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Appendix A Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Evaluating the Value of a Functional

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Solution strategy: Substitute \(f(x) = 2x\) into the functional \(H[f] = \int_0^3 [f(x)]^2\,dx\) and evaluate the definite integral.

Calculation:

\[ H[f] = \int_0^3 (2x)^2\,dx = \int_0^3 4x^2\,dx = 4\left[\frac{x^3}{3}\right]_0^3 = 4 \cdot \frac{27}{3} = 4 \cdot 9 = 36 \]

Final answer:

\[ \boxed{H[f] = 36} \]

Verification: Dimensional check: the integrand \(4x^2\) equals \(4 \times 9 = 36\) at \(x=3\) and \(0\) at \(x=0\). The average value is approximately \(4 \times (9/3) = 12\), and multiplying by the interval width of 3 gives \(36\). This is consistent.

B-2. Basic Calculation of Functional Derivatives

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Solution Strategy: Apply the formula from Calculation Example 2 in the main text, \(\frac{\delta}{\delta f(x)}\int [f(y)]^p\,\varphi(y)\,dy = p[f(x)]^{p-1}\,\varphi(x)\), with \(p=4\) and \(\varphi(y)=1\).

Calculation:

For \(F[f] = \int_0^1 [f(x)]^4\,dx\), we make the replacement \(f(x) \to f(x) + \epsilon\,\delta(x - x_0)\):

\[ F[f + \epsilon\delta] = \int_0^1 [f(x) + \epsilon\,\delta(x-x_0)]^4\,dx \]

Expanding to first order in \(\epsilon\):

\[ [f(x) + \epsilon\,\delta(x-x_0)]^4 \approx [f(x)]^4 + 4[f(x)]^3 \cdot \epsilon\,\delta(x-x_0) \]

Therefore:

\[ \frac{\delta F}{\delta f(x_0)} = \lim_{\epsilon \to 0}\frac{1}{\epsilon}\int_0^1 4[f(x)]^3\,\epsilon\,\delta(x-x_0)\,dx = 4\int_0^1 [f(x)]^3\,\delta(x-x_0)\,dx \]

By the sifting property of the delta function (when \(0 \leq x_0 \leq 1\)):

\[ \frac{\delta F}{\delta f(x_0)} = 4[f(x_0)]^3 \]

Final Answer:

\[ \boxed{\frac{\delta F}{\delta f(x_0)} = 4[f(x_0)]^3} \]

Verification: This follows the same pattern as the ordinary derivative \(\frac{d}{dx}x^4 = 4x^3\). The power is reduced by one and the coefficient 4 is brought out front. This is consistent.

B-3. Weighted Functional Derivative

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Solution strategy: Use the formula from Worked Example 2 with \(p = 2\), \(\varphi(y) = e^{-y^2}\).

Calculation:

For \(G[f] = \int_{-\infty}^{\infty} [f(y)]^2\,e^{-y^2}\,dy\), substitute \(f(y) \to f(y) + \epsilon\,\delta(y - x)\):

\[ [f(y) + \epsilon\,\delta(y-x)]^2 \approx [f(y)]^2 + 2f(y)\cdot\epsilon\,\delta(y-x) \]

Extract the terms first-order in \(\epsilon\):

\[ \frac{\delta G}{\delta f(x)} = \int_{-\infty}^{\infty} 2f(y)\,\delta(y-x)\,e^{-y^2}\,dy \]

Apply the sifting property of the delta function:

\[ \frac{\delta G}{\delta f(x)} = 2f(x)\,e^{-x^2} \]

Final answer:

\[ \boxed{\frac{\delta G}{\delta f(x)} = 2f(x)\,e^{-x^2}} \]

Verification: The weight function \(e^{-y^2}\) remains in the result as \(e^{-x^2}\). Setting \(\varphi(y) = 1\) gives \(2f(x)\), which agrees with D2 when \(p=2\). Consistent.

B-4. Functional Derivative Using the Delta Function

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Solution strategy: Express \(F[f] = f(a) = \int f(y)\,\delta(y-a)\,dy\) in integral form, and compute the functional derivative according to the definition.

Calculation:

Substitute \(f(y) \to f(y) + \epsilon\,\delta(y - x)\):

\[ F[f + \epsilon\delta] = \int [f(y) + \epsilon\,\delta(y-x)]\,\delta(y-a)\,dy \]
\[ = \int f(y)\,\delta(y-a)\,dy + \epsilon\int \delta(y-x)\,\delta(y-a)\,dy \]
\[ = f(a) + \epsilon\,\delta(x - a) \]

Here we used \(\int \delta(y-x)\,\delta(y-a)\,dy = \delta(x-a)\) (the composition property of the delta function).

Therefore:

\[ \frac{\delta F}{\delta f(x)} = \lim_{\epsilon \to 0}\frac{F[f+\epsilon\delta] - F[f]}{\epsilon} = \lim_{\epsilon \to 0}\frac{\epsilon\,\delta(x-a)}{\epsilon} = \delta(x-a) \]

Final answer:

\[ \boxed{\frac{\delta F}{\delta f(x)} = \delta(x - a)} \]

Verification: This reflects the fact that "\(f(a)\) depends only on the value of \(f\) at \(y=a\)." There is sensitivity only at \(x = a\), and zero sensitivity everywhere else. It is natural that this is expressed by a delta function. Furthermore, \(\frac{\delta f(a)}{\delta f(x)} = \delta(x-a)\) is known as a fundamental formula of functional differentiation.

