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Appendix D Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Calculation of Canonical Momentum from the Lagrangian

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Solution Strategy

We directly apply the definition of canonical momentum \(p = \frac{\partial L}{\partial \dot{q}}\). In the partial differentiation, \(q\) is treated as a constant.

(a) \(L = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}kq^2\)

\[ p = \frac{\partial L}{\partial \dot{q}} = \frac{\partial}{\partial \dot{q}}\left(\frac{1}{2}m\dot{q}^2\right) = m\dot{q} \]
\[ \boxed{p = m\dot{q}} \]

(b) \(L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)\)

Momentum conjugate to \(r\):

\[ p_r = \frac{\partial L}{\partial \dot{r}} = m\dot{r} \]

Momentum conjugate to \(\theta\):

\[ p_\theta = \frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta} \]
\[ \boxed{p_r = m\dot{r}, \qquad p_\theta = mr^2\dot{\theta}} \]

Verification: \(p_\theta = mr^2\dot{\theta}\) coincides with the \(z\)-component of angular momentum \(L_z\). Its dimensions are \([\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}]\), which is correct for angular momentum.

(c) \(L = \frac{1}{2}m\dot{q}^2 + e\dot{q}A(q) - e\phi(q)\)

Since \(A(q)\) does not depend on \(\dot{q}\):

\[ p = \frac{\partial L}{\partial \dot{q}} = m\dot{q} + eA(q) \]
\[ \boxed{p = m\dot{q} + eA(q)} \]

Verification: The canonical momentum \(p\) differs from the mechanical momentum \(m\dot{q}\) and includes the contribution from the vector potential \(eA\). This agrees with the well-known result for the canonical momentum of a charged particle in an electromagnetic field.

B-2. Construction of the Hamiltonian via Legendre Transform

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Solution Strategy

Solve the \(p\) found in D1 for \(\dot{q}\), then substitute into \(H = p\dot{q} - L\) to eliminate \(\dot{q}\).

(a) From D1(a), \(p = m\dot{q}\), so \(\dot{q} = p/m\).

\[ H = p\dot{q} - L = p \cdot \frac{p}{m} - \left[\frac{1}{2}m\left(\frac{p}{m}\right)^2 - \frac{1}{2}kq^2\right] \]
\[ = \frac{p^2}{m} - \frac{p^2}{2m} + \frac{1}{2}kq^2 = \frac{p^2}{2m} + \frac{1}{2}kq^2 \]
\[ \boxed{H = \frac{p^2}{2m} + \frac{1}{2}kq^2} \]

Check: This has the form \(T + V\). Setting \(k = m\omega^2\), it agrees with the standard Hamiltonian of the harmonic oscillator.

(b) From D1(b), \(p_r = m\dot{r}\), \(p_\theta = mr^2\dot{\theta}\), so:

\[ \dot{r} = \frac{p_r}{m}, \qquad \dot{\theta} = \frac{p_\theta}{mr^2} \]

The Hamiltonian is \(H = p_r\dot{r} + p_\theta\dot{\theta} - L\):

\[ H = p_r \cdot \frac{p_r}{m} + p_\theta \cdot \frac{p_\theta}{mr^2} - \left[\frac{1}{2}m\left(\frac{p_r^2}{m^2} + r^2 \cdot \frac{p_\theta^2}{m^2 r^4}\right) - V(r)\right] \]
\[ = \frac{p_r^2}{m} + \frac{p_\theta^2}{mr^2} - \frac{p_r^2}{2m} - \frac{p_\theta^2}{2mr^2} + V(r) \]
\[ \boxed{H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r)} \]

Check: The second term \(\frac{p_\theta^2}{2mr^2} = \frac{(mr^2\dot{\theta})^2}{2mr^2} = \frac{1}{2}mr^2\dot{\theta}^2\) corresponds to the rotational kinetic energy. The whole expression has the form \(T + V\).

(c) From D1(c), \(p = m\dot{q} + eA(q)\), so:

\[ \dot{q} = \frac{p - eA(q)}{m} \]
\[ H = p\dot{q} - L = p \cdot \frac{p - eA}{m} - \left[\frac{1}{2}m\left(\frac{p - eA}{m}\right)^2 + e \cdot \frac{p - eA}{m} \cdot A - e\phi\right] \]

Computing each term:

\[ p\dot{q} = \frac{p(p - eA)}{m} \]
\[ \frac{1}{2}m\dot{q}^2 = \frac{(p - eA)^2}{2m} \]
\[ e\dot{q}A = \frac{eA(p - eA)}{m} \]

Therefore:

\[ H = \frac{p(p - eA)}{m} - \frac{(p - eA)^2}{2m} - \frac{eA(p - eA)}{m} + e\phi \]
\[ = \frac{(p - eA)}{m}\left[p - \frac{p - eA}{2} - eA\right] + e\phi \]
\[ = \frac{(p - eA)}{m} \cdot \frac{2p - p + eA - 2eA}{2} + e\phi = \frac{(p - eA)}{m} \cdot \frac{p - eA}{2} + e\phi \]
\[ \boxed{H = \frac{(p - eA)^2}{2m} + e\phi} \]

Check: \(\frac{(p-eA)^2}{2m} = \frac{(m\dot{q})^2}{2m} = \frac{1}{2}m\dot{q}^2\) equals the kinetic energy. \(e\phi\) is the potential energy. The whole expression has the form \(T + V\), which agrees with the well-known Hamiltonian for a charged particle in an electromagnetic field.

B-3. Application of Hamilton's Canonical Equations

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Solution Strategy

For \(H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\), compute \(\dot{q} = \frac{\partial H}{\partial p}\) and \(\dot{p} = -\frac{\partial H}{\partial q}\), then derive the second-order equation of motion.

Hamilton's canonical equations:

\[ \dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m} \]
\[ \dot{p} = -\frac{\partial H}{\partial q} = -m\omega^2 q \]
\[ \boxed{\dot{q} = \frac{p}{m}, \qquad \dot{p} = -m\omega^2 q} \]

Differentiating both sides of \(\dot{q} = p/m\) with respect to time:

\[ \ddot{q} = \frac{\dot{p}}{m} \]

Substituting \(\dot{p} = -m\omega^2 q\):

\[ \ddot{q} = \frac{-m\omega^2 q}{m} = -\omega^2 q \]
\[ \boxed{m\ddot{q} = -m\omega^2 q} \]

This is precisely the equation of motion for the harmonic oscillator (restoring force \(F = -kq = -m\omega^2 q\)).

