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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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Problem:

Substitute \(f(x) = 2x\) into the functional \(H[f] = \int_0^3 [f(x)]^2\,dx\) and find the value of \(H[f]\).

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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Problem:

Find the functional derivative \(\frac{\delta F}{\delta f(x_0)}\) (\(0 \leq x_0 \leq 1\)) of the functional \(F[f] = \int_0^1 [f(x)]^4\,dx\).

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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Problem:

Find the functional derivative \(\frac{\delta G}{\delta f(x)}\) of the functional \(G[f] = \int_{-\infty}^{\infty} [f(y)]^2\,e^{-y^2}\,dy\).

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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Problem:

Write the functional \(F[f] = f(a)\) (the function value at a fixed point \(a\)) in its integral representation \(F[f] = \int f(y)\,\delta(y - a)\,dy\), and compute \(\frac{\delta F}{\delta f(x)}\) according to the definition.

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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Problem:

For the Lagrangian \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}k x^2\), compute the following in order.

  1. \(\frac{\partial L}{\partial \dot{x}}\)
  2. \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)\)
  3. \(\frac{\partial L}{\partial x}\)
  4. Write down the Euler-Lagrange equation and verify the resulting equation of motion.

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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Problem:

For each of the following Lagrangians, find the canonical momentum \(p = \frac{\partial L}{\partial \dot{q}}\).

(a) \(L = \frac{1}{2}m\dot{q}^2 - mg q\) (free fall in a uniform gravitational field)

(b) For \(L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)\) (2-dimensional polar coordinates), find \(p_r\) and \(p_\theta\) respectively.

(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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Problem:

For the Lagrangian of a 1-dimensional harmonic oscillator \(L = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}m\omega^2 q^2\):

  1. Find the canonical momentum \(p\).
  2. Express \(\dot{q}\) in terms of \(p\) and \(m\).
  3. Write down the Hamiltonian \(H = p\dot{q} - L\) as a function of \(q\) and \(p\).

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

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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Problem:

For the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\) (without mass term), apply the field Euler-Lagrange equation and derive the equation of motion.

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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Problem:

Apply the Euler-Lagrange equation for fields to the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{m^2}{2}\phi^2 - \frac{g}{3!}\phi^3\), and derive the equation of motion.

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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Problem:

Compute the functional derivative \(\frac{\delta S}{\delta q(t')}\) (\(t_1 < t' < t_2\)) of the functional \(S[q] = \int_{t_1}^{t_2}\frac{1}{2}m[\dot{q}(t)]^2\,dt\). You may use the boundary conditions \(\delta q(t_1) = \delta q(t_2) = 0\).

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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Problem:

A particle of mass \(m\) moves vertically in a uniform gravitational field. The Lagrangian is

\[ L = \frac{1}{2}m\dot{z}^2 - mgz \]

Show the following.

  1. Compute the variation \(\delta S\) of the action \(S[z] = \int_{t_1}^{t_2} L\,dt\) and arrange it into an integral containing \(\delta z(t)\) (explicitly showing the integration by parts).
  2. Derive the Euler-Lagrange equation from \(\delta S = 0\) and obtain \(m\ddot{z} = -mg\).
  3. Confirm that the resulting equation is consistent with Newton's equation of motion \(F = ma\).

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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Problem:

Given the Lagrangian density of the Klein-Gordon field

\[ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{m^2}{2}\phi^2 = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{m^2}{2}\phi^2 \]

perform the following:

  1. Find the canonical momentum density \(\pi(x) = \frac{\partial\mathcal{L}}{\partial\dot{\phi}}\).
  2. Write down the Hamiltonian density \(\mathcal{H} = \pi\dot{\phi} - \mathcal{L}\) in terms of \(\phi\), \(\pi\), and \(\nabla\phi\).
  3. Verify that the resulting \(\mathcal{H}\) is an energy density (positive definite).

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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Problem:

Consider a one-dimensional particle with Hamiltonian given by \(H(q, p) = \frac{p^2}{2m} + V(q)\). The Poisson bracket is defined as

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

Show the following.

