Paper 2, Section II, 33E

Principles of Quantum Mechanics
Part II, 2013

(i) In units where =1\hbar=1, angular momentum states jm|j m\rangle obey

J2jm=j(j+1)jm,J3jm=mjmJ^{2}|j m\rangle=j(j+1)|j m\rangle, \quad J_{3}|j m\rangle=m|j m\rangle

Use the algebra of angular momentum [Ji,Jj]=iϵijkJk\left[J_{i}, J_{j}\right]=i \epsilon_{i j k} J_{k} to derive the following in terms of J2,J±=J1±iJ2J^{2}, J_{\pm}=J_{1} \pm i J_{2} and J3J_{3} : (a) [J2,Ji]\left[J^{2}, J_{i}\right]; (b) [J3,J±]\left[J_{3}, J_{\pm}\right]; (c) [J2,J±]\left[J^{2}, J_{\pm}\right].

(ii) Find J+JJ_{+} J_{-}in terms of J2J^{2} and J3J_{3}. Thus calculate the quantum numbers of the state J±jmJ_{\pm}|j m\rangle in terms of jj and mm. Derive the normalisation of the state JjmJ_{-}|j m\rangle. Therefore, show that

jj1J+j1Jjjj=A(2j1)!\left\langle j j-1\left|J_{+}^{j-1} J_{-}^{j}\right| j j\right\rangle=\sqrt{A}(2 j-1) !

finding AA in terms of jj.

(iii) Consider the combination of a spinless particle with an electron of spin 1/21 / 2 and orbital angular momentum 1. Calculate the probability that the electron has a spin of +1/2+1 / 2 in the zz-direction if the combined system has an angular momentum of +1/2+1 / 2 in the zz-direction and a total angular momentum of +3/2+3 / 2. Repeat the calculation for a total angular momentum of +1/2+1 / 2.