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Appendix B Solutions

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Basic

Medium

Advanced

Basic

B-1. Antisymmetry of Infinitesimal Lorentz Transformations

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Solution Strategy

Substitute the infinitesimal Lorentz transformation \(\Lambda^\mu{}_\nu = \delta^\mu{}_\nu + \omega^\mu{}_\nu\) into the metric preservation condition and keep terms up to first order in \(\omega\).

Detailed Calculation

The metric preservation condition is

\[ \Lambda^\mu{}_\alpha\,\Lambda^\nu{}_\beta\,\eta^{\alpha\beta} = \eta^{\mu\nu} \]

Substituting \(\Lambda^\mu{}_\alpha = \delta^\mu{}_\alpha + \omega^\mu{}_\alpha\) into the left-hand side:

\[ (\delta^\mu{}_\alpha + \omega^\mu{}_\alpha)(\delta^\nu{}_\beta + \omega^\nu{}_\beta)\,\eta^{\alpha\beta} \]

Expanding gives

\[ = \delta^\mu{}_\alpha\,\delta^\nu{}_\beta\,\eta^{\alpha\beta} + \omega^\mu{}_\alpha\,\delta^\nu{}_\beta\,\eta^{\alpha\beta} + \delta^\mu{}_\alpha\,\omega^\nu{}_\beta\,\eta^{\alpha\beta} + \omega^\mu{}_\alpha\,\omega^\nu{}_\beta\,\eta^{\alpha\beta} \]

Discarding the second-order term in \(\omega\):

\[ \approx \eta^{\mu\nu} + \omega^\mu{}_\alpha\,\eta^{\alpha\nu} + \omega^\nu{}_\beta\,\eta^{\mu\beta} \]

Here we used \(\omega^\mu{}_\alpha\,\eta^{\alpha\nu} = \omega^{\mu\nu}\) (the index-raising operation). Similarly, \(\omega^\nu{}_\beta\,\eta^{\mu\beta} = \omega^{\nu\mu}\).

Therefore

\[ \eta^{\mu\nu} + \omega^{\mu\nu} + \omega^{\nu\mu} = \eta^{\mu\nu} \]

Final Answer

\[ \boxed{\omega^{\mu\nu} + \omega^{\nu\mu} = 0} \]

That is, \(\omega^{\mu\nu}\) is antisymmetric. Writing this with lowered indices as \(\omega_{\mu\nu} = \eta_{\mu\alpha}\omega^{\alpha}{}_\nu\), the relation \(\omega_{\mu\nu} + \omega_{\nu\mu} = 0\) likewise holds.

Verification

The number of independent components of a \(4 \times 4\) antisymmetric matrix is \(\frac{4 \times 3}{2} = 6\). This matches the number of parameters of the Lorentz group (3 rotations + 3 boosts = 6). ✓

B-2. Verifying Matrix Elements of Generators

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Solution Strategy

Set \(\rho = 2, \sigma = 3\) and compute equation (B.10) for each component.

Detailed Calculation

\[ (M^{[23]})^\mu{}_\nu = \eta^{2\mu}\,\delta^3{}_\nu - \eta^{3\mu}\,\delta^3{}_\nu \quad \text{rather} \quad \eta^{2\mu}\,\delta^3{}_\nu - \eta^{3\mu}\,\delta^2{}_\nu \]

\(\eta^{2\mu}\) equals \(+1\) when \(\mu = 2\), and \(0\) otherwise. \(\eta^{3\mu}\) equals \(+1\) when \(\mu = 3\), and \(0\) otherwise.

Computing each component:

  • \((M^{[23]})^0{}_\nu = \eta^{20}\delta^3{}_\nu - \eta^{30}\delta^2{}_\nu = 0 - 0 = 0\) (for all \(\nu\))
  • \((M^{[23]})^1{}_\nu = \eta^{21}\delta^3{}_\nu - \eta^{31}\delta^2{}_\nu = 0 - 0 = 0\) (for all \(\nu\))
  • \((M^{[23]})^2{}_\nu = \eta^{22}\delta^3{}_\nu - \eta^{32}\delta^2{}_\nu = (+1)\delta^3{}_\nu - 0 = \delta^3{}_\nu\)
  • Equals \(+1\) when \(\nu = 3\), and \(0\) otherwise
  • \((M^{[23]})^3{}_\nu = \eta^{23}\delta^3{}_\nu - \eta^{33}\delta^2{}_\nu = 0 - (+1)\delta^2{}_\nu = -\delta^2{}_\nu\)
  • Equals \(-1\) when \(\nu = 2\), and \(0\) otherwise

Final Answer

\[ \boxed{M^{[23]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}} \]

Rows correspond to \(\mu = 0,1,2,3\) and columns to \(\nu = 0,1,2,3\).

Verification

This is the rotation generator in the \(yz\)-plane, where only the \(y\) and \(z\) components mix. The \(x^0, x^1\) components are invariant, and the \(x^2, x^3\) block gives the standard 2-dimensional rotation generator \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\). ✓

B-3. Matrix Representation of Boost Generator \(K^2 = M^{[02]}\)

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Solution Strategy

Set \(\rho = 0, \sigma = 2\) in equation (B.10) to find the matrix and verify the infinitesimal transformation.

Detailed Calculation

\[ (M^{[02]})^\mu{}_\nu = \eta^{0\mu}\,\delta^2{}_\nu - \eta^{2\mu}\,\delta^0{}_\nu \]

\(\eta^{0\mu}\) is \(-1\) when \(\mu = 0\), and \(0\) otherwise. \(\eta^{2\mu}\) is \(+1\) when \(\mu = 2\), and \(0\) otherwise.

Each component:

  • \((M^{[02]})^0{}_\nu = (-1)\delta^2{}_\nu - 0 = -\delta^2{}_\nu\)
  • Equals \(-1\) when \(\nu = 2\)
  • \((M^{[02]})^1{}_\nu = 0 - 0 = 0\) (for all \(\nu\))
  • \((M^{[02]})^2{}_\nu = 0 - (+1)\delta^0{}_\nu = -\delta^0{}_\nu\)
  • Equals \(-1\) when \(\nu = 0\)
  • \((M^{[02]})^3{}_\nu = 0 - 0 = 0\) (for all \(\nu\))

Therefore

\[ M^{[02]} = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Writing the infinitesimal transformation \(x'^\mu = (\delta^\mu{}_\nu + \phi\,(M^{[02]})^\mu{}_\nu)\,x^\nu\) in components:

\[ x'^0 = x^0 + \phi \cdot (-1) \cdot x^2 = t - \phi\, y \]
\[ x'^1 = x^1 \]
\[ x'^2 = x^2 + \phi \cdot (-1) \cdot x^0 = y - \phi\, t \]
\[ x'^3 = x^3 \]

Final Answer

\[ \boxed{M^{[02]} = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}} \]

The infinitesimal transformation gives \(t' \approx t - \phi\,y\), \(y' \approx y - \phi\,t\) (\(x, z\) are unchanged). ✓

Verification

For a finite boost, \(t' = t\cosh\phi - y\sinh\phi\), \(y' = y\cosh\phi - t\sinh\phi\). Taking \(\phi \ll 1\) with \(\cosh\phi \approx 1\), \(\sinh\phi \approx \phi\) reproduces the infinitesimal transformation above. ✓

B-4. Direct Calculation of Commutation Relations for Rotation Generators

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Solution Strategy

We directly compute \([J^3, J^1]\) using the \(4\times4\) matrices of \(J^3 = M^{[12]}\), \(J^1 = M^{[23]}\), \(J^2 = M^{[31]}\).

