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Ch. 6 Solutions

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Basic

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Basic

B-1. Antisymmetry of the Field Strength Tensor and Verification of Components

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(a) Calculation of \(F_{01}\)

Substituting \(\mu=0\), \(\nu=1\) into the definition \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\):

\[ F_{01} = \partial_0 A_1 - \partial_1 A_0 = \frac{\partial A_1}{\partial t} - \frac{\partial A_0}{\partial x} \]

(b) Calculation of \(F_{10}\) and verification of antisymmetry

Substituting \(\mu=1\), \(\nu=0\):

\[ F_{10} = \partial_1 A_0 - \partial_0 A_1 = \frac{\partial A_0}{\partial x} - \frac{\partial A_1}{\partial t} \]

Comparing with (a):

\[ F_{10} = -\left(\frac{\partial A_1}{\partial t} - \frac{\partial A_0}{\partial x}\right) = -F_{01} \quad \checkmark \]

(c) Calculation of \(F_{12}\)

Substituting \(\mu=1\), \(\nu=2\):

\[ F_{12} = \partial_1 A_2 - \partial_2 A_1 = \frac{\partial A_2}{\partial x} - \frac{\partial A_1}{\partial y} \]

Meanwhile, the \(z\)-component of the magnetic field is \(B_z = (\nabla \times \mathbf{A})_z = \partial_1 A^2 - \partial_2 A^1\). Since \(F^{12} = -B_z\) according to Eq. (6.4), and \(F^{12} = \eta^{1\alpha}\eta^{2\beta}F_{\alpha\beta} = (-1)(-1)F_{12} = F_{12}\), so \(F^{12} = F_{12} = -B_z\)...

In fact, Eq. (6.4) is the matrix for \(F^{\mu\nu}\), and the \((1,2)\) component is written as \(-B_z\). Using the metric \(\eta^{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\):

\[ F^{12} = \eta^{1\alpha}\eta^{2\beta}F_{\alpha\beta} = (-1)(-1)F_{12} = F_{12} \]

Therefore \(F_{12} = F^{12} = -B_z\).

However, from \(B_z = \partial_1 A_2 - \partial_2 A_1\) and \(F_{12} = \partial_1 A_2 - \partial_2 A_1\), it would seem that \(F_{12} = B_z\).

This apparent contradiction depends on how one reads the sign conventions of Eq. (6.4). Looking at Eq. (6.4), the \((0,1)\) component of \(F^{\mu\nu}\) is \(E_x\), so \(F^{01} = E_x\). Meanwhile, \(F_{01} = \partial_0 A_1 - \partial_1 A_0\), and from Eq. (6.3), \(E_x = -\partial_x A_0 - \partial_t A_1\), so \(F_{01} = \partial_0 A_1 - \partial_1 A_0 = -E_x\). Then \(F^{01} = \eta^{00}\eta^{11}F_{01} = (+1)(-1)(-E_x) = E_x\). This is consistent with Eq. (6.4).

Similarly, \(F_{12} = \partial_1 A_2 - \partial_2 A_1\) and \(F^{12} = \eta^{11}\eta^{22}F_{12} = (-1)(-1)F_{12} = F_{12}\).

Re-checking the \((1,2)\) component of Eq. (6.4), it is written as \(-B_z\). This means \(F^{12} = -B_z\) in the textbook's sign convention.

Conclusion: In the textbook's Eq. (6.4) convention, \((\nabla \times \mathbf{A})_z = \partial_x A_y - \partial_y A_x\) gives \(F_{12} = \partial_1 A_2 - \partial_2 A_1 = B_z\), and \(F^{12} = (-1)(-1)F_{12} = B_z\) should hold, but the \(-B_z\) in Eq. (6.4) is related to the convention for raising/lowering spatial component indices (\(A^i = -A_i\), etc.).

Following the most standard convention:

\[ F_{12} = \partial_1 A_2 - \partial_2 A_1 = B_z \]

and the \((1,2)\) component being \(-B_z\) in Eq. (6.4) means \(F^{12} = -B_z\), corresponding to the relation \(F^{ij} = -\epsilon^{ijk}B_k\) (so \(F^{12} = -B_3 = -B_z\)). In this case, \(F_{12} = \eta_{1\alpha}\eta_{2\beta}F^{\alpha\beta} = (-1)(-1)F^{12} = F^{12} = -B_z\).

Final reconciliation: Following Eq. (6.4), \(F^{12} = -B_z\), and \(F_{12} = F^{12} = -B_z\) (since raising/lowering spatial indices gives \(\eta_{11}\eta_{22} = (-1)(-1) = 1\), so \(F_{12} = F^{12}\)).

Meanwhile, \(F_{12} = \partial_1 A_2 - \partial_2 A_1\), and \(B_z = \partial_x A_y - \partial_y A_x = \partial_1 A_2 - \partial_2 A_1\) would give \(F_{12} = B_z\), which seems contradictory.

The resolution lies in raising/lowering indices: \(F_{ij} = \eta_{i\alpha}\eta_{j\beta}F^{\alpha\beta}\). Only spatial indices contribute, so \(F_{12} = \eta_{11}\eta_{22}F^{12} = (-1)(-1)F^{12} = F^{12} = -B_z\).

Meanwhile, in \(F_{12} = \partial_1 A_2 - \partial_2 A_1\), here \(A_i\) has a lower index. Since \(A_i = \eta_{i\mu}A^\mu = -A^i\), and \(\mathbf{A} = (A^1, A^2, A^3)\), we have \(A_i = -A^i\).

The \(z\)-component of \(\mathbf{B} = \nabla \times \mathbf{A}\) is \(B_z = \partial_x A^y - \partial_y A^x = \partial_1 A^2 - \partial_2 A^1\).

\[ F_{12} = \partial_1 A_2 - \partial_2 A_1 = \partial_1(-A^2) - \partial_2(-A^1) = -\partial_1 A^2 + \partial_2 A^1 = -(\partial_1 A^2 - \partial_2 A^1) = -B_z \]

This is now consistent! \(F_{12} = -B_z\), which agrees with the \((1,2)\) component \(-B_z\) of Eq. (6.4) where \(F^{12} = -B_z = F_{12}\).

Final answer:

\[ F_{12} = \partial_1 A_2 - \partial_2 A_1 = -(\partial_1 A^2 - \partial_2 A^1) = -B_z \]

This is consistent with the \((1,2)\) component \(-B_z\) of the matrix in Eq. (6.4). \(\square\)

B-2. Expansion of the Lagrangian

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Solution Strategy

Compute \(F_{\mu\nu}F^{\mu\nu}\) by separating it into \((0,i)\) components and \((i,j)\) components.

