Classically, the angular momentum of a compound is the corner products of its po-Sition vector r = (x; y; z) and its momentum vector p = (PX; py; pz):
L = r £ p:
Orbital angular momentum solutions with physics assignment help
The huge technical orbital angular momentum owner is de¯ned in the same way with p changed by the momentum owner p! ¡i¹hr. Thus, the Cartesian elements of L are
Lx = ¹hi³y @@z ¡ z @@y´; Ly = ¹hi³z @@x ¡ x @@z´; LZ = ¹hi³x @@y ¡ y @@x´: (1.1)
With the aid of the commutation interaction between p and r:
[PX; x] = ¡i¹h; [py; y] = ¡i¹h; [pz; z] = ¡i¹h; (1.2)
One quickly ensures the following commutation interaction for the Cartesian solutions with physics assignment help
Components of the huge technical angular momentum operator:
LxLy ¡ LyLx = i¹hLz; LyLz ¡ LzLy = i¹hLx; LzLx ¡ LxLz = i¹hLy :( 1.3)
Since the elements of L do not travel with each other, it is not possible to ¯nd multiple eigenstates of any two of these three employees.
y +L2 operator physics assignment help
L2 = L2
Z, however, trips with each part of L. It is, there-
Fore, possible to ¯nd a multiple eigenstate of L2 and any one part of. It is traditional to search for eigenstates of L2 and LZ.