B-5. Application of the Euler-Lagrange Equation (1D Harmonic Oscillator)

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Solution strategy: For \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\), compute each partial derivative in sequence.

Calculation:

1.

\[ \frac{\partial L}{\partial \dot{x}} = \frac{\partial}{\partial \dot{x}}\left(\frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\right) = m\dot{x} \]

2.

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = \frac{d}{dt}(m\dot{x}) = m\ddot{x} \]

3.

\[ \frac{\partial L}{\partial x} = \frac{\partial}{\partial x}\left(\frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\right) = -kx \]

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

\[ m\ddot{x} - (-kx) = 0 \quad \Longrightarrow \quad m\ddot{x} + kx = 0 \]

That is:

\[ \boxed{m\ddot{x} = -kx} \]

Verification: This is the equation of motion for a harmonic oscillator, describing simple harmonic motion with angular frequency \(\omega = \sqrt{k/m}\). It is consistent with Newton's second law \(F = -kx = ma\).

B-6. Calculation of Canonical Momenta

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(a) Free Fall in a Uniform Gravitational Field

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

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

(b) 2-Dimensional Polar Coordinates

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

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

Verification: \(p_r = m\dot{r}\) is the radial linear momentum. \(p_\theta = mr^2\dot{\theta}\) corresponds to the angular momentum \(L_z\). Since \(V(r)\) does not depend on \(\theta\), \(\theta\) is a cyclic coordinate, and \(p_\theta\) is a conserved quantity (conservation of angular momentum). This is physically correct.

B-7. Construction of the Hamiltonian

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

1. Canonical momentum:

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

2. Express \(\dot{q}\) in terms of \(p\):

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

3. Construct the Hamiltonian \(H = p\dot{q} - L\):

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

Final answer:

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

Verification: This has the form \(H = T + V\) (total energy). That the Legendre transform of \(L = T - V\) yields \(H = T + V\) is a general result. Furthermore, this agrees with the classical version of the quantum mechanical harmonic oscillator Hamiltonian \(\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{q}^2\).

B-8. Application of the Field Euler-Lagrange Equation

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Solution strategy: Apply the field Euler-Lagrange equation \(\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0\) to \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\).

Calculation:

Since \(\mathcal{L}\) does not contain \(\phi\) itself:

\[ \frac{\partial\mathcal{L}}{\partial\phi} = 0 \]

Partial derivative with respect to \(\partial_\mu\phi\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \frac{\partial}{\partial(\partial_\mu\phi)}\left(\frac{1}{2}\partial_\nu\phi\,\partial^\nu\phi\right) = \partial^\mu\phi \]

(Here, differentiating \(\frac{1}{2}\partial_\nu\phi\,g^{\nu\rho}\partial_\rho\phi\) with respect to \(\partial_\mu\phi\) gives \(g^{\mu\rho}\partial_\rho\phi = \partial^\mu\phi\).)

Substituting into the field Euler-Lagrange equation:

\[ \partial_\mu(\partial^\mu\phi) - 0 = 0 \]
\[ \boxed{\partial_\mu\partial^\mu\phi = \Box\phi = 0} \]

Final answer: This is the wave equation (massless Klein-Gordon equation). Written in components:

\[ \frac{\partial^2\phi}{\partial t^2} - \nabla^2\phi = 0 \]

Verification: Since there is no mass term \(\frac{m^2}{2}\phi^2\), we should obtain the equation for a massless free field. \(\Box\phi = 0\) is indeed the massless Klein-Gordon equation (= wave equation). This is consistent.

B-9. Equation of Motion for \(\phi^3\) Theory

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Solution strategy: Apply the field Euler-Lagrange equation to \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{m^2}{2}\phi^2 - \frac{g}{3!}\phi^3\).

Calculation:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi \]
\[ \frac{\partial\mathcal{L}}{\partial\phi} = -m^2\phi - \frac{g}{3!}\cdot 3\phi^2 = -m^2\phi - \frac{g}{2}\phi^2 \]

Field Euler-Lagrange equation \(\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0\):

\[ \partial_\mu\partial^\mu\phi - \left(-m^2\phi - \frac{g}{2}\phi^2\right) = 0 \]
\[ \boxed{\left(\Box + m^2\right)\phi + \frac{g}{2}\phi^2 = 0} \]

Or in an equivalent form:

\[ (\partial_\mu\partial^\mu + m^2)\phi = -\frac{g}{2}\phi^2 \]

Final answer: The equation of motion for \(\phi^3\) theory is as given above. The left-hand side is the free Klein-Gordon equation operator, and the right-hand side is the nonlinear interaction term.

Verification: Setting \(g = 0\) reduces to the free Klein-Gordon equation \((\Box + m^2)\phi = 0\). Also, comparing with the \(\phi^4\) theory example in the text, where \(\frac{\lambda}{4!}\phi^4\) gives \(\frac{\lambda}{3!}\phi^3\) on the right-hand side following the same pattern, \(\frac{g}{3!}\phi^3\) yields \(\frac{g}{2}\phi^2\). This is consistent.

