David Tong Standard Model Solutions — Sheet 1: The Basics

These prompts paraphrase David Tong's The Standard Model, Example Sheet 1 (January 2026). Problem numbers, starred status, equations, and guidance follow the official sheet.

1 Sheet 1, Problem 1: Lorentz Generators for Weyl Spinors

Define \[\sigma^{\mu\nu} =\frac{i}{4}(\sigma^\mu\bar\sigma^\nu -\sigma^\nu\bar\sigma^\mu).\tag{1.1}\] Verify that these matrices obey \[[\sigma^{\mu\nu},\sigma^{\rho\sigma}] =i\left(\eta^{\nu\rho}\sigma^{\mu\sigma} -\eta^{\nu\sigma}\sigma^{\mu\rho} +\eta^{\mu\sigma}\sigma^{\nu\rho} -\eta^{\mu\rho}\sigma^{\nu\sigma}\right).\tag{1.2}\] Hint: Rewrite $\sigma^{\mu\nu} =\tfrac i2(\eta^{\mu\nu}-\sigma^\nu\bar\sigma^\mu)$ and establish $\sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu=2\eta^{\mu\nu}\mathbf1_2$.

Recall $\{\sigma_i,\sigma_j\} = 2\delta_{ij}$, which follows \[\sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu=2\eta^{\mu\nu}\mathbf1_2. \tag{1.3}\] Hence, \[\sigma^{\mu\nu} =\tfrac i2(\eta^{\mu\nu}-\sigma^\nu\bar\sigma^\mu), \tag{1.4}\] which gives \[[\sigma^{\mu\nu},\sigma^{\rho\sigma}] = -\frac{1}{4}(\sigma^\nu\bar\sigma^\mu\sigma^\sigma\bar\sigma^\rho - \sigma^\sigma\bar\sigma^\rho\sigma^\nu\bar\sigma^\mu).\tag{1.5}\] Now the strategy is to permute the indices using (1.3) so that the 4 $\sigma$ term eventually cancels. Carrying out the calculation, we find \[[\sigma^{\mu\nu},\sigma^{\rho\sigma}] = -\frac{1}{4}\left(-2\eta^{\rho\nu}\sigma^\sigma \bar \sigma^\mu + 2\eta^{\mu\rho}\sigma^\sigma \bar \sigma^\nu - 2\eta^{\nu\sigma}\sigma^\mu \bar \sigma^\rho + 2\eta^{\mu\sigma}\sigma^\nu \bar \sigma^\rho\right).\tag{1.6}\] Then, plugging (1.4) in, we find \[[\sigma^{\mu\nu},\sigma^{\rho\sigma}] =i\left(\eta^{\nu\rho}\sigma^{\mu\sigma} -\eta^{\nu\sigma}\sigma^{\mu\rho} +\eta^{\mu\sigma}\sigma^{\nu\rho} -\eta^{\mu\rho}\sigma^{\nu\sigma}\right).\tag{1.7}\]

2 Sheet 1, Problem 2: Raised Weyl Indices and Lorentz Scalars

Let $S\in SL(2,\mathbb C)$ act on left- and right-handed spinors by \[(\psi_L)_\alpha\longmapsto S_\alpha{}^\beta(\psi_L)_\beta, \qquad (\psi_R)_{\dot\alpha} \longmapsto (S^*)_{\dot\alpha}{}^{\dot\beta} (\psi_R)_{\dot\beta}.\tag{2.1}\] The dotted index distinguishes the two representations. Prove the following statements in the sheet's epsilon convention:

