Section 5-4 Notes

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Section 5-4

<html><body>Section 5-4 Slater Determinants and the Pauli Principle 5-4 Slater Determinants and the Pauli Principle

It was pointed out by Slater [5] that there is a simple way to write wavefunctions guaranteeing that they will be antisymmetric for interchange of electronic space and spin coordinates: one writes the wavefunction as a determinant. For the 1s22s conﬁguration of lithium, one would write

1s

(1)α(1)

1s(2)α(2)

1s(3)α(3)

ψ = 1

√ 1s(1)β(1) 1s(2)β(2) 1s(3)β(3)

(5-38)

6

2s(1)α(1)

2s(2)α(2)

2s(3)α(3)

Expanding this according to the usual rules governing determinants (see Appendix 2)

gives

ψ = 1

√ [1s(1)α(1)1s(2)β(2)2s(3)α(3) + 2s(1)α(1)1s(2)α(2)1s(3)β(3)

6

+1s(1)β(1)2s(2)α(2)1s(3)α(3) − 2s(1)α(1)1s(2)β(2)1s(3)α(3) −1s(1)β(1)1s(2)α(2)2s(3)α(3) − 1s(1)α(1)2s(2)α(2)1s(3)β(3)] (5-39)

This can be factored and shown to be identical to wavefunction (5-37) of the preceding section.

A simplifying notation in common usage is to delete the α, β symbols of the spin

orbitals and to let a bar over the space orbital signify β spin, absence of a bar being understood to signify α spin. In this notation, Eq. (5-38) would be written

1s(1) 1s(2) 1s(3)

ψ = 1

√ 1¯s(1) 1¯s(2) 1¯s(3)

(5-40)

6

2s(1)

2s(2)

2s(3)

The general prescription to follow in writing a Slater determinantal wavefunction is very simple: 1. Choose the conﬁguration to be represented. 1s1¯s2s was used above. (Here we write

1s1¯s2s rather than 1s22s to emphasize that the two 1s electrons occupy different spinorbitals.) For our general example, we will let Ui stand for a general spin-orbital and take a four-electron example of conﬁguration U1U2U3U4. 2. For n electrons, set up an n × n determinant with (n!)−1/2 as normalizing factor.

Every position in the ﬁrst row should be occupied by the ﬁrst spin-orbital of the conﬁguration; every position in the second row by the second spin-orbital, etc. Now put in electron indices so that all positions in column 1 are occupied by electron 1, column 2 by electron 2, etc.

In the case of our four-electron conﬁguration, the recipe gives

U

1(1) U1(2) U1(3) U1(4)

U2(1) U2(2) U2(3) U2(4)

ψ = 1

√

(5-41)

4! U3(1) U3(2) U3(3) U3(4)

U

4(1)

U4(2) U4(3) U4(4)

Chapter 5 Many-Electron Atoms

Notice that the principal diagonal (top left to bottom right) contains our original conﬁguration U1U2U3U4. Often, the Slater determinant is represented in a space-saving way by simply writing the principal diagonal between short vertical bars. The normalizing factor is deleted. Thus, Eq. (5-41) would be symbolized as |U1(1)U2(2)

U3(3)U4(4)|.

We have indicated the general recipe for writing down a Slater determinant, and we

have seen that, for the conﬁguration 1s1¯s2s, this gives an antisymmetric wavefunction. Now we will give a general proof of the antisymmetry of such wavefunctions for exchange of electrons. We have already seen that interchanging the space and spin coordinates of electrons 1 and 2 corresponds to going through the wavefunction and changing all the 1s to 2s and vice versa; i.e., electron labels denote coordinates. In a Slater determinant, interchanging electron labels 1 and 2 is the same thing as interchanging columns 1 and 2 of the determinant. [See Eq. (5-41) and note that columns 1 and 2 differ only in electron index.] But a determinant reverses sign upon interchange of two rows or columns. (See Appendix 2 for a summary of the properties of determinants.) Hence, any Slater determinant reverses sign (i.e., is antisymmetric) upon the interchange of space and spin coordinates of any two electrons.

Suppose we tried to put two electrons into the same space-orbital with the same spin.

This would require that the same spin-orbital be written twice in the conﬁguration, causing two rows of the Slater determinant to be identical. [If both 1s electrons in Eq. (5-40) had α spin, the bars would be absent from row 2.] We just stated that the determinant must reverse sign upon interchange of two rows. If we interchange two identical rows, we change nothing yet the sign must reverse: the determinant must be equal to zero. Thus, the determinantal wavefunction vanishes when we try to put more than one electron into the same spin-orbital, indicating that this is not a physically allowed situation. This is a generalization of our earlier discovery that no 1s3 conﬁguration is allowed by the exclusion principle, such a conﬁguration requiring at least two electrons to have the same space and spin functions.

