Poission bracket Quiz

Explore the essential properties and applications of the Poisson bracket in classical mechanics, including its definition, algebraic properties, and role in dynamical systems. This quiz helps reinforce your understanding of key Poisson bracket concepts with practical and theoretical questions.

1. Definition of the Poisson Bracket

   Which of the following best defines the Poisson bracket {f, g} of two functions f and g in canonical coordinates (q, p)?

   1. {f, g} = f g − g f
   2. {f, g} = ∂^2f/∂q^2 ∂^2g/∂p^2
   3. {f, g} = ∂f/∂q + ∂g/∂p
   4. {f, g} = ∂f/∂q ∂g/∂p − ∂f/∂p ∂g/∂q

   Explanation: The Poisson bracket {f, g} in canonical coordinates is defined as ∂f/∂q ∂g/∂p minus ∂f/∂p ∂g/∂q. Option B is simply the commutator from linear algebra, which is not applicable to functions in classical mechanics. Option C adds derivatives incorrectly and does not represent the bracket. Option D involves second derivatives but the correct definition uses first derivatives. Thus, only the first option accurately represents the standard definition.

2. Antisymmetry Property

   What property of the Poisson bracket is demonstrated by the equation {f, g} = −{g, f} for any two differentiable functions f and g?

   1. Symplecticity
   2. Associativity
   3. Linearity
   4. Antisymmetry

   Explanation: Antisymmetry means switching the order of the functions in the Poisson bracket changes the sign, which is exactly what the equation shows. Associativity relates to how operations group, and does not describe this property. Linearity means distributing over addition, not inversion of sign. Symplecticity is a property of transformations, not the bracket itself. Therefore, antisymmetry is the only correct answer.

3. Poisson Brackets of Canonical Variables

   Given canonical coordinates q and p, what is the value of the Poisson bracket {q, p}?

   1. -1
   2. q p
   3. 0
   4. 1

   Explanation: The Poisson bracket {q, p} equals 1 by the fundamental canonical relations in Hamiltonian mechanics. Zero is incorrect as only the bracket of a coordinate with itself is zero. Negative one is the value for {p, q}, showing the antisymmetry. The product q p is not related to the bracket's value here. The correct numerical value is one.

4. Poisson Bracket and Constants of Motion

   If the Poisson bracket {f, H} = 0 for a function f and Hamiltonian H, what does this imply about f in a Hamiltonian system?

   1. f is a constant of motion
   2. f equals Hamiltonian H
   3. f is an explicit function only of time
   4. f is always zero

   Explanation: When the Poisson bracket with the Hamiltonian vanishes, f is conserved and does not change during the system's evolution, thus making it a constant of motion. It does not mean f equals H, since many different functions can commute with H. f being zero is not implied by the bracket vanishing. The last option confuses time dependence, but the bracket only involves phase space variables, not explicit time.

5. Leibniz Rule (Derivation Property)

   Which statement correctly describes how the Poisson bracket acts on a product of functions: {f, g h}?

   1. {f, g h} = {f, g} + {f, h}
   2. {f, g h} = {h, g} f + {f, h} g
   3. {f, g h} = {f, g} {f, h}
   4. {f, g h} = {f, g} h + g {f, h}

   Explanation: The Leibniz rule states that the Poisson bracket acts on products like a derivation, yielding {f, g} h + g {f, h}. Multiplying the brackets as in option B is incorrect. Option C mixes the order of functions and associates incorrectly. Option D simply adds the brackets without considering multiplication. The first option is the correct statement of the derivation property.