Verification: From Newton's equation of motion \(F = ma\) with \(F = -\frac{dV}{dq} = -m\omega^2 q\), we directly obtain \(m\ddot{q} = -m\omega^2 q\), which is consistent with the result above.

B-4. Direct Application of the Euler-Lagrange Equation

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Solution Strategy

For each Lagrangian, compute \(\frac{\partial L}{\partial \dot{q}}\) and \(\frac{\partial L}{\partial q}\), then substitute into \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0\).

(a) \(L = \frac{1}{2}m\dot{q}^2\)

\[ \frac{\partial L}{\partial \dot{q}} = m\dot{q}, \qquad \frac{\partial L}{\partial q} = 0 \]
\[ \frac{d}{dt}(m\dot{q}) - 0 = 0 \quad \Longrightarrow \quad \boxed{m\ddot{q} = 0} \]

A free particle moves with constant velocity in a straight line.

(b) \(L = \frac{1}{2}m\dot{q}^2 - mgq\)

\[ \frac{\partial L}{\partial \dot{q}} = m\dot{q}, \qquad \frac{\partial L}{\partial q} = -mg \]
\[ \frac{d}{dt}(m\dot{q}) - (-mg) = 0 \quad \Longrightarrow \quad m\ddot{q} + mg = 0 \]
\[ \boxed{m\ddot{q} = -mg} \]

Uniform acceleration in a gravitational field (free fall with \(q\) taken as positive upward).

(c) \(L = \frac{1}{2}m\dot{q}^2 - V(q)\)

\[ \frac{\partial L}{\partial \dot{q}} = m\dot{q}, \qquad \frac{\partial L}{\partial q} = -\frac{dV}{dq} \]
\[ \frac{d}{dt}(m\dot{q}) - \left(-\frac{dV}{dq}\right) = 0 \]
\[ \boxed{m\ddot{q} = -\frac{dV}{dq}} \]

This is Newton's second law \(F = ma\) (with \(F = -dV/dq\)) itself.

Verification: (a) is the special case \(V = 0\), and (b) is the special case \(V = mgq\), both of which are correctly reproduced from the general result in (c).

B-5. Direct Verification of Energy Conservation

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Solution Strategy

Expand the total time derivative of \(H(q,p)\) using the chain rule, then substitute Hamilton's canonical equations.

Since \(H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\) does not depend explicitly on time:

\[ \frac{dH}{dt} = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} \]

Computing each partial derivative:

\[ \frac{\partial H}{\partial q} = m\omega^2 q, \qquad \frac{\partial H}{\partial p} = \frac{p}{m} \]

Substituting the results from D3, \(\dot{q} = \frac{p}{m}\), \(\dot{p} = -m\omega^2 q\):

\[ \frac{dH}{dt} = (m\omega^2 q)\cdot\frac{p}{m} + \frac{p}{m}\cdot(-m\omega^2 q) \]
\[ = \omega^2 qp - \omega^2 qp = 0 \]
\[ \boxed{\frac{dH}{dt} = 0} \]

Verification: This result holds in general. For any system where \(H\) does not depend explicitly on time, \(\frac{dH}{dt} = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} = \frac{\partial H}{\partial q}\frac{\partial H}{\partial p} + \frac{\partial H}{\partial p}\left(-\frac{\partial H}{\partial q}\right) = 0\). This follows automatically from the structure of Hamilton's canonical equations.

B-6. Concrete Calculation of the Action

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Solution Strategy

Since the velocity of uniform linear motion \(q(t) = \frac{d}{T}t\) is the constant \(\dot{q} = d/T\), the integrand is also constant.

\[ \dot{q}(t) = \frac{d}{T} \quad (\text{constant}) \]
\[ S[q] = \int_0^T \frac{1}{2}m\dot{q}^2\, dt = \int_0^T \frac{1}{2}m\left(\frac{d}{T}\right)^2 dt = \frac{1}{2}m\frac{d^2}{T^2} \cdot T \]
\[ \boxed{S = \frac{md^2}{2T}} \]

Verification (dimensional analysis): \([m][d^2]/[T] = \text{kg}\cdot\text{m}^2/\text{s} = \text{J}\cdot\text{s}\). The dimensions of action are \([\text{energy}]\times[\text{time}]\), which is correct.

Verification (special cases): If \(d = 0\) (no motion), then \(S = 0\). If \(T \to \infty\) (very slow motion), then \(S \to 0\). If \(T \to 0\) (instantaneous motion), then \(S \to \infty\). All of these are physically reasonable.

B-7. Practice Calculating Variations

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Solution Strategy

Calculate the velocity of the displaced path and expand the action in powers of \(\epsilon\).

(a) Verification of endpoint conditions:

\[ \delta q(0) = \epsilon \sin\!\left(\frac{\pi \cdot 0}{T}\right) = \epsilon \sin 0 = 0 \quad \checkmark \]
\[ \delta q(T) = \epsilon \sin\!\left(\frac{\pi T}{T}\right) = \epsilon \sin \pi = 0 \quad \checkmark \]
\[ \boxed{\delta q(0) = \delta q(T) = 0 \text{ is satisfied}} \]

(b) Velocity of the displaced path:

\[ \dot{q}(t) = \dot{q}_0 + \epsilon\frac{\pi}{T}\cos\!\left(\frac{\pi t}{T}\right) = \frac{d}{T} + \epsilon\frac{\pi}{T}\cos\!\left(\frac{\pi t}{T}\right) \]

Computing the action:

\[ S[q_0 + \delta q] = \int_0^T \frac{1}{2}m\dot{q}^2\, dt = \frac{m}{2}\int_0^T \left[\frac{d}{T} + \epsilon\frac{\pi}{T}\cos\!\left(\frac{\pi t}{T}\right)\right]^2 dt \]

Expanding:

\[ \dot{q}^2 = \frac{d^2}{T^2} + 2\frac{d}{T}\cdot\epsilon\frac{\pi}{T}\cos\!\left(\frac{\pi t}{T}\right) + \epsilon^2\frac{\pi^2}{T^2}\cos^2\!\left(\frac{\pi t}{T}\right) \]

Integrating each term:

Zeroth-order term (\(\epsilon^0\)):

\[ \int_0^T \frac{d^2}{T^2}\, dt = \frac{d^2}{T} \]

First-order term (\(\epsilon^1\)):

\[ 2\frac{d\pi}{T^2}\epsilon \int_0^T \cos\!\left(\frac{\pi t}{T}\right) dt = 2\frac{d\pi}{T^2}\epsilon \left[\frac{T}{\pi}\sin\!\left(\frac{\pi t}{T}\right)\right]_0^T = 2\frac{d\pi}{T^2}\epsilon \cdot \frac{T}{\pi}(\sin\pi - \sin 0) = 0 \]