B-10. Chain Rule for Functional Derivatives

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Solution strategy: Substitute \(q(t) \to q(t) + \epsilon\,\delta(t-t')\) into \(S[q] = \int_{t_1}^{t_2}\frac{1}{2}m[\dot{q}(t)]^2\,dt\) and extract the terms first order in \(\epsilon\).

Calculation:

Under \(q(t) \to q(t) + \epsilon\,\delta(t-t')\), we have \(\dot{q}(t) \to \dot{q}(t) + \epsilon\,\frac{d}{dt}\delta(t-t')\).

\[ S[q + \epsilon\delta] = \int_{t_1}^{t_2}\frac{1}{2}m\left[\dot{q}(t) + \epsilon\,\frac{d}{dt}\delta(t-t')\right]^2 dt \]

Expanding to first order in \(\epsilon\):

\[ \approx \int_{t_1}^{t_2}\frac{1}{2}m\left[\dot{q}^2 + 2\dot{q}(t)\cdot\epsilon\,\frac{d}{dt}\delta(t-t')\right]dt \]

The first-order term in \(\epsilon\):

\[ \frac{\delta S}{\delta q(t')} = \int_{t_1}^{t_2} m\dot{q}(t)\,\frac{d}{dt}\delta(t-t')\,dt \]

Integrating by parts (\(u = m\dot{q}(t)\), \(dv = \frac{d}{dt}\delta(t-t')\,dt\)):

\[ = \left[m\dot{q}(t)\,\delta(t-t')\right]_{t_1}^{t_2} - \int_{t_1}^{t_2} m\ddot{q}(t)\,\delta(t-t')\,dt \]

Corresponding to the endpoint conditions \(\delta q(t_1) = \delta q(t_2) = 0\), the surface term vanishes when \(t'\) lies in the interior of the interval (since \(\delta(t-t')\) is zero at \(t = t_1, t_2\)).

Applying the sifting property of the delta function:

\[ \frac{\delta S}{\delta q(t')} = -m\ddot{q}(t') \]

Final answer:

\[ \boxed{\frac{\delta S}{\delta q(t')} = -m\ddot{q}(t')} \]

Verification: The Euler-Lagrange equation for \(L = \frac{1}{2}m\dot{q}^2\) (with \(V = 0\)) is \(m\ddot{q} = 0\). Setting the functional derivative \(\frac{\delta S}{\delta q(t')} = 0\) gives \(-m\ddot{q}(t') = 0\), i.e., \(m\ddot{q} = 0\), which is consistent.

Medium

M-1. Deriving Newton's Gravitational Equation of Motion from the Action Principle

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Solution strategy: Compute the variation of the action for \(L = \frac{1}{2}m\dot{z}^2 - mgz\), and derive the Euler-Lagrange equation through integration by parts.

1. Variation of the Action

The action is \(S[z] = \int_{t_1}^{t_2}\left(\frac{1}{2}m\dot{z}^2 - mgz\right)dt\).

We displace the path as \(z(t) \to z(t) + \delta z(t)\) (with boundary conditions \(\delta z(t_1) = \delta z(t_2) = 0\)).

\[ \delta S = \int_{t_1}^{t_2}\left[\frac{\partial L}{\partial z}\delta z + \frac{\partial L}{\partial \dot{z}}\delta\dot{z}\right]dt \]

Computing each partial derivative:

\[ \frac{\partial L}{\partial z} = -mg, \qquad \frac{\partial L}{\partial \dot{z}} = m\dot{z} \]

Substituting:

\[ \delta S = \int_{t_1}^{t_2}\left[(-mg)\,\delta z + m\dot{z}\,\delta\dot{z}\right]dt \]

We integrate the second term by parts. Since \(\delta\dot{z} = \frac{d}{dt}(\delta z)\):

\[ \int_{t_1}^{t_2} m\dot{z}\,\frac{d(\delta z)}{dt}\,dt = \left[m\dot{z}\,\delta z\right]_{t_1}^{t_2} - \int_{t_1}^{t_2} m\ddot{z}\,\delta z\,dt \]

The boundary conditions \(\delta z(t_1) = \delta z(t_2) = 0\) cause the surface term to vanish:

\[ \left[m\dot{z}\,\delta z\right]_{t_1}^{t_2} = m\dot{z}(t_2)\cdot 0 - m\dot{z}(t_1)\cdot 0 = 0 \]

Therefore:

\[ \delta S = \int_{t_1}^{t_2}\left[-mg - m\ddot{z}\right]\delta z\,dt \]

2. Euler-Lagrange Equation

For \(\delta S = 0\) to hold for arbitrary \(\delta z(t)\), the integrand must vanish:

\[ -mg - m\ddot{z} = 0 \]
\[ \boxed{m\ddot{z} = -mg} \]

3. Agreement with Newton's Equation of Motion

The force acting on a particle of mass \(m\) in a uniform gravitational field is \(F = -mg\) (taking the vertically upward direction as positive). Newton's second law \(F = ma\) gives:

\[ ma = -mg \quad \Longrightarrow \quad m\ddot{z} = -mg \]

This is in complete agreement with the result obtained above.

