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Poisson brackets

Poisson brackets provide us an elegant and powerful way to rewrite Hamilton's equations. If we have nn degrees of freedom and two functions ff and gg that depend on pip_i and qiq_i:[1]

{f,g}=∑i=1n(∂f∂qi∂g∂pi−∂f∂pi∂g∂qi).\begin{align} \{f, g\} &= \sum_{i=1}^n \left( \dfrac{\partial f}{\partial q_i} \dfrac{\partial g}{\partial p_i} - \dfrac{\partial f}{\partial p_i}\dfrac{\partial g}{\partial q_i} \right). \tag{1} \end{align}

Poisson brackets also have several interesting properties, such as antisymmetry:

{A,B}=−{B,A}.\begin{align} \{A, B\} &= -\{B, A\}. \tag{2} \end{align}

Further, assuming each of the coordinates of the phase space are independent, it follows that:

{qk,pk}=∑i=1n(∂qk∂qi∂pk∂pi−∂qk∂pi∂pk∂qi)=∑i=1nδikδik=1.\begin{align} \{q_k, p_k\} &= \sum_{i=1}^n \left( \dfrac{\partial q_k}{\partial q_i} \dfrac{\partial p_k}{\partial p_i} - \dfrac{\partial q_k}{\partial p_i} \dfrac{\partial p_k}{\partial q_i} \right) \notag \\ &= \sum_{i=1}^n \delta_{ik}\delta_{ik} \notag \\ &= 1. \tag{3} \end{align}

Where δik\delta_{ik} is the Kronecker delta function. It is equal to zero if i≠ki\neq k and one if i=ki=k.

{qj,qk}=∑i=1n(∂qj∂qi∂qk∂pi−∂qj∂pi∂qk∂qi)=0{pj,pk}=∑i=1n(∂pj∂qi∂pk∂pi−∂pj∂pi∂pk∂qi)=0.\begin{align} \{q_j, q_k\} &= \sum_{i=1}^n \left( \dfrac{\partial q_j}{\partial q_i} \dfrac{\partial q_k}{\partial p_i} - \dfrac{\partial q_j}{\partial p_i} \dfrac{\partial q_k}{\partial q_i} \right) \notag \\ &= 0 \tag{4} \\ \{p_j, p_k\} &= \sum_{i=1}^n \left(\dfrac{\partial p_j}{\partial q_i}\dfrac{\partial p_k}{\partial p_i} - \dfrac{\partial p_j}{\partial p_i}\dfrac{\partial p_k}{\partial q_i}\right) \notag \\ &= 0. \tag{5} \end{align}

The quality of two functions, say ff and gg, having a Poisson bracket that equals zero is phrased as ff and gg commuting under the Poisson bracket.

If a function f(qi,pi)f(q_i, p_i) is not explicitly dependent on tt and only depends on tt indirectly through qiq_i and pip_i, then, via the chain rule:

dfdt=∑i=1n∂f∂qidqidt+∂f∂pidpidt.\begin{align} \dfrac{df}{dt} &= \sum_{i=1}^n \dfrac{\partial f}{\partial q_i}\dfrac{d q_i}{dt} + \dfrac{\partial f}{\partial p_i}\dfrac{dp_i}{dt}. \tag{6} \end{align}

We can use Hamilton's equations to replace dqidt\dfrac{dq_i}{dt} and dpidt\dfrac{dp_i}{dt}:

dfdt=∑i=1n∂f∂qi∂H∂pi−∂f∂pi∂H∂qi.\begin{align} \dfrac{df}{dt} &= \sum_{i=1}^n \dfrac{\partial f}{\partial q_i}\dfrac{\partial H}{\partial p_i} - \dfrac{\partial f}{\partial p_i}\dfrac{\partial H}{\partial q_i}. \tag{7} \end{align}

Interestingly, the right-hand side is equal to the Poisson bracket of ff with HH, in other words:

dfdt={f,H}.\begin{align} \dfrac{df}{dt} &= \{f, H\}. \tag{8} \end{align}

Hence commuting with the Hamiltonian under the Poisson bracket is a sufficient condition for a quantity that does not explicitly depend on time to have a zero time derivative. A zero time derivative means the quantity is static over time which means it is conserved.

This quality makes it unsurprising that Hamilton's equations can be rewritten as:

q˙k={qk,H}p˙k=−{pk,H}.\begin{align} \dot{q}_k &= \{q_k, H\} \tag{9} \\ \dot{p}_k &= -\{p_k, H\}. \tag{10} \end{align}
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Last modified: October 31, 2026.