Nonabelian Symmetries — Srednicki Ch. 24

Source: M. Srednicki, Quantum Field Theory, Chapter 24 (Nonabelian Symmetries), Problems 24.1–24.4.

1 Antisymmetry of the Generator, Q24.1

Show that $\theta_{ij}$ in eq. (24.4) must be antisymmetric if $R$ is orthogonal.

2 Structure Constants from SO$(N)$ Transformations, Q24.2

By considering the $\mathrm{SO}(N)$ transformation $R'^{-1}R^{-1}R'R$, where $R$ and $R'$ are independent infinitesimal $\mathrm{SO}(N)$ transformations, prove eq. (24.7).

3 Noether Current and Charge Algebra, Q24.3

Find the Noether current $j^{a\mu}$ for the transformation of eq. (24.6).
Show that $[\phi_i, Q^a] = (T^a)_{ij}\phi_j$, where $Q^a$ is the Noether charge.
Use this result, eq. (24.7), and the Jacobi identity (see problem 2.8) to show that $[Q^a, Q^b] = if^{abc}Q^c$.

For scalar field theories that is invariant under $SO(N)$ ($\delta \mathcal{L} = 0$). The infinitesimal transformation is given by \[\phi_i(x) \to \phi'_i(x) = \phi_i(x) -i\theta^a(T^a)_{ij}\phi_j(x).\tag{3.1}\] Therefore, the Noether current is given by \[j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_i(x))}\delta\phi_i(x) = i\theta^a\partial^\mu \phi_i(x)(T^a)_{ij}\phi_j(x).\tag{3.2}\] Getting rid of the parameter, we find \[j^{a\mu} = i\partial^\mu \phi_i(x)(T^a)_{ij}\phi_j(x).\tag{3.3}\]

Then, we can defined the conserved charge \[Q^a = \int_{\Sigma}d^{d-1}x\, j^{a0} = -i(T^a)_{ij}\int_{\Sigma}d^{d-1}x\,\pi_i(x)\phi_j(x).\tag{3.4}\] Using the equal time commutation relation, \[[\phi_i(\mathbf{x},t), \pi_j(\mathbf{y},t)] = i\delta_{ij}\delta^{d-1}(\mathbf{x}-\mathbf{y}),\quad [\phi_i(\mathbf{x},t), \phi_j(\mathbf{y},t)] = 0,\tag{3.5}\] we find \[[\phi_i(\mathbf{x},t), Q^a] = (T^a)_{ij}\phi_j(\mathbf{x},t),\tag{3.6}\] where we have used the fact that $Q_a$ is conserved, and therefore the same for any $t$. Similarly \[[\pi_i(\mathbf{x},t), Q^a] = - (T^a)_{ij}\pi_i(\mathbf{x},t).\tag{3.7}\] Then, we consider \[\begin{aligned} [Q^a,Q^b] &= i(T^a)_{ij}\int_{\Sigma}d^{d-1}x\,[\pi_i(x)\phi_j(x), Q^b]\\ &= i(T^a)_{ij}\int_{\Sigma}d^{d-1}x\,\left(\pi_i(x)[\phi_j(x), Q^b] +[\pi_i(x), Q^b]\phi_j(x)\right)\\ &=i(T^a)_{ij}\int_{\Sigma}d^{d-1}x\,(\pi_i(x)(T^b)_{jk}\phi_k(x) - \pi_k(T^b)_{ki}\phi_j(x))\\ &=i[T^a,T^b]_{ik}\int_{\Sigma}d^{d-1}x\,\pi_i(x)\phi_k(x) = if^{abc}Q^c. \end{aligned}\tag{3.8}\]

4 Generators of Sp$(2N)$, Q24.4

The elements of the group $\mathrm{SO}(N)$ can be defined as $N\times N$ matrices $R$ that satisfy \[R_{ii'}R_{jj'}\delta_{i'j'} = \delta_{ij}.\tag{4.1}\] The elements of the symplectic group $\mathrm{Sp}(2N)$ can be defined as $2N\times 2N$ matrices $S$ that satisfy \[S_{ii'}S_{jj'}\eta_{i'j'} = \eta_{ij},\tag{4.2}\] where the symplectic metric $\eta_{ij}$ is antisymmetric, $\eta_{ij}=-\eta_{ji}$, and squares to minus the identity: $\eta^2 = -I$. One way to write $\eta$ is \[\eta = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix},\tag{4.3}\] where $I$ is the $N\times N$ identity matrix. Find the number of generators of $\mathrm{Sp}(2N)$.