Correct zeroth-order functions

This integral vanishes because of the orthogonality of the spherical harmonics. Hence (4.38) are the correct zeroth-order functions. 

The functions (6.85) each correspond to a specific pair of the normal-mode vibrational quantum numbers v2a and v2b, whereas the correct zeroth-order functions (6.87) do not. Hence the quantum numbers v2a and v2b are not particularly physically significant. The quantum numbers of physical significance are v2 = v2a -I- v2b, which (together with c, and t>3) specifies which degenerate vibrational level we are dealing with, and the vibrational angular-momentum quantum number. 

Our first task is thus to determine the correct zeroth-order wave functions (9.75) for the perturbation H. Calling these correct zeroth-order functions f we have... 

Having found the d first-order energy corrections, we go back to the set of equations (9.84) to find the unknowns C , which determine the correct zeroth-order wave functions. To find the correct zeroth-order function... 

The converse is also true. If the initially assumed functions are the correct zeroth-order functions, then the secular determinant is in diagonal form. This is seen as follows. From we know that the coefficients in the expansion are... 

The secular determinant is in block-diagonal form and factors into four determinants, each of second order. We conclude that the correct zeroth-order functions have the form... 

We cannot say which orbital electron 1 is in for either or 4. This property of the wave functions of systems containing more than one electron results fi-om the indistinguishability of identical particles in quantum mechanics and will be discussed further in Chapter 10. Since the functions and have different energies, the exchange degeneracy is removed when the correct zeroth-order functions are used. 

What now Recall that we ran into essentially the same situation in treating the helium excited states (Section 9.7), where we started with the functions ls(l)2s(2) and 2s(l)ls(2). We found that these two functions, which distinguished between electrons 1 and 2, are not the correct zeroth-order functions and that the correct zeroth-order functions are 2 [ls(l)2s(2) 2s(l)ls(2)]. This result suggests pretty strongly that instead of a(l)j8(2) and /3(l)a(2), we use... 

We might compare the preceding treatment of H2 with the perturbation treatment of the helium ls2s levels (Section 9.7). There we started with the degenerate functions ls (l) (2) and ls(2)2s(l). Because of the symmetry of the Hamiltonian with respect to interchange of the identical electrons, we found the correct zeroth-order functions to be [ls (l)2s(2) ls(2)2s(l)]/V2. For H2,we started with the degenerate functions Is and Isj. Because of the symmetry of the electronic Hamiltonian with respect to the identical nuclei, we found the correct zeroth-order functions to be Is, ls,)/V2(l ... 

Applying the same reasoning to the remaining functions we conclude that = 0 for i m. Hence, use of the correct zeroth-order functions makes the secular determinant diagonal. Note also that the first-order corrections to the energy can be found by averaging the perturbation over the correct zeroth-order wave functions ... 

For a configuration of closed subshells (for example, the helium ground state), we cm write only a single Slater determinant. This determinant is an eigenfunction of I and and is the correct zeroth-order function for the nondegenerate S term. A configuration... 

The preceding discussion is oversimplified. For the hydrogen atom, the 2s and 2p AOs are degenerate, and so we can expect the correct zeroth-order functions for the (Tg2s, (t 2s, (Tg2p, and (r 2j9 MOs of Hj to each be mixtures of 2s and 2p AOs rather than containing only 2s or 2p character. [In the i limit, Hj consists of an H atom pertnrbed by the essentially uniform electric field of a far-distant proton. Problem 9.23... 

Higher-order effects and overlap corrections may be treated as in earlier sections, using the correct zeroth-order functions defined in (14.6.5). The energy formula (14.6.6) will then assume the general form (14.5.18), except for the presence of an extra term, (res), obtained above. 

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