Detailed Calculation

From D2, \(J^1 = M^{[23]}\):

\[ J^1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \]

From equation (B.11) in the text, \(J^3 = M^{[12]}\):

\[ J^3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

We compute \(M^{[31]}\) using equation (B.10) with \(\rho = 3, \sigma = 1\):

\[ (M^{[31]})^\mu{}_\nu = \eta^{3\mu}\delta^1{}_\nu - \eta^{1\mu}\delta^3{}_\nu \]
  • \(\mu = 1\): \(0 - (+1)\delta^3{}_\nu = -\delta^3{}_\nu\) → \(-1\) at \(\nu = 3\)
  • \(\mu = 3\): \((+1)\delta^1{}_\nu - 0 = \delta^1{}_\nu\) → \(+1\) at \(\nu = 1\)
  • All others are \(0\)
\[ J^2 = M^{[31]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \]

We compute the matrix product \(J^3 \cdot J^1\). The non-zero columns of \(J^1\) are the \(\nu = 2\) column \((0,0,0,-1)^T\) and the \(\nu = 3\) column \((0,0,1,0)^T\).

\(J^3 \cdot J^1\): rows of \(J^3\) \(\times\) columns of \(J^1\)

\[ J^3 \cdot J^1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \]

Row 1: \((0,0,0,0) \cdot \text{each column} = (0,0,0,0)\)

Row 2: \((0,0,1,0)\) - Row 2 \(\times\) column 1: \(0\) - Row 2 \(\times\) column 2: \(0\) - Row 2 \(\times\) column 3: \(0 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 + 0 \cdot (-1) = 0\) - Row 2 \(\times\) column 4: \(0 + 0 + 1 \cdot 1 + 0 = 1\)

Row 3: \((0,-1,0,0)\) - All \(0\) (since rows 1 and 2 of \(J^1\) are all zeros)

Row 4: \((0,0,0,0) \cdot \text{each column} = (0,0,0,0)\)

\[ J^3 \cdot J^1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
\[ J^1 \cdot J^3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Row 1: all \(0\)

Row 2: all \(0\) (row 2 of \(J^1\) is all zeros)

Row 3: \((0,0,0,1)\) - Row 3 \(\times\) each column: row 4 of \(J^3\) = \((0,0,0,0)\), so all \(0\)

Row 4: \((0,0,-1,0)\) - Row 4 \(\times\) column 1: \(0\) - Row 4 \(\times\) column 2: \((-1)(-1) = 1\) - Row 4 \(\times\) column 3: \((-1)(0) = 0\) - Row 4 \(\times\) column 4: \(0\)

\[ J^1 \cdot J^3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \]

Therefore

\[ [J^3, J^1] = J^3 J^1 - J^1 J^3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \]

On the other hand, we compute \(iJ^2\):

\[ iJ^2 = i\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \end{pmatrix} \]

Note: A problem arises here. \([J^3, J^1]\) is a matrix with real components, while \(iJ^2\) contains imaginary components.

This is a matter of the conventions in the text. The \(M^{[\rho\sigma]}\) defined by equation (B.10) in the text are used with the convention of equation (B.14): \(\Lambda = \exp(i\boldsymbol{\theta}\cdot\mathbf{J} + i\boldsymbol{\phi}\cdot\mathbf{K})\). In the four-vector representation, the generators are real matrices rather than purely imaginary, and the commutation relations hold in the form

\[ [J^i, J^j] = i\varepsilon^{ijk}J^k \]

Let us verify this. The computed result is:

\[ [J^3, J^1] = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \]

And

\[ iJ^2 = i \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \]

These do not match. In fact, the commutation relations for generators in the four-vector representation, in the convention \(\Lambda = e^{\omega_{\rho\sigma}\mathcal{J}^{\rho\sigma}/2}\) (without \(i\)), take the form

\[ [\mathcal{J}^i, \mathcal{J}^j] = \varepsilon^{ijk}\mathcal{J}^k \]

In the text's convention \(\Lambda = e^{i\theta_i J^i + i\phi_i K^i}\), the generators in the four-vector representation become \(J^i_{\text{vec}} = -iM^{[jk]}_{\text{text}}\) (with appropriate correspondence).

In practice, using the matrices from equation (B.10) directly, we should verify the commutation relation in the form without \(i\):

\[ [M^{[12]}, M^{[23]}] = M^{[13]} \]

Let us compute \(M^{[13]}\) (\(\rho=1, \sigma=3\)):

\[ (M^{[13]})^\mu{}_\nu = \eta^{1\mu}\delta^3{}_\nu - \eta^{3\mu}\delta^1{}_\nu \]
  • \(\mu=1\): \((+1)\delta^3{}_\nu - 0 = \delta^3{}_\nu\) → \(+1\) at \(\nu=3\)
  • \(\mu=3\): \(0 - (+1)\delta^1{}_\nu = -\delta^1{}_\nu\) → \(-1\) at \(\nu=1\)
\[ M^{[13]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \]

Final Result

\[ [M^{[12]}, M^{[23]}] = M^{[13]} \]

Translating this into the language of \(J^i\): since \(J^3 = M^{[12]}\), \(J^1 = M^{[23]}\), \(J^2 = M^{[31]} = -M^{[13]}\), we have

\[ [J^3, J^1] = M^{[13]} = -M^{[31]} = -(-J^2) \]

Wait, since \(J^2 = M^{[31]}\), we get \(M^{[13]} = -M^{[31]} = -J^2\)...

Let us reorganize. Since \(M^{[31]} = -M^{[13]}\), we have \(J^2 = M^{[31]} = -M^{[13]}\).

Therefore \(M^{[13]} = -J^2\), and \([J^3, J^1] = M^{[13]} = -J^2\)?

This contradicts \([J^3, J^1] = iJ^2\). The issue is whether the generator matrices in the four-vector representation include a factor of \(i\).

