Question Statement:
A system with time independent Hamiltonian $H(q, p)$ has two constants of motion $f(q, p)$ and $g(q, p)$. Then which of the following Poisson brackets are always zero?
Options:
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(A) $\{H, f + g\}$
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(B) $\{H, \{f, g\}\}$
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(C) $\{H + f, g\}$
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(D) $\{H, H + fg\}$
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.54)**
**1. Classical Mechanics (Poisson Brackets & Constants of Motion)**:
- Let $H$ be time-independent. By Poisson's theorem, if $f$ and $g$ are constants of motion:
$\{H, f\} = 0, \quad \{H, g\} = 0$
**2. Detailed Step-by-Step Evaluation**:
- **(A) $\{H, f + g\}$**:
$\{H, f + g\} = \{H, f\} + \{H, g\} = 0 + 0 = 0$
Hence $f + g$ is a constant of motion, meaning **(A) vanishes**.
- **(B) $\{H, \{f, g\}\}$**:
By Jacobi's Identity:
$\{H, \{f, g\}\} + \{f, \{g, H\}\} + \{g, \{H, f\}\} = 0$
Since $\{g, H\} = 0$ and $\{H, f\} = 0$:
$\{H, \{f, g\}\} = 0$
This is Poisson's Theorem: the Poisson bracket of two constants of motion is itself a constant of motion. **(B) vanishes**.
- **(D) $\{H, H + fg\}$**:
$\{H, H + fg\} = \{H, H\} + \{H, fg\} = 0 + f\{H, g\} + g\{H, f\} = 0 + 0 + 0 = 0$
**(D) vanishes**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, B, D)**.