  1. Verify that \(\{q, p\}_{\mathrm{PB}} = 1\).
  2. Compute Hamilton's equations of motion \(\dot{q} = \{q, H\}_{\mathrm{PB}}\), \(\dot{p} = \{p, H\}_{\mathrm{PB}}\), and obtain \(\dot{q} = p/m\) and \(\dot{p} = -\frac{dV}{dq}\), respectively.
  3. Using the canonical quantization prescription "\(\{A, B\}_{\mathrm{PB}} \to \frac{1}{i\hbar}[\hat{A}, \hat{B}]\)", confirm that \([\hat{q}, \hat{p}] = i\hbar\) is obtained.

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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Problem:

For a system with \(N\) generalized coordinates \(q_1, \ldots, q_N\) and Lagrangian \(L(q_i, \dot{q}_i)\):

  1. Define the canonical momenta \(p_i = \frac{\partial L}{\partial \dot{q}_i}\) and construct the Hamiltonian as
\[ H = \sum_{i=1}^N p_i \dot{q}_i - L \]

Show that \(H\) does not contain \(\dot{q}_i\) and is a function only of \((q_i, p_i)\), by computing the total differential \(dH\).

  1. From the expression for \(dH\), derive Hamilton's canonical equations
\[ \dot{q}_i = \frac{\partial H}{\partial p_i}, \qquad \dot{p}_i = -\frac{\partial H}{\partial q_i} \]

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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Problem:

For the action

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

compute the functional derivative \(\frac{\delta S}{\delta q(t')}\) and show that the result is

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

Using this, reconfirm in the language of functional derivatives that \(\delta S = 0\) is equivalent to the Euler-Lagrange equation.

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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Problem:

The Lagrangian of a charged particle (charge \(e\), mass \(m\)) in an electromagnetic field \((V, \mathbf{A})\) is given by

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

Carry out the following.

  1. Find the canonical momentum \(\mathbf{p} = \frac{\partial L}{\partial \dot{\mathbf{r}}}\) and show that it differs from the kinetic momentum \(m\dot{\mathbf{r}}\).
  2. Write down the Hamiltonian \(H = \mathbf{p}\cdot\dot{\mathbf{r}} - L\) in terms of \((\mathbf{r}, \mathbf{p})\).
  3. Show how the canonical momentum transforms under the gauge transformation \(\mathbf{A} \to \mathbf{A} + \nabla\chi\), \(V \to V - \frac{\partial\chi}{\partial t}\), and verify that the Hamiltonian (and hence the equations of motion) is gauge invariant.
  4. Discuss how this result serves as the classical origin of the prescription (minimal coupling) of "replacing \(\hat{\mathbf{p}}\) with \(\hat{\mathbf{p}} - e\hat{\mathbf{A}}\)" in Ch. 6 (quantization of QED) in the main text.

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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Problem:

For a scalar field \(\phi(\mathbf{x}, t)\) and the canonical momentum density \(\pi(\mathbf{x}, t) = \dot{\phi}(\mathbf{x}, t)\), the field Poisson bracket is defined as

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

Show the following.

  1. Verify that \(\{\phi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^3(\mathbf{x} - \mathbf{y})\).
  2. Verify that \(\{\phi(\mathbf{x}), \phi(\mathbf{y})\}_{\mathrm{PB}} = 0\) and \(\{\pi(\mathbf{x}), \pi(\mathbf{y})\}_{\mathrm{PB}} = 0\).
  3. For the Hamiltonian \(H = \int d^3x\,\mathcal{H}\) (where \(\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2}\phi^2\)), compute Hamilton's equation of motion \(\dot{\phi}(\mathbf{x}) = \{\phi(\mathbf{x}), H\}_{\mathrm{PB}}\) and obtain \(\dot{\phi} = \pi\).
  4. Similarly, compute \(\dot{\pi}(\mathbf{x}) = \{\pi(\mathbf{x}), H\}_{\mathrm{PB}}\) and obtain \(\dot{\pi} = \nabla^2\phi - m^2\phi\). Confirm that combining these reproduces the Klein-Gordon equation.
  5. Apply the canonical quantization prescription \(\{\cdot, \cdot\}_{\mathrm{PB}} \to \frac{1}{i\hbar}[\cdot, \cdot]\) and verify that the equal-time commutation relation \([\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = i\hbar\,\delta^3(\mathbf{x} - \mathbf{y})\) from Ch. 4 is obtained.

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).