Correct correspondence: In the text's convention \(\Lambda = e^{i\boldsymbol{\theta}\cdot\mathbf{J}}\), the matrix representation of \(\mathbf{J}\) in the four-vector representation is

\[ (J^i)^\mu{}_\nu = -\frac{i}{2}\varepsilon^{ijk}(M^{[jk]})^\mu{}_\nu \]

rather than the \(M^{[jk]}\) matrices themselves. However, since the problem statement says "using the \(4\times4\) matrix representation (B.11)," the text uses the \(M^{[\rho\sigma]}\) matrices themselves as \(J^i\) (with \(J^3 = M^{[12]}\), etc.).

In this case, the commutation relation for the matrices holds in the form without \(i\):

\[ [M^{[12]}, M^{[23]}] = M^{[13]} \]

where \(M^{[13]} = -M^{[31]}\).

The simplest interpretation: using the matrix from (B.11) as \(\mathcal{J}^3\) and the matrix from D2 as \(\mathcal{J}^1\), the commutation relation holds as

\[ [\mathcal{J}^3, \mathcal{J}^1] = \mathcal{J}^2 \]

(without \(i\)), where \(\mathcal{J}^2 = M^{[13]}\) (\(= -M^{[31]}\)).

The actual computation confirms:

\[ \boxed{[M^{[12]}, M^{[23]}] = M^{[13]} = -M^{[31]}} \]

This is consistent with the Lorentz algebra commutation relation \([J^3, J^1] = iJ^2\) in the four-vector representation (once one accounts for the factor of \(-i\) contained in the generator matrices of the four-vector representation).

Verification (Alternative Method)

The non-zero components of \(M^{[13]}\) are \(+1\) at position \((1,3)\) and \(-1\) at position \((3,1)\). This is the matrix generating rotations in the \(xz\)-plane, which is consistent with the physical expectation that \([J^3, J^1]\) should be proportional to \(J^2\) (rotation in the \(xz\)-plane). ✓

B-5. Recovering Rotation Generators Using the Levi-Civita Symbol

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Strategy

Expand \(J^i = \frac{1}{2}\varepsilon^{ijk}M^{[jk]}\) for \(i = 1, 2, 3\).

Detailed Calculation

\(J^1\):

\[ J^1 = \frac{1}{2}\varepsilon^{1jk}M^{[jk]} \]

\(\varepsilon^{1jk} \neq 0\) for \((j,k) = (2,3)\) and \((3,2)\): - \(\varepsilon^{123} = +1\) - \(\varepsilon^{132} = -1\)

\[ J^1 = \frac{1}{2}\left(\varepsilon^{123}M^{[23]} + \varepsilon^{132}M^{[32]}\right) = \frac{1}{2}\left(M^{[23]} + (-1)(-M^{[23]})\right) = \frac{1}{2}(M^{[23]} + M^{[23]}) = M^{[23]} \]

\(J^2\):

\[ J^2 = \frac{1}{2}\varepsilon^{2jk}M^{[jk]} \]

\(\varepsilon^{2jk} \neq 0\) for \((j,k) = (3,1)\) and \((1,3)\): - \(\varepsilon^{231} = +1\) - \(\varepsilon^{213} = -1\)

\[ J^2 = \frac{1}{2}\left(\varepsilon^{231}M^{[31]} + \varepsilon^{213}M^{[13]}\right) = \frac{1}{2}\left(M^{[31]} + (-1)(-M^{[31]})\right) = M^{[31]} \]

\(J^3\):

\[ J^3 = \frac{1}{2}\varepsilon^{3jk}M^{[jk]} \]

\(\varepsilon^{3jk} \neq 0\) for \((j,k) = (1,2)\) and \((2,1)\): - \(\varepsilon^{312} = +1\) - \(\varepsilon^{321} = -1\)

\[ J^3 = \frac{1}{2}\left(\varepsilon^{312}M^{[12]} + \varepsilon^{321}M^{[21]}\right) = \frac{1}{2}\left(M^{[12]} + (-1)(-M^{[12]})\right) = M^{[12]} \]

Final Answer

\[ \boxed{J^1 = M^{[23]}, \quad J^2 = M^{[31]}, \quad J^3 = M^{[12]}} \]

Verification

\(J^3 = M^{[12]}\) is the generator of rotations in the \(xy\) plane, corresponding to rotation about the \(z\)-axis. This is physically correct. ✓

B-6. Expressing \(\mathbf{J}\) and \(\mathbf{K}\) in terms of \(\mathbf{J}_+\) and \(\mathbf{J}_-\)

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Calculation Details

Restating equation (B.18):

\[ \mathbf{J}_+ = \frac{\mathbf{J} + i\mathbf{K}}{2}, \qquad \mathbf{J}_- = \frac{\mathbf{J} - i\mathbf{K}}{2} \]

Adding the two equations:

\[ \mathbf{J}_+ + \mathbf{J}_- = \frac{\mathbf{J} + i\mathbf{K}}{2} + \frac{\mathbf{J} - i\mathbf{K}}{2} = \mathbf{J} \]

Subtracting the two equations:

\[ \mathbf{J}_+ - \mathbf{J}_- = \frac{\mathbf{J} + i\mathbf{K}}{2} - \frac{\mathbf{J} - i\mathbf{K}}{2} = i\mathbf{K} \]

Final Answer

\[ \boxed{\mathbf{J} = \mathbf{J}_+ + \mathbf{J}_-, \qquad \mathbf{K} = -i(\mathbf{J}_+ - \mathbf{J}_-)} \]

Verification

Substituting \(\mathbf{K} = -i(\mathbf{J}_+ - \mathbf{J}_-)\) back into the original definition to confirm: \(\frac{\mathbf{J} + i\mathbf{K}}{2} = \frac{(\mathbf{J}_+ + \mathbf{J}_-) + i(-i)(\mathbf{J}_+ - \mathbf{J}_-)}{2} = \frac{(\mathbf{J}_+ + \mathbf{J}_-) + (\mathbf{J}_+ - \mathbf{J}_-)}{2} = \mathbf{J}_+\). ✓

B-7. Calculating Dimensions of Representations

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Final Answer

The dimension of the representation \((j_+, j_-)\) is \((2j_+ + 1)(2j_- + 1)\).

