An infinitesimal <Lorentz transformation> is $\delta X^\mu=\omega^\mu{}_{\nu}X^\nu$ with $\omega_{\mu\nu}=-\omega_{\nu\mu}$. Applying <Noether theorem> to this continuous symmetry gives the conserved <Lorentz current>
$$
J_\alpha^{\mu\nu}=\frac1{2\pi\alpha'}
\left(X^\mu\partial_\alpha X^\nu-X^\nu\partial_\alpha X^\mu\right),
\qquad \partial^\alpha J_\alpha^{\mu\nu}=0.
$$
Its <Noether charge> is $M^{\mu\nu}=\int_0^\pi d\sigma\,J_\tau^{\mu\nu}$. Substituting the <open-string mode expansion> and using <orthogonality> of the cosine modes yields
$$
M^{\mu\nu}=x^\mu p^\nu-x^\nu p^\mu
-i\sum_{n=1}^{\infty}\frac1n
\left(\alpha_{-n}^\mu\alpha_n^\nu-
\alpha_{-n}^\nu\alpha_n^\mu\right).
$$
The first term is orbital angular momentum; the sum is the contribution of the <string oscillators>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-306/1/b/solution