Calculation

\[ F_{\mu\nu}F^{\mu\nu} = \sum_{\mu,\nu=0}^{3} F_{\mu\nu}F^{\mu\nu} \]

Due to antisymmetry, the \(\mu = \nu\) terms are zero. The independent components are those with \(\mu < \nu\), and by symmetry each pair contributes twice:

\[ F_{\mu\nu}F^{\mu\nu} = 2\sum_{\mu < \nu} F_{\mu\nu}F^{\mu\nu} \]

\((0,i)\) components (\(i = 1,2,3\)):

\(F_{0i} = -E_i\) (from the discussion in D1(c), \(F_{0i} = \partial_0 A_i - \partial_i A_0\), and \(E_i = -\partial_0 A^i - \partial_i A^0 = \partial_0 A_i + \partial_i A_0\)... )

Using the results from D1 above: \(F^{0i} = E^i = E_i\) (spatial components of the electric field). \(F_{0i} = \eta_{00}\eta_{ii}F^{0i} = (1)(-1)E_i = -E_i\).

Therefore:

\[ F_{0i}F^{0i} = (-E_i)(E_i) = -E_i^2 \]

Summing over the three values of \(i\):

\[ \sum_{i=1}^{3} F_{0i}F^{0i} = -\mathbf{E}^2 \]

Combining the contributions from \((0,i)\) and \((i,0)\) using antisymmetry:

\[ 2\sum_{i=1}^{3} F_{0i}F^{0i} = -2\mathbf{E}^2 \]

Let us redo this more carefully.

\[ F_{\mu\nu}F^{\mu\nu} = \sum_{\mu=0}^{3}\sum_{\nu=0}^{3} F_{\mu\nu}F^{\mu\nu} \]

Here \(F^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta}\), and writing the index contraction of \(F_{\mu\nu}F^{\mu\nu}\) correctly:

\[ F_{\mu\nu}F^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}F_{\mu\nu}F_{\alpha\beta} \]

This is cumbersome, so we compute directly in components.

\[ F_{\mu\nu}F^{\mu\nu} = F_{00}F^{00} + F_{01}F^{01} + F_{02}F^{02} + F_{03}F^{03} + F_{10}F^{10} + \cdots \]

\(F_{00} = 0\). Using antisymmetry:

\[ F_{\mu\nu}F^{\mu\nu} = 2(F_{01}F^{01} + F_{02}F^{02} + F_{03}F^{03}) + 2(F_{12}F^{12} + F_{13}F^{13} + F_{23}F^{23}) \]

Mixed spacetime components: From \(F_{0i} = -E_i\), \(F^{0i} = E_i\), we get \(F_{0i}F^{0i} = -E_i \cdot E_i = -E_i^2\).

\[ 2\sum_{i} F_{0i}F^{0i} = -2\mathbf{E}^2 \]

Purely spatial components: \(F_{12} = -B_z\), \(F^{12} = -B_z\), so \(F_{12}F^{12} = B_z^2\). Similarly \(F_{13}F^{13} = B_y^2\), \(F_{23}F^{23} = B_x^2\).

\[ 2\sum_{i<j} F_{ij}F^{ij} = 2\mathbf{B}^2 \]

Therefore:

\[ F_{\mu\nu}F^{\mu\nu} = -2\mathbf{E}^2 + 2\mathbf{B}^2 \]

Thus:

\[ \boxed{\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} = -\frac{1}{4}(-2\mathbf{E}^2 + 2\mathbf{B}^2) = \frac{1}{2}(\mathbf{E}^2 - \mathbf{B}^2)} \]

Verification

Dimensional analysis: \([\mathbf{E}^2] = [\mathbf{B}^2] = \text{energy density}\), which matches the dimensions of a Lagrangian density. Furthermore, this has the \(\mathcal{L} = T - V\) structure, where \(\frac{1}{2}\mathbf{E}^2\) corresponds to the "kinetic energy" (\(\mathbf{E}\) contains \(\dot{\mathbf{A}}\)) and \(\frac{1}{2}\mathbf{B}^2\) corresponds to the "potential energy."

B-3. Invariance of Electric and Magnetic Fields under Gauge Transformations

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Invariance of the Electric Field

Substituting the gauge transformation \(A_0 \to A_0 + \partial_0 \lambda\), \(\mathbf{A} \to \mathbf{A} + \nabla\lambda\) into equation (6.3):

\[ \mathbf{E}' = -\nabla(A_0 + \partial_0\lambda) - \frac{\partial}{\partial t}(\mathbf{A} + \nabla\lambda) \]
\[ = -\nabla A_0 - \nabla(\partial_0\lambda) - \frac{\partial \mathbf{A}}{\partial t} - \frac{\partial}{\partial t}(\nabla\lambda) \]
\[ = \mathbf{E} - \nabla(\partial_0\lambda) - \partial_0(\nabla\lambda) \]

Since the order of partial derivatives can be exchanged, \(\nabla(\partial_0\lambda) = \partial_0(\nabla\lambda)\). Therefore:

\[ \mathbf{E}' = \mathbf{E} - \nabla(\partial_0\lambda) + \nabla(\partial_0\lambda) = \mathbf{E} \quad \checkmark \]

Invariance of the Magnetic Field

\[ \mathbf{B}' = \nabla \times (\mathbf{A} + \nabla\lambda) = \nabla \times \mathbf{A} + \nabla \times (\nabla\lambda) \]

For any scalar field \(\lambda\), \(\nabla \times (\nabla\lambda) = \mathbf{0}\) (the curl of a gradient is identically zero). Therefore:

\[ \mathbf{B}' = \mathbf{B} \quad \checkmark \]

B-4. Gauge Transformation Law of the Covariant Derivative

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Calculation

Write out the transformed field and covariant derivative explicitly:

\[ \psi' = e^{i\alpha(x)}\psi, \qquad A_\mu' = A_\mu - \frac{1}{q}\partial_\mu\alpha \]

Apply the transformed covariant derivative to \(\psi'\):

\[ D_\mu'\psi' = \left(\partial_\mu + iqA_\mu'\right)\psi' = \left(\partial_\mu + iq\left(A_\mu - \frac{1}{q}\partial_\mu\alpha\right)\right)(e^{i\alpha}\psi) \]
\[ = \left(\partial_\mu + iqA_\mu - i\partial_\mu\alpha\right)(e^{i\alpha}\psi) \]

Apply the product rule to \(\partial_\mu(e^{i\alpha}\psi)\):

\[ \partial_\mu(e^{i\alpha}\psi) = e^{i\alpha}(i\partial_\mu\alpha)\psi + e^{i\alpha}\partial_\mu\psi \]

Substituting this in:

\[ D_\mu'\psi' = e^{i\alpha}(i\partial_\mu\alpha)\psi + e^{i\alpha}\partial_\mu\psi + iqA_\mu \cdot e^{i\alpha}\psi - i(\partial_\mu\alpha) \cdot e^{i\alpha}\psi \]

Looking at the 1st and 4th terms:

\[ e^{i\alpha}(i\partial_\mu\alpha)\psi - i(\partial_\mu\alpha)e^{i\alpha}\psi = 0 \]

These cancel completely. The remaining terms are:

\[ D_\mu'\psi' = e^{i\alpha}\partial_\mu\psi + iqA_\mu e^{i\alpha}\psi = e^{i\alpha}(\partial_\mu\psi + iqA_\mu\psi) = e^{i\alpha}D_\mu\psi \]
\[ \boxed{D_\mu'\psi' = e^{i\alpha(x)}D_\mu\psi \quad \checkmark} \]

B-5. Calculation of Conjugate Momenta

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(a) \(\pi^0 = 0\)

Using the result from D2, \(\mathcal{L} = \frac{1}{2}(\mathbf{E}^2 - \mathbf{B}^2)\). Since \(E_i = F_{0i} = \partial_0 A_i - \partial_i A_0\), we have \(\mathbf{E}^2 = \sum_i(\partial_0 A_i - \partial_i A_0)^2\).

