Symmetries in phase space

Symmetry in phase space

We defined a symmetry in coordinate space, $q_i$, as a transformation that doesn’t change the Lagrangian, and hence doesn't change the action.

A symmetry in phase space, $q_i,p_i$, is a transformation that doesn’t change the energy, or Hamiltonian, associate with the system.

In order to consider such transformations, we need to determine the conditions for any transformation to be allowable in phase space.

Canonical transformations

We say that a change of (phase space) coordinates, \begin{align*}Q_i &= Q_i(q_i,p_i)\\P_i &= P_i(q_i,p_i)\end{align*} is called a canonical transformation if it preserves Poisson brackets. That is, if $$\left\{ Q_i, P_i \right\} = \left\{ q_i, p_i \right\}$$

Example 1

In one dimension, suppose \begin{align*}Q &= \alpha q\\P &= \beta p\end{align*}

We want to show that the new coordinates still satisfy the axioms, and in particular, that $$\left\{ Q, P \right\} = 1$$

But, \begin{align*}\left\{ Q, P \right\} &= \left\{ \alpha q, \beta p \right\}\\&= \alpha \beta \left\{ q, p \right\}\\&= \alpha \beta\\&= 1 \quad \Leftrightarrow \quad \beta = \alpha^{-1}\end{align*}

So, $$(Q, P) \to \left( \alpha q, \alpha^{-1} p \right)$$ is a canonical transformation.

Example 2

We can mix up the (Lagrangian) coordinates and momenta and still have a canonical transformation. Again in one dimension, suppose we rotate the coordinate-momentum, $q, p$, pair about a fixed angle $\theta$, \begin{align*}Q &= q \cos \theta + p \sin \theta\\P &= -q \sin \theta + p \cos \theta\end{align*}

This is also a canonical transformation, since \begin{align*}\left\{Q, P \right\} &= \left\{ q \cos \theta + p \sin \theta, -q \sin \theta + p \cos \theta \right\}\\&= \left\{ q \cos \theta, -q \sin \theta \right\} + \left\{ q \cos \theta, p \cos \theta \right\}+ \left\{ p \sin \theta, -q \sin \theta \right\} + \left\{ p \sin \theta, p \cos \theta \right\}\\&= \left\{ q, q \right\} \left( -\cos \theta \sin \theta \right) + \left\{ q, p \right\} \left( \cos^2 \theta \right)+ \left\{ p, q\right\} \left( - \sin^2 \theta \right) + \left\{ p, p \right\} \left( \sin \theta \cos \theta \right)\\&= \left(0 \right)\left( -\cos \theta \sin \theta \right) + \left(1 \right) \left(\cos^2 \theta \right)+ \left(-1 \right) \left( - \sin^2 \theta \right) + \left(0 \right) \sin \theta \cos \theta\\&= \cos^2 \theta + \sin^2 \theta\\&= 1\end{align*}

Infinitesimal transformations

We consider general transformations as being built up of a series of infinitesimal transformations, \begin{align*}Q_i &= q_i + \delta q_i\\P_i &= p_i + \delta p_i\end{align*}

For this to be a canonical transformation, we require $$\left\{Q_i, P_i\right\} = 1$$

That is, we want \begin{align*}1 &= \left\{Q_i, P_i\right\}\\&= \left\{q_i + \delta q_i, p_i + \delta p_i\right\}\\&= \left\{q_i, p_i\right\}+ \left\{q_i, \delta p_i\right\}+ \left\{\delta q_i, p_i\right\}+ \left\{\delta q_i, \delta p_i\right\}\\&= 1 + \left\{q_i, \delta p_i\right\} + \left\{\delta q_i, p_i\right\} + 0\end{align*}

So, the condition for a infinitesimal canonical transformation is $$\left\{q_i, \delta p_i\right\} + \left\{\delta q_i, p_i\right\} = 0$$

Generator functions

Suppose we define the variations, $\delta q_i, \delta p_i$, in the (phase space) coordinates in terms of a function $G = G(q_1, \cdots q_N, p_1, \cdots p_N)$, which, in general, is a function of all of the coordinates.

That is, write \begin{align*}\delta q_i &= \epsilon \left\{q_i, G\right\} = \epsilon \frac{\partial G}{\partial p_i}\\\delta p_i &= \epsilon \left\{p_i, G\right\} = -\epsilon \frac{\partial G}{\partial q_i}\end{align*} with $\epsilon$ assumed to be infinitesimal.

Nb. This is analagous to how we defined infinitesimal variations in the context of establishing the Euler-Lagrange equations from the principle of least action.