Verification: Dimensional analysis: \([m\ddot{z}] = \text{kg}\cdot\text{m/s}^2 = \text{N}\), \([mg] = \text{kg}\cdot\text{m/s}^2 = \text{N}\). Consistent. Furthermore, taking \(g \to 0\) yields \(m\ddot{z} = 0\) (uniform rectilinear motion), which is physically correct.

M-2. Canonical Momentum and Hamiltonian Density of a Field

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1. Canonical Momentum Density

\[ \mathcal{L} = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{m^2}{2}\phi^2 \]
\[ \pi(x) = \frac{\partial\mathcal{L}}{\partial\dot{\phi}} = \frac{\partial}{\partial\dot{\phi}}\left(\frac{1}{2}\dot{\phi}^2\right) = \dot{\phi} \]
\[ \boxed{\pi(x) = \dot{\phi}(x)} \]

2. Hamiltonian Density

\[ \mathcal{H} = \pi\dot{\phi} - \mathcal{L} \]

Substituting \(\dot{\phi} = \pi\):

\[ \mathcal{H} = \pi \cdot \pi - \left[\frac{1}{2}\pi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{m^2}{2}\phi^2\right] \]
\[ = \pi^2 - \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2}\phi^2 \]
\[ \boxed{\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2}\phi^2} \]

3. Verification of Positive Definiteness

\(\mathcal{H}\) is the sum of three terms:

  • \(\frac{1}{2}\pi^2 \geq 0\) (square of a real number)
  • \(\frac{1}{2}(\nabla\phi)^2 \geq 0\) (squared norm of a vector)
  • \(\frac{m^2}{2}\phi^2 \geq 0\) (\(m^2 > 0\) and square of a real number)

Therefore \(\mathcal{H} \geq 0\), and the energy density is positive definite (strictly speaking, non-negative definite). The equality \(\mathcal{H} = 0\) holds only when \(\pi = 0\), \(\nabla\phi = 0\), and \(\phi = 0\).

Consistency check: This has the same structure as the harmonic oscillator in particle mechanics, where \(H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\) (positive definite). This is consistent with the fact that each mode of the field is an independent harmonic oscillator. Furthermore, it agrees with the classical version of the Hamiltonian \(\hat{H} = \int d^3x\,\hat{\mathcal{H}}\) obtained when quantizing the Klein-Gordon field in Ch. 4.

M-3. Poisson Brackets and Hamilton's Equations of Motion

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1. Verification of \(\{q, p\}_{\mathrm{PB}} = 1\)

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

2. Hamilton's Equations of Motion

For \(H = \frac{p^2}{2m} + V(q)\):

Equation for \(\dot{q}\):

\[ \{q, H\}_{\mathrm{PB}} = \frac{\partial q}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial H}{\partial q} \]
\[ = 1 \cdot \frac{p}{m} - 0 \cdot \frac{dV}{dq} = \frac{p}{m} \]
\[ \boxed{\dot{q} = \{q, H\}_{\mathrm{PB}} = \frac{p}{m}} \]

Equation for \(\dot{p}\):

\[ \{p, H\}_{\mathrm{PB}} = \frac{\partial p}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial p}{\partial p}\frac{\partial H}{\partial q} \]
\[ = 0 \cdot \frac{p}{m} - 1 \cdot \frac{dV}{dq} = -\frac{dV}{dq} \]
\[ \boxed{\dot{p} = \{p, H\}_{\mathrm{PB}} = -\frac{dV}{dq}} \]

3. Canonical Quantization Prescription

Applying the prescription \(\{A, B\}_{\mathrm{PB}} \to \frac{1}{i\hbar}[\hat{A}, \hat{B}]\) to \(\{q, p\}_{\mathrm{PB}} = 1\):

\[ \frac{1}{i\hbar}[\hat{q}, \hat{p}] = 1 \]
\[ \boxed{[\hat{q}, \hat{p}] = i\hbar} \]

Verification: This is the fundamental commutation relation of quantum mechanics, and served as the starting point for canonical quantization of fields in Ch. 4. Combining \(\dot{q} = p/m\) and \(\dot{p} = -dV/dq\) reproduces \(m\ddot{q} = -dV/dq = F\) (Newton's second law).

M-4. Legendre Transform for Systems with Multiple Degrees of Freedom

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1. Proof that \(H\) is a function of \((q_i, p_i)\) only

We compute the total differential of \(H = \sum_{i=1}^N p_i\dot{q}_i - L(q_i, \dot{q}_i)\):

\[ dH = \sum_{i=1}^N \left(\dot{q}_i\,dp_i + p_i\,d\dot{q}_i\right) - \sum_{i=1}^N\left(\frac{\partial L}{\partial q_i}dq_i + \frac{\partial L}{\partial \dot{q}_i}d\dot{q}_i\right) \]