(a) \((1, 0)\): \((2 \cdot 1 + 1)(2 \cdot 0 + 1) = 3 \times 1 = \boxed{3}\)

(b) \((1, 1)\): \((2 \cdot 1 + 1)(2 \cdot 1 + 1) = 3 \times 3 = \boxed{9}\)

(c) \((3/2, 0)\): \((2 \cdot \frac{3}{2} + 1)(2 \cdot 0 + 1) = 4 \times 1 = \boxed{4}\)

(d) \((1/2, 1)\): \((2 \cdot \frac{1}{2} + 1)(2 \cdot 1 + 1) = 2 \times 3 = \boxed{6}\)

Verification

(a) corresponds to a self-dual antisymmetric tensor (3 components). (b) corresponds to part of a symmetric traceless rank-2 tensor (9 components). (c) is related to part of a Rarita-Schwinger field (4 components). (d) has 6 components and corresponds to part of a vector-spinor. The dimension counting is consistent. ✓

B-8. Verification of \([K^1, K^2] = -iJ^3\)

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Solution Strategy

Compute the product of the 4×4 matrices for \(K^1 = M^{[01]}\) and \(K^2 = M^{[02]}\).

Detailed Calculation

From equation (B.12):

\[ M^{[01]} = \begin{pmatrix} 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

From D3:

\[ M^{[02]} = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

\(M^{[01]} \cdot M^{[02]}\):

Row 1 \((0,-1,0,0)\) × each column: - Column 1: \(0 + 0 + 0 + 0 = 0\) - Column 2: \(0 + 0 + 0 + 0 = 0\) - Column 3: \(0 + 0 + 0 + 0 = 0\) - Column 4: \(0\)

→ \((0, 0, 0, 0)\)

Row 2 \((-1,0,0,0)\) × each column: - Column 1: \((-1)(0) = 0\) - Column 2: \((-1)(0) = 0\) - Column 3: \((-1)(-1) = 1\) - Column 4: \(0\)

→ \((0, 0, 1, 0)\)

Row 3 \((0,0,0,0)\) → \((0,0,0,0)\)

Row 4 \((0,0,0,0)\) → \((0,0,0,0)\)

\[ M^{[01]} \cdot M^{[02]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

\(M^{[02]} \cdot M^{[01]}\):

Row 1 \((0,0,-1,0)\) × each column: - Column 1: \(0 + 0 + 0 + 0 = 0\) (Row 3 of \(M^{[01]}\) is all zeros) - Column 2: \((-1)(0) = 0\) - Column 3: \(0\) - Column 4: \(0\)

→ \((0, 0, 0, 0)\)

Row 2 \((0,0,0,0)\) → \((0,0,0,0)\)

Row 3 \((-1,0,0,0)\) × each column: - Column 1: \((-1)(0) = 0\) - Column 2: \((-1)(-1) = 1\) - Column 3: \(0\) - Column 4: \(0\)

→ \((0, 1, 0, 0)\)

Row 4 \((0,0,0,0)\) → \((0,0,0,0)\)

\[ M^{[02]} \cdot M^{[01]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Commutator:

\[ [M^{[01]}, M^{[02]}] = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Comparing this with \(M^{[12]}\) (equation (B.11)):

\[ M^{[12]} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Therefore

\[ [M^{[01]}, M^{[02]}] = M^{[12]} \]

Final Answer

\[ \boxed{[K^1, K^2] = [M^{[01]}, M^{[02]}] = M^{[12]} = J^3} \]

Regarding consistency with the abstract commutation relation \([K^i, K^j] = -i\varepsilon^{ijk}J^k\): \([K^1, K^2] = -i\varepsilon^{123}J^3 = -iJ^3\).

In the four-vector representation, with the convention \(\Lambda = e^{i\phi_i K^i}\), the matrix representation of the generators is \(K^i_{\text{rep}} = -iM^{[0i]}\). Then

\[ [K^1_{\text{rep}}, K^2_{\text{rep}}] = [-iM^{[01]}, -iM^{[02]}] = -[M^{[01]}, M^{[02]}] = -M^{[12]} \]

On the other hand, \(-iJ^3_{\text{rep}} = -i(-iM^{[12]}) = -M^{[12]}\). ✓

This confirms that \([K^1, K^2] = -iJ^3\) is correctly realized in the four-vector representation.

Verification

Physically, performing successive boosts in the \(x\)-direction and \(y\)-direction produces a rotation in the \(xy\)-plane (Thomas precession). The relation \([K^1, K^2] \propto J^3\) precisely reflects this effect. ✓

Medium

M-1. Complete Derivation of \([J^i_+, J^j_-] = 0\)

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Strategy

Substitute definition (B.18), expand, and show that all terms cancel using the commutation relations of the Lorentz algebra.

Detailed Calculation

\[ [J^i_+, J^j_-] = \left[\frac{J^i + iK^i}{2},\, \frac{J^j - iK^j}{2}\right] \]
\[ = \frac{1}{4}\left[J^i + iK^i,\, J^j - iK^j\right] \]

Expanding using the linearity of the commutator:

\[ = \frac{1}{4}\left([J^i, J^j] - i[J^i, K^j] + i[K^i, J^j] - i^2[K^i, K^j]\right) \]
\[ = \frac{1}{4}\left([J^i, J^j] - i[J^i, K^j] + i[K^i, J^j] + [K^i, K^j]\right) \]

Substituting the commutation relations (B.15)–(B.17) into each term:

Term 1: \([J^i, J^j] = i\varepsilon^{ijk}J^k\)

Term 2: \(-i[J^i, K^j] = -i \cdot i\varepsilon^{ijk}K^k = \varepsilon^{ijk}K^k\)

Term 3: \(i[K^i, J^j]\)

\([K^i, J^j] = -[J^j, K^i]\). Substituting \(i \to j, j \to i\) in equation (B.16) gives \([J^j, K^i] = i\varepsilon^{jik}K^k\).

Since \(\varepsilon^{jik} = -\varepsilon^{ijk}\), we have \([J^j, K^i] = -i\varepsilon^{ijk}K^k\).

Therefore \([K^i, J^j] = -[J^j, K^i] = i\varepsilon^{ijk}K^k\).

Thus \(i[K^i, J^j] = i \cdot i\varepsilon^{ijk}K^k = -\varepsilon^{ijk}K^k\).

Term 4: \([K^i, K^j] = -i\varepsilon^{ijk}J^k\)

Combining everything:

\[ [J^i_+, J^j_-] = \frac{1}{4}\left(i\varepsilon^{ijk}J^k + \varepsilon^{ijk}K^k - \varepsilon^{ijk}K^k - i\varepsilon^{ijk}J^k\right) \]
\[ = \frac{1}{4}\left(i\varepsilon^{ijk}J^k - i\varepsilon^{ijk}J^k + \varepsilon^{ijk}K^k - \varepsilon^{ijk}K^k\right) \]

Final Answer

\[ \boxed{[J^i_+, J^j_-] = \frac{1}{4}(0 + 0) = 0} \]

Terms 1 and 4 cancel, and terms 2 and 3 cancel.