Second-order term (\(\epsilon^2\)):

\[ \frac{\pi^2}{T^2}\epsilon^2 \int_0^T \cos^2\!\left(\frac{\pi t}{T}\right) dt = \frac{\pi^2}{T^2}\epsilon^2 \cdot \frac{T}{2} = \frac{\pi^2\epsilon^2}{2T} \]

Therefore:

\[ S[q_0 + \delta q] = \frac{m}{2}\left(\frac{d^2}{T} + 0 + \frac{\pi^2\epsilon^2}{2T}\right) = \frac{md^2}{2T} + \frac{m\pi^2\epsilon^2}{4T} \]
\[ S[q_0 + \delta q] - S[q_0] = \frac{m\pi^2\epsilon^2}{4T} \]
\[ \boxed{S[q_0 + \delta q] - S[q_0] = \frac{m\pi^2}{4T}\epsilon^2} \]

The first-order term in \(\epsilon\) is zero, and the difference begins at order \(\epsilon^2\). This means that \(q_0(t)\) is a stationary point of the action.

Consistency check: Since \(\Delta S > 0\), the uniform rectilinear motion path gives a minimum of the action (a true minimum for the free particle case). This is physically reasonable.

B-8. Calculation of Poisson Brackets

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Solution Strategy

Definition of the Poisson bracket for 1 degree of freedom:

\[ \{A, B\}_{\mathrm{PB}} = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q} \]

(a) \(\{q, p\}_{\mathrm{PB}}\)

\[ \{q, p\}_{\mathrm{PB}} = \frac{\partial q}{\partial q}\frac{\partial p}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial p}{\partial q} = 1 \cdot 1 - 0 \cdot 0 = 1 \]
\[ \boxed{\{q, p\}_{\mathrm{PB}} = 1} \]

(b) \(\{q, q\}_{\mathrm{PB}}\) and \(\{p, p\}_{\mathrm{PB}}\)

\[ \{q, q\}_{\mathrm{PB}} = \frac{\partial q}{\partial q}\frac{\partial q}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial q}{\partial q} = 1 \cdot 0 - 0 \cdot 1 = 0 \]
\[ \{p, p\}_{\mathrm{PB}} = \frac{\partial p}{\partial q}\frac{\partial p}{\partial p} - \frac{\partial p}{\partial p}\frac{\partial p}{\partial q} = 0 \cdot 1 - 1 \cdot 0 = 0 \]
\[ \boxed{\{q, q\}_{\mathrm{PB}} = 0, \qquad \{p, p\}_{\mathrm{PB}} = 0} \]

Verification: For any \(A\), \(\{A, A\}_{\mathrm{PB}} = 0\) follows automatically from the antisymmetry of the Poisson bracket \(\{A, B\} = -\{B, A\}\).

(c) \(\{q^2, p\}_{\mathrm{PB}}\)

\[ \{q^2, p\}_{\mathrm{PB}} = \frac{\partial(q^2)}{\partial q}\frac{\partial p}{\partial p} - \frac{\partial(q^2)}{\partial p}\frac{\partial p}{\partial q} = 2q \cdot 1 - 0 \cdot 0 = 2q \]
\[ \boxed{\{q^2, p\}_{\mathrm{PB}} = 2q} \]

(d) \(\{q, p^2\}_{\mathrm{PB}}\)

\[ \{q, p^2\}_{\mathrm{PB}} = \frac{\partial q}{\partial q}\frac{\partial(p^2)}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial(p^2)}{\partial q} = 1 \cdot 2p - 0 \cdot 0 = 2p \]
\[ \boxed{\{q, p^2\}_{\mathrm{PB}} = 2p} \]

Verification: Using the general Poisson bracket identity \(\{A, BC\} = \{A, B\}C + B\{A, C\}\), we get \(\{q, p^2\} = \{q, p\}p + p\{q, p\} = p + p = 2p\), which agrees. Similarly, \(\{q^2, p\} = \{q, p\}q + q\{q, p\} = q + q = 2q\) also agrees.

Medium

M-1. Derivation of Euler-Lagrange Equations in 2D Polar Coordinates

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Solution Strategy

Apply the Euler-Lagrange equation for each of \(r\) and \(\theta\) to \(L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)\).

(a) Euler-Lagrange equation for \(r\)

Computing the necessary partial derivatives:

\[ \frac{\partial L}{\partial \dot{r}} = m\dot{r} \]
\[ \frac{\partial L}{\partial r} = mr\dot{\theta}^2 - V'(r) \]

Substituting into the Euler-Lagrange equation \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right) - \frac{\partial L}{\partial r} = 0\):

\[ m\ddot{r} - mr\dot{\theta}^2 + V'(r) = 0 \]
\[ \boxed{m\ddot{r} = mr\dot{\theta}^2 - V'(r)} \]

Physical interpretation: The left-hand side \(m\ddot{r}\) is the mass times the radial acceleration. The first term on the right-hand side \(mr\dot{\theta}^2\) is the centrifugal force (a fictitious force) that acts radially outward due to rotational motion. The second term \(-V'(r)\) is the force from the central potential. This equation represents the equation of motion in the radial direction.

(b) Euler-Lagrange equation for \(\theta\)

\[ \frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta} \]
\[ \frac{\partial L}{\partial \theta} = 0 \]

\(L\) does not depend explicitly on \(\theta\) (since \(V\) is a function of \(r\) only). Substituting into the Euler-Lagrange equation:

\[ \frac{d}{dt}(mr^2\dot{\theta}) - 0 = 0 \]
\[ \boxed{\frac{d}{dt}(mr^2\dot{\theta}) = 0} \]

This means that \(p_\theta = mr^2\dot{\theta}\) is a constant independent of time. Since \(mr^2\dot{\theta}\) is nothing other than the \(z\)-component of angular momentum \(L_z\), this equation expresses conservation of angular momentum.

In general, when the Lagrangian does not explicitly depend on a generalized coordinate \(q_j\) (i.e., \(\frac{\partial L}{\partial q_j} = 0\)), that coordinate is called a cyclic coordinate, and the corresponding canonical momentum \(p_j\) becomes a conserved quantity.

