David Tong Standard Model Solutions — Sheet 2: Symmetry Breaking and QCD

These prompts paraphrase David Tong's The Standard Model, Example Sheet 2 (February 2026). Problem numbers, starred status, equations, and guidance follow the official sheet.

1 Sheet 2, Problem 1: Broken Discrete Symmetry

Construct a polynomial potential for a complex scalar $\phi$ whose $\mathbb Z_N$ symmetry breaks spontaneously. Identify every ground state.

2 Sheet 2, Problem 2: Matrix Order Parameters and Goldstone Modes

Let $M(x)$ be a complex $N\times N$ matrix with $\lambda>0$ and \[S=\int d^4x\,\operatorname{Tr}\left[ \partial^\mu M^\dagger\partial_\mu M -kM^\dagger M-\frac\lambda2M^\dagger M M^\dagger M\right].\tag{2.1}\]

Show invariance under $M\mapsto AMB^\dagger$ for $A,B\in U(N)$ and identify the $U(1)$ subgroup of $U(N)\times U(N)$ that acts trivially.
When $k<0$, show that a vacuum satisfies $M_0^\dagger M_0=v^2\mathbf1$. Find the unbroken group, describe the vacuum manifold as a coset, and count Goldstone bosons.
Add $\int d^4x\,h(\det M+\det M^\dagger)$. Determine the new symmetry group and the Goldstone count if the vacuum still has $M_0^\dagger M_0=v^2\mathbf1$ with $v\ne0$.

The invariance is trivial to see. For $A = B \in U(1)$, the transformation is trivial.
The differentiation of the potential with respect to $M^\dagger$, we find \[\frac{\partial}{\partial M^\dagger}\left(\frac{\lambda}{2}M_0^\dagger M_0 M_0^\dagger M_0 + k M_0^\dagger M_0\right) = M_0 (\lambda M_0^\dagger M_0 + k ) = 0 \implies M_0^\dagger M_0 = -\frac{k}{\lambda}\mathbf{1}.\tag{2.2}\] Therefore, the vacuum must take the form $M_0 = v U$ with $U \in U(N)$. We can choose $U = \mathbf{1}$. Under $U(N) \times U(N)$, the vacuum transforms as \[v\mathbf{1} \to A v\mathbf{1} B^\dagger.\tag{2.3}\] Therefore, the vacuum is invariant for $A = B \in U(N)$ . Therefore, the stability group is given by $H = U(N)$, while the coset space is $G = (U(N)\times U(N))/U(N) $. Therefore, the number of Goldstone bosons is $\dim G = N^2$.
With this additional term, the symmetry reduces to $A, B \in SU(N)$, namely $SU(N) \times SU(N)$. If the vacuum still takes the same form, then the stability group becomes $H = SU(N)$. Hence, the number of Goldstone bosons becomes $\dim SU(N) = N^2 -1$.

3 Sheet 2, Problem 3*: Fundamental Higgs Fields in $SU(2)$ and $SU(N)$

For a fundamental $SU(2)$ scalar $\phi_a$ ($a=1,2$), use \[S=\int d^4x\left[-\frac12\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu}) +D_\mu\phi^\dagger D^\mu\phi -\frac\lambda2(\phi^\dagger\phi-v^2)^2\right],\tag{3.1}\] where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu -ig[A_\mu,A_\nu]$, $D_\mu\phi=\partial_\mu\phi-igA_\mu\phi$, and $T^a=\sigma^a/2$ in the fundamental representation. Find the particle masses. Generalise to a fundamental scalar in $SU(N)$: after condensation, what symmetry survives, how many gauge bosons become massive, and how many real scalar components are eaten?

For $SU(2)$, we can choose the vacuum to be \[\phi_0 = \begin{pmatrix} 0 \\ v \end{pmatrix}.\tag{3.2}\] Therefore, all of the symmetries are broken, and we can reparameterise the scalar field as \[\phi(x) = \begin{pmatrix} 0 \\ v + \sigma(x) \end{pmatrix}e^{i\pi^a(x)T^a/ F_\pi},\tag{3.3}\] where $F_\pi $ is chosen to by $v$ so that the kinetic term of $\pi$ is canonically normalised. The gauge boson acquires a mass through the term \[g^2A_\mu A^\mu |\phi|^2 = v^2g^2A_\mu A^\mu + \cdots,\tag{3.4}\] where $m_A = \sqrt{2}vg$. The $\pi(x)$ field is eaten by the gauge bosons. The mass of $\sigma(x)$ is given by \[\lambda v^2 |\phi|^2 = \lambda v^2 \sigma^2 + \cdots,\tag{3.5}\] which gives $m_\sigma = \sqrt{2\lambda}v$.