$(S^{-1})_\beta{}^\alpha =\epsilon^{\alpha\gamma}S_\gamma{}^\lambda \epsilon_{\lambda\beta}$.
$(\psi_L)^\alpha=\epsilon^{\alpha\beta}(\psi_L)_\beta$ transforms with $(S^{-1})_\beta{}^\alpha$ on its right.
$\psi_L\chi_L=(\psi_L)^\alpha(\chi_L)_\alpha$ is Lorentz invariant.
$(\psi_R)^{\dot\alpha} =\epsilon^{\dot\alpha\dot\beta}(\psi_R)_{\dot\beta}$ transforms with $S^{-1\dagger}$.
$\bar\psi_R\psi_L =(\psi_R^*)^\alpha(\psi_L)_\alpha$ is also a Lorentz scalar.

\[S_\alpha{}^\gamma S_\beta{}^\delta \epsilon_{\gamma \delta} = \begin{pmatrix} a & b\\ c &d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 &0 \end{pmatrix} \begin{pmatrix} a & c\\ b &d \end{pmatrix} = \begin{pmatrix} 0 & ad-bc \\ -ad +bc &0 \end{pmatrix} = \det S \epsilon_{\alpha \beta}.\tag{2.2}\] Since $S \in SL(2,\mathbb{C})$, $\det S = 1$, we find \[S_\alpha{}^\gamma S_\beta{}^\delta \epsilon_{\gamma \delta} = \epsilon_{\alpha \beta} \implies \epsilon^{\beta \lambda}S_\alpha{}^\gamma S_\lambda{}^\delta \epsilon_{\gamma \delta} = \delta_\alpha{}^\beta \implies (S^{-1})_\beta{}^\alpha =\epsilon^{\alpha\gamma}S_\gamma{}^\lambda \epsilon_{\lambda\beta}.\tag{2.3}\]
\[(\psi_L)^\alpha = \epsilon^{\alpha \beta } (\psi_L)_\beta \to \epsilon^{\alpha\beta}S_\beta{}^\gamma(\psi_L)_\gamma = \epsilon^{\alpha\beta}S_\beta{}^\gamma\epsilon_{\lambda\gamma}(\psi_L)^\gamma = (S^{-1})_\gamma{}^\alpha (\psi_L)^\gamma.\tag{2.4}\]
\[\psi_L \chi_L \to (\psi_L)^\gamma(S^{-1})_\gamma{}^\alpha S_\alpha{}^\gamma(\chi_L)_\beta = \psi_L \chi_L .\tag{2.5}\]
Completely analogous to the left-handed case, we find \[(\psi_R)^{\dot \alpha} \to (S^{-1\dagger})^{\dot \alpha}{}_{\dot \beta} (\psi_R)^{\dot \beta}.\tag{2.6}\]
\[\bar \psi_R \psi_L \to (\psi_R^*)^{\dot \alpha}(S^{-1})_{\dot \alpha}{}^\alpha S_\alpha{}^\beta(\psi_L)_\beta = \bar \psi_R \psi_L.\tag{2.7}\]

3 Sheet 1, Problem 3: The Right-Handed Lorentz Representation

In addition to $\sigma^{\mu\nu}$ above, define \[\bar\sigma^{\mu\nu} =\frac i4(\bar\sigma^\mu\sigma^\nu -\bar\sigma^\nu\sigma^\mu).\tag{3.1}\] Show $\bar\sigma^{\mu\nu}=(\sigma^{\mu\nu})^\dagger$. If \[S=\exp\!\left(-\frac i2\omega_{\mu\nu} \sigma^{\mu\nu}\right),\tag{3.2}\] show that \[S^{-1\dagger} =\exp\!\left(-\frac i2\omega_{\mu\nu} \bar\sigma^{\mu\nu}\right).\tag{3.3}\] It is enough to verify the infinitesimal transformation and then use the group law. Relate this to the dotted and undotted components of a Dirac spinor.

4 Sheet 1, Problem 4: Dirac and Majorana Mass Mixing

Two Weyl spinors have a real Dirac mass $M$ and complex Majorana masses $m_1,m_2$: \[\mathcal L_{\rm mass} =-M(\bar\psi_L\psi_R+\bar\psi_R\psi_L) +\frac{m_1}{2}\psi_L\psi_L +\frac{m_1^*}{2}\bar\psi_L\bar\psi_L +\frac{m_2}{2}\psi_R\psi_R +\frac{m_2^*}{2}\bar\psi_R\bar\psi_R.\tag{4.1}\] Determine the physical fermion masses.