Using the definition of canonical momentum \(p_i = \frac{\partial L}{\partial \dot{q}_i}\), the terms containing \(d\dot{q}_i\) are:

\[ \sum_{i=1}^N p_i\,d\dot{q}_i - \sum_{i=1}^N \frac{\partial L}{\partial \dot{q}_i}d\dot{q}_i = \sum_{i=1}^N (p_i - p_i)\,d\dot{q}_i = 0 \]

Therefore the \(d\dot{q}_i\) terms cancel completely:

\[ dH = \sum_{i=1}^N \dot{q}_i\,dp_i - \sum_{i=1}^N \frac{\partial L}{\partial q_i}dq_i \]

Since \(dH\) is written solely in terms of \(dp_i\) and \(dq_i\), \(H\) is a function of \((q_i, p_i)\) only. It does not depend on \(\dot{q}_i\). \(\square\)

2. Derivation of Hamilton's canonical equations

Since \(H\) is a function of \((q_i, p_i)\), its total differential is:

\[ dH = \sum_{i=1}^N \frac{\partial H}{\partial p_i}dp_i + \sum_{i=1}^N \frac{\partial H}{\partial q_i}dq_i \]

We compare this with the expression obtained above:

\[ dH = \sum_{i=1}^N \dot{q}_i\,dp_i - \sum_{i=1}^N \frac{\partial L}{\partial q_i}dq_i \]

Comparing the coefficients of \(dp_i\):

\[ \frac{\partial H}{\partial p_i} = \dot{q}_i \]

Comparing the coefficients of \(dq_i\):

\[ \frac{\partial H}{\partial q_i} = -\frac{\partial L}{\partial q_i} \]

From the Euler-Lagrange equations, \(\frac{\partial L}{\partial q_i} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = \dot{p}_i\), so:

\[ \frac{\partial H}{\partial q_i} = -\dot{p}_i \]

Summarizing, we obtain Hamilton's canonical equations:

\[ \boxed{\dot{q}_i = \frac{\partial H}{\partial p_i}, \qquad \dot{p}_i = -\frac{\partial H}{\partial q_i}} \]

Verification: We check with \(N=1\), \(H = \frac{p^2}{2m} + V(q)\): \(\dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m}\), \(\dot{p} = -\frac{\partial H}{\partial q} = -\frac{dV}{dq}\). This agrees with the results from S3.

M-5. Relationship Between Functional Derivatives and the Euler-Lagrange Equation

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Solution strategy: Substitute \(q(t) \to q(t) + \epsilon\,\delta(t-t')\) and extract the terms first-order in \(\epsilon\) from the action.

Calculation:

\[ S[q] = \int_{t_1}^{t_2} L(q(t), \dot{q}(t))\,dt \]

Under \(q(t) \to q(t) + \epsilon\,\delta(t-t')\), we have \(\dot{q}(t) \to \dot{q}(t) + \epsilon\,\frac{d}{dt}\delta(t-t')\).

Expanding \(L\) to first order in \(\epsilon\):

\[ L(q + \epsilon\delta(t-t'),\, \dot{q} + \epsilon\dot{\delta}(t-t')) \approx L(q,\dot{q}) + \frac{\partial L}{\partial q}\epsilon\,\delta(t-t') + \frac{\partial L}{\partial \dot{q}}\epsilon\,\frac{d}{dt}\delta(t-t') \]

Extracting the terms first-order in \(\epsilon\):

\[ \frac{\delta S}{\delta q(t')} = \int_{t_1}^{t_2}\left[\frac{\partial L}{\partial q}\,\delta(t-t') + \frac{\partial L}{\partial \dot{q}}\,\frac{d}{dt}\delta(t-t')\right]dt \]

The first term, by the sifting property of the delta function:

\[ \int_{t_1}^{t_2}\frac{\partial L}{\partial q}\,\delta(t-t')\,dt = \frac{\partial L}{\partial q}\bigg|_{t=t'} \]

Integrating the second term by parts:

\[ \int_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}}\,\frac{d}{dt}\delta(t-t')\,dt = \left[\frac{\partial L}{\partial \dot{q}}\,\delta(t-t')\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\delta(t-t')\,dt \]

When \(t'\) is in the interior of the interval, the boundary terms vanish (\(\delta(t_1 - t') = \delta(t_2 - t') = 0\)). Applying the sifting property of the delta function to the remaining term:

\[ = -\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\bigg|_{t=t'} \]

Combining everything:

\[ \boxed{\frac{\delta S}{\delta q(t')} = \frac{\partial L}{\partial q}\bigg|_{t=t'} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\bigg|_{t=t'}} \]

Therefore, \(\frac{\delta S}{\delta q(t')} = 0\) (for all \(t'\)) is equivalent to:

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

That is, the Euler-Lagrange equation.

Verification: In D10, for the case \(L = \frac{1}{2}m\dot{q}^2\), we obtained \(\frac{\delta S}{\delta q(t')} = -m\ddot{q}(t')\). Checking with the formula above: \(\frac{\partial L}{\partial q} = 0\), \(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = m\ddot{q}\). Therefore \(\frac{\delta S}{\delta q(t')} = 0 - m\ddot{q}(t') = -m\ddot{q}(t')\). This agrees.