For $SU(N)$, a choice of vacuum could be \[\phi_0 = \begin{pmatrix} 0 \\ \vdots \\ v \end{pmatrix}.\tag{3.6}\] Therefore, the stability group is $H = SU(N-1)$, and hence the coset space is $G = SU(N) / SU(N-1)$. Therefore, we can reparameterise the scalar field as \[\phi(x) = mm\begin{pmatrix} 0 \\\vdots \\ v + \sigma(x) \end{pmatrix}e^{i\pi^a(x)T^a/ F_\pi},\tag{3.7}\] where $T^a$ are the generators of $G$ and $a = 1, \cdots , \dim G$. For consistency, $\phi(x)$ has $2N$ degrees of freedom, while $\pi(x)$ has $\dim G = 2N-1$ degrees of freedom and $\sigma(x)$ has $1$ degree of freedom. All of the $\pi$ are eaten by the gauge fields. Therefore, there will be $2N-1$ massive bosons.

4 Sheet 2, Problem 4: Adjoint Higgs Breaking Patterns

An $SU(N)$ adjoint scalar is a traceless matrix $\phi=\phi^AT^A$ with $D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi]$. Explain why a potential minimum can be represented by \[\phi_0=\operatorname{diag}(v_1,\ldots,v_N), \qquad \sum_{a=1}^Nv_a=0,\quad v_a\leq v_{a+1}.\tag{4.1}\] Classify the residual gauge groups according to coincidences among the $v_a$.

Given a vacuum $\phi_0$, we can always diagonalise it by $U\phi_0 U^\dagger$, where $U \in SU(N)$. Therefore, we can choose the vacuum to be \[\phi_0 = \operatorname{diag}(v_1,\ldots,v_N),\quad v_a\leq v_{a+1}.\tag{4.2}\] Since $T^A$ are traceless, we must have $\Tr \phi_0 = 0$, which implies $\sum_a v_a = 0$. For a gauge transformation generated by $T^A$ to be unbroken, we must have \[[T^A, \phi_0] = 0.\tag{4.3}\] In general, we have \[[T^A,\phi_0]_{ik} = \sum_j T^A_{ij}\delta_{jk}v_k - \sum_j \delta_{ij}v_j T^A_{jk} = (v_k -v_i) T^A_{ik}.\tag{4.4}\] Therefore, there are no restrictions on $T_{ii}$, which gives $N-1$ unbroken generators. Meanwhile, we must either have $T_{ik} = 0$ for all $i\neq k$, or $v_i - v_k = 0$ for some $i,k$. Therefore, the number of unbroken generator is equal to the number of identical eigenvalue pairs. Suppose each distinct eigenvalues have multiplicity $n_i$, they give \[2\sum_{j}\begin{pmatrix} n_j \\ 2 \end{pmatrix} = \sum_j(n_j^2 -n_j) = \sum_jn_j^2 - N.\tag{4.5}\] Therefore, the total number of unbroken generators is \[\sum_jn_j^2 - 1.\tag{4.6}\]

5 Sheet 2, Problem 5*: One-Loop QCD Scale

For $SU(N_c)$ with $N_f$ massless fundamental Dirac fermions, use \[\frac1{g^2(\mu)}=\frac1{g_0^2} -\frac{11N_c-2N_f}{3(4\pi)^2} \log\frac{\Lambda_{\rm UV}^2}{\mu^2}.\tag{5.1}\] Taking $\alpha_s=g_s^2/(4\pi)\simeq0.12$ at $\mu=M_Z\simeq90\,\mathrm{GeV}$, estimate $\Lambda_{\rm QCD}$ if every quark lighter than $M_Z$ is treated as massless. How can the estimate be made more realistic?