(c) Comparison with the Newtonian formulation

To write Newton's equations of motion in polar coordinates, one must separately derive the polar coordinate components of acceleration:

\[ a_r = \ddot{r} - r\dot{\theta}^2, \qquad a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} \]

Radial direction: \(m(\ddot{r} - r\dot{\theta}^2) = F_r = -V'(r)\)

Angular direction: \(m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) = F_\theta = 0\)

These are consistent with the results of (a) and (b), but in the Newtonian formulation: - The polar coordinate components of acceleration (the centrifugal term \(-r\dot{\theta}^2\) and the Coriolis term \(2\dot{r}\dot{\theta}\)) must be derived beforehand - The expression for acceleration must be rederived separately for each coordinate system

On the other hand, the advantages of the Lagrangian formulation are: 1. One only needs to write \(L = T - V\) in the chosen coordinates; the form of the Euler-Lagrange equations is the same regardless of the coordinate system 2. Centrifugal and Coriolis forces appear automatically without needing to be introduced separately 3. The relationship between cyclic coordinates and conserved quantities is immediately apparent

Verification: Expanding the result of (b) gives \(mr^2\ddot{\theta} + 2mr\dot{r}\dot{\theta} = 0\), i.e., \(m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) = 0\), which agrees with the angular equation of motion in the Newtonian formulation (for the case \(F_\theta = 0\)).

M-2. Inverse of the Legendre Transform

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Solution Strategy

Starting from the Hamiltonian, we recover the Lagrangian and demonstrate the involutive property of the Legendre transformation (applying it twice returns to the original function).

(a) Procedure for the inverse transformation

Given a Hamiltonian \(H(q, p)\):

  1. Compute Hamilton's first canonical equation \(\dot{q} = \frac{\partial H}{\partial p}\)
  2. Solve this equation for \(p\) to obtain \(p = p(q, \dot{q})\)
  3. Substitute \(p = p(q, \dot{q})\) into the inverse Legendre transformation \(L(q, \dot{q}) = p\dot{q} - H(q, p)\) to eliminate \(p\)

This procedure recovers \(L(q, \dot{q})\) from \(H(q, p)\). Note that step 2 requires that \(p\) can be uniquely solved for (i.e., \(\frac{\partial^2 H}{\partial p^2} \neq 0\), the convexity condition for the Legendre transformation).

(b) Explicit recovery for the harmonic oscillator

Starting from \(H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\).

Step 1:

\[ \dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m} \]

Step 2:

\[ p = m\dot{q} \]

Step 3:

\[ L = p\dot{q} - H = m\dot{q} \cdot \dot{q} - \left[\frac{(m\dot{q})^2}{2m} + \frac{1}{2}m\omega^2 q^2\right] \]
\[ = m\dot{q}^2 - \frac{m\dot{q}^2}{2} - \frac{1}{2}m\omega^2 q^2 \]
\[ \boxed{L = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}m\omega^2 q^2} \]

The original Lagrangian is correctly recovered.

(c) General proof of involutivity

Starting from \(L(q, \dot{q})\), we apply the Legendre transformation twice and show that it returns to the original function.

First Legendre transformation: \(L(q, \dot{q}) \to H(q, p)\)

\[ p = \frac{\partial L}{\partial \dot{q}}, \qquad H(q, p) = p\dot{q} - L(q, \dot{q}) \]

where \(\dot{q} = \dot{q}(q, p)\) is obtained by solving \(p = \frac{\partial L}{\partial \dot{q}}\) for \(\dot{q}\).

Second Legendre transformation: \(H(q, p) \to L'(q, \dot{q}')\)

Define the new variable as \(\dot{q}' = \frac{\partial H}{\partial p}\). Computing the partial derivative of \(H = p\dot{q} - L\) with respect to \(p\):

\[ \frac{\partial H}{\partial p} = \dot{q} + p\frac{\partial \dot{q}}{\partial p} - \frac{\partial L}{\partial \dot{q}}\frac{\partial \dot{q}}{\partial p} = \dot{q} + p\frac{\partial \dot{q}}{\partial p} - p\frac{\partial \dot{q}}{\partial p} = \dot{q} \]

(where we used \(\frac{\partial L}{\partial \dot{q}} = p\).)

Therefore \(\dot{q}' = \dot{q}\), and the new variable is identical to the original velocity.

The function obtained by the inverse Legendre transformation is:

\[ L'(q, \dot{q}') = p\dot{q}' - H(q, p) = p\dot{q} - (p\dot{q} - L) = L(q, \dot{q}) \]
\[ \boxed{L' = L} \]

Thus, applying the Legendre transformation twice returns to the original function (involutivity). This means that the Legendre transformation is a reversible transformation that preserves information.

Verification: Confirmed with the concrete example in (b). The round trip \(L \to H \to L\) correctly recovers \(L = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}m\omega^2 q^2\).

M-3. Trajectory of a Harmonic Oscillator in Phase Space

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Solution Strategy

Solve the canonical equations to obtain the general solution, then eliminate \(t\) using trigonometric identities to find the trajectory in phase space.

(a) General solution of the system of differential equations

Differentiating \(\dot{q} = p/m\) with respect to time and substituting \(\dot{p} = -m\omega^2 q\):

\[ \ddot{q} = \frac{\dot{p}}{m} = -\omega^2 q \]

The general solution is:

\[ q(t) = C_1 \cos\omega t + C_2 \sin\omega t \]

From the initial condition \(q(0) = q_0\), we get \(C_1 = q_0\).

\(p(t) = m\dot{q}(t) = m(-C_1\omega\sin\omega t + C_2\omega\cos\omega t)\)

From the initial condition \(p(0) = p_0\), we get \(m\omega C_2 = p_0\), i.e., \(C_2 = \frac{p_0}{m\omega}\).

\[ \boxed{q(t) = q_0\cos\omega t + \frac{p_0}{m\omega}\sin\omega t} \]
\[ \boxed{p(t) = -m\omega q_0\sin\omega t + p_0\cos\omega t} \]

Verification: At \(t = 0\), \(q(0) = q_0\), \(p(0) = p_0\) ✓. \(\dot{q}(t) = -q_0\omega\sin\omega t + \frac{p_0}{m}\cos\omega t = p(t)/m\) ✓.

