Discussion
For a system with n degrees of freedom, the Poisson's bracket \( [x_{i}, p_{j}] \) is :
Answer: B
The Poisson bracket is \( \delta_{ij} \).
By definition, for canonical coordinates \( q_{k}, p_{k} \):
\[ \begin{aligned} [x_{i}, p_{j}] = \sum_{k} \Big( &\frac{\partial x_{i}}{\partial q_{k}} \frac{\partial p_{j}}{\partial p_{k}} \\ - &\frac{\partial x_{i}}{\partial p_{k}} \frac{\partial p_{j}}{\partial q_{k}} \Big) \end{aligned} \]
The second term vanishes and the first is \( \sum_{k} \delta_{ik}\delta_{jk} = \delta_{ij} \): 1 when the coordinate and momentum are conjugate, 0 otherwise. This is the classical counterpart of \( [x_{i}, p_{j}] = i\hbar\delta_{ij} \).
Zero holds only for \( i \ne j \), \( -\delta_{ij} \) is the bracket taken in the reverse order, \( [p_{j}, x_{i}] \), and the option with \( p_{k} \) is not a constant at all.
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