The mass Lagrangian can be written as \[\mathcal L_{\rm mass} = -\frac{1}{2}\begin{pmatrix} \bar \psi _L & \psi_R \end{pmatrix}\begin{pmatrix} -m_1^* & M\\ M & -m_2^* \end{pmatrix}\begin{pmatrix} \bar \psi _L \\ \psi_R \end{pmatrix} + h.c. \equiv -\frac{1}{2} \chi_i \mathcal{M}_{ij}\chi_j + h.c.,\tag{4.2}\] where $\chi^T \equiv (\bar \psi_L, \psi_R)$. The kinetic part of the Lagrangian can be written as \[\bar \psi_L i\bar \sigma^\mu \partial_\mu + \bar\psi_R i\sigma^\mu \partial_\mu \psi_R = \frac{1}{2}( \bar\chi_i i\bar\sigma^\mu\partial_\mu\chi_i + \chi_i i\sigma^\mu\partial_\mu\bar\chi_i).\tag{4.3}\] The equation of motion is given by \[i \sigma^\mu \partial_\mu \chi_i -\mathcal{M}_{ij}^*\bar \chi_j = 0\tag{4.4}\] and its conjugate \[i\bar \sigma^\mu \partial_\mu \bar\chi_i - \mathcal{M}_{ij}\chi_j = 0.\tag{4.5}\] Applying $i\bar\sigma^\nu \partial_\nu$ to the first equation, we find \[-\bar\sigma^\nu \sigma^\mu \partial_\mu \partial_\nu \chi_i - \mathcal{M}_{ij}^*i\bar\sigma^\nu \partial_\nu \chi_j =-\bar\sigma^{(\nu} \sigma^{\mu)} \partial_\mu \partial_\nu \chi_i - \mathcal{M}_{ij}^*\mathcal{M}_{jk}\chi_k = 0 \implies (\partial^2 + \mathcal{M}^\dagger\mathcal{M}) \chi = 0.\tag{4.6}\] To progress, let us diagonalise \[\mathcal{M}^\dagger \mathcal{M} = U D U^\dagger.\tag{4.7}\] Then, the equation of motion becomes \[(\partial^2 + UDU^\dagger) \chi = 0 \implies (\partial^2 + D)(U^\dagger\chi) = 0.\tag{4.8}\] Therefore, the physical masses squared are the eigenvalues of $\mathcal{M}^\dagger\mathcal{M}$. The mass eigenvalues are found to be \[m_\pm^2 = \frac{1}{2}\left[ |m_1|^2+|m_2|^2+2M^2 \pm \sqrt{ \left(|m_1|^2+|m_2|^2+2M^2\right)^2 -4|m_1m_2-M^2|^2 } \right].\tag{4.9}\]

5 Sheet 1, Problem 6*: Discrete Symmetries of Gauge Fields

Use the chiral-basis matrices \[\gamma^\mu=\begin{pmatrix}0&\sigma^\mu\\ \bar\sigma^\mu&0\end{pmatrix}, \qquad \gamma^5=\begin{pmatrix}\mathbf1&0\\0&-\mathbf1\end{pmatrix}, \qquad \sigma^\mu=(\mathbf1,\sigma^i),\quad \bar\sigma^\mu=(\mathbf1,-\sigma^i).\tag{5.1}\] For a Dirac fermion coupled to $U(1)$, the gauge-field equation is $\partial_\nu F^{\mu\nu}=e\bar\psi\gamma^\mu\psi$. Starting from the spinor actions of $C,P,T$, work out how $A_\mu$ must transform so that this equation remains invariant. Then determine how the Maxwell theta term \[S_\theta=\int d^4x\,F_{\mu\nu}{}^\star F^{\mu\nu}\tag{5.2}\] transforms under each of $C,P,T$.