(b) Elliptical trajectory in phase space

We eliminate \(t\) from \(q(t)\) and \(p(t)\). Rewriting:

\[ q = q_0\cos\omega t + \frac{p_0}{m\omega}\sin\omega t \]
\[ \frac{p}{m\omega} = -q_0\sin\omega t + \frac{p_0}{m\omega}\cos\omega t \]

We treat these as simultaneous equations in \(\cos\omega t\) and \(\sin\omega t\). Multiplying the first equation by \(\frac{p_0}{m\omega}\) and the second by \(q_0\), then adding:

\[ \frac{p_0}{m\omega}q + q_0\frac{p}{m\omega} = \left(q_0\frac{p_0}{m\omega} + \frac{p_0}{m\omega}q_0\right)\cos\omega t \cdot (\text{...}) \]

More directly, we use \(\cos^2\omega t + \sin^2\omega t = 1\). Writing the system in matrix form:

\[ \begin{pmatrix} q \\ p/(m\omega) \end{pmatrix} = \begin{pmatrix} \cos\omega t & \sin\omega t \\ -\sin\omega t & \cos\omega t \end{pmatrix} \begin{pmatrix} q_0 \\ p_0/(m\omega) \end{pmatrix} \]

Since the rotation matrix is orthogonal, the squared magnitude of the vector is conserved:

\[ q^2 + \frac{p^2}{m^2\omega^2} = q_0^2 + \frac{p_0^2}{m^2\omega^2} \]

The right-hand side can be written as \(\frac{2E}{m\omega^2}\) (confirmed in part (c) below). This is the equation of an ellipse in phase space.

(c) Expression in terms of energy

The initial energy is:

\[ E = H(q_0, p_0) = \frac{p_0^2}{2m} + \frac{1}{2}m\omega^2 q_0^2 \]

Therefore:

\[ q_0^2 + \frac{p_0^2}{m^2\omega^2} = \frac{2E}{m\omega^2} \]

Substituting into the result from (b):

\[ q^2 + \frac{p^2}{m^2\omega^2} = \frac{2E}{m\omega^2} \]

Dividing both sides by \(\frac{2E}{m\omega^2}\):

\[ \boxed{\frac{m\omega^2 q^2}{2E} + \frac{p^2}{2mE} = 1} \]

This is an ellipse in standard form \(\frac{q^2}{a^2} + \frac{p^2}{b^2} = 1\), where:

  • Semi-axis along the \(q\)-direction (semi-major axis): \(a = \sqrt{\frac{2E}{m\omega^2}}\)
  • Semi-axis along the \(p\)-direction (semi-minor axis): \(b = \sqrt{2mE}\)
\[ \boxed{a = \sqrt{\frac{2E}{m\omega^2}}, \qquad b = \sqrt{2mE}} \]

Verification: The area of the ellipse is \(\pi ab = \pi\sqrt{\frac{2E}{m\omega^2}}\cdot\sqrt{2mE} = \pi\cdot\frac{2E}{\omega} = \frac{2\pi E}{\omega}\). Since the period is \(T = 2\pi/\omega\), the area \(= ET\). This agrees with the well-known result for the adiabatic invariant \(J = \oint p\,dq = ET\).

Dimensional check: \(a = \sqrt{2E/(m\omega^2)}\): \([E/(m\omega^2)] = \text{J}/(\text{kg}\cdot\text{s}^{-2}) = \text{m}^2\), so \([a] = \text{m}\) ✓. \(b = \sqrt{2mE}\): \([mE] = \text{kg}\cdot\text{J} = \text{kg}^2\cdot\text{m}^2\cdot\text{s}^{-2}\), so \([b] = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}\) (dimensions of momentum) ✓.

M-4. Poisson Brackets and Hamilton's Equations of Motion

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Solution Strategy

Expand the total time derivative of the dynamical variable \(A(q, p, t)\) using the chain rule, then substitute Hamilton's canonical equations to express the result in terms of the Poisson bracket.

Writing the total time derivative of \(A(q_j, p_j, t)\) using the chain rule (employing Einstein's summation convention):

\[ \frac{dA}{dt} = \frac{\partial A}{\partial q_j}\dot{q}_j + \frac{\partial A}{\partial p_j}\dot{p}_j + \frac{\partial A}{\partial t} \]

Substituting Hamilton's canonical equations \(\dot{q}_j = \frac{\partial H}{\partial p_j}\), \(\dot{p}_j = -\frac{\partial H}{\partial q_j}\):

\[ \frac{dA}{dt} = \frac{\partial A}{\partial q_j}\frac{\partial H}{\partial p_j} + \frac{\partial A}{\partial p_j}\left(-\frac{\partial H}{\partial q_j}\right) + \frac{\partial A}{\partial t} \]
\[ = \sum_j\left(\frac{\partial A}{\partial q_j}\frac{\partial H}{\partial p_j} - \frac{\partial A}{\partial p_j}\frac{\partial H}{\partial q_j}\right) + \frac{\partial A}{\partial t} \]

Comparing with the definition of the Poisson bracket:

\[ \boxed{\frac{dA}{dt} = \{A, H\}_{\mathrm{PB}} + \frac{\partial A}{\partial t}} \]

Case \(A = q_j\): Since \(q_j\) does not depend explicitly on time, \(\frac{\partial q_j}{\partial t} = 0\).

\[ \frac{dq_j}{dt} = \{q_j, H\}_{\mathrm{PB}} = \frac{\partial q_j}{\partial q_k}\frac{\partial H}{\partial p_k} - \frac{\partial q_j}{\partial p_k}\frac{\partial H}{\partial q_k} \]

Since \(\frac{\partial q_j}{\partial q_k} = \delta_{jk}\) and \(\frac{\partial q_j}{\partial p_k} = 0\):

\[ \dot{q}_j = \frac{\partial H}{\partial p_j} \quad \checkmark \]

Case \(A = p_j\): Similarly, \(\frac{\partial p_j}{\partial t} = 0\).

\[ \frac{dp_j}{dt} = \{p_j, H\}_{\mathrm{PB}} = \frac{\partial p_j}{\partial q_k}\frac{\partial H}{\partial p_k} - \frac{\partial p_j}{\partial p_k}\frac{\partial H}{\partial q_k} \]

Since \(\frac{\partial p_j}{\partial q_k} = 0\) and \(\frac{\partial p_j}{\partial p_k} = \delta_{jk}\):

\[ \dot{p}_j = -\frac{\partial H}{\partial q_j} \quad \checkmark \]
\[ \boxed{\dot{q}_j = \frac{\partial H}{\partial p_j}, \qquad \dot{p}_j = -\frac{\partial H}{\partial q_j}} \]

Hamilton's canonical equations are reproduced as special cases of the Poisson bracket.

Verification: Substituting \(A = H\) (when \(\frac{\partial H}{\partial t} = 0\)) gives \(\frac{dH}{dt} = \{H, H\}_{\mathrm{PB}} = 0\) (by antisymmetry). This is the law of energy conservation, consistent with the result of D5.

