Operators Linear momentum

From the definition of the translational linear momentum operator / (in (eqnation Al.4,97)) we see that... 

Here we have uncovered an interesting connection between the linear displacement operator D and the linear momentum operator P ... 

Linear algebraic problem, 53 Linear displacement operator, 392 Linear manifolds in Hilbert space, 429 Linear momentum operator, 392 Linear operators in Hilbert space, 431 Linear programming, 252,261 diet problem, 294 dual problem, 304 evaluation of methods, 302 in matrix notation, simplex method, 292... 

The frozen-core (fc) approach is not restricted to spin-independent electronic interactions the spin-orbit (SO) interaction between core and valence electrons can be expressed by a sum of Coulomb- and exchange-type operators. The matrix element formulas can be derived in a similar way as the Sla-ter-Condon rules.27 Here, it is not important whether the Breit-Pauli spin-orbit operators or their no-pair analogs are employed as these are structurally equivalent. Differences with respect to the Slater-Condon rules occur due to the symmetry properties of the angular momentum operators and because of the presence of the spin-other-orbit interaction. It is easily shown by partial integration that the linear momentum operator p is antisymmetric with respect to orbital exchange, and the same applies to t = r x p. Therefore, spin-orbit... 

From the above derivation it is seen that after the series expansion of the exponential in the space part of the vector potential, the transition moment operator involves the linear momentum operator fij or the gradient operator Equation (1.32) is obtained from Equation (1.30) in the following way From the commutation relation [A, rj = Htj - rfi = fij, we have... 

The use of the dipole velocity formulation requires the matrix elements of the linear momentum operator p. From the commutation relation of H with r, Linderberg (l%7) showed that the following expression is obtained in the ZDO approximation ... 

Here, e, m, A, and c are well-known natural constants, pj = -iftV, is the linear momentum operator, and r, is the position operator of electron /. For... 

Using the expressions for the linear momentum operators (2.24), the angular momentum operators are obtained ... 

It should however be noted that the eigenfunctions of the hamiltonian are not eigenfunctions of the linear momentum operator. Accordingly, a measurement of the momentum does not lead to p = 2mE , but to a probability distribution as shown in Fig. 2.3 (refs. 18 and 19). It can be noted that the most probable momentum is not p , when the particle is in a state except when n becomes large. Then the quantum and classical descriptions are similar. 

Light, circularly polaridzed, 41, 139-44. 154, 158, 162-63 elUplically polarized, 139-43 linearly polarized, 1-3, 5, 38-41, 139-41 Light-gathering antennae, 473 Linear momentum operator. 22. 24, 56, 145 Line-shape function, 156 Liquid crystals, 272 Localized orbital model, 115-16... 

Of course, the connection between Sections 4.1 and 4.2.2 is provided by Euler s formula = cos mtp isin mtp. Thus, the simultaneous raising and lowering actions of the linear momentum operators on the order of the angular momentum and radial eigenfunctions is established for cartesian, spherical, and spheroconal representations. 

The linear momentum operator and its connection with the angular momentum operator in spheroconal and cartesian components ... 

In Section 3.3 we found the eigenfunctions and eigenvalues for the linear-momentum operator p. In this section we consider the same problem for the angular momentum of a particle. Angular momentum is important in the quantum mechanics of atomic structure. We begin by reviewing the classical mechanics of angular momentum. 

There is a close connection between symmetry and the constants of the motion (these are properties whose operators commute with the Hamiltonian H). For a system whose Hamiltonian is invariant (that is, doesn t change) under any translation of spatial coordinates, the linear-momentum operator p will commute with H, and p can be assigned a definite value in a stationary state. An example is the free particle. For a system with H invariant under any rotation of coordinates, the operators for the angular-momentum components commute with H, and the toted angular momentum and one of its components are specifiable. An example is an atom. A linear molecule has axial symmetry, rather than the spherical synunetry of an atom here only the axial component of angular momentum can be specified (Chapter 13). 

Laplacian, symmetry of, 13 transformation properties of, 9 Legendre polynomials, 144 Linear equations, solutions of, 42 Linear momentum operator, symmetry of, 167... 

Tlie linear momentum operators Pi = ihd/dx, p = ihd/dy,p = ihd/dij, transform like polar vectors, since... 

The calculation of the magnetic transition dipoles requires a preamble. The magnetic moment was already defined in Eq. (4.128) of Chap. 4. By explicitly writing the angular momentum operator in terms of the linear momentum operator sr x p one obtains ... 

Considering the derivation of DKH Hamiltonians so far, we are facing the problem to express all operators in momemtum space, which is somewhat unpleasant for most molecular quantum chemical calculations which employ atom-centered position-space basis functions of the Gaussian type as explained in section 10.3. The origin of the momentum-space presentation of the DKH method is traced back to the square-root operator in Sq of Eq. (12.54). This square root requires the evaluation of the square root of the momentum operator as already discussed in the context of the Klein-Gordon equation in chapter 5. Such a square-root expression can hardly be evaluated in a position-space formulation with linear momentum operators as differential operators. In a momentum-space formulation, however, the momentum operator takes a... 

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