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.

Q: For a particle moving in an inverse square force field, \(V(r) = -\frac{K}{r}\), the action angle variable analysis gives :

On the key, the action-angle analysis gives two similar frequencies of oscillation.Motion under the inverse-square force is planar, with two degrees of freedom, r and \(\theta\). Their actions \(J_{r}\) and \(J_{\theta}\) give the energy\[ E = -\frac{2\pi^{2}mK^{2}}{(J_{r} + J_{\theta})^{2}} \]which depends only on the sum \(J_{r} + J_{\theta}\). The two frequencies \(\nu = \partial E/\partial J\) are therefore equal: the radial oscillation and the angular revolution have the same period, which is why bound Kepler orbits are closed ellipses. This degeneracy is the point of the question.Two dissimilar frequencies would give non-closed, precessing orbits, as in a perturbed potential. 'Only one frequency' misses that there are two degrees of freedom whose frequencies coincide.

Q: If \(p_{j}\) and \(p_{j}'\) are the components of the generalised momenta corresponding to the Lagrangians L and L' respectively, where\[ L' = L + \frac{d}{dt}F(q_{j}, t), \]then :

The momenta are related by \(p_{j}' = p_{j} + \frac{\partial F}{\partial q_{j}}\).Expand the total derivative:\[ \begin{aligned} \frac{dF}{dt} &= \sum_{k} \frac{\partial F}{\partial q_{k}}\dot{q}_{k} \\ &\quad + \frac{\partial F}{\partial t} \end{aligned} \]Only the first sum contains the velocities, and it is linear in them, so\[ \begin{aligned} p_{j}' &= \frac{\partial L'}{\partial \dot{q}_{j}} \\ &= p_{j} + \frac{\partial F}{\partial q_{j}} \end{aligned} \]Adding a total time derivative leaves the equations of motion unchanged but shifts the canonical momenta, so \(p_{j}' = p_{j}\) is wrong. \(\partial F / \partial t\) has no velocity dependence and drops out, and option D differentiates once too often.

Q: The in order to have following transformation to be canonical\[ \begin{aligned} Q &= q^{1/\alpha}\cos(\beta p) \\ P &= q^{1/\alpha}\sin(\beta p) \end{aligned} \]what should be the value of \(\alpha\) and \(\beta\) ?

The transformation is canonical for \(\alpha = 2, \beta = 2\).A transformation is canonical when the Poisson bracket \(\{Q, P\}_{q,p} = 1\). With \(Q = q^{1/\alpha}\cos\beta p\) and \(P = q^{1/\alpha}\sin\beta p\):\[ \begin{aligned} \{Q,P\} &= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1}(\cos^{2}\beta p \\ &\qquad + \sin^{2}\beta p) \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1} \end{aligned} \]For this to equal 1 for every \(q\), the power of \(q\) must vanish, so \(2/\alpha - 1 = 0\) and \(\alpha = 2\); then \(\beta/\alpha = 1\) gives \(\beta = 2\).\(\alpha = \beta = 1\) leaves a factor \(q\); \(\alpha = 1/2\) leaves \(q^{3}\); and \(\alpha = 2, \beta = 1/2\) gives the bracket 1/4.

Q: Which identity of the following does not hold true for Poisson bracket ?(Where all the symbols have their usual meaning)

The identity that does not hold is \([L_{i}, L_{j}] = \delta_{ij}\).The components of angular momentum do not have a Kronecker-delta bracket with each other; they form the rotation algebra:\[ [L_{i}, L_{j}] = \epsilon_{ijk}L_{k} \]In particular \([L_{i}, L_{i}] = 0\) for any single component, whereas \(\delta_{ii} = 1\), so the stated identity fails even for \(i = j\).The other three are the standard results: \([x_{i}, p_{j}] = \delta_{ij}\) is the fundamental bracket, and \([L_{i}, p_{j}] = \epsilon_{ijk}p_{k}\) and \([x_{i}, L_{j}] = \epsilon_{ijk}x_{k}\) express that \(\vec{p}\) and \(\vec{r}\) transform as vectors under rotations.

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.

Discuss in Forum

For a particle moving in an inverse square force field, \(V(r) = -\frac{K}{r}\), the action angle variable analysis gives :

A.only one frequency of oscillation
B.two similar frequencies of oscillation
C.two dissimilar frequencies of oscillation
D.more than two dissimilar frequencies of oscillation

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

On the key, the action-angle analysis gives two similar frequencies of oscillation.

Motion under the inverse-square force is planar, with two degrees of freedom, r and \(\theta\). Their actions \(J_{r}\) and \(J_{\theta}\) give the energy

\[ E = -\frac{2\pi^{2}mK^{2}}{(J_{r} + J_{\theta})^{2}} \]

which depends only on the sum \(J_{r} + J_{\theta}\). The two frequencies \(\nu = \partial E/\partial J\) are therefore equal: the radial oscillation and the angular revolution have the same period, which is why bound Kepler orbits are closed ellipses. This degeneracy is the point of the question.

Two dissimilar frequencies would give non-closed, precessing orbits, as in a perturbed potential. 'Only one frequency' misses that there are two degrees of freedom whose frequencies coincide.

Discuss in Forum

The condition for generating function F, defining canonical transformations, is that the differential dF is exact. Which of the following relations does not satisfy this condition ?