M-5. Canonical Quantization: Verification of Commutation Relations

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Solution Strategy

Apply the replacement rule from Poisson brackets to commutators, and derive the operator version of the classical canonical equations from Heisenberg's equations of motion.

(a) Derivation of canonical commutation relations

Applying the canonical quantization recipe to the result \(\{q, p\}_{\mathrm{PB}} = 1\) from D8(a):

\[ \{q, p\}_{\mathrm{PB}} = 1 \quad \longrightarrow \quad \frac{1}{i\hbar}[\hat{q}, \hat{p}] = 1 \]
\[ \boxed{[\hat{q}, \hat{p}] = i\hbar} \]

Similarly, from the results \(\{q, q\}_{\mathrm{PB}} = 0\), \(\{p, p\}_{\mathrm{PB}} = 0\) of D8(b):

\[ [\hat{q}, \hat{q}] = 0, \qquad [\hat{p}, \hat{p}] = 0 \]

These are consistent with the canonical commutation relations introduced in Ch. 10.

(b) Application of Heisenberg's equations of motion

We apply Heisenberg's equations of motion to \(\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{q}^2\).

Case \(\hat{A} = \hat{q}\):

\[ \frac{d\hat{q}}{dt} = \frac{1}{i\hbar}[\hat{q}, \hat{H}] = \frac{1}{i\hbar}\left[\hat{q},\, \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{q}^2\right] \]

Since \([\hat{q}, \hat{q}^2] = 0\) (\(\hat{q}\) commutes with itself):

\[ \frac{d\hat{q}}{dt} = \frac{1}{i\hbar}\cdot\frac{1}{2m}[\hat{q}, \hat{p}^2] \]

Using the commutator identity \([\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}]\):

\[ [\hat{q}, \hat{p}^2] = [\hat{q}, \hat{p}]\hat{p} + \hat{p}[\hat{q}, \hat{p}] = i\hbar\hat{p} + \hat{p}\cdot i\hbar = 2i\hbar\hat{p} \]

Substituting:

\[ \frac{d\hat{q}}{dt} = \frac{1}{i\hbar}\cdot\frac{1}{2m}\cdot 2i\hbar\hat{p} = \frac{\hat{p}}{m} \]
\[ \boxed{\frac{d\hat{q}}{dt} = \frac{\hat{p}}{m}} \]

Case \(\hat{A} = \hat{p}\):

\[ \frac{d\hat{p}}{dt} = \frac{1}{i\hbar}[\hat{p}, \hat{H}] = \frac{1}{i\hbar}\left[\hat{p},\, \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{q}^2\right] \]

Since \([\hat{p}, \hat{p}^2] = 0\):

\[ \frac{d\hat{p}}{dt} = \frac{1}{i\hbar}\cdot\frac{1}{2}m\omega^2[\hat{p}, \hat{q}^2] \]

Using the same identity:

\[ [\hat{p}, \hat{q}^2] = [\hat{p}, \hat{q}]\hat{q} + \hat{q}[\hat{p}, \hat{q}] = (-i\hbar)\hat{q} + \hat{q}(-i\hbar) = -2i\hbar\hat{q} \]

Substituting:

\[ \frac{d\hat{p}}{dt} = \frac{1}{i\hbar}\cdot\frac{1}{2}m\omega^2\cdot(-2i\hbar\hat{q}) = -m\omega^2\hat{q} \]
\[ \boxed{\frac{d\hat{p}}{dt} = -m\omega^2\hat{q}} \]

From the above, we have obtained the operator version of Hamilton's canonical equations:

\[ \frac{d\hat{q}}{dt} = \frac{\hat{p}}{m}, \qquad \frac{d\hat{p}}{dt} = -m\omega^2\hat{q} \]

These have exactly the same form as the classical canonical equations (results of D3).

Verification: By Ehrenfest's theorem, quantum mechanical expectation values obey the classical equations of motion. Taking expectation values of the operator equations above gives \(\frac{d\langle\hat{q}\rangle}{dt} = \frac{\langle\hat{p}\rangle}{m}\), \(\frac{d\langle\hat{p}\rangle}{dt} = -m\omega^2\langle\hat{q}\rangle\), which is consistent with the classical results. For the harmonic oscillator, since the potential is quadratic, Ehrenfest's theorem yields equations that are exactly the same form as the classical equations.

Advanced

A-1. Canonical Quantization of a Charged Particle in an Electromagnetic Field

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Solution Strategy

Derive the canonical momentum and Hamiltonian from the 3-dimensional Lagrangian, obtain the Schrödinger equation through canonical quantization, and then verify covariance under gauge transformations.

(a) Canonical momentum

\[ L = \frac{1}{2}m\dot{\mathbf{r}}^2 + e\dot{\mathbf{r}}\cdot\mathbf{A}(\mathbf{r}, t) - e\phi(\mathbf{r}, t) \]

The canonical momentum for the \(i\)-th component is:

\[ p_i = \frac{\partial L}{\partial \dot{r}_i} = m\dot{r}_i + eA_i(\mathbf{r}, t) \]

In vector notation:

\[ \boxed{\mathbf{p} = m\dot{\mathbf{r}} + e\mathbf{A}(\mathbf{r}, t)} \]

The difference from the kinetic momentum \(m\dot{\mathbf{r}}\):

\[ \mathbf{p} - m\dot{\mathbf{r}} = e\mathbf{A} \neq \mathbf{0} \quad (\text{when } \mathbf{A} \neq \mathbf{0}) \]

The canonical momentum includes the contribution from the vector potential and generally differs from the kinetic momentum \(m\dot{\mathbf{r}}\).