Taking the physical scale, we have \[\frac{1}{g^2(M_Z)} = \frac{1}{g_0^2} -\frac{11N_c-2N_f}{3(4\pi)^2} \log\frac{\Lambda_{\rm UV}^2}{M_Z^2},\tag{5.2}\] while the Landau pole is at \[0 = \frac{1}{g^2(\Lambda_\mathrm{QCD})} = \frac{1}{g_0^2} -\frac{11N_c-2N_f}{3(4\pi)^2} \log\frac{\Lambda_{\rm UV}^2}{\Lambda_\mathrm{QCD}^2}.\tag{5.3}\] Combining the two, we find \[\frac{11N_c-2N_f}{3(4\pi)^2} \log\frac{M_Z^2}{\Lambda_\mathrm{QCD}^2} = \frac{1}{g^2(M_Z)} \implies \Lambda_{\mathrm{QCD}}= M_Z\exp\left(-\frac{6\pi}{(11N_c - 2N_f)\alpha_s}\right). \tag{5.4}\] Five quark flavours are lighter than $M_Z$: $u,d,s,c,b$. The top quark is heavier, so take $N_c=3$ and $N_f=5$. This gives $\Lambda_\mathrm{QCD} \approx 0.097\,\mathrm{GeV}$.

For a more realistic estimation, we can run the coupling towards the IR. When we go across the boundary of the bottom quark mass, we take $N_f$ to be $4$. Then, we match the value at $\alpha(m_b)$ in both regime. We repeat this procedure until ran into the landau pole.

6 Sheet 2, Problem 6: Casimirs and Dynkin Indices

For a representation $R$, define \[T^AT^A=C(R)\mathbf1, \qquad \operatorname{Tr}(T^AT^B)=\frac12I(R)\delta^{AB}.\tag{6.1}\] Prove $2C(R)\dim(R)=I(R)\dim(G)$. Find $C(R)$ for the fundamental and anti-fundamental of $SU(N)$, and calculate the adjoint Casimir of $SU(2)$.

7 Sheet 2, Problem 7: Expanding the Chiral Lagrangian

For $U=e^{2i\pi/f_\pi}$ with $\pi\in su(N_f)$, expand \[\mathcal L_{\rm pion} =\frac{f_\pi^2}{4} \operatorname{tr}(\partial^\mu U^\dagger\partial_\mu U)\tag{7.1}\] through fourth order in $\pi$. Show that the quadratic term is $\operatorname{tr}(\partial_\mu\pi)^2$ and that the quartic term can be written \[-\frac{2}{3f_\pi^2}\operatorname{tr}\left[ \pi^2(\partial_\mu\pi)^2 -(\pi\partial_\mu\pi)^2\right].\tag{7.2}\] For $N_f=2$ and $T^a=\sigma^a/2$, reduce the interaction to \[\mathcal L_{\rm int}=-\frac1{6f_\pi^2} \left[\pi^a\pi^a\partial_\mu\pi^b\partial^\mu\pi^b -\pi^a\partial_\mu\pi^a\, \pi^b\partial^\mu\pi^b\right].\tag{7.3}\]

Expanding $U$, we find \[U = 1+ \frac{2i}{f_\pi}\pi - \frac{2}{f_\pi^2}\pi^2 - \frac{4i}{3f_\pi^3}\pi^3 + \cdots,\tag{7.4}\] and hence \[\partial_\mu U =\frac{2i}{f_\pi}\partial_\mu\pi -\frac{2}{f_\pi^2}\bigl((\partial_\mu\pi)\pi+\pi(\partial_\mu\pi)\bigr) -\frac{4i}{3f_\pi^3}\bigl((\partial_\mu\pi)\pi^2+\pi(\partial_\mu\pi)\pi+\pi^2(\partial_\mu\pi)\bigr) +\cdots.\tag{7.5}\] Expanding the Lagrangian, we have \[\begin{aligned} \mathcal L_{\rm pion} &=\operatorname{tr}(\partial_\mu \pi\partial^\mu \pi)-\frac{2}{3f_\pi^2}\operatorname{tr}\left[ \pi^2(\partial_\mu\pi)^2 -(\pi\partial_\mu\pi)^2\right] +\cdots. \end{aligned}\tag{7.6}\]