This restriction on allowable electronic conﬁgurations is more familiar to chemists as

the Pauli principle: In assigning electrons to atomic orbitals in the independent electron scheme, no two electrons are allowed to have all four quantum numbers (n, l, m, spin) the same. The Pauli principle is a restatement of the exclusion principle as it applies in the special case of an orbital approximation to the wavefunction. 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

We showed in Section 5-2 that two space functions having proper space symmetry

could be written for the conﬁguration 1s2s. One was symmetric (Eq. 5-15) and one was antisymmetric (Eq. 5-16). Now we ﬁnd that spin functions must be included in our wavefunctions, and in a way that makes the ﬁnal result antisymmetric when space and spin coordinates are interchanged. We can accomplish this by multiplying the symmetric space function by an antisymmetric spin function, calling the result ψs,a. Thus,

√

ψs,a(1, 2) = (1/ 2) 1s(1)2s(2) + 2s(1)1s(2) (1/ 2) α(1)β(2) − β(1)α(2) (5-42)

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

Alternatively, we can multiply the antisymmetric space term by any one of the three

possible symmetric spin terms:



√



 α(1)α(2)

(5-43a)

√

ψa,s(1, 2) = (1/ 2) 1s(1)2s(2) − 2s(1)1s(2) (1/ 2) α(1)β(2)+β(1)α(2)

(5-43b)

 β(1)β(2)

(5-43c)

All four of these wavefunctions satisfy the exclusion principle and each is linearly

independent of the others, indicating that four distinct physical states arise from the conﬁguration 1s2s.

There are a number of important points that can be illustrated using these wave

functions. The ﬁrst has to do with Slater determinants. Let us write down a Slater determinantal expression corresponding to wavefunction (5-43a). The conﬁguration is 1s(1)α(1)2s(2)α(2), giving the Slater determinant (where absence of a bar indicates α spin)

1s(1)

1s(2)

ψa,s(1, 2) = 1

√

2 2s(1)

2s(2)

(5-44)

which, upon expansion, gives us Eq. (5-43a). If we attempt the same process to obtain Eq. (5-43b), we encounter a difﬁculty. The conﬁguration 1s(1)α(1)2s(2)β(2) leads to a 2 × 2 determinant, which, upon expansion, gives two product terms, whereas Eq. (5-43b) involves four product terms. The Slater determinantal functions corresponding to Eqs. (5-42) and (5-43b) are, in fact,

1 1s(1) 1s(2)

1¯s(1) 1¯s(2)

ψs,a(1, 2) = 1

√

√

∓ 1

√

(5-45)

a,s

2

2 2¯s(1)

2¯s(2)

2 2s(1)

2s(2)

The lesson to be gained from this is that a single Slater determinant does not always display all of the symmetry possessed by the correct wavefunction. (In this particular case, a single determinant restricts one of the AOs to α spin and the other to β, which is an artiﬁcial limitation.)

Next we will investigate the energies of the states as they are described by these

wavefunctions. We have already pointed out that they are degenerate eigenfunctions of Happrox, but we will now examine their interactions with the full hamiltonian (5-2). Since our wavefunctions are not eigenfunctions of this hamiltonian, we cannot compare eigenvalues. Instead we must calculate the average values of the energy for each wavefunction, using the formula

¯

ψ*H ψ dτ

E =

(5-46)

ψ*ψ dτ

The symbol “dτ ” stands for integration over space and spin coordinates of the electrons: dτ = dv dω. Since both space and spin parts of our wavefunctions are normalized [cf. Eqs. (5-25) and (5-26)], the denominator of Eq. (5-46) is unity and may be ignored. The energy thus is given by the expression

¯E = ψ* −1∇2 − 1∇2 − (2/r1) − (2/r2) + (1/r12) ψ dτ

(5-47)

2 1

2 2 <img src="./tmp/raw_deb0d1ce85a5c5c4d1ce09953a8c9ef8-163_1.png"/>

Chapter 5 Many-Electron Atoms

Notice that the energy operator H contains no terms that would interact with spin functions α and β. (Such terms do arise at higher levels of reﬁnement, but we ignore them for now.) Hence, the spin terms of ψ can be integrated separately, and, since all spin factors in Eqs. (5-42) and (5-43) are normalized, this gives a factor of unity in all four cases. This means that the average energies will be entirely determined by the space parts of the wavefunctions. This, in turn, means that all three states (5-43), which have the same space term, will have the same energy but that the state approximated by the function (5-42) may have a different energy. If our approximate representation of the exact eigenfunctions is physically realistic, we expect helium to display two excited state energies in the energy range consistent with a 1s2s conﬁguration. Furthermore, we expect one of these state energies to be triply degenerate.