Verification

This result is the essential condition for the decomposition of the Lorentz algebra into \(\mathfrak{su}(2)_+ \oplus \mathfrak{su}(2)_-\). The fact that \(\mathbf{J}_+\) and \(\mathbf{J}_-\) commute with each other means that the two \(\mathfrak{su}(2)\) algebras act independently, which provides the basis for classifying representations by pairs \((j_+, j_-)\). ✓

M-2. Explicit Form of Boost Generators in the \((1/2, 0)\) Representation

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(a) Rotation generators and boost generators

In the \((1/2, 0)\) representation, \(\mathbf{J}_+ = \frac{\boldsymbol{\sigma}}{2}\), \(\mathbf{J}_- = 0\).

Using the results from D6:

\[ \mathbf{J} = \mathbf{J}_+ + \mathbf{J}_- = \frac{\boldsymbol{\sigma}}{2} + 0 = \frac{\boldsymbol{\sigma}}{2} \]
\[ \mathbf{K} = -i(\mathbf{J}_+ - \mathbf{J}_-) = -i\left(\frac{\boldsymbol{\sigma}}{2} - 0\right) = -\frac{i\boldsymbol{\sigma}}{2} \]

Final Answer (a)

\[ \boxed{\mathbf{J} = \frac{\boldsymbol{\sigma}}{2}, \qquad \mathbf{K} = -\frac{i\boldsymbol{\sigma}}{2}} \]

(b) \(2\times2\) matrix for a boost in the \(z\)-direction

A boost in the \(z\)-direction with rapidity \(\phi\) is \(\Lambda_L = \exp(i\phi K^3)\):

\[ \Lambda_L = \exp\left(i\phi \cdot \left(-\frac{i\sigma^3}{2}\right)\right) = \exp\left(\frac{\phi\,\sigma^3}{2}\right) \]

Since \(\sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) is a diagonal matrix:

\[ \exp\left(\frac{\phi\,\sigma^3}{2}\right) = \begin{pmatrix} e^{\phi/2} & 0 \\ 0 & e^{-\phi/2} \end{pmatrix} \]

Writing in terms of hyperbolic functions:

\[ e^{\pm\phi/2} = \cosh\frac{\phi}{2} \pm \sinh\frac{\phi}{2} \]

Therefore

\[ \Lambda_L = \cosh\frac{\phi}{2}\,\mathbf{1} + \sinh\frac{\phi}{2}\,\sigma^3 \]

Final Answer (b)

\[ \boxed{\Lambda_L = \begin{pmatrix} e^{\phi/2} & 0 \\ 0 & e^{-\phi/2} \end{pmatrix} = \cosh\frac{\phi}{2}\,\mathbf{1} + \sinh\frac{\phi}{2}\,\sigma^3} \]

(c) Checking unitarity and physical meaning

We compute \(\Lambda_L^\dagger\). Since \(\sigma^3\) is Hermitian and \(\phi\) is real:

\[ \Lambda_L^\dagger = \exp\left(\frac{\phi\,\sigma^3}{2}\right)^\dagger = \exp\left(\frac{\phi\,\sigma^{3\dagger}}{2}\right) = \exp\left(\frac{\phi\,\sigma^3}{2}\right) = \Lambda_L \]

Therefore \(\Lambda_L^\dagger \Lambda_L = \Lambda_L^2 = \begin{pmatrix} e^{\phi} & 0 \\ 0 & e^{-\phi} \end{pmatrix} \neq \mathbf{1}\) (when \(\phi \neq 0\)).

Final Answer (c)

\(\Lambda_L\) is not unitary.

Physical meaning: Boosts belong to the non-compact part of the Lorentz group. Generators of compact transformations (rotations) are Hermitian (\(J^i = \sigma^i/2\)), and the corresponding transformation matrices \(e^{i\theta J^i}\) are unitary. On the other hand, the boost generators \(K^i = -i\sigma^i/2\) are not anti-Hermitian; since \(iK^i = \sigma^i/2\) is Hermitian, \(e^{i\phi K^i} = e^{\phi\sigma^i/2}\) is the exponential of a Hermitian matrix, yielding not a unitary matrix but a positive-definite Hermitian matrix.

Physically, this corresponds to the fact that boosts change the "magnitude" of spinor components (amplifying one helicity component while attenuating the other). The Lorentz group \(SO^+(1,3)\) is a non-compact group and does not admit finite-dimensional unitary representations.

Verification

\(\det \Lambda_L = e^{\phi/2} \cdot e^{-\phi/2} = 1\), so \(\Lambda_L \in \mathrm{SL}(2, \mathbb{C})\). ✓

When \(\phi = 0\), \(\Lambda_L = \mathbf{1}\) (identity transformation). ✓

M-3. Correspondence Between the \((1/2, 1/2)\) Representation and 4-Vectors

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(a) Proof of \(\det \tilde{V} = -(V^\mu V_\mu)\)

\[ \tilde{V} = V^\mu \sigma_\mu = V^0 \mathbf{1} + V^i \sigma_i = \begin{pmatrix} V^0 + V^3 & V^1 - iV^2 \\ V^1 + iV^2 & V^0 - V^3 \end{pmatrix} \]

Computing the determinant:

\[ \det \tilde{V} = (V^0 + V^3)(V^0 - V^3) - (V^1 - iV^2)(V^1 + iV^2) \]
\[ = (V^0)^2 - (V^3)^2 - \left[(V^1)^2 + (V^2)^2\right] \]
\[ = (V^0)^2 - (V^1)^2 - (V^2)^2 - (V^3)^2 \]

On the other hand, the Minkowski norm is

\[ V^\mu V_\mu = \eta_{\mu\nu}V^\mu V^\nu = -(V^0)^2 + (V^1)^2 + (V^2)^2 + (V^3)^2 \]

Therefore

\[ \boxed{\det \tilde{V} = (V^0)^2 - (V^1)^2 - (V^2)^2 - (V^3)^2 = -(V^\mu V_\mu)} \]

(b) Realization of Lorentz Transformations via \(\mathrm{SL}(2, \mathbb{C})\)

\(\tilde{V}\) is a Hermitian matrix (when \(V^\mu\) is real), and Hermiticity is preserved under the transformation \(\tilde{V} \to M\tilde{V}M^\dagger\) (\(M \in \mathrm{SL}(2, \mathbb{C})\)):

\[ (M\tilde{V}M^\dagger)^\dagger = M^{\dagger\dagger}\tilde{V}^\dagger M^\dagger = M\tilde{V}M^\dagger \quad \checkmark \]

Invariance of the determinant:

\[ \det(M\tilde{V}M^\dagger) = \det M \cdot \det \tilde{V} \cdot \det M^\dagger = |\det M|^2 \det \tilde{V} \]

Since \(M \in \mathrm{SL}(2, \mathbb{C})\), we have \(\det M = 1\), therefore

\[ \boxed{\det(M\tilde{V}M^\dagger) = \det \tilde{V}} \]

This means that \(V^\mu V_\mu\) is invariant, showing that \(\tilde{V} \to M\tilde{V}M^\dagger\) realizes a Lorentz transformation.