Then, \begin{align*}\left\{q_i, \delta p_i\right\} + \left\{\delta q_i, p_i\right\}&= \left\{q_i, -\epsilon \frac{\partial G}{\partial q_i}\right\} + \left\{\epsilon \frac{\partial G}{\partial p_i}, p_i\right\}\\&= \epsilon \left\{q_i, -\frac{\partial G}{\partial q_i}\right\} -\, \epsilon \left\{p_i, \frac{\partial G}{\partial p_i}\right\}\\&= \epsilon \left(\frac{\partial}{\partial p_i} \left( -\frac{\partial G}{\partial q_i} \right) \right) - \epsilon \left(-\frac{\partial}{\partial q_i} \left( \frac{\partial G}{\partial p_i} \right) \right)\\&=\epsilon \left(-\frac{\partial^2 G}{\partial p_i \partial q_i} + \frac{\partial^2 G}{\partial q_i \partial p_i} \right)\\& = 0\end{align*}

So these transformations are canonical. The function, $G$, is called a generator function. We characterise flows in phase space by generator functions. These flows do not have to represent trajectories, but we can note that, by the time postulate, $$\dot{q} = \left\{q, H \right\}$$ which is the same as saying $$\delta q = \delta t \left\{q, H \right\}$$

With $\epsilon = \delta t$, we see that the Hamiltonian is the generator of the trajectory - the generator of the flow with respect to time.

Change along generators

We can ask how a general function, $A = A(q_i, p_i)$, changes along a flow characterised by a generator, $G$, \begin{align*}\delta A &= \sum_i \left( \frac{\partial A}{\partial q_i}\delta q_i + \frac{\partial A}{\partial p_i}\delta p_i \right)\\&= \sum_i \left( \frac{\partial A}{\partial q_i}\left(\epsilon \left\{ q_i, G \right\} \right) + \frac{\partial A}{\partial p_i} \left(\epsilon \left\{ p_i, G \right\} \right) \right)\\&= \epsilon \sum_i \left( \frac{\partial A}{\partial q_i}\left( \frac{\partial G}{\partial p_i}\right) + \frac{\partial A}{\partial p_i} \left(- \frac{\partial G}{\partial q_i}\right) \right)\\&= \epsilon \sum_i \left( \frac{\partial A}{\partial q_i} \frac{\partial G}{\partial p_i} -\, \frac{\partial A}{\partial p_i} \frac{\partial G}{\partial q_i} \right)\\&= \epsilon \left\{ A, G \right\}\end{align*}

So, the condition for $A$ not changing along $G$ is $$\left\{ A, G \right\} = 0$$

Symmetry

By definition, a symmetry is a canonical transformation, $G$, (which preserves Poisson brackets) that doesn't change the energy, $H$, of the system. That is, we want $$\left\{ H, G \right\} = 0$$

But, $$\dot{G} = \left\{ G, H \right\} = -\left\{ H, G \right\}$$ thus we can conclude that $G$ is a symmetry in phase space if $$\dot{G} = 0$$

An example

Suppose we have a system, in the plane $x, y$, subject to the potential, $U = U(x^2 + y^2)$, so that the Hamiltonian is $$H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + U(x^2 + y^2)$$

We can write $\lambda = x^2 + y^2$, so that \begin{align*}\frac{\partial H}{\partial x} &= \frac{\partial U}{\partial x} = 2x \frac{\partial U}{\partial \lambda}\\\frac{\partial H}{\partial y} &= \frac{\partial U}{\partial y} = 2y \frac{\partial U}{\partial \lambda}\end{align*}

We can show that a rotation of the phase space coordinates, defined by $$G(x, y,p_x, p_y) = x p_y - y p_x$$ is a symmetry of the system, since \begin{align*}\left\{ G, H \right\} &= \left(\frac{\partial G}{\partial x}\frac{\partial H}{\partial p_x} -\, \frac{\partial G}{\partial p_x}\frac{\partial H}{\partial x} \right) + \left(\frac{\partial G}{\partial y}\frac{\partial H}{\partial p_y} -\, \frac{\partial G}{\partial p_y}\frac{\partial H}{\partial y} \right)\\&= \left( p_y \right) \left(\frac{p_x}{m} \right) - \left( -y \right) \left(2x \frac{\partial U}{\partial \lambda} \right)+\left( -p_x \right) \left(\frac{p_y}{m} \right) - \left( x \right) \left(2y\frac{\partial U}{\partial \lambda} \right)\\&= \frac{p_x p_y}{m} + 2xy \frac{\partial U}{\partial \lambda} -\, \frac{p_x p_y}{m} - 2xy\frac{\partial U}{\partial \lambda}\\&= 0\end{align*}

As a final note, there is certain symmetry (in a different sense to the symmetry we have been discussing) between the Hamiltonian and symmetric quantities. Given that $$\left\{ G, H \right\} = 0$$ then, $$\left\{ H, G \right\} = 0$$ so, in principle, we can reverse the roles of $G$ and $H$. That is, we can consider $G$ as the Hamiltonian and $H$ as a conserved quantity of the $H$-system.

With the above example in mind, if a system has a Hamiltonian, $$x p_y - y p_x$$ then $$\frac{p_x^2}{2m} + \frac{p_y^2}{2m} + U(x^2 + y^2)$$ will be a conserved quantity, for any $U$.