Advanced

A-1. Charged Particle in an Electromagnetic Field and Gauge Dependence of Canonical Momentum

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1. Canonical Momentum

\[ L = \frac{1}{2}m\dot{\mathbf{r}}^2 - eV + e\dot{\mathbf{r}}\cdot\mathbf{A} \]

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

\[ p_i = \frac{\partial L}{\partial \dot{r}_i} = m\dot{r}_i + eA_i \]

In vector notation:

\[ \boxed{\mathbf{p} = m\dot{\mathbf{r}} + e\mathbf{A}} \]

This differs from the mechanical momentum \(m\dot{\mathbf{r}}\). The difference is \(e\mathbf{A}\), which depends on the vector potential.

2. Construction of the Hamiltonian

Using \(\dot{\mathbf{r}} = \frac{\mathbf{p} - e\mathbf{A}}{m}\):

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

Computing each term:

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

Therefore:

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

Defining \(\boldsymbol{\Pi} \equiv \mathbf{p} - e\mathbf{A}\):

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

3. Verification of Gauge Invariance

Gauge transformation:

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

Change in canonical momentum:

\[ \mathbf{p}' = m\dot{\mathbf{r}} + e\mathbf{A}' = m\dot{\mathbf{r}} + e(\mathbf{A} + \nabla\chi) = \mathbf{p} + e\nabla\chi \]

Therefore the canonical momentum is gauge-dependent:

\[ \boxed{\mathbf{p} \to \mathbf{p}' = \mathbf{p} + e\nabla\chi} \]

However, the combination \(\mathbf{p} - e\mathbf{A}\) appearing in the Hamiltonian transforms as:

\[ \mathbf{p}' - e\mathbf{A}' = (\mathbf{p} + e\nabla\chi) - e(\mathbf{A} + \nabla\chi) = \mathbf{p} - e\mathbf{A} \]

Gauge invariant! Furthermore:

\[ H' = \frac{(\mathbf{p}' - e\mathbf{A}')^2}{2m} + eV' = \frac{(\mathbf{p} - e\mathbf{A})^2}{2m} + e\left(V - \frac{\partial\chi}{\partial t}\right) \]

At first glance it appears that \(H\) changes, but this corresponds to the fact that \(H\) transforms under a canonical transformation in the case of a time-dependent gauge transformation. The equations of motion (the physical equations written in terms of \(\dot{\mathbf{r}}\) and \(\ddot{\mathbf{r}}\)) are gauge invariant. Indeed, the mechanical momentum \(m\dot{\mathbf{r}} = \mathbf{p} - e\mathbf{A}\) is gauge invariant, and the Lorentz force equation:

\[ m\ddot{\mathbf{r}} = e(\mathbf{E} + \dot{\mathbf{r}}\times\mathbf{B}) \]

is gauge invariant because \(\mathbf{E} = -\nabla V - \frac{\partial\mathbf{A}}{\partial t}\) and \(\mathbf{B} = \nabla\times\mathbf{A}\) are gauge invariant.

4. Relation to Minimal Coupling

Upon quantization, the canonical momentum is promoted to an operator: \(\mathbf{p} \to \hat{\mathbf{p}}\). The Hamiltonian becomes:

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

This is nothing other than the minimal coupling prescription: "in the free-particle Hamiltonian \(\frac{\hat{\mathbf{p}}^2}{2m}\), replace \(\hat{\mathbf{p}} \to \hat{\mathbf{p}} - e\mathbf{A}\)."

In Ch. 6 (quantization of QED) of the main text, we introduced the covariant derivative \(\partial_\mu \to D_\mu = \partial_\mu + ieA_\mu\) in covariant form. Looking at the spatial components, this corresponds to \(-i\hbar\nabla \to -i\hbar\nabla - e\mathbf{A}\), i.e., \(\hat{\mathbf{p}} \to \hat{\mathbf{p}} - e\mathbf{A}\).

In other words, the minimal coupling prescription of QED originates from the distinction between canonical momentum and mechanical momentum in classical analytical mechanics. Gauge-invariant physical quantities always appear in the combination \(\mathbf{p} - e\mathbf{A}\) (mechanical momentum), and this naturally requires gauge-covariant coupling in quantum theory as well.

Consistency checks: - Dimensions: \([e\mathbf{A}] = \text{C}\cdot\text{V·s/m} = \text{kg·m/s}\), which has dimensions of momentum. Consistent. - In the limit \(\mathbf{A} = 0\), we recover \(H = \frac{p^2}{2m} + eV\) (a particle in an electrostatic potential). - Lorentz covariance: In four-vector form, \(p^\mu - eA^\mu\) is the covariant combination.

A-2. From Field Poisson Brackets to Canonical Quantization

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1. Verification of \(\{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^3(\mathbf{x} - \mathbf{y})\)

Definition of the Poisson bracket for fields:

\[ \{A, B\}_{\mathrm{PB}} = \int d^3z\left(\frac{\delta A}{\delta\phi(\mathbf{z})}\frac{\delta B}{\delta\pi(\mathbf{z})} - \frac{\delta A}{\delta\pi(\mathbf{z})}\frac{\delta B}{\delta\phi(\mathbf{z})}\right) \]

Let \(A = \phi(\mathbf{x})\), \(B = \pi(\mathbf{y})\).

