Physics
Physics, 10.04.2020 15:48, azalialujan4634

In this guided problem, you will explore some ideas from a general topic of "addition of angular momenta". Let Ji and J2 be two angular momentum operators (I am dropping hats on operators). They may refer to angular momenta of two different particles or to the orbital angular momentum and to the spin of the same particle; that is, they act on different "coordinates" in the description of our system. It means that these operators commute with each other: (Jli, J2k] = 0, for all Cartesian indices i, k. These momenta "interact", e. g. due to the associated magnetic momenta, the relevant part of the system Hamiltonian is their scalar product ("energy is scalar"): (1) H = BJ1 J2 (2) where B is an appropriate coefficient. Hamiltonian (2) could, e. g., describe an important phenomenon of the so-called spin-orbit interaction. We now also define the total angular momentum of the system: J=J1 +J2. (3) (c) Show that all three squares: J, Jz, and J2 are conserved. In addition, these three squares and operator J, let us consider its J, component as usual, all commute with each other. That is, these four operators represent compatible observables. An important consequence of the statements above is that we can label the station- ary eigenstates as jmjija) by the respective eigenvalues of the conserved compatible observables: J?|jmjij2) = hạj(j + 1).jmj1j2), Jz[jmjij2) = ħm|jmjija), J{jmjija) = j1(01 + 1).jmj1j2), Jž jmj1j2) = t´ja(j2 +1)|jm|1j2). (d) You should be able now to show that \jmj1j2) is indeed the stationary state of (2): Hjmj1j2) = E|jmj1j2), (8) with energy Bha +1) - jiji + 1) - 12(12 + 1)]. (9) All you need to do for this is square (3) and then follow our already established rela- tionships (4)-(7).

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