Canonical Transformations and Poisson Brackets Questions and Answers

Q: If \(F_{1}(Q, q) = Qq\) is the generating function of a canonical transformation from \((p, q)\) to \((P, Q)\), then which one of the following relations is correct?

The correct relation is \(\dfrac{p}{P} = -\dfrac{Q}{q}\).A type-1 generating function F1(q, Q) of the old and new coordinates gives the momenta by\[ \begin{aligned} p &= \frac{\partial F_{1}}{\partial q} \\ P &= -\frac{\partial F_{1}}{\partial Q} \end{aligned} \]With \(F_{1} = Qq\): \(p = Q\) and \(P = -q\). Dividing, \(\dfrac{p}{P} = \dfrac{Q}{-q} = -\dfrac{Q}{q}\).The options without the minus sign miss the sign in P = -∂F1/∂Q. The forms with P/p invert the ratio: P/p = -q/Q, not -Q/q.

Q: For a system with n degrees of freedom, the Poisson's bracket \( [x_{i}, p_{j}] \) is :

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.

If \(F_{1}(Q, q) = Qq\) is the generating function of a canonical transformation from \((p, q)\) to \((P, Q)\), then which one of the following relations is correct?

A.\(\dfrac{p}{P} = \dfrac{Q}{q}\)
B.\(\dfrac{P}{p} = \dfrac{Q}{q}\)
C.\(\dfrac{p}{P} = -\dfrac{Q}{q}\)
D.\(\dfrac{P}{p} = -\dfrac{Q}{q}\)

Exam Year(s)

GATE Physics - 2024,

Answer: C

The correct relation is \(\dfrac{p}{P} = -\dfrac{Q}{q}\).

A type-1 generating function F₁(q, Q) of the old and new coordinates gives the momenta by

\[ \begin{aligned} p &= \frac{\partial F_{1}}{\partial q} \\ P &= -\frac{\partial F_{1}}{\partial Q} \end{aligned} \]

With \(F_{1} = Qq\): \(p = Q\) and \(P = -q\). Dividing, \(\dfrac{p}{P} = \dfrac{Q}{-q} = -\dfrac{Q}{q}\).

The options without the minus sign miss the sign in P = -∂F₁/∂Q. The forms with P/p invert the ratio: P/p = -q/Q, not -Q/q.

Discuss in Forum

For a system with n degrees of freedom, the Poisson's bracket \( [x_{i}, p_{j}] \) is :

A.Zero
B.\( \delta_{ij} \)
C.\( \delta_{ij}\delta_{jk}p_{k} \)
D.\( -\delta_{ij} \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Canonical Transformations and Poisson Brackets Questions and Answers

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