Computing the functional derivatives:

\[ \frac{\delta\phi(\mathbf{x})}{\delta\phi(\mathbf{z})} = \delta^3(\mathbf{x} - \mathbf{z}), \qquad \frac{\delta\phi(\mathbf{x})}{\delta\pi(\mathbf{z})} = 0 \]
\[ \frac{\delta\pi(\mathbf{y})}{\delta\pi(\mathbf{z})} = \delta^3(\mathbf{y} - \mathbf{z}), \qquad \frac{\delta\pi(\mathbf{y})}{\delta\phi(\mathbf{z})} = 0 \]

Substituting:

\[ \{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \int d^3z\left[\delta^3(\mathbf{x}-\mathbf{z})\cdot\delta^3(\mathbf{y}-\mathbf{z}) - 0\cdot 0\right] \]
\[ = \int d^3z\,\delta^3(\mathbf{x}-\mathbf{z})\,\delta^3(\mathbf{y}-\mathbf{z}) \]

Using the sifting property of the delta function (the \(\mathbf{z}\) integration picks out \(\mathbf{z} = \mathbf{x}\)):

\[ = \delta^3(\mathbf{y} - \mathbf{x}) = \delta^3(\mathbf{x} - \mathbf{y}) \]
\[ \boxed{\{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^3(\mathbf{x} - \mathbf{y})} \]

2. \(\{\phi(\mathbf{x}), \phi(\mathbf{y})\}_{\mathrm{PB}} = 0\) and \(\{\pi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = 0\)

\(\phi\)-\(\phi\):

\[ \{\phi(\mathbf{x}), \phi(\mathbf{y})\}_{\mathrm{PB}} = \int d^3z\left[\frac{\delta\phi(\mathbf{x})}{\delta\phi(\mathbf{z})}\frac{\delta\phi(\mathbf{y})}{\delta\pi(\mathbf{z})} - \frac{\delta\phi(\mathbf{x})}{\delta\pi(\mathbf{z})}\frac{\delta\phi(\mathbf{y})}{\delta\phi(\mathbf{z})}\right] \]
\[ = \int d^3z\left[\delta^3(\mathbf{x}-\mathbf{z})\cdot 0 - 0\cdot\delta^3(\mathbf{y}-\mathbf{z})\right] = 0 \]

\(\pi\)-\(\pi\):

\[ \{\pi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \int d^3z\left[\frac{\delta\pi(\mathbf{x})}{\delta\phi(\mathbf{z})}\frac{\delta\pi(\mathbf{y})}{\delta\pi(\mathbf{z})} - \frac{\delta\pi(\mathbf{x})}{\delta\pi(\mathbf{z})}\frac{\delta\pi(\mathbf{y})}{\delta\phi(\mathbf{z})}\right] \]
\[ = \int d^3z\left[0\cdot\delta^3(\mathbf{y}-\mathbf{z}) - \delta^3(\mathbf{x}-\mathbf{z})\cdot 0\right] = 0 \]
\[ \boxed{\{\phi(\mathbf{x}), \phi(\mathbf{y})\}_{\mathrm{PB}} = 0, \qquad \{\pi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = 0} \]

3. Derivation of \(\dot{\phi} = \pi\)

With \(H = \int d^3y\,\mathcal{H}\) where \(\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2}\phi^2\):

\[ \dot{\phi}(\mathbf{x}) = \{\phi(\mathbf{x}), H\}_{\mathrm{PB}} = \int d^3z\left[\frac{\delta\phi(\mathbf{x})}{\delta\phi(\mathbf{z})}\frac{\delta H}{\delta\pi(\mathbf{z})} - \frac{\delta\phi(\mathbf{x})}{\delta\pi(\mathbf{z})}\frac{\delta H}{\delta\phi(\mathbf{z})}\right] \]
\[ = \int d^3z\,\delta^3(\mathbf{x}-\mathbf{z})\frac{\delta H}{\delta\pi(\mathbf{z})} \]

Computing \(\frac{\delta H}{\delta\pi(\mathbf{z})}\). The only term in \(H\) containing \(\pi\) is \(\int d^3y\,\frac{1}{2}\pi(\mathbf{y})^2\):

\[ \frac{\delta H}{\delta\pi(\mathbf{z})} = \frac{\delta}{\delta\pi(\mathbf{z})}\int d^3y\,\frac{1}{2}\pi(\mathbf{y})^2 = \pi(\mathbf{z}) \]

Therefore:

\[ \dot{\phi}(\mathbf{x}) = \int d^3z\,\delta^3(\mathbf{x}-\mathbf{z})\,\pi(\mathbf{z}) = \pi(\mathbf{x}) \]
\[ \boxed{\dot{\phi}(\mathbf{x}) = \pi(\mathbf{x})} \]

4. Derivation of \(\dot{\pi} = \nabla^2\phi - m^2\phi\) and the Klein-Gordon Equation

\[ \dot{\pi}(\mathbf{x}) = \{\pi(\mathbf{x}), H\}_{\mathrm{PB}} = \int d^3z\left[\frac{\delta\pi(\mathbf{x})}{\delta\phi(\mathbf{z})}\frac{\delta H}{\delta\pi(\mathbf{z})} - \frac{\delta\pi(\mathbf{x})}{\delta\pi(\mathbf{z})}\frac{\delta H}{\delta\phi(\mathbf{z})}\right] \]