A.\(dF = \sum_{i}p_{i}\,\delta q_{i} - \sum_{i}P_{i}\,\delta Q_{i}\)
B.\(dF = \sum_{i}q_{i}\,\delta p_{i} + \sum_{i}Q_{i}\,\delta P_{i}\)
C.\(dF = \sum_{i}p_{i}\,\delta q_{i} + \sum_{i}Q_{i}\,\delta P_{i}\)
D.\(dF = \sum_{i}q_{i}\,\delta p_{i} + \sum_{i}P_{i}\,\delta Q_{i}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The relation that does not satisfy the condition is \(dF = \sum_{i}q_{i}\,\delta p_{i} + \sum_{i}Q_{i}\,\delta P_{i}\).

A transformation is canonical if \(\sum p\,dq - \sum P\,dQ\) is an exact differential \(dF_{1}\). Legendre transformations give the other standard forms:

\(\sum p\,dq + \sum Q\,dP = d(F_{1} + \sum PQ)\): exact (type 2).
\(-\sum q\,dp - \sum P\,dQ = d(F_{1} - \sum pq)\): exact (type 3), so its negative \(\sum q\,dp + \sum P\,dQ\) is exact too.
\(-\sum q\,dp + \sum Q\,dP\) is the type 4 form.

\(\sum q\,dp + \sum Q\,dP\) has the sign of the \(Q\,dP\) term reversed relative to type 4; it differs from an exact form by \(2\sum Q\,dP\), which is not exact in general, so it does not define a canonical transformation.

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If \(p_{j}\) and \(p_{j}'\) are the components of the generalised momenta corresponding to the Lagrangians L and L' respectively, where

\[ L' = L + \frac{d}{dt}F(q_{j}, t), \]

then :

A.\(p_{j}' = p_{j}\)
B.\(p_{j}' = p_{j} + \frac{\partial F}{\partial q_{j}}\)
C.\(p_{j}' = p_{j} + \frac{\partial F}{\partial t}\)
D.\(p_{j}' = p_{j} + \frac{d}{dt}\left(\frac{\partial F}{\partial q_{j}}\right)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The momenta are related by \(p_{j}' = p_{j} + \frac{\partial F}{\partial q_{j}}\).

Expand the total derivative:

\[ \begin{aligned} \frac{dF}{dt} &= \sum_{k} \frac{\partial F}{\partial q_{k}}\dot{q}_{k} \\ &\quad + \frac{\partial F}{\partial t} \end{aligned} \]

Only the first sum contains the velocities, and it is linear in them, so

\[ \begin{aligned} p_{j}' &= \frac{\partial L'}{\partial \dot{q}_{j}} \\ &= p_{j} + \frac{\partial F}{\partial q_{j}} \end{aligned} \]

Adding a total time derivative leaves the equations of motion unchanged but shifts the canonical momenta, so \(p_{j}' = p_{j}\) is wrong. \(\partial F / \partial t\) has no velocity dependence and drops out, and option D differentiates once too often.

Discuss in Forum

The in order to have following transformation to be canonical

\[ \begin{aligned} Q &= q^{1/\alpha}\cos(\beta p) \\ P &= q^{1/\alpha}\sin(\beta p) \end{aligned} \]

what should be the value of \(\alpha\) and \(\beta\) ?

A.\(\alpha = 1, \beta = 1\)
B.\(\alpha = \frac{1}{2}, \beta = 2\)
C.\(\alpha = 2, \beta = \frac{1}{2}\)
D.\(\alpha = 2, \beta = 2\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The transformation is canonical for \(\alpha = 2, \beta = 2\).

A transformation is canonical when the Poisson bracket \(\{Q, P\}_{q,p} = 1\). With \(Q = q^{1/\alpha}\cos\beta p\) and \(P = q^{1/\alpha}\sin\beta p\):

\[ \begin{aligned} \{Q,P\} &= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1}(\cos^{2}\beta p \\ &\qquad + \sin^{2}\beta p) \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1} \end{aligned} \]

For this to equal 1 for every \(q\), the power of \(q\) must vanish, so \(2/\alpha - 1 = 0\) and \(\alpha = 2\); then \(\beta/\alpha = 1\) gives \(\beta = 2\).

\(\alpha = \beta = 1\) leaves a factor \(q\); \(\alpha = 1/2\) leaves \(q^{3}\); and \(\alpha = 2, \beta = 1/2\) gives the bracket 1/4.

Discuss in Forum

Which identity of the following does not hold true for Poisson bracket ?

(Where all the symbols have their usual meaning)

A.\([x_{i}, p_{j}] = \delta_{ij}\)
B.\([L_{i}, L_{j}] = \delta_{ij}\)
C.\([L_{i}, p_{j}] = \epsilon_{ijk}\, p_{k}\)
D.\([x_{i}, L_{j}] = \epsilon_{ijk}\, x_{k}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The identity that does not hold is \([L_{i}, L_{j}] = \delta_{ij}\).

The components of angular momentum do not have a Kronecker-delta bracket with each other; they form the rotation algebra:

\[ [L_{i}, L_{j}] = \epsilon_{ijk}L_{k} \]

In particular \([L_{i}, L_{i}] = 0\) for any single component, whereas \(\delta_{ii} = 1\), so the stated identity fails even for \(i = j\).

The other three are the standard results: \([x_{i}, p_{j}] = \delta_{ij}\) is the fundamental bracket, and \([L_{i}, p_{j}] = \epsilon_{ijk}p_{k}\) and \([x_{i}, L_{j}] = \epsilon_{ijk}x_{k}\) express that \(\vec{p}\) and \(\vec{r}\) transform as vectors under rotations.

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