7.7 Poisson Brackets

The Poisson Bracket between two functions of the canonical variable, \(U(q_i,p_i,t)\), and \(V(q_i, p_i, t)\) is defined as:

\begin{align} \{U,V\}\equiv\sum_i^n\left(\die{U}{q_i}\die{V}{p_i}-\die{U}{p_i}\die{V}{q_i}\right) \tag{7.36}\end{align}

The Poisson Bracket has several properties that are easily verified (in the following, capital letters denote functions of \(q_i\) and \(p_i\), whereas \(k\) is a constant):

\begin{align} \{U,V\}&=-\{V,U\}\\ \{U,U\}&=0\\ \{kU,V\}&=k\{U,V\}\\ \{A+B,V\}&=\{A,V\}+\{B,V\}\\ \{AB,V\}&=A\{B,V\}+B\{A,V\}\\ \{q_i,q_j\}=\{p_i,p_j\}&=0\\ \{q_i,p_j\}&=\delta_{ij}\\ \{q_i^n,p_j\}&=nq_i^{n-1}\delta_{ij}\\ \{U,p_i\}&=\die{U}{q_i}\\ \{U,q_i\}&=-\die{U}{p_i}\\ \{U,\{V,W\}\}+\{V,\{W,U\}\}+\{W,\{U,V\}\}&=0 \tag{7.37}\end{align}

This last relation is called “Jacobi’s identity”. \(\delta_{ij}\) is the Kronecker delta, and is equal to zero unless \(i=j\). It seems a little strange to introduce the Poisson Brackets as a mathematical artefact, but they are related to the commutators that appear in Quantum Mechanics, so it is worthwhile to explore them further. Consider for example the Poisson Bracket of a canonical variable with the Hamiltonian:

\begin{align} \{q_j,H\}&=\sum_i^n\left(\die{q_j}{q_i}\die{H}{p_i}-\die{q_j}{p_i}\die{H}{q_i}\right)\\ &=\die{H}{p_j}\\ \{p_j,H\}&=\sum_i^n\left(\die{p_j}{q_i}\die{H}{p_i}-\die{p_j}{p_i}\die{H}{q_i}\right)\\ &=-\die{H}{q_j}\end{align}

where the terms \(\die{q_j}{q_i}\) and \(\die{p_j}{p_i}\) are zero unless \(i=j\), and the terms \(\die{q_j}{p_i}\) and \(\die{p_j}{q_i}\) are zero. Identifying this with Hamilton’s canonical equations, we can re-write the canonical equations as:

\begin{align} \dot q_i&=\{q_i,H\}\\ \dot p_i&=\{p_i,H\}\end{align}

More generally, consider the total time derivative of a function, \(F(q_i,p_i,t)\), of the canonical variables:

\begin{align} \frac{dF}{dt}&=\sum_i\left(\die{F}{q_i}\dot q_i+\die{F}{p_i}\dot p_i\right)+\die{F}{t}\\ &=\sum_i\left(\die{F}{q_i}\die{H}{p_i}-\die{F}{p_i}\die{H}{q_i}\right)+\die{F}{t}\\ &=\{F,H\}+\die{F}{t} \tag{7.40}\end{align}

We can see that the time derivative of a quantity is given by its Poisson Bracket with the Hamiltonian. If \(F\) does not depend explicitly on time, it is a constant of motion if its Poisson Bracket with the Hamiltonian is zero:

\begin{align} \{F(q_i,p_i),H\}=0 \to F=\text{const.} \tag{7.41}\end{align}

Poisson Brackets and symmetries

Recall that we showed in Chapter ? that there exists a conserved quantity, \(Q\), for each axis of rotation about which the Lagrangian was invariant. For an infinitesimal rotation of angle \(\delta \epsilon\) about the z-axis:

\begin{align} x' &= x+f_x\delta\epsilon = x-y\delta \epsilon\\ y &= y+f_y\delta\epsilon = y+x\delta \epsilon\\ z &= z+f_z\delta\epsilon = z \\ Q&=\sum_i f_ip_i=(xp_y - yp_x)=L_z\end{align}

and the conserved quantity was the z-component of angular momentum, \(L_z\).

Now consider the Poisson Bracket:

\begin{align} \{x,L_z\}&=\{x,L_z\}\\ &=\{x,xp_y - yp_x\}\\ &=\{x,xp_y\} - \{x,yp_x\}\\ &=x\{x,p_y\}+p_y\{x,x\} - y\{x,p_x\}-p_x\{x,y\}\\ &=-y \tag{7.43}\end{align}

where we have made use of the properties from equations 7.37. One can easily show that:

\begin{align} \{x,L_z\}&=-y=f_x\\ \{y,L_z\}&=x=f_y\\ \{z,L_z\}&=0=f_z \tag{7.44}\end{align}

Thus, the Poisson Bracket of a coordinate with a component of angular momentum gives the coefficient (\(f_i\)) corresponding to the transformation of that coordinate with respect to infinitesimal rotations.