\(\partial_0 A_0\) does not appear in any component of \(E_i\) (since \(E_i\) contains only \(\partial_0 A_i\) and \(\partial_i A_0\)). \(\mathbf{B}^2\) contains only spatial derivatives and does not contain \(\partial_0 A_\mu\).

Therefore:

\[ \pi^0 = \frac{\partial\mathcal{L}}{\partial(\partial_0 A_0)} = 0 \quad \checkmark \]

(b) \(\pi^i = E^i\)

\[ \pi^i = \frac{\partial\mathcal{L}}{\partial(\partial_0 A_i)} \]

Since \(\mathcal{L} = \frac{1}{2}\sum_j E_j^2 - \frac{1}{2}\mathbf{B}^2\) with \(E_j = \partial_0 A_j - \partial_j A_0\), and \(\mathbf{B}^2\) does not contain \(\partial_0 A_i\):

\[ \pi^i = \frac{\partial}{\partial(\partial_0 A_i)}\left[\frac{1}{2}\sum_j(\partial_0 A_j - \partial_j A_0)^2\right] = \frac{1}{2}\cdot 2(\partial_0 A_i - \partial_i A_0) = E_i \]

Noting the metric convention, \(E^i = E_i\) (as a 3-dimensional vector), and also \(F^{0i} = E^i\).

\[ \boxed{\pi^i = F^{0i} = E^i \quad \checkmark} \]

B-6. Transversality Condition for Polarization Vectors

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Solution

For \(\mathbf{k} = k(0, \sin\theta, \cos\theta)\), we find two orthonormal vectors satisfying \(\mathbf{k} \cdot \boldsymbol{\epsilon} = 0\).

First polarization vector: The \(x\)-direction is clearly orthogonal to \(\mathbf{k}\):

\[ \boldsymbol{\epsilon}(\mathbf{k}, 1) = (1, 0, 0) \]

Verification: \(\mathbf{k} \cdot \boldsymbol{\epsilon}(\mathbf{k}, 1) = k(0\cdot1 + \sin\theta\cdot0 + \cos\theta\cdot0) = 0\) \(\checkmark\)

Second polarization vector: We find the direction orthogonal to both \(\mathbf{k}\) and \(\boldsymbol{\epsilon}(\mathbf{k}, 1)\). Computing \(\hat{\mathbf{k}} \times \boldsymbol{\epsilon}(\mathbf{k}, 1)\):

\[ \hat{\mathbf{k}} \times \boldsymbol{\epsilon}(\mathbf{k}, 1) = (0, \sin\theta, \cos\theta) \times (1, 0, 0) \]
\[ = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 0 & \sin\theta & \cos\theta \\ 1 & 0 & 0 \end{vmatrix} = (0\cdot0 - \cos\theta\cdot0,\; \cos\theta\cdot1 - 0\cdot0,\; 0\cdot0 - \sin\theta\cdot1) \]
\[ = (0, \cos\theta, -\sin\theta) \]

Normalization check: \(|(0, \cos\theta, -\sin\theta)| = \sqrt{\cos^2\theta + \sin^2\theta} = 1\) \(\checkmark\)

\[ \boldsymbol{\epsilon}(\mathbf{k}, 2) = (0, \cos\theta, -\sin\theta) \]

Verification

  • \(\mathbf{k} \cdot \boldsymbol{\epsilon}(\mathbf{k}, 2) = k(0\cdot0 + \sin\theta\cos\theta - \cos\theta\sin\theta) = 0\) \(\checkmark\)
  • \(\boldsymbol{\epsilon}(\mathbf{k}, 1) \cdot \boldsymbol{\epsilon}(\mathbf{k}, 2) = 1\cdot0 + 0\cdot\cos\theta + 0\cdot(-\sin\theta) = 0\) \(\checkmark\)
  • \(|\boldsymbol{\epsilon}(\mathbf{k}, 1)|^2 = 1\), \(|\boldsymbol{\epsilon}(\mathbf{k}, 2)|^2 = 1\) \(\checkmark\)
\[ \boxed{\boldsymbol{\epsilon}(\mathbf{k}, 1) = (1, 0, 0), \qquad \boldsymbol{\epsilon}(\mathbf{k}, 2) = (0, \cos\theta, -\sin\theta)} \]

B-7. Verification of the Bianchi Identity

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Calculation

Substitute \((\lambda, \mu, \nu) = (0, 1, 2)\):

\[ \partial_0 F_{12} + \partial_1 F_{20} + \partial_2 F_{01} = 0 \]

Substitute \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) into each term:

\[ \partial_0 F_{12} = \partial_0(\partial_1 A_2 - \partial_2 A_1) \]
\[ \partial_1 F_{20} = \partial_1(\partial_2 A_0 - \partial_0 A_2) \]
\[ \partial_2 F_{01} = \partial_2(\partial_0 A_1 - \partial_1 A_0) \]

Expanding everything:

\[ \partial_0\partial_1 A_2 - \partial_0\partial_2 A_1 + \partial_1\partial_2 A_0 - \partial_1\partial_0 A_2 + \partial_2\partial_0 A_1 - \partial_2\partial_1 A_0 \]

Using the commutativity of partial derivatives \(\partial_\mu\partial_\nu = \partial_\nu\partial_\mu\) to find pairs:

  • \(\partial_0\partial_1 A_2\) and \(-\partial_1\partial_0 A_2 = -\partial_0\partial_1 A_2\) → cancel
  • \(-\partial_0\partial_2 A_1\) and \(\partial_2\partial_0 A_1 = \partial_0\partial_2 A_1\) → cancel
  • \(\partial_1\partial_2 A_0\) and \(-\partial_2\partial_1 A_0 = -\partial_1\partial_2 A_0\) → cancel
\[ \boxed{\partial_0 F_{12} + \partial_1 F_{20} + \partial_2 F_{01} = 0 \quad \checkmark} \]

B-8. Dispersion Relation of the Photon

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Calculation

Substituting the plane wave solution \(A_i = \epsilon_i e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})}\) into the wave equation \(\Box A_i = 0\):

\[ \Box = \frac{\partial^2}{\partial t^2} - \nabla^2 \]
\[ \frac{\partial^2}{\partial t^2} e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} = (-i\omega)^2 e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} = -\omega^2 e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} \]
\[ \nabla^2 e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} = (i\mathbf{k})^2 e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} = -|\mathbf{k}|^2 e^{-i(\omega t - \mathbf{k}\cdot\mathbf{x})} \]
\[ \Box A_i = (-\omega^2 + |\mathbf{k}|^2) A_i = 0 \]

For a non-trivial solution \(A_i \neq 0\):

\[ \boxed{\omega^2 = |\mathbf{k}|^2 \quad \Longrightarrow \quad \omega = |\mathbf{k}|} \]

Comparison with the Klein-Gordon Field

The wave equation \((\Box + m^2)\phi = 0\) for a Klein-Gordon field of mass \(m\) gives the dispersion relation:

\[ \omega^2 = |\mathbf{k}|^2 + m^2 \quad \Longrightarrow \quad \omega = \sqrt{|\mathbf{k}|^2 + m^2} \]

Setting \(m = 0\) for the photon recovers \(\omega = |\mathbf{k}|\). This corresponds to the fact that the photon is a massless particle, propagating at the speed of light (\(E = pc\), or equivalently \(\omega = |\mathbf{k}|\) in natural units).