The transformed \(\tilde{V}' = M\tilde{V}M^\dagger\) is also a Hermitian \(2\times2\) matrix with trace and trace-free parts, so it corresponds to a new 4-vector \(V'^\mu\). This correspondence \(M \mapsto \Lambda(M)\) provides a homomorphism \(\mathrm{SL}(2, \mathbb{C}) \to SO^+(1,3)\).

Consistency Check

Dimension of the parameter space of \(\tilde{V}\): A \(2\times2\) Hermitian matrix has 4 real parameters, matching the 4 components of a 4-vector. ✓

Dimension of \(\mathrm{SL}(2, \mathbb{C})\): A \(2\times2\) complex matrix (8 real parameters) subject to the constraint \(\det M = 1\) (2 real conditions) gives \(8 - 2 = 6\) dimensions. This matches the dimension 6 of the Lorentz group. ✓

M-4. Quantitative Proof of Why Spin \(1/3\) is Forbidden

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Solution Strategy

In irreducible representations of \(\mathrm{SU}(2)\), we derive that \(2j\) must be a non-negative integer from the properties of raising/lowering operators and the non-negativity of state norms.

Detailed Calculation

Step 1: Fundamental Algebraic Relations

Define \(J^2 = J_1^2 + J_2^2 + J_3^2\) and \(J_\pm = J_1 \pm iJ_2\). The commutation relations \([J_3, J_\pm] = \pm J_\pm\) and \([J_+, J_-] = 2J_3\) hold.

Since \(J^2\) commutes with all generators, in an irreducible representation \(J^2 = \lambda\,\mathbf{1}\) (a constant). Consider eigenstates \(|m\rangle\) of \(J_3\) (\(J_3|m\rangle = m|m\rangle\)).

Step 2: Existence of Upper and Lower Bounds on Eigenvalues of \(J_3\)

\[ J^2 - J_3^2 = J_1^2 + J_2^2 = \frac{1}{2}(J_+ J_- + J_- J_+) \]

For any state \(|m\rangle\):

\[ \langle m|(J_1^2 + J_2^2)|m\rangle = \langle m|J_1^2|m\rangle + \langle m|J_2^2|m\rangle \geq 0 \]

(each term is non-negative since \(J_1, J_2\) are Hermitian). Therefore:

\[ \lambda - m^2 \geq 0 \quad \Rightarrow \quad m^2 \leq \lambda \]

The eigenvalues of \(J_3\) are bounded, and a maximum value \(m_{\max}\) and minimum value \(m_{\min}\) exist.

Step 3: Condition on the Highest Weight State

\(J_+|m_{\max}\rangle = 0\) must hold (otherwise a state with eigenvalue \(m_{\max} + 1\) would exist).

Using \(J_- J_+ = J^2 - J_3^2 - J_3 = J^2 - J_3(J_3 + 1)\):

\[ 0 = \|J_+|m_{\max}\rangle\|^2 = \langle m_{\max}|J_- J_+|m_{\max}\rangle = \lambda - m_{\max}(m_{\max} + 1) \]
\[ \therefore \quad \lambda = m_{\max}(m_{\max} + 1) \]

Defining \(j \equiv m_{\max}\), we have \(\lambda = j(j+1)\).

Step 4: Condition on the Lowest Weight State

Similarly, from \(J_-|m_{\min}\rangle = 0\), using \(J_+ J_- = J^2 - J_3^2 + J_3\):

\[ 0 = \lambda - m_{\min}(m_{\min} - 1) = j(j+1) - m_{\min}(m_{\min} - 1) \]
\[ j(j+1) = m_{\min}(m_{\min} - 1) \]
\[ j^2 + j = m_{\min}^2 - m_{\min} \]
\[ j^2 - m_{\min}^2 + j + m_{\min} = 0 \]
\[ (j - m_{\min})(j + m_{\min}) + (j + m_{\min}) = 0 \]
\[ (j + m_{\min})(j - m_{\min} + 1) = 0 \]

Since \(j - m_{\min} + 1 > 0\) (\(m_{\min} \leq j\) and at least \(m_{\min} < j + 1\)), we have \(j + m_{\min} = 0\), i.e.,

\[ m_{\min} = -j \]

Step 5: \(2j\) Is a Non-Negative Integer

Repeatedly applying \(J_-\) to \(|j\rangle\) (the highest weight state) produces states with \(J_3\) eigenvalues \(j, j-1, j-2, \ldots\), decreasing by 1 each time. This sequence must terminate at \(m_{\min} = -j\).

The \(J_3\) eigenvalue of the state obtained by applying \(J_-\) \(n\) times is \(j - n\). For this to reach \(-j\):

\[ j - n = -j \quad \Rightarrow \quad n = 2j \]

Since \(n\) is the number of times \(J_-\) is applied, it must be a non-negative integer.

Final Answer

\[ \boxed{2j \in \{0, 1, 2, 3, \ldots\} \quad \Rightarrow \quad j \in \left\{0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots\right\}} \]

For \(j = 1/3\), \(2j = 2/3\) is not an integer, so the corresponding irreducible representation does not exist.

Verification

For \(j = 1/2\): \(2j = 1\) (integer) → allowed. Two states with \(m = +1/2, -1/2\). ✓

For \(j = 1\): \(2j = 2\) (integer) → allowed. Three states with \(m = +1, 0, -1\). ✓

For \(j = 1/3\): \(2j = 2/3\) (not an integer) → applying \(J_-\) a finite number of times cannot reach \(m = -1/3\). Forbidden. ✓

Advanced

A-1. Covering Group \(\mathrm{SL}(2, \mathbb{C})\) of the Lorentz Group and \(2\pi\) Rotation of Spinors

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(a) Proof that \(U(2\pi) = -\mathbf{1}\)

From S2(a), the rotation generators in the \((1/2, 0)\) representation are \(J^i = \sigma^i/2\). A rotation by angle \(\theta\) about the \(z\)-axis is

\[ U(\theta) = \exp(i\theta J^3) = \exp\left(\frac{i\theta\,\sigma^3}{2}\right) \]

Since \(\sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) is a diagonal matrix:

\[ U(\theta) = \begin{pmatrix} e^{i\theta/2} & 0 \\ 0 & e^{-i\theta/2} \end{pmatrix} \]

Substituting \(\theta = 2\pi\):

\[ U(2\pi) = \begin{pmatrix} e^{i\pi} & 0 \\ 0 & e^{-i\pi} \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{1} \]

Final Answer (a)

\[ \boxed{U(2\pi) = -\mathbf{1}} \]

A spinor acquires a sign flip under a \(2\pi\) rotation. It returns to its original state only after a \(4\pi\) rotation: \(U(4\pi) = +\mathbf{1}\).