Under parity, we have \[P: \quad \psi(t,\mathbf{x}) \to \gamma^0\psi(t,-\mathbf{x}), \quad \bar \psi \gamma^\mu \psi(t,\mathbf{x}) \to \bar \psi\gamma^\mu \psi(t,-\mathbf{x}),\tag{5.3}\] which implies \[P: \quad \partial_\mu F^{\mu 0}(t,\mathbf{x}) \to \partial_\mu F^{\mu 0}(t,-\mathbf{x}), \quad \partial_\mu F^{\mu i}(t,\mathbf{x}) \to - \partial_\mu F^{\mu i}(t,-\mathbf{x}).\tag{5.4}\] This follows \[P: \quad \partial_i\partial^i A^0 - \partial_i \partial^0 A^i \to \partial_i\partial^i A'^0 + \partial_i \partial^0 A'^i \implies P:\quad A^0(t,\mathbf{x}) \to A^0(t,-\mathbf{x}), \quad A^i(t,\mathbf{x}) \to -A^i(t,-\mathbf{x}).\tag{5.5}\] One can check that the spatial components also transform as desired.

Under charge conjugation, we have \[C: \quad \psi(t,\mathbf{x}) \to \pm i \gamma^2 \psi^* (t,\mathbf{x}), \quad \bar \psi \gamma^\mu \psi(t,\mathbf{x}) \to -\psi^T\gamma^2 \gamma^0 \gamma^\mu \gamma^2 \psi^* (t,\mathbf{x}) = -\bar \psi \gamma^\mu \psi (t,\mathbf{x}),\tag{5.6}\] where we have used $\gamma^2 \gamma^\mu \gamma^2 = - \gamma^2$. Therefore, we must have \[C: \quad A^\mu (t,\mathbf{x}) \to -A^\mu (t,\mathbf{x}) .\tag{5.7}\]

Under time reversal, we have \[\begin{aligned} T: \quad \psi(t,\mathbf{x}) \to \gamma^1\gamma^3 \psi^*(-t,\mathbf{x}), \quad \bar \psi\gamma^0 \psi (t,\mathbf{x}) &\to \psi^T (\gamma^0)^*\psi^*(-t,\mathbf{x}) = \bar \psi \gamma^0\psi(-t,\mathbf{x}),\\ \bar \psi\gamma^i \psi (t,\mathbf{x}) &\to -\psi^T (\gamma^i)^*\psi^*(-t,\mathbf{x}) = -\bar \psi \gamma^i\psi(-t,\mathbf{x}). \end{aligned}\tag{5.8}\] The LHS of the gauge-field equation transforms as \[T: \quad \partial_\mu F^{\mu0}(t,\mathbf{x}) \to (\partial_j \partial^j A'^0 + \partial_j \partial^0 A'^j)(-t,\mathbf{x}), \quad \partial_\mu F^{\mu i}(t,\mathbf{x}) \to (\partial^2A'^i + \partial_0 \partial^j A'^0 - \partial_i \partial^jA'^j)(-t,\mathbf{x}).\tag{5.9}\] For the theory to be invariant under time reversal, we must have \[T: \quad A^0(t,\mathbf{x}) \to A^0(-t,\mathbf{x}), \quad A^i(t,\mathbf{x}) \to - A^i(-t,\mathbf{x}).\tag{5.10}\]

The theta term is proportional to \[\int d^4x \mathbf{E}\cdot \mathbf{B},\tag{5.11}\] whereas the fields transform as \[P:\quad \mathbf{E}(t,\mathbf{x}) \to -\mathbf{E}(t,-\mathbf{x}), \quad \mathbf{B}(t,\mathbf{x}) \to \mathbf{B}(t,-\mathbf{x})\tag{5.12}\] \[C:\quad \mathbf{E}(t,\mathbf{x}) \to -\mathbf{E}(t,\mathbf{x}), \quad \mathbf{B}(t,\mathbf{x}) \to - \mathbf{B}(t,\mathbf{x})\tag{5.13}\] \[P:\quad \mathbf{E}(t,\mathbf{x}) \to \mathbf{E}(-t,\mathbf{x}), \quad \mathbf{B}(t,\mathbf{x}) \to - \mathbf{B}(-t,\mathbf{x}).\tag{5.14}\] Therefore, the theta term is invariant under C, and odd under P and T.

