Z-component of the spin operator

In this Hamiltonian, the a/ are the shieldings, or chemical shifts, of the nuclei, J / the indirect spin-spin coupling constants between pairs of nuclei, IZi the z component of the spin operator 7, y the gyromagnetic ratic and H0 the strength of the static applied field. Within the experimental accuracy of measurements achieved thus far, no additional terms are required in the Hamiltonian to attain an exact fit of theoretical and calcu-... 

The mean value of the z-component of the spin operator is calculated as a Boltzmann average... 

To determine the first-order spin-orbit splitting pattern of an orbitally degenerate electronic state, we shall make use of the energy expression obtained from the phenomenological operator, which in this case reduces to Aso A S because only the z component of the spin-orbit operator is involved. 

In addition we require the electronic wavefunction to be an eigenfunction of the operator for the square of the total spin, , and also of the z component of the spin,... 

Notice that in the transformed basis set the z-component of the Zeeman operator is no longer diagonal there are off-diagonal matrix elements of the type P i Hf Pj).) The full spin Hamiltonian matrix Hy is then diag-onalised. The obtained eigenvalues ,-(/ ) are expanded in powers of B, yielding the van Vleck coefficients. 

Here, both CSFs possess a zero quantum number for the z-component of the spin angular momentum operator (Ms = 0). The corresponding excitation energies are derived as follows ... 

Jt is identified with the total angular momentum along the z axis, and St and Mt are the contributions from spin and orbital motion. The operator for the z component of the spin angular momentum St depends on the representation S, implicit in Zj, under which the function values are invariant. If the value of 0(ri, r, . .., t) is independent of the direction, ri, r2,. . . , and dependent only on their magnitudes, 0 is clearly a scalar function. In that event the function value space has all the symmetry of the sphere and s is the totally symmetric representation of the group. Zi V is then equal to zero, and J,0 = A/,0. The spin infinitesimal operator 1, only appears in the problem when the values of the function 0(ri, r, . .., 1) at various points of application form a basis for a representation other than the totally symmetric rep of the... 

Similarly the operator for the z component of the spin takes a singlet spin function into a triplet spin function... 

The usual choice of the basis functions in the spin space is a and JS, These functions are adapted to the z axis in real space in the sense of being the eigenfunctions of the z component of the spin angular momentum operator ha /2, with eigenvalues +h/2 and - /2, respectively. The usual choice of spin functions similarly adapted to the orthogonal 7 and x axes in real space is (a i )/j2 and (a P)/j2 respectively. These are eigenfunctions of the spin angular momentum operators hGy/2 and fta /2, respectively. 

An angular momentum is a vector, which also pertains to spin angular momenta (see Chapter 1). The spin angular momentum of a certain number of elementary particles is a sum of their spin vectors. To obtain the total spin vector, therefore, we have to add the x-, y-, and z-components of the spins of the particles, and to construct from them the total vector. Then we might he interested in the corresponding spin operators. These operators will he created using the Pauli matrices. ... 

Like the angular momentum of an electron in its orbit, there are two measurables for spin that can be observed simultaneously the square of the total spin and the z component of the spin. Because spin is an angular momentum, there are eigenvalue equations for the spin observables that are the same as for and L, except we use the operators S and to indicate the spin observables. We also introduce the quantum numbers s and m to represent the quantized values of the spin of the particles. (Do not confuse s, the symbol for the spin angular momentum, with s, an orbital that has = 0.) The eigenvalue equations are therefore... 

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