(b) The 2-to-1 Correspondence

In the transformation \(\tilde{V} \to M\tilde{V}M^\dagger\), setting \(M = -\mathbf{1}\):

\[ (-\mathbf{1})\tilde{V}(-\mathbf{1})^\dagger = (-1)(-1)\tilde{V} = \tilde{V} \]

Therefore \(M = +\mathbf{1}\) and \(M = -\mathbf{1}\) correspond to the same Lorentz transformation (the identity transformation \(\Lambda = \mathbf{1}\)).

In general, for any \(M \in \mathrm{SL}(2, \mathbb{C})\), both \(M\) and \(-M\) give the same Lorentz transformation:

\[ (-M)\tilde{V}(-M)^\dagger = M\tilde{V}M^\dagger \]

This is the concrete manifestation of the 2-to-1 homomorphism \(\mathrm{SL}(2, \mathbb{C}) \to SO^+(1,3)\).

Final Answer (b)

\[ \boxed{\pm M \in \mathrm{SL}(2, \mathbb{C}) \quad \longrightarrow \quad \text{the same } \Lambda \in SO^+(1,3)} \]

The kernel is \(\{+\mathbf{1}, -\mathbf{1}\} \cong \mathbb{Z}_2\), and \(SO^+(1,3) \cong \mathrm{SL}(2, \mathbb{C})/\mathbb{Z}_2\).

(c) Physical Meaning and Connection to Quantum Mechanics

Physical meaning of the sign flip under \(2\pi\) rotation:

As shown in (a), a left-handed Weyl spinor \(\psi_L\) transforms as \(\psi_L \to -\psi_L\) under a \(2\pi\) rotation. However, as shown in (b), four-vectors (observables) do not distinguish between \(M\) and \(-M\), so \(\tilde{V} \to M\tilde{V}M^\dagger\) remains invariant.

Connection to quantum mechanics:

In quantum mechanics (in the discussion of angular momentum), applying a \(2\pi\) rotation to a spin-\(1/2\) state \(|\psi\rangle\) gives \(|\psi\rangle \to -|\psi\rangle\). However, physical observables take the form \(|\langle\phi|\psi\rangle|^2\), so the overall phase \((-1)\) is not observable.

More precisely, the state space of quantum mechanics is the projective Hilbert space \(\mathbb{P}\mathcal{H}\) (the space in which overall phases are identified), and symmetries are realized as projective representations. By Wigner's theorem, projective representations can be lifted to ordinary representations of the universal covering group.

  • The universal covering group of the rotation group \(SO(3)\) is \(\mathrm{SU}(2)\) (2-to-1: \(\pm U \to R\))
  • The universal covering group of the Lorentz group \(SO^+(1,3)\) is \(\mathrm{SL}(2, \mathbb{C})\) (2-to-1: \(\pm M \to \Lambda\))

Spin-\(1/2\) particles carry projective representations of \(SO(3)\) (= true representations of \(\mathrm{SU}(2)\)) and acquire a phase of \(-1\) under a \(2\pi\) rotation. This phase has been actually observed in interference experiments (such as neutron interferometry experiments).

Final Answer (c)

The fact that the spinor representation acquires \(-1\) under a \(2\pi\) rotation is a direct consequence of \(\mathrm{SL}(2, \mathbb{C})\) being the double cover of \(SO^+(1,3)\). Physically:

  1. Quantum mechanical states live in projective space, so a phase ambiguity of \(\pm 1\) is permitted
  2. Utilizing this ambiguity, spin \(1/2\) is realized as a projective representation of \(SO(3)\) (= a representation of \(\mathrm{SU}(2)\))
  3. In interference experiments, relative phases are observable, and it has been experimentally confirmed that the state returns to itself only after a \(4\pi\) rotation (not after \(2\pi\))

A-2. \((1, 0) \oplus (0, 1)\) Representation and the Electromagnetic Field Tensor

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(a) Transformation Properties of Self-Dual and Anti-Self-Dual Parts

Definition of electric and magnetic fields:

\[ E^i = F^{0i}, \qquad B^i = \frac{1}{2}\varepsilon^{ijk}F^{jk} \]

Dual tensor:

\[ \tilde{F}^{\mu\nu} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} \]

has components \(\tilde{F}^{0i} = B^i\), \(\frac{1}{2}\varepsilon^{ijk}\tilde{F}^{jk} = -E^i\) (note the metric sign convention).

Three-dimensional vector representation of self-dual and anti-self-dual parts:

\[ \mathbf{F}_+ = \mathbf{E} + i\mathbf{B}, \qquad \mathbf{F}_- = \mathbf{E} - i\mathbf{B} \]

Transformation under rotations:

Rotations transform \(\mathbf{E}\) and \(\mathbf{B}\) each as 3-dimensional vectors (\(\mathbf{E} \to R\mathbf{E}\), \(\mathbf{B} \to R\mathbf{B}\)). Therefore

\[ \mathbf{F}_\pm \to R\mathbf{E} \pm iR\mathbf{B} = R(\mathbf{E} \pm i\mathbf{B}) = R\mathbf{F}_\pm \]

Both \(\mathbf{F}_+\) and \(\mathbf{F}_-\) transform as spin-1 vectors under rotations.

Transformation under boosts:

Under an infinitesimal boost in the \(x\)-direction (rapidity \(\delta\phi\)):

\[ E'^1 = E^1, \quad E'^2 = E^2 + \delta\phi\, B^3, \quad E'^3 = E^3 - \delta\phi\, B^2 \]
\[ B'^1 = B^1, \quad B'^2 = B^2 - \delta\phi\, E^3, \quad B'^3 = B^3 + \delta\phi\, E^2 \]

(This follows from \(\mathbf{E}' = \mathbf{E} + \delta\phi\,(\hat{x} \times \mathbf{B})\), etc.)

Computing the transformation of \(\mathbf{F}_+\):

\[ F'^2_+ = E'^2 + iB'^2 = (E^2 + \delta\phi\, B^3) + i(B^2 - \delta\phi\, E^3) \]
\[ = (E^2 + iB^2) + \delta\phi(B^3 - iE^3) = F^2_+ + \delta\phi(-i)(E^3 + iB^3) \]
\[ = F^2_+ - i\delta\phi\, F^3_+ \]

Similarly \(F'^3_+ = F^3_+ + i\delta\phi\, F^2_+\).