The first term vanishes since \(\frac{\delta\pi(\mathbf{x})}{\delta\phi(\mathbf{z})} = 0\). The second term gives:

\[ \dot{\pi}(\mathbf{x}) = -\int d^3z\,\delta^3(\mathbf{x}-\mathbf{z})\frac{\delta H}{\delta\phi(\mathbf{z})} = -\frac{\delta H}{\delta\phi(\mathbf{x})} \]

Computing \(\frac{\delta H}{\delta\phi(\mathbf{x})}\). The terms in \(H\) containing \(\phi\) are:

\[ \int d^3y\left[\frac{1}{2}(\nabla\phi(\mathbf{y}))^2 + \frac{m^2}{2}\phi(\mathbf{y})^2\right] \]

Functional derivative of the second term:

\[ \frac{\delta}{\delta\phi(\mathbf{x})}\int d^3y\,\frac{m^2}{2}\phi(\mathbf{y})^2 = m^2\phi(\mathbf{x}) \]

Functional derivative of the first term. For \((\nabla\phi)^2 = \nabla_i\phi\,\nabla_i\phi\):

\[ \frac{\delta}{\delta\phi(\mathbf{x})}\int d^3y\,\frac{1}{2}\nabla_i\phi(\mathbf{y})\,\nabla_i\phi(\mathbf{y}) \]

Substituting \(\phi(\mathbf{y}) \to \phi(\mathbf{y}) + \epsilon\,\delta^3(\mathbf{y}-\mathbf{x})\) gives \(\nabla_i\phi(\mathbf{y}) \to \nabla_i\phi(\mathbf{y}) + \epsilon\,\nabla_i^{(y)}\delta^3(\mathbf{y}-\mathbf{x})\). The first-order term in \(\epsilon\):

\[ \int d^3y\,\nabla_i\phi(\mathbf{y})\,\nabla_i^{(y)}\delta^3(\mathbf{y}-\mathbf{x}) \]

Integrating by parts (the surface term vanishes at infinity):

\[ = -\int d^3y\,\nabla_i^2\phi(\mathbf{y})\,\delta^3(\mathbf{y}-\mathbf{x}) = -\nabla^2\phi(\mathbf{x}) \]

Therefore:

\[ \frac{\delta H}{\delta\phi(\mathbf{x})} = -\nabla^2\phi(\mathbf{x}) + m^2\phi(\mathbf{x}) \]

Hence:

\[ \dot{\pi}(\mathbf{x}) = -\frac{\delta H}{\delta\phi(\mathbf{x})} = \nabla^2\phi(\mathbf{x}) - m^2\phi(\mathbf{x}) \]
\[ \boxed{\dot{\pi}(\mathbf{x}) = \nabla^2\phi(\mathbf{x}) - m^2\phi(\mathbf{x})} \]

Recovery of the Klein-Gordon equation:

From \(\dot{\phi} = \pi\), we have \(\dot{\pi} = \ddot{\phi}\). Therefore:

\[ \ddot{\phi} = \nabla^2\phi - m^2\phi \]
\[ \ddot{\phi} - \nabla^2\phi + m^2\phi = 0 \]
\[ \boxed{(\partial_\mu\partial^\mu + m^2)\phi = (\Box + m^2)\phi = 0} \]

This is the Klein-Gordon equation.

5. The Canonical Quantization Prescription

Applying the prescription \(\{A, B\}_{\mathrm{PB}} \to \frac{1}{i\hbar}[\hat{A}, \hat{B}]\).

Applying it to the result of part 1, \(\{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^3(\mathbf{x}-\mathbf{y})\):

\[ \frac{1}{i\hbar}[\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = \delta^3(\mathbf{x}-\mathbf{y}) \]
\[ \boxed{[\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = i\hbar\,\delta^3(\mathbf{x}-\mathbf{y})} \]

Similarly, from the results of part 2:

\[ [\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})] = 0, \qquad [\hat{\pi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = 0 \]

These are precisely the equal-time commutation relations introduced in Ch. 4 of the main text.

Consistency checks: - Correspondence with particle mechanics: \(\{q, p\}_{\mathrm{PB}} = 1 \to [\hat{q}, \hat{p}] = i\hbar\) and \(\{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^3(\mathbf{x}-\mathbf{y}) \to [\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = i\hbar\,\delta^3(\mathbf{x}-\mathbf{y})\) represent a natural extension replacing the discrete index \(i\) with the continuous index \(\mathbf{x}\) (Kronecker delta \(\delta_{ij}\) → Dirac delta \(\delta^3(\mathbf{x}-\mathbf{y})\)). - The fact that Hamilton's equations of motion reproduce the Klein-Gordon equation confirms the equivalence between the Hamiltonian and Lagrangian formulations. - Lorentz covariance: The equal-time commutation relations single out a particular time slice, but the Klein-Gordon equation itself is Lorentz covariant. In the quantum theory, these are extended to covariant commutation relations (the Pauli-Jordan function) (see Ch. 4 in the main text).