11.2 Properties of the Poisson bracket
(a) Algebraic properties
Directly from the definition, for any dynamical variables \(A\), \(B\), \(C\), any constants \(a\), \(b\) and any differentiable function \(f\), the Poisson bracket has the following properties. It is antisymmetric, \(\{A,B\}=-\{B,A\}\), so that \(\{A,A\}=0\); more generally \(\{A,f(A)\}=0\), since \(\partial f(A)/\partial q_i=f'(A)\,\partial A/\partial q_i\) and the two terms of the bracket cancel. It is linear in each argument, \(\{A,bB+cC\}=b\{A,B\}+c\{A,C\}\), and a constant has a vanishing bracket with everything. It obeys the product (Leibniz) rule
\[\{A,BC\}=\{A,B\}C+B\{A,C\},\qquad \{AB,C\}=\{A,C\}B+A\{B,C\},\]
which follows from the product rule for derivatives. Finally it satisfies the Jacobi identity
\[\boxed{\{A,\{B,C\}\}+\{B,\{C,A\}\}+\{C,\{A,B\}\}=0}.\]
The Jacobi identity can be verified by expanding all twelve terms of second derivatives: each product of a second derivative of one function with first derivatives of the other two appears twice with opposite signs. Together with antisymmetry and linearity it makes the dynamical variables a Lie algebra under the Poisson bracket. We also note the useful rule for partial derivatives with respect to a parameter, \(\dfrac{\partial}{\partial t}\{A,B\}=\left\{\dfrac{\partial A}{\partial t},B\right\}+\left\{A,\dfrac{\partial B}{\partial t}\right\}\).
(b) The fundamental Poisson brackets
The brackets of the canonical variables among themselves are
\[\boxed{\{q_i,q_j\}=0,\qquad \{p_i,p_j\}=0,\qquad \{q_i,p_j\}=\delta_{ij}},\]
where \(\delta_{ij}=1\) for \(i=j\) and \(0\) otherwise. These fundamental brackets are the starting point of most calculations, because the product rule reduces any bracket of polynomials to them. Two consequences are used constantly: for any function \(f(q)\) of the coordinates and \(g(p)\) of the momenta,
\[\{q_i,g(p)\}=\frac{\partial g}{\partial p_i},\qquad \{p_i,f(q)\}=-\frac{\partial f}{\partial q_i}.\]
For a particle with position \(\mathbf r=(x,y,z)\) and momentum \(\mathbf p\), the angular momentum \(\mathbf L=\mathbf r\times\mathbf p\) has components such as \(L_z=xp_y-yp_x\). Using the product rule,
\[\{x,L_z\}=\{x,xp_y\}-\{x,yp_x\}=-y\{x,p_x\}=-y,\qquad \{y,L_z\}=x\{y,p_y\}=x,\]
and in the same way \(\{p_x,L_z\}=-p_y\) and \(\{p_y,L_z\}=p_x\). Writing \(x_1,x_2,x_3\) for \(x,y,z\), the general results are
\[\{x_i,L_j\}=\epsilon_{ijk}x_k,\qquad \{p_i,L_j\}=\epsilon_{ijk}p_k,\qquad \boxed{\{L_i,L_j\}=\epsilon_{ijk}L_k},\]
with summation over \(k\) and \(\epsilon_{ijk}\) the Levi-Civita symbol. For example \(\{L_x,L_y\}=L_z\). Since the bracket of \(L^2\) with each component vanishes, \(\{L^2,L_i\}=0\).
(c) Poisson’s theorem
If \(f\) and \(g\) are two constants of the motion, then their Poisson bracket \(\{f,g\}\) is also a constant of the motion. This is Poisson’s theorem. For time-independent \(f\) and \(g\) the proof is a single application of the Jacobi identity:
\[\{\{f,g\},H\}=-\{\{g,H\},f\}-\{\{H,f\},g\}=-\{0,f\}-\{0,g\}=0.\]
If \(f\) and \(g\) depend explicitly on time, we combine the Jacobi identity with the rule for \(\partial/\partial t\) of a bracket:
\[\frac{d}{dt}\{f,g\}=\{\{f,g\},H\}+\frac{\partial}{\partial t}\{f,g\}=\left\{\frac{df}{dt},g\right\}+\left\{f,\frac{dg}{dt}\right\}=0.\]
The theorem sometimes yields new integrals of the motion. For instance, if \(L_x\) and \(L_y\) are conserved, then so is \(\{L_x,L_y\}=L_z\). Often, however, the bracket is merely a constant, a function of the integrals already known, or zero.
(d) Poisson brackets and quantum commutators
The algebra of Poisson brackets carries over to quantum mechanics. In Dirac’s correspondence the classical bracket is replaced by the commutator of the corresponding operators according to
\[\{A,B\}\;\longrightarrow\;\frac{1}{i\hbar}[\hat A,\hat B],\]
so that \(\{x,p\}=1\) becomes \([\hat x,\hat p]=i\hbar\) and the classical equation \(df/dt=\{f,H\}+\partial f/\partial t\) becomes the Heisenberg equation of motion. For polynomials of low degree, symmetrized in \(x\) and \(p\) where necessary, a classical bracket calculation and the corresponding commutator calculation therefore give the same result apart from the factor \(i\hbar\). Several examination questions exploit this parallel.