Poisson Brackets

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, \(U(t) = U(0) + t\frac{dU}{dt}(0) + \mathcal{O}(t^2) = U(0) + t[U,H](0) + \mathcal{O}(t^2)\).

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

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*}