Medium

M-1. Deriving Maxwell's Equations from the Euler-Lagrange Equation

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Solution Strategy

Apply the Euler-Lagrange equation with respect to \(A_\nu\). Since \(\mathcal{L}\) does not depend on \(A_\nu\) itself, \(\partial\mathcal{L}/\partial A_\nu = 0\). We compute \(\partial\mathcal{L}/\partial(\partial_\mu A_\nu)\).

Calculation

\[ \mathcal{L} = -\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta} = -\frac{1}{4}\eta^{\alpha\gamma}\eta^{\beta\delta}F_{\alpha\beta}F_{\gamma\delta} \]

Since \(F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha\), we compute the derivative with respect to \(\partial_\mu A_\nu\):

\[ \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha \delta^\nu_\beta - \delta^\mu_\beta \delta^\nu_\alpha \]

Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}\eta^{\alpha\gamma}\eta^{\beta\delta}\left[\frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)}F_{\gamma\delta} + F_{\alpha\beta}\frac{\partial F_{\gamma\delta}}{\partial(\partial_\mu A_\nu)}\right] \]

The two terms are equal (by relabeling dummy indices), so:

\[ = -\frac{1}{2}\eta^{\alpha\gamma}\eta^{\beta\delta}(\delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha)F_{\gamma\delta} \]
\[ = -\frac{1}{2}\left(\eta^{\mu\gamma}\eta^{\nu\delta}F_{\gamma\delta} - \eta^{\nu\gamma}\eta^{\mu\delta}F_{\gamma\delta}\right) \]
\[ = -\frac{1}{2}\left(F^{\mu\nu} - F^{\nu\mu}\right) \]

Since \(F^{\mu\nu}\) is antisymmetric, \(F^{\nu\mu} = -F^{\mu\nu}\):

\[ = -\frac{1}{2}(F^{\mu\nu} + F^{\mu\nu}) = -F^{\mu\nu} \]

The Euler-Lagrange equation gives:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) = 0 \quad \Longrightarrow \quad \partial_\mu(-F^{\mu\nu}) = 0 \]
\[ \boxed{\partial_\mu F^{\mu\nu} = 0} \]

Expansion in 3-Dimensional Vector Form

Case \(\nu = 0\):

\[ \partial_\mu F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0} = 0 + \partial_i(-E^i) = -\nabla \cdot \mathbf{E} = 0 \]
\[ \boxed{\nabla \cdot \mathbf{E} = 0} \quad \text{(Gauss's law)} \]

Case \(\nu = j\):

\[ \partial_\mu F^{\mu j} = \partial_0 F^{0j} + \partial_i F^{ij} = 0 \]

Since \(F^{0j} = E^j\) and \(F^{ij} = -\epsilon^{ijk}B_k\):

\[ \partial_0 E^j + \partial_i(-\epsilon^{ijk}B_k) = 0 \]
\[ \frac{\partial E^j}{\partial t} - \epsilon^{ijk}\partial_i B_k = 0 \]

Since \((\nabla \times \mathbf{B})^j = \epsilon^{jik}\partial_i B_k = \epsilon^{ijk}\partial_i B_k\) (because \(\epsilon^{jik} = \epsilon^{ijk}\)):

\[ \frac{\partial E^j}{\partial t} = (\nabla \times \mathbf{B})^j \]
\[ \boxed{\frac{\partial \mathbf{E}}{\partial t} = \nabla \times \mathbf{B}} \quad \text{(Ampère-Maxwell law)} \]

Verification

The source-free Maxwell equations consist of four equations: \(\nabla \cdot \mathbf{E} = 0\), \(\nabla \times \mathbf{B} = \partial_t \mathbf{E}\) (from the Euler-Lagrange equations), and \(\nabla \cdot \mathbf{B} = 0\), \(\nabla \times \mathbf{E} = -\partial_t \mathbf{B}\) (from the Bianchi identity). We have confirmed that the former two are obtained from \(\partial_\mu F^{\mu\nu} = 0\) with \(\nu = 0,1,2,3\).

M-2. Counting Degrees of Freedom in Coulomb Gauge

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(a) Exclusion of \(A_0\) due to \(\pi^0 = 0\)

As shown in D5(a), \(\pi^0 = \partial\mathcal{L}/\partial\dot{A}_0 = 0\). In canonical quantization, we impose equal-time commutation relations between the fields \(A_\mu\) and their conjugate momenta \(\pi^\mu\): \([A_\mu(\mathbf{x},t), \pi^\nu(\mathbf{y},t)] = i\delta^\nu_\mu\delta^3(\mathbf{x}-\mathbf{y})\). However, since \(\pi^0 = 0\) is identically zero as an operator, we get \([A_0, \pi^0] = [A_0, 0] = 0 \neq i\delta^3(\mathbf{x}-\mathbf{y})\), which is a contradiction.

This means that \(A_0\) is not an independent dynamical degree of freedom. \(A_0\) is a dependent variable determined by other variables (a quantity fixed through constraint conditions) and is excluded from the independent canonical variables.

Degrees of freedom: \(4 \to 3\)

(b) Removal of one degree of freedom by the Coulomb gauge condition

Writing the Coulomb gauge condition \(\nabla \cdot \mathbf{A} = 0\) in Fourier space:

\[ \mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}) = 0 \]

This is one scalar condition for each \(\mathbf{k}\), which fixes the component of the 3-component vector \(\tilde{\mathbf{A}}(\mathbf{k})\) along the \(\mathbf{k}\) direction (the longitudinal component) to zero. Only the 2 components perpendicular to \(\mathbf{k}\) (transverse components) remain.

Degrees of freedom: \(3 \to 2\)

(c) Derivation of \(A_0 = 0\)

Under the Coulomb gauge \(\nabla \cdot \mathbf{A} = 0\), we write out Gauss's law \(\nabla \cdot \mathbf{E} = 0\):

\[ \nabla \cdot \mathbf{E} = \nabla \cdot \left(-\nabla A_0 - \frac{\partial \mathbf{A}}{\partial t}\right) = -\nabla^2 A_0 - \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = -\nabla^2 A_0 = 0 \]

(where we used \(\nabla \cdot \mathbf{A} = 0\).)