(b) Derivation of the Hamiltonian

From \(\mathbf{p} = m\dot{\mathbf{r}} + e\mathbf{A}\):

\[ \dot{\mathbf{r}} = \frac{\mathbf{p} - e\mathbf{A}}{m} \]

The Hamiltonian is:

\[ H = \mathbf{p}\cdot\dot{\mathbf{r}} - L \]

Computing each term:

\[ \mathbf{p}\cdot\dot{\mathbf{r}} = \mathbf{p}\cdot\frac{\mathbf{p} - e\mathbf{A}}{m} \]
\[ \frac{1}{2}m\dot{\mathbf{r}}^2 = \frac{(\mathbf{p} - e\mathbf{A})^2}{2m} \]
\[ e\dot{\mathbf{r}}\cdot\mathbf{A} = \frac{e\mathbf{A}\cdot(\mathbf{p} - e\mathbf{A})}{m} \]

Substituting:

\[ H = \frac{\mathbf{p}\cdot(\mathbf{p} - e\mathbf{A})}{m} - \frac{(\mathbf{p} - e\mathbf{A})^2}{2m} - \frac{e\mathbf{A}\cdot(\mathbf{p} - e\mathbf{A})}{m} + e\phi \]
\[ = \frac{(\mathbf{p} - e\mathbf{A})\cdot(\mathbf{p} - e\mathbf{A} + e\mathbf{A})}{m} - \frac{(\mathbf{p} - e\mathbf{A})^2}{2m} - \frac{e\mathbf{A}\cdot(\mathbf{p} - e\mathbf{A})}{m} + e\phi \]

More concisely, defining \(\boldsymbol{\pi} \equiv \mathbf{p} - e\mathbf{A}\) (kinetic momentum) so that \(\dot{\mathbf{r}} = \boldsymbol{\pi}/m\):

\[ H = (\boldsymbol{\pi} + e\mathbf{A})\cdot\frac{\boldsymbol{\pi}}{m} - \frac{\boldsymbol{\pi}^2}{2m} - \frac{e\mathbf{A}\cdot\boldsymbol{\pi}}{m} + e\phi \]
\[ = \frac{\boldsymbol{\pi}^2}{m} + \frac{e\mathbf{A}\cdot\boldsymbol{\pi}}{m} - \frac{\boldsymbol{\pi}^2}{2m} - \frac{e\mathbf{A}\cdot\boldsymbol{\pi}}{m} + e\phi = \frac{\boldsymbol{\pi}^2}{2m} + e\phi \]
\[ \boxed{H = \frac{(\mathbf{p} - e\mathbf{A})^2}{2m} + e\phi} \]

Verification: When \(\mathbf{A} = \mathbf{0}\) and \(\phi = 0\), we get \(H = p^2/(2m)\), which reduces to the free-particle Hamiltonian ✓.

(c) Schrödinger equation

Applying the canonical quantization prescription \(\mathbf{r} \to \hat{\mathbf{r}}\), \(\mathbf{p} \to \hat{\mathbf{p}} = -i\hbar\nabla\) to the Hamiltonian:

\[ \hat{H} = \frac{(\hat{\mathbf{p}} - e\mathbf{A})^2}{2m} + e\phi = \frac{(-i\hbar\nabla - e\mathbf{A})^2}{2m} + e\phi \]

Substituting into the Schrödinger equation \(i\hbar\frac{\partial}{\partial t}\Psi = \hat{H}\Psi\):

\[ \boxed{i\hbar\frac{\partial}{\partial t}\Psi = \frac{1}{2m}(-i\hbar\nabla - e\mathbf{A})^2\Psi + e\phi\,\Psi} \]

(d) Gauge invariance

Under the gauge transformation:

\[ \mathbf{A} \to \mathbf{A}' = \mathbf{A} + \nabla\chi, \qquad \phi \to \phi' = \phi - \frac{\partial\chi}{\partial t} \]

we set \(\Psi' = e^{ie\chi/\hbar}\Psi\). We show that \(\Psi'\) satisfies the Schrödinger equation after the gauge transformation.

Step 1: Compute \((-i\hbar\nabla - e\mathbf{A}')\Psi'\).

\[ \nabla\Psi' = \nabla(e^{ie\chi/\hbar}\Psi) = \frac{ie}{\hbar}(\nabla\chi)e^{ie\chi/\hbar}\Psi + e^{ie\chi/\hbar}\nabla\Psi \]
\[ = e^{ie\chi/\hbar}\left(\frac{ie}{\hbar}\nabla\chi + \nabla\right)\Psi \]

Therefore:

\[ (-i\hbar\nabla - e\mathbf{A}')\Psi' = e^{ie\chi/\hbar}\left(-i\hbar\nabla + e\nabla\chi - e\mathbf{A} - e\nabla\chi\right)\Psi \]
\[ = e^{ie\chi/\hbar}(-i\hbar\nabla - e\mathbf{A})\Psi \]

Step 2: Apply the same operator a second time.

\[ (-i\hbar\nabla - e\mathbf{A}')^2\Psi' = (-i\hbar\nabla - e\mathbf{A}')\left[e^{ie\chi/\hbar}(-i\hbar\nabla - e\mathbf{A})\Psi\right] \]

This has exactly the same structure as Step 1 (with \(\Psi\) replaced by \((-i\hbar\nabla - e\mathbf{A})\Psi\)), so:

\[ = e^{ie\chi/\hbar}(-i\hbar\nabla - e\mathbf{A})^2\Psi \]

Step 3: Compute the time derivative.

\[ i\hbar\frac{\partial\Psi'}{\partial t} = i\hbar\left(\frac{ie}{\hbar}\frac{\partial\chi}{\partial t}e^{ie\chi/\hbar}\Psi + e^{ie\chi/\hbar}\frac{\partial\Psi}{\partial t}\right) \]
\[ = e^{ie\chi/\hbar}\left(-e\frac{\partial\chi}{\partial t}\Psi + i\hbar\frac{\partial\Psi}{\partial t}\right) \]

Step 4: The right-hand side of the Schrödinger equation after the gauge transformation is:

\[ \frac{1}{2m}(-i\hbar\nabla - e\mathbf{A}')^2\Psi' + e\phi'\Psi' \]
\[ = e^{ie\chi/\hbar}\left[\frac{1}{2m}(-i\hbar\nabla - e\mathbf{A})^2\Psi + e\left(\phi - \frac{\partial\chi}{\partial t}\right)\Psi\right] \]

Since the original \(\Psi\) satisfies the original Schrödinger equation, \(i\hbar\frac{\partial\Psi}{\partial t} = \frac{1}{2m}(-i\hbar\nabla - e\mathbf{A})^2\Psi + e\phi\Psi\). Substituting this, the right-hand side becomes:

\[ = e^{ie\chi/\hbar}\left[i\hbar\frac{\partial\Psi}{\partial t} - e\phi\Psi + e\phi\Psi - e\frac{\partial\chi}{\partial t}\Psi\right] \]
\[ = e^{ie\chi/\hbar}\left[i\hbar\frac{\partial\Psi}{\partial t} - e\frac{\partial\chi}{\partial t}\Psi\right] \]

This agrees with the left-hand side \(i\hbar\frac{\partial\Psi'}{\partial t}\) computed in Step 3.

\[ \boxed{\Psi' = e^{ie\chi/\hbar}\Psi \text{ satisfies the Schrödinger equation after the gauge transformation}} \]

Gauge invariance of physical observables:

Probability density:

\[ |\Psi'|^2 = |e^{ie\chi/\hbar}|^2|\Psi|^2 = |\Psi|^2 \]

Since \(e^{ie\chi/\hbar}\) is a phase factor with unit modulus, the probability density is invariant under gauge transformations.