Now consider the Poisson Bracket with the components of momentum:

\begin{align} \{p_x,L_z\}&=-p_y=f_x\\ \{p_y,L_z\}&=p_x=f_y\\ \{p_z,L_z\}&=0=f_z \tag{7.45}\end{align}

which are easily demonstrated. The momentum vector transforms the same way as coordinates under a rotation, so the Poisson Brackets also correspond to the correct coefficients for rotations of the momentum vector. We find that the \(f_i\) corresponding to how a particular quantity is rotated about the z-axis are given by the Poisson Bracket of the quantity with the z-component of angular momentum. We say that angular momentum is the “generator” of rotations.

Example 7-2

Show that a charged sphere rotating about some arbitrary axis precesses about an axis parallel to a uniform magnetic field. Assume the magnetic field is in the z-direction

The Hamiltonian for the sphere is simply the total energy of the rotating sphere. The kinetic energy is a constant and will thus not influence the motion, and we can ignore it. The potential energy is related to the magnetic dipole of the sphere:

\begin{align*} V=\vec\mu\cdot\vec B \end{align*}

The magnetic moment will be proportional to the angular momentum of the sphere, and will point in the direction of the angular momentum:

\begin{align*} \vec\mu&\propto \vec L\\ \therefore H=V&\propto \vec L\cdot \vec B=\omega L_z \end{align*}

where \(\omega\) is a constant including the magnitude of the magnetic field, the charge and the radius of the sphere, and we used the fact that \(B\) points in the z-direction. Now consider how the components of angular momentum change with time (using Poisson Brackets):

\begin{align} \dot L_z=\{L_z,H\}=\{L_z,\omega L_z\}=0\tag{7.46}\\ \dot L_x=\{L_x,H\}=\{L_x,\omega L_z\}=-kL_y\tag{7.47}\\ \dot L_y=\{L_y,H\}=\{L_y,\omega L_z\}=kL_x \tag{7.48}\end{align}

Thus the z-component of the angular momentum does not change. The equations for the \(x\) and \(y\) components correspond to the equations for a particle rotating around a circle with angular speed \(\omega\). Indeed, consider a particle rotating on a circle:

\begin{align*} x&=\cos(\omega t)\\ y&=\sin(\omega t)\\ \dot x &=-\omega\sin(\omega t)=-\omega y\\ \dot y &=\omega\cos(\omega t)=\omega x \end{align*}

Thus the tip of the angular momentum vector of the charged sphere rotates at an angular speed \(\omega\) in a circle that is perpendicular to the \(z\) axis. In other words, the charged sphere precesses about the axis of the magnetic field.

Let’s examine the relation between Poisson Brackets, symmetries and conserved quantities further. Recall that we showed that if the Lagrangian is invariant under translation in a direction, then momentum in that direction is conserved. The infinitesimal translation about the x-axis transformation equations and the associated conserved quantity are given by:

\begin{align} x'&=x+f_x\delta\epsilon =x+\delta\epsilon\\ y'&=y+f_y\delta\epsilon =y\\ z'&=z+f_z\delta\epsilon =z\\ Q &=p_x \tag{7.49}\end{align}

And we find that the Poisson Brackets are evaluated trivially:

\begin{align} \{x,p_x\}&=1=f_x\\ \{y,p_x\}&=0=f_y\\ \{z,p_x\}&=0=f_z \tag{7.50}\end{align}

More generally, given a function, \(F\), we have:

\begin{align} \{F,p_x\}=\die{F}{x} \tag{7.51}\end{align}

Under an infinitesimal translation a distance \(\delta x=\delta\epsilon\) in the x-direction, the variation of \(F\) is:

\begin{align} F'&=F+\die{F}{x}\delta\epsilon =F+\{F,p_x\}\delta\epsilon\\ \delta F&=\{F,p_x\}\delta\epsilon \tag{7.52}\end{align}

Thus, taking the Poisson Bracket of a conserved quantity and a coordinate, tells us how that coordinate transforms under the symmetry that corresponds to that conserved quantity.

Recall that if the Hamiltonian does not depend on time, then energy is conserved. Energy in this case is the Hamiltonian (so the statement is a little redundant). We can thus find how a quantity transforms under a translation in time by taking the Poisson Bracket with the Hamiltonian. Under a time translation, \(\delta t = \delta \epsilon\):

\begin{align} F'=F+\{F,H\}\delta\epsilon \tag{7.53}\end{align}

Now, consider a general function, \(G(q_i,p_i)\) of the \(2n\) coordinates. Let’s define a transformation of the coordinates given by:

\begin{align} q_i'&=q_i+\{q_i,G\}\delta\epsilon=q_i+\die{G}{p_i}\delta\epsilon \\ p_i'&=p_i+\{p_i,G\}\delta\epsilon=p_i-\die{G}{q_i}\delta\epsilon \tag{7.54}\end{align}

We will call \(G\) the “generator” for this transformation. The Hamiltonian will transform just as any other function of \(q_i\) and \(p_i\). If the Hamiltonian is invariant under the transformation generated by \(G\):

\begin{align} H'&=H+\{H,G\}\delta\epsilon\\ &=H \\ \therefore \{H,G\}&=0 \tag{7.55}\end{align}

That is, if the Poisson Bracket \(\{H,G\}=0\), then the Hamiltonian in invariant under the coordinate transformation generated by \(G\). But this also means that \(\{G,H\}=0\), hence that \(G\) does not change with time (or is a constant). Here, the Poisson Brackets give us this insight into how a conserved quantity is related to the coordinates transformations under which the Hamiltonian is invariant.