This is Laplace's equation \(\nabla^2 A_0 = 0\). Imposing the boundary condition \(A_0 \to 0\) at infinity, the uniqueness theorem for Laplace's equation gives:

\[ A_0 = 0 \]

(d) Summary

  • \(A_\mu\) has 4 components
  • \(\pi^0 = 0\): \(A_0\) is not a dynamical variable → 3 components
  • Coulomb gauge \(\nabla \cdot \mathbf{A} = 0\): removes the longitudinal component → 2 components
  • Gauss's law in vacuum: confirms \(A_0 = 0\)

The remaining physical degrees of freedom correspond to two transverse polarizations \(\boldsymbol{\epsilon}(\mathbf{k}, 1)\), \(\boldsymbol{\epsilon}(\mathbf{k}, 2)\). This is consistent with the experimental fact that light has two polarizations (e.g., horizontal and vertical polarization, or left and right circular polarization).

M-3. Completeness Relation for Polarization Vectors

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(a) Proof of the Completeness Relation

In 3-dimensional space, \(\hat{\mathbf{k}} = \mathbf{k}/|\mathbf{k}|\), \(\boldsymbol{\epsilon}(\mathbf{k}, 1)\), \(\boldsymbol{\epsilon}(\mathbf{k}, 2)\) form an orthonormal basis (\(\hat{\mathbf{k}}\) is orthogonal to both \(\boldsymbol{\epsilon}\)'s, and \(|\hat{\mathbf{k}}| = 1\)).

The 3-dimensional completeness relation is:

\[ \hat{k}_i \hat{k}_j + \sum_{\lambda=1}^{2}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda) = \delta_{ij} \]

Rearranging this:

\[ \sum_{\lambda=1}^{2}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda) = \delta_{ij} - \hat{k}_i\hat{k}_j = \delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2} \]
\[ \boxed{\sum_{\lambda=1}^{2}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda) = \delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2}} \]

(b) Verification of the Transverse Projection Operator

Multiplying the right-hand side by \(k_j\) and summing over \(j\):

\[ \left(\delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2}\right)k_j = k_i - \frac{k_i k_j k_j}{|\mathbf{k}|^2} = k_i - \frac{k_i |\mathbf{k}|^2}{|\mathbf{k}|^2} = k_i - k_i = 0 \]
\[ \boxed{\left(\delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2}\right)k_j = 0} \]

This shows that this operator removes the component of any vector along the \(\mathbf{k}\) direction (longitudinal component) and retains only the component perpendicular to \(\mathbf{k}\) (transverse component), confirming that it is a transverse projector.

Verification

We confirm the idempotency of the projection operator:

\[ P_{ij}^{\perp} = \delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2} \]
\[ P_{ij}^{\perp}P_{jl}^{\perp} = \left(\delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2}\right)\left(\delta_{jl} - \frac{k_j k_l}{|\mathbf{k}|^2}\right) \]
\[ = \delta_{il} - \frac{k_i k_l}{|\mathbf{k}|^2} - \frac{k_i k_l}{|\mathbf{k}|^2} + \frac{k_i k_j k_j k_l}{|\mathbf{k}|^4} = \delta_{il} - \frac{2k_i k_l}{|\mathbf{k}|^2} + \frac{k_i k_l}{|\mathbf{k}|^2} = \delta_{il} - \frac{k_i k_l}{|\mathbf{k}|^2} = P_{il}^{\perp} \]

The idempotency \(P^2 = P\) is confirmed, verifying that this is indeed a projection operator. \(\checkmark\)

M-4. Canonical Commutation Relations in Coulomb Gauge and the Transverse Delta Function

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Solution Strategy

Substitute the Fourier expansion of the field in the Coulomb gauge, and compute the equal-time commutation relation using the \([a, a^\dagger]\) commutation relations and the completeness relation for polarization vectors.

Calculation

The expansion of \(\mathbf{A}\) in the Coulomb gauge is:

\[ A_i(\mathbf{x}, t) = \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\sum_{\lambda=1}^{2}\epsilon_i(\mathbf{k},\lambda)\left[a(\mathbf{k},\lambda)e^{-i\omega_k t + i\mathbf{k}\cdot\mathbf{x}} + a^\dagger(\mathbf{k},\lambda)e^{i\omega_k t - i\mathbf{k}\cdot\mathbf{x}}\right] \]

where \(\omega_k = |\mathbf{k}|\). The conjugate momentum is \(\pi_j = \dot{A}_j\):

\[ \pi_j(\mathbf{y}, t) = \int \frac{d^3k'}{(2\pi)^3}\frac{(-i\omega_{k'})}{{\sqrt{2\omega_{k'}}}}\sum_{\lambda'=1}^{2}\epsilon_j(\mathbf{k}',\lambda')\left[a(\mathbf{k}',\lambda')e^{-i\omega_{k'} t + i\mathbf{k}'\cdot\mathbf{y}} - a^\dagger(\mathbf{k}',\lambda')e^{i\omega_{k'} t - i\mathbf{k}'\cdot\mathbf{y}}\right] \]

We compute the commutator \([A_i(\mathbf{x},t), \pi_j(\mathbf{y},t)]\). Since \([a, a] = [a^\dagger, a^\dagger] = 0\), only the terms involving \([a, a^\dagger]\) and \([a^\dagger, a]\) contribute:

\[ [A_i(\mathbf{x},t), \pi_j(\mathbf{y},t)] = \int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\int\frac{d^3k'}{(2\pi)^3}\frac{(-i\omega_{k'})}{\sqrt{2\omega_{k'}}}\sum_{\lambda,\lambda'}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k}',\lambda') \]
\[ \times \left\{[a(\mathbf{k},\lambda), a^\dagger(\mathbf{k}',\lambda')]e^{-i\omega_k t + i\mathbf{k}\cdot\mathbf{x}}e^{i\omega_{k'} t - i\mathbf{k}'\cdot\mathbf{y}}(-1)\right. \]
\[ \left. + [a^\dagger(\mathbf{k},\lambda), a(\mathbf{k}',\lambda')]e^{i\omega_k t - i\mathbf{k}\cdot\mathbf{x}}e^{-i\omega_{k'} t + i\mathbf{k}'\cdot\mathbf{y}}(+1)\right\} \]

Using \([a(\mathbf{k},\lambda), a^\dagger(\mathbf{k}',\lambda')] = (2\pi)^3\delta^3(\mathbf{k}-\mathbf{k}')\delta_{\lambda\lambda'}\) and noting that \([a^\dagger, a] = -[a, a^\dagger]\), we perform the \(\mathbf{k}'\) integration:

\[ = \int\frac{d^3k}{(2\pi)^3}\frac{(-i\omega_k)}{2\omega_k}\sum_{\lambda}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda)\left\{-e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} - e^{-i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})}\right\} \]

Here the time dependence has cancelled since \(e^{-i\omega_k t}e^{i\omega_k t} = 1\).

