Free Trial

Safari Books Online is a digital library providing on-demand subscription access to thousands of learning resources.

Share this Page URL
Help

Chapter 6. Quantum Physics > Angular Momentum - Pg. 639

6.5 Angular Momentum 1 2 NIntegrateAu@xD , 8x, 0, 4<E 2 639 normalizationConstant = ; normalizedu@x_D = normalizationConstant u@xD; PlotBnormalizedu@xD, 8x, -3, 3<, Epilog ® 8Line@881, -1<, 81, 1<<D, Line@88-1, -1<, 8-1, 1<<D<, Frame ® True, FrameLabel ® :' (a)' ' 'x ', 'u(x) (1/ a)' '>, ImageSize -> 84 ´ 72, 3 ´ 72<, PlotLabel -> ' = ' < > ToString@E@nD /. V 0 ® 1D <> ' 0 ' 'E ' 'V 'EF Column@Table@plotFunc@u@nDD, 8n, 1, 3<DD We have also omitted the graphs of the eigenfunctions because they appear identical to those from the analytical solution in Section 6.4.3.1. Our numerical method, a form of what is known as the shooting method, can be adapted for other one-dimensional potentials, and it can still be used when the time-independent Schr¨ dinger equation cannot be solved by even the most powerful analytical tools. However, o a few words of caution are in order. With this numerical method, it is quite possible to miss some energy eigenvalues, but this pitfall can be avoided. A general property of one-dimensional bound states can serve as a safeguard. If the bound states are arranged in the order of increas- ing energies E 1 , E 2 , . . . , E n , . . . , the nth eigenfunction has n - 1 nodes (see [Mes00]). Thus, if, for example, the eigenfunctions corresponding to two consecutive energies in our spectrum have, respectively, n and n + 2 nodes, we must have missed an energy eigenvalue. In determining the appropriate starting values r 0 and r 1 with initialValues@ e i , e f , n , eqn D , consider making n bigger or reducing e f - e i to scan regions that are likely to have roots. (For a discussion of the shooting method, see [Pat94].) In[46]:= ClearAll@l, F, e, sol, eqnEven, eqnOdd, initialValues, evenvalues, oddvalues, E, u, V, plotFuncD 6.5 ANGULAR MOMENTUM 6.5.1 The Problem The Hamiltonian of a force-free rigid rotator is H = 1 2 1 2 1 2 L x + L y + L 2I 1 2I 2 2I 3 z (6.5.1) where I 1 , I 2 , and I 3 are the principal moments of inertia, and L x , L y , and L z are the components of the total angular momentum operator along the principal axes. Find the eigenvalues of H when the total angular momentum quantum number equals 1, 2, and 3. 6.5.2 Physics of the Problem 6.5.2.1 Angular Momentum in Quantum Mechanics In classical mechanics, angular momentum of a particle is a vector defined in the coordinate space in terms of the position and momentum vectors of the particle. Corresponding to this