For the probability current density as well, since \((-i\hbar\nabla - e\mathbf{A}')\Psi' = e^{ie\chi/\hbar}(-i\hbar\nabla - e\mathbf{A})\Psi\):

\[ \mathbf{j}' = \frac{1}{m}\mathrm{Re}\left[\Psi'^*(-i\hbar\nabla - e\mathbf{A}')\Psi'\right] = \frac{1}{m}\mathrm{Re}\left[\Psi^*(-i\hbar\nabla - e\mathbf{A})\Psi\right] = \mathbf{j} \]

Therefore, the probability current density is also gauge invariant.

Verification: A gauge transformation does not change the physical electric field \(\mathbf{E} = -\nabla\phi - \frac{\partial\mathbf{A}}{\partial t}\) or the magnetic field \(\mathbf{B} = \nabla\times\mathbf{A}\). The gauge invariance of physical observables is consistent with this fact.

A-2. Noether's Theorem: From Symmetry to Conservation Laws

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Solution Strategy

Derive conserved quantities by combining the invariance of the Lagrangian with the Euler-Lagrange equations.

(a) Derivation of Noether charge conservation

Under the infinitesimal transformation \(q_j \to q_j + \epsilon\,\eta_j\), the variation of the Lagrangian is:

\[ \delta L = \frac{\partial L}{\partial q_j}\epsilon\eta_j + \frac{\partial L}{\partial \dot{q}_j}\epsilon\dot{\eta}_j \]

(using the fact that \(\dot{q}_j \to \dot{q}_j + \epsilon\dot{\eta}_j\).)

Under the condition \(\delta L = 0\) (invariance of the Lagrangian):

\[ \frac{\partial L}{\partial q_j}\eta_j + \frac{\partial L}{\partial \dot{q}_j}\dot{\eta}_j = 0 \tag{*} \]

Substituting the Euler-Lagrange equation \(\frac{\partial L}{\partial q_j} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\) into the first term of \((*)\):

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\eta_j + \frac{\partial L}{\partial \dot{q}_j}\dot{\eta}_j = 0 \]

The left-hand side, by the product rule of differentiation, gives:

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\eta_j\right) = 0 \]

Defining the Noether charge \(Q = \sum_j \frac{\partial L}{\partial \dot{q}_j}\eta_j = \sum_j p_j\eta_j\):

\[ \boxed{\frac{dQ}{dt} = 0} \]

(b) Conservation of angular momentum

\(L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - V(r)\), \(r = \sqrt{x^2 + y^2 + z^2}\)

Infinitesimal rotation about the \(z\)-axis: \(\eta_x = -y\), \(\eta_y = x\), \(\eta_z = 0\)

Verification of Lagrangian invariance:

Variation of kinetic energy: \(\delta T = m(\dot{x}\delta\dot{x} + \dot{y}\delta\dot{y} + \dot{z}\delta\dot{z})\)

Since \(\delta\dot{x} = -\epsilon\dot{y}\), \(\delta\dot{y} = \epsilon\dot{x}\), \(\delta\dot{z} = 0\):

\[ \delta T = m\epsilon(-\dot{x}\dot{y} + \dot{y}\dot{x}) = 0 \]

Variation of potential energy: \(\delta V = V'(r)\delta r\)

\[ \delta r = \frac{\partial r}{\partial x}\delta x + \frac{\partial r}{\partial y}\delta y = \frac{x}{r}(-\epsilon y) + \frac{y}{r}(\epsilon x) = \frac{\epsilon}{r}(-xy + yx) = 0 \]

Therefore \(\delta L = \delta T - \delta V = 0\) ✓

Applying the result of (a), the Noether charge is:

\[ Q = p_x\eta_x + p_y\eta_y + p_z\eta_z = m\dot{x}(-y) + m\dot{y}(x) + m\dot{z}(0) \]
\[ \boxed{Q = m(x\dot{y} - y\dot{x}) = L_z} \]

From \(\frac{dQ}{dt} = 0\), the \(z\)-component of angular momentum \(L_z\) is conserved.

Verification: Since the central force potential \(V(r)\) is rotationally symmetric, conservation of angular momentum is physically obvious. Noether's theorem rigorously proves this intuition.

(c) Conservation of energy

Consider the case \(\frac{\partial L}{\partial t} = 0\) (the Lagrangian has no explicit time dependence).

The coordinate change corresponding to time translation \(t \to t + \epsilon\) is \(\delta q_j = \dot{q}_j\epsilon\) (when time advances by \(\epsilon\), the coordinates change by \(\dot{q}_j\epsilon\)).

Computing the total time derivative of \(L\):

\[ \frac{dL}{dt} = \frac{\partial L}{\partial q_j}\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j + \frac{\partial L}{\partial t} \]

Under the assumption \(\frac{\partial L}{\partial t} = 0\), substituting the Euler-Lagrange equation \(\frac{\partial L}{\partial q_j} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\):

\[ \frac{dL}{dt} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\dot{q}_j\right) \]

(using the product rule in reverse.)

Rearranging:

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\dot{q}_j - L\right) = 0 \]
\[ \frac{d}{dt}\left(\sum_j p_j\dot{q}_j - L\right) = 0 \]
\[ \boxed{\frac{dH}{dt} = 0, \qquad H = \sum_j p_j\dot{q}_j - L} \]

The conserved quantity corresponding to time translation symmetry is the Hamiltonian (energy).

Verification: This result is consistent with \(\frac{dH}{dt} = 0\) computed directly in D5. In D5, it was shown using Hamilton's canonical equations, whereas here it is derived from the general framework of Noether's theorem.

Remark: In the formulation of (a), we assumed \(\delta L = 0\), but for time translation the situation is slightly different. The time translation \(t \to t + \epsilon\) induces the coordinate transformation \(\delta q_j = \dot{q}_j\epsilon\), but \(L\) itself undergoes a change \(\delta L = \frac{dL}{dt}\epsilon\) (variation as a total derivative). When \(\frac{\partial L}{\partial t} = 0\), \(\delta L = \frac{dL}{dt}\epsilon\) takes the form of a total derivative, which does not contribute to the variation of the action (it becomes a boundary term). For this reason, the derivation above employs the method of directly computing \(\frac{dL}{dt}\).