnothing$$. This will be important to find the constants of motion.

Looking ahead: $$[U,H] = 0$$ and $$U$$ does not explicitly depend on time, $$U$$ is a constant of the motion. Note: $$\frac{dU}{dt} = \frac{\partial U}{\partial\eta}^TJ\frac{\partial H}{\partial\eta} + \frac{\partial U}{\partial t} = [U,H] + \frac{\partial U}{\partial t}$$.

For, $$U\to\eta_i$$, $$\dot{\eta}_i = [\eta_i, H]\Rightarrow \dot{\eta} = [\eta, H] = \left(\frac{\partial\eta}{\partial\eta}\right)^TJ\frac{\partial H}{\partial\eta} = J\frac{\partial H}{\partial\eta}$$. For $$U\to H$$, $$\frac{dH}{dt} = \frac{\partial H}{\partial t}$$. Thus, if $$H$$ does not explicitly depend on time, $$H$$ is a constant of the motion.

If we have $$U(q,p,t)$$ how do we determine if it is a constant of the motion? So, $$\frac{dU}{dt} = 0\Rightarrow [U,H] = -\frac{\partial U}{\partial t}$$.

Then, if $$U$$ and $$V$$ are both constants of the motion, then $$\alpha U + \beta V$$ is a constant of the motion. Further, if $$U,V$$ do not depend on time explicitly, $$[U,V]=\mathbb{C}$$. Hence, $$[U,V]$$ is a constant of the motion.

Additionally, you can show that $$U,V$$ are both constants of the motion, this still holds. I.e. $$[U,V]=\mathbb{C}$$ is still a constant of the motion.

\begin{align*} \frac{d[U,V]}{dt} &= [[U,V], H] + \frac{\partial[U,V]}{\partial t} \\ &= -[H,[U,V]] + \frac{\partial[U,V]}{\partial t} \\ &= [U,[V,H]] + [V, [H,U]] + \frac{\partial[U,V]}{\partial t} \\ &= \left[U,-\frac{\partial V}{\partial t}\right] + \left[V, \frac{\partial U}{\partial t}\right] + \frac{\partial[U,V]}{\partial t} \\ &= -\left[U,\frac{\partial V}{\partial t}\right] + \left[V, \frac{\partial U}{\partial t}\right] + \frac{\partial[U,V]}{\partial t} \\ &= -\left[U,\frac{\partial V}{\partial t}\right] + \left[V, \frac{\partial U}{\partial t}\right] + [U_t,V] + [U,V_t] \\ &= 0. \end{align*}

Suppose we have a function driven by a single parameter in the phase space: $$U(\alpha)$$. Suppose $$A$$ and $$B$$ are two points in the phase space with canonical coordinates. Thus, we can think of this as a finite canonical transformation. So, you can take infinitesimal steps of this parameter $$\alpha$$ to reach $$B$$ from $$A$$. So, $$\zeta = \eta + \delta \eta$$. Hence, we have some generator $$\delta\eta = \varepsilon J\frac{\partial G}{\partial \eta}$$. Recall, $$[\eta, H] = J\frac{\partial H}{\partial\eta}$$. So, $$\delta\eta = \varepsilon [\eta, G]$$. Consider $$\varepsilon\to dt$$ and $$G\to H$$. So, $$\delta\eta = dt[\eta, H] = dt\dot{\eta} = d\eta$$. Thus, the Hamiltonian drives the system along its trajectory. Hence, the motion of the system is a canonical transformation.

Define $$dU = U(B) - U(A)$$ as the change in the system under the canonical transformation, for $$B$$ an infinitesimal point away from $$A$$. So, $$dU = U(\eta + \delta\eta) - U(\eta)$$. Then, $$\frac{\partial U}{\partial U}^T\delta\eta = \frac{\partial U}{\partial\eta}^T\varepsilon J\frac{\partial G}{\partial \eta}$$. Remark: in this case we can change $$\varepsilon\to\alpha$$. So, $$dU = = \varepsilon\frac{\partial U}{\partial\eta}^T J\frac{\partial G}{\partial \eta} = \varepsilon[U,G]$$. So, for $$\alpha\to t$$, $$\frac{dU}{dt} = [U,G]$$. In general, $$\frac{dU}{d\alpha} = [U,G]$$. $$U(\alpha) = U(A) + \alpha\frac{dU}{d\alpha}(A) + \mathcal{O}(\alpha^2)$$. So, $$\alpha\to t$$ and $$G\to H$$. Then, $$U(t) \Rightarrow \frac{dU}{dt} = [U,H]$$. Hence, $$U(t) = U(0) + t\frac{dU}{dt}(0) + \mathcal{O}(t^2) = U(0) + t[U,H](0) + \mathcal{O}(t^2)$$.

## Example

$$H = \frac{p^2}{2m} - max$$.

Let $$U = x$$. Then,

\begin{align*} x(t) &= x(0) + t\frac{dx}{dt}(0) + \frac{t^2}{2}\frac{dx^2}{dt^2} + \mathcal{O}(t^3) \\ &= x(0) + t[x,H](0) + \frac{t^2}{2}\frac{dx^2}{dt^2} + \mathcal{O}(t^3) \\ &= x(0) + t\left(\frac{p}{m}\right)(0) + \frac{t^2}{2}\frac{dx^2}{dt^2} + \mathcal{O}(t^3) \\ &= x(0) + v(0)t + \frac{t^2}{2m}\frac{dp}{dt} + \mathcal{O}(t^3) \\ &= x(0) + v(0)t + \frac{t^2}{2m}\left(0 - (-ma)\right) + \mathcal{O}(t^3) \\ &= x(0) + v(0)t + \frac{at^2}{2} + \mathcal{O}(t^3) \\ &= x(0) + v(0)t + \frac{at^2}{2}. \end{align*}

Created: 2024-05-30 Thu 21:21

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