PHYS 6572 - Quantum Mechanics I - Fall 2011 Problem Set 7 — Solutions Joe

• Using the ladder operators, Jx = 1 2(J+ + J−) and Jy = 1 2i(J+ − J−). When Jx (resp. Jy) acts on | j,m〉, it produces a linear combination of two ket states | j,m+ 1〉 and | j,m− 1〉, both of which are orthogonal to | j,m〉. So if we put the bra 〈j,m | with the ket Jx| j,m〉 (resp. Jy| j,m〉) together, the bracket must vanish: that is, the expectation value of Jx (resp. Jy) in the state | j,m〉 is 0. • It is also possible to exploit the commutation relations of the Ji alone without invoking the ladder operators. Since [Jy, Jz] = i~Jx, and Jz| j,m〉 = m~| j,m〉, we can compute the expectation value of Jx in | j,m〉 as follows: 〈j,m | Jx | j,m〉 = 1 i~ 〈j,m | [Jy, Jz] | j,m〉