6 Sheet 1, Problem 7: Parity with Scalar and Pseudoscalar Masses

Prove that $\bar\psi\psi$ and $i\bar\psi\gamma^5\psi$ are real. For \[\mathcal L_{\rm mass} =m_1\bar\psi\psi+im_2\bar\psi\gamma^5\psi,\tag{6.1}\] rewrite the mass term in left- and right-handed Weyl fields using the chiral gamma matrices. Find an action of parity on $\psi_L,\psi_R$ that preserves the theory, and express it on the Dirac spinor.

7 Sheet 1, Problem 8: Parity and Mass in $2+1$ Dimensions

In signature $(+,-,-)$ use imaginary matrices $\gamma^\mu=(\sigma^2,i\sigma^1,i\sigma^3)$ and the action \[S=-\int d^dx\left(i\bar\psi\gamma^\mu\partial_\mu\psi -M\bar\psi\psi\right).\tag{7.1}\] Explain why parity reverses only $x^1$ and leaves $x^0,x^2$ unchanged, instead of sending the whole spatial vector to its negative. For $M=0$, find parity, charge-conjugation, and time-reversal actions that preserve $S$. Which are broken by $M\ne0$?

We wish to use the parity action to enlarge $SO(1,2)$ to $O(1,2)$. To do that, we need the determinant of parity to be negative. Therefore, this can only be achieved with one coordinate flipped in two spatial dimensions.

For parity, the massless Lagrangian transform as \[P:\quad i\bar \psi \sigma^\mu \partial_\mu \psi \to i\psi^\dagger P^\dagger\gamma^0(\gamma^0\partial_0 - \gamma^1 \partial_1 + \gamma^2 \partial_2)P\psi.\tag{7.2}\] Therefore, by choosing $P: \psi \to \gamma^0 \psi$, the massless theory remains invariant.

For charge conjugation, the Dirac field transforms as \[C: \quad \psi \to C\psi^*,\tag{7.3}\] while the kinetic term transforms as \[C: \quad i\bar\psi\gamma^\mu \partial_\mu \psi \to i \psi^TC^\dagger \gamma^0 \gamma^\mu C\partial_\mu\psi^* = -i(\partial_\mu \psi^\dagger)C^T (\gamma^\mu)^T (\gamma^0)^TC^*\psi = i\psi^\dagger C^T\gamma^\mu \gamma^0 C^* \psi,\tag{7.4}\] where we have used $(\gamma^\mu)^T = -\gamma^\mu$. Therefore, we must have $C: \psi \to e^{i\alpha}\psi^*$.

For time reversal, we have \[T: \quad i \to -i, \quad \psi \to T \psi^*,\tag{7.5}\] and hence \[\begin{aligned} T: \quad i\bar \psi \gamma^\mu \partial_\mu \psi &\to -i \psi^T T^\dagger \gamma^0 (-\gamma^0 T \partial_0 -\gamma^1 T \partial_1-\gamma^2 T \partial_2)\psi^*\\ &= -i\psi^\dagger T^T (-\gamma^0 \gamma^0 T^* \partial_0 - \gamma^1 \gamma^0 T^* \partial_1 - \gamma^2 \gamma^0 T^* \partial_2) \psi. \end{aligned}\tag{7.6}\] For the kinetic term to be invariant, we must have $T:\psi \to \gamma^1\gamma^2\psi^* = \gamma^0 \psi^*$.

Now it is clear that the mass term is even under C, and odd under T and P .