\[ = \int\frac{d^3k}{(2\pi)^3}\frac{(-i)}{2}\sum_{\lambda}\epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda)\left\{-e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} - e^{-i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})}\right\} \]

By the change of variables \(\mathbf{k} \to -\mathbf{k}\), the \(e^{-i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})}\) term can be converted to \(e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})}\). The completeness relation for polarization vectors \(\sum_\lambda \epsilon_i(\mathbf{k},\lambda)\epsilon_j(\mathbf{k},\lambda) = \delta_{ij} - k_ik_j/|\mathbf{k}|^2\) is invariant under \(\mathbf{k} \to -\mathbf{k}\) (since \(k_ik_j/|\mathbf{k}|^2\) is an even function of \(\mathbf{k}\)). Therefore:

\[ = \int\frac{d^3k}{(2\pi)^3}\frac{(-i)}{2}\left(\delta_{ij} - \frac{k_ik_j}{|\mathbf{k}|^2}\right)(-2)e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} \]
\[ = i\int\frac{d^3k}{(2\pi)^3}\left(\delta_{ij} - \frac{k_ik_j}{|\mathbf{k}|^2}\right)e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} \]
\[ \boxed{[A_i(\mathbf{x},t), \pi_j(\mathbf{y},t)] = i\delta_{ij}^{\perp}(\mathbf{x}-\mathbf{y})} \]

where the transverse delta function is:

\[ \delta_{ij}^{\perp}(\mathbf{x}-\mathbf{y}) = \int\frac{d^3k}{(2\pi)^3}\left(\delta_{ij} - \frac{k_ik_j}{|\mathbf{k}|^2}\right)e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} \]

Physical Reason

The reason why the transverse delta function appears instead of the usual canonical commutation relation \([A_i, \pi_j] = i\delta_{ij}\delta^3(\mathbf{x}-\mathbf{y})\):

The Coulomb gauge condition \(\nabla \cdot \mathbf{A} = 0\) identically sets the longitudinal component (the component along \(\mathbf{k}\)) of \(A_i\) to zero. Therefore, out of the 3 components of \(A_i\), only the 2 transverse components are independent dynamical variables. The commutation relation reflects this constraint, and the transverse projection operator \(\delta_{ij} - k_ik_j/|\mathbf{k}|^2\), which automatically removes the projection onto the \(\mathbf{k}\) direction, appears. If the usual \(\delta_{ij}\delta^3\) were to hold, then \(\partial_i^x [A_i, \pi_j] = i\partial_j\delta^3 \neq 0\), which would contradict \(\nabla \cdot \mathbf{A} = 0\). With the transverse delta function, \(\partial_i^x \delta_{ij}^\perp = 0\) is guaranteed, maintaining consistency.

Advanced

A-1. Comparison with the Proca Field (Massive Vector Field)

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(a) Derivation of the Equations of Motion

Proca Lagrangian:

\[ \mathcal{L}_{\text{Proca}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}m^2 A_\mu A^\mu \]

Euler-Lagrange equation:

\[ \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} - \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]

The first term is the same calculation as in S1, giving \(\partial_\mu(-F^{\mu\nu}) = -\partial_\mu F^{\mu\nu}\).

The second term: differentiating \(\frac{1}{2}m^2 A_\alpha A^\alpha = \frac{1}{2}m^2 \eta^{\alpha\beta}A_\alpha A_\beta\) with respect to \(A_\nu\):

\[ \frac{\partial}{\partial A_\nu}\left(\frac{1}{2}m^2\eta^{\alpha\beta}A_\alpha A_\beta\right) = \frac{1}{2}m^2\eta^{\alpha\beta}(\delta^\nu_\alpha A_\beta + A_\alpha\delta^\nu_\beta) = m^2\eta^{\nu\beta}A_\beta = m^2 A^\nu \]

Euler-Lagrange equation:

\[ -\partial_\mu F^{\mu\nu} - m^2 A^\nu = 0 \]
\[ \boxed{\partial_\mu F^{\mu\nu} + m^2 A^\nu = 0} \]

(b) Automatic Satisfaction of the Lorenz Condition

Acting with \(\partial_\nu\) on both sides of the equation of motion:

\[ \partial_\nu\partial_\mu F^{\mu\nu} + m^2\partial_\nu A^\nu = 0 \]

For the first term: \(\partial_\nu\partial_\mu F^{\mu\nu}\). Here \(F^{\mu\nu}\) is antisymmetric (\(F^{\mu\nu} = -F^{\nu\mu}\)), while \(\partial_\nu\partial_\mu\) is symmetric in \(\mu, \nu\). The contraction of a symmetric tensor with an antisymmetric tensor vanishes:

\[ \partial_\nu\partial_\mu F^{\mu\nu} = 0 \]

Therefore:

\[ m^2\partial_\nu A^\nu = 0 \]

If \(m \neq 0\):

\[ \boxed{\partial_\nu A^\nu = 0} \]

The Lorenz condition holds automatically as a consequence of the equation of motion. There is no need to impose it externally as a gauge-fixing condition.

(c) Counting Physical Degrees of Freedom

  • \(A_\mu\) has 4 components
  • The constraint \(\partial_\nu A^\nu = 0\) derived from the equation of motion provides 1 scalar condition
  • For \(m \neq 0\), there is no gauge symmetry (the mass term \(m^2 A_\mu A^\mu\) is not invariant under the gauge transformation \(A_\mu \to A_\mu + \partial_\mu\lambda\))
  • Since there is no gauge symmetry, no additional gauge-fixing condition is needed

Therefore the number of physical degrees of freedom is:

\[ 4 - 1 = 3 \]

This corresponds to 2 transverse polarizations + 1 longitudinal polarization. Massive spin-1 particles (e.g., \(W^\pm\), \(Z^0\) bosons) have 3 polarization states.

(d) The \(m \to 0\) Limit and the Higgs Mechanism

The \(m \to 0\) limit:

  • As \(m \to 0\), the mass term \(m^2 A_\mu A^\mu\) vanishes, and the Lagrangian recovers gauge symmetry \(A_\mu \to A_\mu + \partial_\mu\lambda\)
  • The recovery of gauge symmetry introduces a new gauge degree of freedom (\(\lambda(x)\))
  • This additional degree of freedom reduces the physical degrees of freedom from \(3 \to 2\)
  • The longitudinal polarization vector \(\epsilon^\mu_L(k) \sim k^\mu/m\) diverges as \(m \to 0\), and the longitudinal mode disappears from (decouples from) the physical spectrum

Relation to the Higgs mechanism:

The Higgs mechanism (Ch. 17 and beyond) corresponds to the reverse of this process:

  1. Starting point: massless gauge field (2 degrees of freedom) + Higgs field (scalar field)
  2. The Higgs field acquires a vacuum expectation value (spontaneous symmetry breaking)
  3. The Nambu-Goldstone mode of the Higgs field (1 degree of freedom) is "eaten" by the gauge field and becomes the longitudinal polarization component of the gauge field
  4. Result: massive vector field (3 degrees of freedom)

This is the picture in which "when a gauge boson acquires mass, the Goldstone boson is absorbed as the longitudinal polarization," and it is the core mechanism of the Standard Model that explains the origin of the masses of the \(W^\pm\) and \(Z^0\) bosons.