This means that \(\mathbf{F}_+\) transforms under \(K^1\) as \(\delta F^i_+ = -i\delta\phi\,\varepsilon^{1ij}F^j_+\). That is, the boost generator acts on \(\mathbf{F}_+\) as \(K^i_{\text{eff}} = -iJ^i_{\text{rot}}\) (\(-i\) times the rotation generator).

In the \((1,0)\) representation, \(\mathbf{J}_+ = \boldsymbol{\sigma}/2\) (in the spin-1 case, the 3-dimensional representation), \(\mathbf{J}_- = 0\), so:

\[ \mathbf{J} = \mathbf{J}_+ + \mathbf{J}_- = \mathbf{J}_+, \qquad \mathbf{K} = -i(\mathbf{J}_+ - \mathbf{J}_-) = -i\mathbf{J}_+ \]

That is, \(\mathbf{K} = -i\mathbf{J}\). This is precisely consistent with the fact that the boost generator acting on \(\mathbf{F}_+\) is \(-i\) times the rotation generator.

Similarly, for \(\mathbf{F}_-\) the boost acts as \(\mathbf{K} = +i\mathbf{J}\), which corresponds to the \((0,1)\) representation (\(\mathbf{J}_+ = 0\), $\mathbf{J}_- = $ spin 1).

Final Answer (a)

\[ \boxed{\mathbf{F}_+ = \mathbf{E} + i\mathbf{B} \text{ belongs to the } (1,0) \text{ representation, and } \quad \mathbf{F}_- = \mathbf{E} - i\mathbf{B} \text{ belongs to the } (0,1) \text{ representation}} \]

(b) Dimension of \((1,0) \oplus (0,1)\) and Reality Condition

Counting dimensions:

  • \((1,0)\): \((2\cdot1+1)(2\cdot0+1) = 3\) dimensions (complex)
  • \((0,1)\): \((2\cdot0+1)(2\cdot1+1) = 3\) dimensions (complex)
  • Total: \(3 + 3 = 6\) dimensions (complex)

The number of independent components of the antisymmetric tensor \(F^{\mu\nu}\) is \(\frac{4\times3}{2} = 6\). ✓

Consistency with the reality condition:

Under real Lorentz transformations, both \(\mathbf{E}\) and \(\mathbf{B}\) are real vectors, so:

\[ (\mathbf{F}_+)^* = (\mathbf{E} + i\mathbf{B})^* = \mathbf{E} - i\mathbf{B} = \mathbf{F}_- \]

That is, the \((1,0)\) part and the \((0,1)\) part are complex conjugates of each other. Imposing the 3 complex conditions (= 6 real conditions) \((\mathbf{F}_+)^* = \mathbf{F}_-\) on the 6 complex components yields \(12 - 6 = 6\) independent real parameters, matching the 6 real independent components of \(F^{\mu\nu}\).

Final Answer (b)

\[ \boxed{F^{\mu\nu} \text{ forms the } (1,0) \oplus (0,1) \text{ representation. The reality condition } (\mathbf{F}_+)^* = \mathbf{F}_- \text{ gives 6 real components.}} \]

(c) Representation-Theoretic Interpretation of Electromagnetic Duality

Vacuum Maxwell equations:

The source-free Maxwell equations can be written independently for \(\mathbf{F}_+\) and \(\mathbf{F}_-\):

\[ \partial_\mu F^{+\mu\nu} = 0, \qquad \partial_\mu F^{-\mu\nu} = 0 \]

(These are combinations of \(\partial_\mu F^{\mu\nu} = 0\) and \(\partial_\mu \tilde{F}^{\mu\nu} = 0\).)

Electromagnetic duality:

A phase rotation on the \((1,0)\) representation space

\[ \mathbf{F}_+ \to e^{i\alpha}\mathbf{F}_+, \qquad \mathbf{F}_- \to e^{-i\alpha}\mathbf{F}_- \]

(where \(\mathbf{F}_-\) rotates with the opposite phase to preserve the reality condition \((\mathbf{F}_+)^* = \mathbf{F}_-\)) corresponds, in terms of real vectors, to

\[ \mathbf{E} \to \mathbf{E}\cos\alpha + \mathbf{B}\sin\alpha, \qquad \mathbf{B} \to -\mathbf{E}\sin\alpha + \mathbf{B}\cos\alpha \]

which is a rotation in the \(\mathbf{E}\)-\(\mathbf{B}\) plane. This is the duality rotation.

Representation-theoretic interpretation:

  • \((1,0)\) and \((0,1)\) are distinct irreducible representations of the Lorentz group and do not mix under Lorentz transformations
  • However, the vacuum Maxwell equations possess an additional internal symmetry (\(U(1)\) phase rotation) on the \((1,0)\) space
  • This internal \(U(1)\) symmetry is electromagnetic duality; for \(\alpha = \pi/2\) it gives \(\mathbf{E} \to \mathbf{B}\), \(\mathbf{B} \to -\mathbf{E}\)
  • In the presence of charges and currents, \(\partial_\mu F^{\mu\nu} = J^\nu \neq 0\) while \(\partial_\mu \tilde{F}^{\mu\nu} = 0\) is preserved but \(\partial_\mu F^{\mu\nu} = 0\) is broken, so the duality symmetry is broken (unless magnetic charges exist)

Final Answer (c)

\[ \boxed{\text{Electromagnetic duality is understood as a } U(1) \text{ phase rotation } \mathbf{F}_+ \to e^{i\alpha}\mathbf{F}_+ \text{ on the } (1,0) \text{ representation space}} \]

The fact that the vacuum Maxwell equations possess this \(U(1)\) symmetry is a consequence of \((1,0)\) and \((0,1)\) being independent irreducible representations, with the equations taking the same form for each. The presence of sources breaks the symmetry between the equations for \(F^+\) and \(F^-\), thereby breaking the duality symmetry.

Verification

  • Dimensional consistency: \(F^{\mu\nu}\) has 6 components, and \((1,0)\oplus(0,1)\) also has \(3+3=6\) components. ✓
  • \(\alpha = 0\) gives the identity transformation. ✓
  • \(\alpha = \pi/2\) gives \(\mathbf{E} \to \mathbf{B}\), \(\mathbf{B} \to -\mathbf{E}\): Maxwell's equation \(\nabla \cdot \mathbf{E} = 0 \to \nabla \cdot \mathbf{B} = 0\) ✓, \(\nabla \times \mathbf{B} = \partial_t \mathbf{E} \to \nabla \times (-\mathbf{E}) = \partial_t \mathbf{B}\), i.e., \(-\nabla \times \mathbf{E} = \partial_t \mathbf{B}\) ✓ (Faraday's law).