For $N_f = 2$, we need to evaluate \[\operatorname{tr}(\sigma^a \sigma^b \sigma^c \sigma^d ) = \operatorname{tr}(\sigma^a \sigma^b \{\sigma^c, \sigma^d \}) - \operatorname{tr}(\sigma^a \{\sigma^b, \sigma^d\} \sigma^c ) + \operatorname{tr}(\{\sigma^a ,\sigma^d\} \sigma^b \sigma^c ) + \operatorname{tr}( \sigma^d \sigma^a \sigma^b \sigma^c).\tag{7.7}\] Using $\{\sigma^a,\sigma^b\} = 2\delta^{ab}$ and $\operatorname{Tr}\!\left(\sigma^a \sigma^b\right) = 2\delta^{ab}$, we find \[\operatorname{tr}(\sigma^a \sigma^b \sigma^c \sigma^d ) = 2(\delta^{ab}\delta^{cd}-\delta^{bd}\delta^{ac} + \delta^{ad}\delta^{bc}).\tag{7.8}\] Substituting into the Lagrangian, the interaction reduces to \[\mathcal L_{\rm int}=-\frac1{6f_\pi^2} \left[\pi^a\pi^a\partial_\mu\pi^b\partial^\mu\pi^b -\pi^a\partial_\mu\pi^a\, \pi^b\partial^\mu\pi^b\right].\tag{7.9}\]

8 Sheet 2, Problem 8: Meson Mass Relations

For $N_f=3$, use the pion, kaon, and eta matrix \[\pi=\frac1{\sqrt2} \begin{pmatrix} \pi^0/\sqrt2+\eta/\sqrt6&\pi^+&K^+\\ \pi^-&-\pi^0/\sqrt2+\eta/\sqrt6&K^0\\ K^-&\bar K^0&-2\eta/\sqrt6 \end{pmatrix}.\tag{8.1}\] With $M=\operatorname{diag}(m_u,m_d,m_s)$ and a mass term proportional to $f_\pi\operatorname{tr}[(M+M^\dagger)\pi^2]$, show \[\frac{m_{K^+}^2-m_{K^0}^2}{m_\pi^2} =\frac{m_u-m_d}{m_u+m_d}.\tag{8.2}\] Then take $m_u\simeq m_d$ to derive the Gell-Mann--Okubo relation $4m_K^2\simeq3m_\eta^2+m_\pi^2$ and compare it with measured meson masses.

\[(\pi^2)_{11}=\frac12\left[\left(\frac{\pi^0}{\sqrt2}+\frac{\eta}{\sqrt6}\right)^2+\pi^+\pi^-+K^+K^-\right],\]\[(\pi^2)_{22}=\frac12\left[\left(-\frac{\pi^0}{\sqrt2}+\frac{\eta}{\sqrt6}\right)^2+\pi^-\pi^++K^0\bar K^0\right],\]\[(\pi^2)_{33}=\frac12\left[K^-K^++\bar K^0K^0+\frac{2\eta^2}{3}\right].\] Hence, the mass term is proportional to \[m_u (\pi^2)_{11} + m_d (\pi^2)_{22} + m_s (\pi^2)_{33}.\tag{8.3}\] Then, we can read off the masses of the Mesons, which are \[m_{\pi}^2 = m_u + m_d, \quad m_{K^0}^2 = m_d + m_s, \quad m_{K^+}^2 = m_u + m_s,\quad m_\eta^2 = \frac{m_u + m_d + 4m_s}{3},\tag{8.4}\] up to an overall constant. Therefore, we obtain the relation \[\frac{m_{K^+}^2-m_{K^0}^2}{m_\pi^2} =\frac{m_u-m_d}{m_u+m_d}.\tag{8.5}\] Take $m_u \simeq m_d$, we immediately find $m_{K}^2 \simeq m_{K^0}^2 \simeq m_{K^+}^2\simeq m_u + m_s$. Then, it follows \[4m_K^2\simeq3m_\eta^2+m_\pi^2.\tag{8.6}\] Using the PDG measured masses and averaging the charged and neutral squared masses within each pion and kaon multiplet gives \[4m_K^2 \simeq 0.983\,\mathrm{GeV}^2.\]\[3m_\eta^2+m_\pi^2 \simeq 0.920\,\mathrm{GeV}^2.\]