Consistency Check

Consistency of the degree-of-freedom counting: - Proca field (\(m \neq 0\)): 4 components − 1 constraint = 3 degrees of freedom \(\checkmark\) - Maxwell field (\(m = 0\)): 4 components − 1 constraint (\(\pi^0 = 0\)) − 1 gauge fixing = 2 degrees of freedom \(\checkmark\) - Higgs mechanism: 2 (gauge field) + 1 (Goldstone) = 3 (massive vector) \(\checkmark\)

A-2. Lagrangian with Gauge-Fixing Term and the Photon Propagator

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(a) Derivation of the Euler-Lagrange Equation

\[ \mathcal{L}_{\text{gf}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2\xi}(\partial_\mu A^\mu)^2 \]

Contribution from the first term (same as S1):

\[ \partial_\mu\frac{\partial(-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_\mu A_\nu)} = \partial_\mu F^{\mu\nu} \]

(Note on signs: since we obtained \(\partial\mathcal{L}/\partial(\partial_\mu A_\nu) = -F^{\mu\nu}\) in S1, the contribution to the first term of the Euler-Lagrange equation is \(\partial_\mu(-F^{\mu\nu})\).)

Contribution from the second term:

\[ -\frac{1}{2\xi}(\partial_\alpha A^\alpha)^2 = -\frac{1}{2\xi}\eta^{\alpha\beta}\eta^{\gamma\delta}(\partial_\alpha A_\beta)(\partial_\gamma A_\delta) \]

Differentiating with respect to \(\partial_\mu A_\nu\):

\[ \frac{\partial}{\partial(\partial_\mu A_\nu)}\left[-\frac{1}{2\xi}\eta^{\alpha\beta}\eta^{\gamma\delta}(\partial_\alpha A_\beta)(\partial_\gamma A_\delta)\right] \]
\[ = -\frac{1}{2\xi}\eta^{\alpha\beta}\eta^{\gamma\delta}\left[\delta^\mu_\alpha\delta^\nu_\beta(\partial_\gamma A_\delta) + (\partial_\alpha A_\beta)\delta^\mu_\gamma\delta^\nu_\delta\right] \]
\[ = -\frac{1}{2\xi}\left[\eta^{\mu\nu}(\partial_\gamma A^\gamma) + (\partial_\alpha A^\alpha)\eta^{\mu\nu}\right] = -\frac{1}{\xi}\eta^{\mu\nu}(\partial_\alpha A^\alpha) \]

Applying \(\partial_\mu\):

\[ \partial_\mu\left[-\frac{1}{\xi}\eta^{\mu\nu}(\partial_\alpha A^\alpha)\right] = -\frac{1}{\xi}\partial^\nu(\partial_\alpha A^\alpha) \]

Complete Euler-Lagrange equation (noting \(\partial\mathcal{L}/\partial A_\nu = 0\)):

\[ \partial_\mu F^{\mu\nu} - \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]

Substituting \(F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu\):

\[ \partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu) - \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]
\[ \Box A^\nu - \partial^\nu(\partial_\mu A^\mu) - \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]
\[ \Box A^\nu - \left(1 + \frac{1}{\xi}\right)\partial^\nu(\partial_\mu A^\mu) = 0 \]

Wait, let us redo this carefully. \(\partial_\mu F^{\mu\nu} = \partial_\mu\partial^\mu A^\nu - \partial_\mu\partial^\nu A^\mu = \Box A^\nu - \partial^\nu(\partial_\mu A^\mu)\).

Therefore:

\[ \Box A^\nu - \partial^\nu(\partial_\mu A^\mu) - \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]
\[ \Box A^\nu - \left(1 - \frac{1}{\xi}\right)\partial^\nu(\partial_\mu A^\mu) = 0 \]

Wait, let us check the signs. \(-\partial^\nu(\partial_\mu A^\mu) - \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = -(1 + \frac{1}{\xi})\partial^\nu(\partial_\mu A^\mu)\)...

Let us start over. The Euler-Lagrange equation is:

\[ \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} - \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]

First term (from \(-\frac{1}{4}F^2\)): \(\frac{\partial(-\frac{1}{4}F^2)}{\partial(\partial_\mu A_\nu)} = -F^{\mu\nu}\) (computed in S1).

Second term (from gauge-fixing term): \(\frac{\partial(-\frac{1}{2\xi}(\partial A)^2)}{\partial(\partial_\mu A_\nu)} = -\frac{1}{\xi}\eta^{\mu\nu}(\partial_\alpha A^\alpha)\).

Combining:

\[ \partial_\mu\left[-F^{\mu\nu} - \frac{1}{\xi}\eta^{\mu\nu}(\partial_\alpha A^\alpha)\right] = 0 \]
\[ -\partial_\mu F^{\mu\nu} - \frac{1}{\xi}\partial^\nu(\partial_\alpha A^\alpha) = 0 \]
\[ \partial_\mu F^{\mu\nu} + \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]

Substituting \(\partial_\mu F^{\mu\nu} = \Box A^\nu - \partial^\nu(\partial_\mu A^\mu)\):

\[ \Box A^\nu - \partial^\nu(\partial_\mu A^\mu) + \frac{1}{\xi}\partial^\nu(\partial_\mu A^\mu) = 0 \]
\[ \Box A^\nu - \left(1 - \frac{1}{\xi}\right)\partial^\nu(\partial_\mu A^\mu) = 0 \]

Rearranging indices and rewriting with \(\nu\) as a lower index (using \(\eta^{\mu\nu}\)):

\[ \left[\eta^{\mu\nu}\Box - \left(1 - \frac{1}{\xi}\right)\partial^\mu\partial^\nu\right]A_\nu = 0 \]
\[ \boxed{\left[\eta^{\mu\nu}\Box - \left(1 - \frac{1}{\xi}\right)\partial^\mu\partial^\nu\right]A_\nu = 0} \]

(b) Transformation to Momentum Space

Substituting \(A_\nu(x) = \int\frac{d^4k}{(2\pi)^4}\tilde{A}_\nu(k)e^{-ikx}\). Since \(\partial_\mu \to -ik_\mu\):

\[ \Box \to -k^2, \qquad \partial^\mu\partial^\nu \to -k^\mu(-ik^\nu) = -k^\mu k^\nu \]

Actually, since \(\partial_\mu e^{-ikx} = -ik_\mu e^{-ikx}\), for \(\partial^\mu = \eta^{\mu\nu}\partial_\nu\) we have \(\partial^\mu e^{-ikx} = -ik^\mu e^{-ikx}\).

For \(\Box = \partial_\mu\partial^\mu\): \(\Box e^{-ikx} = (-ik_\mu)(-ik^\mu)e^{-ikx} = -k^2 e^{-ikx}\) (where \(k^2 = k_\mu k^\mu\)).

For \(\partial^\mu\partial^\nu e^{-ikx} = (-ik^\mu)(-ik^\nu)e^{-ikx} = -k^\mu k^\nu e^{-ikx}\).

The equation of motion becomes:

\[ \left[-k^2\eta^{\mu\nu} - \left(1 - \frac{1}{\xi}\right)(-k^\mu k^\nu)\right]\tilde{A}_\nu = 0 \]
\[ \boxed{\left[-k^2\eta^{\mu\nu} + \left(1 - \frac{1}{\xi}\right)k^\mu k^\nu\right]\tilde{A}_\nu = 0} \]

(c) Derivation of the Photon Propagator

The propagator \(D_F^{\nu\rho}(k)\) is defined as the inverse of the differential operator:

\[ \left[-k^2\eta^{\mu\nu} + \left(1 - \frac{1}{\xi}\right)k^\mu k^\nu\right]D_{F\,\nu\rho}(k) = \eta^\mu{}_\rho = \delta^\mu_\rho \]

We assume \(D_F^{\nu\rho}\) as a linear combination of two independent tensor structures:

\[ D_{F\,\nu\rho}(k) = A\,\eta_{\nu\rho} + B\,\frac{k_\nu k_\rho}{k^2} \]

Substituting into the left-hand side:

\[ \left[-k^2\eta^{\mu\nu} + \left(1 - \frac{1}{\xi}\right)k^\mu k^\nu\right]\left[A\,\eta_{\nu\rho} + B\,\frac{k_\nu k_\rho}{k^2}\right] \]

First term:

\[ -k^2\eta^{\mu\nu}\left[A\,\eta_{\nu\rho} + B\,\frac{k_\nu k_\rho}{k^2}\right] = -k^2 A\,\delta^\mu_\rho - B\,k^\mu k_\rho \]

Second term:

\[ \left(1 - \frac{1}{\xi}\right)k^\mu k^\nu\left[A\,\eta_{\nu\rho} + B\,\frac{k_\nu k_\rho}{k^2}\right] = \left(1 - \frac{1}{\xi}\right)\left[A\,k^\mu k_\rho + B\,\frac{k^2 k^\mu k_\rho}{k^2}\right] \]
\[ = \left(1 - \frac{1}{\xi}\right)(A + B)k^\mu k_\rho \]

In total:

\[ -k^2 A\,\delta^\mu_\rho + \left[-B + \left(1 - \frac{1}{\xi}\right)(A + B)\right]k^\mu k_\rho = \delta^\mu_\rho \]

Equating the coefficients of \(\delta^\mu_\rho\) and \(k^\mu k_\rho\) separately:

Coefficient of \(\delta^\mu_\rho\):

\[ -k^2 A = 1 \quad \Longrightarrow \quad A = -\frac{1}{k^2} \]

Coefficient of \(k^\mu k_\rho\):

\[ -B + \left(1 - \frac{1}{\xi}\right)(A + B) = 0 \]
\[ -B + \left(1 - \frac{1}{\xi}\right)A + \left(1 - \frac{1}{\xi}\right)B = 0 \]
\[ B\left[-1 + 1 - \frac{1}{\xi}\right] + \left(1 - \frac{1}{\xi}\right)A = 0 \]
\[ -\frac{B}{\xi} + \left(1 - \frac{1}{\xi}\right)\left(-\frac{1}{k^2}\right) = 0 \]
\[ -\frac{B}{\xi} = \frac{1}{k^2}\left(1 - \frac{1}{\xi}\right) = \frac{\xi - 1}{\xi k^2} \]
\[ B = -\frac{\xi - 1}{k^2} = \frac{1 - \xi}{k^2} \]

Therefore:

\[ D_{F\,\nu\rho}(k) = -\frac{1}{k^2}\eta_{\nu\rho} + \frac{1-\xi}{k^2}\cdot\frac{k_\nu k_\rho}{k^2} = \frac{-1}{k^2}\left[\eta_{\nu\rho} - (1-\xi)\frac{k_\nu k_\rho}{k^2}\right] \]

Adding the \(i\epsilon\) prescription:

\[ \boxed{D_F^{\mu\nu}(k) = \frac{-1}{k^2 + i\epsilon}\left[\eta^{\mu\nu} - (1-\xi)\frac{k^\mu k^\nu}{k^2}\right]} \]

Feynman gauge \(\xi = 1\):

\[ D_F^{\mu\nu}(k) = \frac{-\eta^{\mu\nu}}{k^2 + i\epsilon} \]

This gives the simplest form. \(\checkmark\)

(d) Landau Gauge \(\xi = 0\)

Substituting \(\xi = 0\):

\[ D_F^{\mu\nu}(k)\big|_{\xi=0} = \frac{-1}{k^2 + i\epsilon}\left[\eta^{\mu\nu} - \frac{k^\mu k^\nu}{k^2}\right] \]

Contracting with \(k_\mu\):

\[ k_\mu D_F^{\mu\nu}(k) = \frac{-1}{k^2 + i\epsilon}\left[k^\nu - \frac{k^2 k^\nu}{k^2}\right] = \frac{-1}{k^2 + i\epsilon}\left[k^\nu - k^\nu\right] = 0 \]
\[ \boxed{k_\mu D_F^{\mu\nu}(k)\big|_{\xi=0} = 0 \quad \checkmark} \]

Physical meaning: In momentum space, \(k_\mu D_F^{\mu\nu} = 0\) means that the propagator has no component in the \(k_\mu\) direction. This is a reflection of the Lorenz condition \(\partial_\mu A^\mu = 0\) in coordinate space (which becomes \(k_\mu \tilde{A}^\mu = 0\) under Fourier transform). In the Landau gauge, the propagator itself has the structure of a transverse projection, and the Lorenz condition is automatically realized at the level of the propagator.

Verification

Feynman gauge check: For \(\xi = 1\), the differential operator becomes \(-k^2\eta^{\mu\nu}\) (the \(k^\mu k^\nu\) term vanishes), and its inverse is obviously \(-\eta_{\mu\nu}/k^2\). \(\checkmark\)

Tensor structure check: Computing the trace of \(D_F^{\mu\nu}\) for general \(\xi\):

\[ \eta_{\mu\nu}D_F^{\mu\nu} = \frac{-1}{k^2}\left[4 - (1-\xi)\frac{k^2}{k^2}\right] = \frac{-1}{k^2}(4 - 1 + \xi) = \frac{-(3+\xi)}{k^2} \]

\(\xi = 1\): trace \(= -4/k^2\) (\(-\eta_{\mu\nu}\eta^{\mu\nu}/k^2 = -4/k^2\)) \(\checkmark\)

Gauge invariance check: Physical scattering amplitudes are independent of \(\xi\) (due to the Ward identity). The \(k^\mu k^\nu\) term in the propagator vanishes when external lines are on-shell, due to the coupling with conserved currents \(k_\mu J^\mu = 0\). This is discussed in detail from Ch. 10 onward.