Which of these two state energies should be higher? To determine this requires that

we expand our energy expression (5-47) for each of the two space functions (5-42) and (5-43).

¯E1 = 1

[1s*(1)2s*(2) ± 2s*(1)1s*(2)] −1∇2 − 1∇2 −

1

2

(2/r1)

3

2

2

− (2/r2) + (1/r12) [1s(1)2s(2) ± 2s(1)1s(2)]dv(1)dv(2) (5-48)

(The subscript on ¯

E refers to the degeneracy of whichever energy level we are consider

ing.) This expands into a large number of terms. Integrals over one-electron operators may be written as products of two integrals, each over a different electron.6 Thus, the expansion over the kinetic energy operators gives

(5-49)

6We have already shown that, if the 1/r12 term is absent, the energy is equal to E1s + E2s, for He+, which is

equal to −2.5 a.u. Therefore, the detailed breakdown leading to Eqs. (5-49)–(5-51) is not necessary. However, we will present it in detail in the belief that some students will beneﬁt from another speciﬁc example of integration of two-electron products over one-electron operators.

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

The orthogonality of the 1s and 2s orbitals causes the terms preceded by ± to vanish. Furthermore, integrals that differ only in the variable label [such as those in the second and third terms of (5-49)] are equal, so that this expansion becomes

1s*(1) − 1 ∇2 1s(1)dv(1) +

2s*(1) − 1 ∇2 2s(1) dv(1)

(5-50)

2 1

2 1

Expansion of Eq. (5-48) over (−2/r1 − 2/r2) proceeds analogously to give

1s*(1)(−2/r1)1s(1) dv(1) +

2s*(1) (−2/r1) 2s(1) dv(1)

(5-51)

The ﬁnal term in the hamiltonian, 1/r12, occurs in four two-electron integrals:

1

1s*(1)2s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

2

+

2s*(1)1s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

±

1s*(1)2s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

±

2s*(1)1s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

(5-52)

The ﬁrst two integrals of (5-52) differ only by an interchange of labels “1” and “2 ,” and so they are equal to each other. The same is true of the second pair. Thus, the average value of the energy is

¯E1 =

1s*(1) − 1 ∇21 1s(1) dv(1) +

1s*(1) [−2/r1] 1s(1) dv(1)

3

2

+ 2s*(1) −1∇2 2s(1) dv(1) + 2s*(1) [−2/r1] 2s(1) dv(1)

2 1

+

1s*(1)2s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

±

1s*(1)2s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

(5-53)

Notice that, since − 1 ∇2 − 2/r is the hamiltonian for He+, the ﬁrst two integrals of

2

Eq. (5-33) combine to give the average energy of He+ in its 1s state. The second pair gives the energy for He+ in the 2s state. Thus, Eq. (5-53) can be written

¯E1 = E1s + E2s + J ± K

(5-54)

3

where J and K represent the last two integrals in Eq. (5-53). No bars appear on E1s or E2s because these “average energies” are identical to the eigenvalues for the He+ hamiltonian (Problem 5-15).

The integral J denotes electrons 1 and 2 as being in “charge clouds” described by

1s1s and 2s2s, respectively. The operator 1/r12 gives the electrostatic repulsion energy between these two charge clouds. Since these charge clouds are everywhere negatively charged, all the interactions are repulsive, and it is necessary that this “coulomb

Chapter 5 Many-Electron Atoms Figure 5-2 The function produced by multiplying together hydrogenlike 1s and 2s orbitals. R is the radius of the spherical nodal surface.

integral” J be positive. Alternatively, we can argue that the functions 1s1s, 2s2s, and 1/r12 are everywhere positive, so the integrand of J is everywhere positive and J must be positive.

The integral K is called an “exchange integral” because the two product functions in

the integrand differ by an exchange of electrons. This integral gives the net interaction between an electron “distribution” described by 1s2s, and another electron in the same distribution. (These distributions are mathematical functions, not physically realizable electron distributions.) The 1s2s function is sketched in Fig. 5-2. Because the 2s orbital has a radial node, the 1s2s function (which is the same as 1s2s since the 1s function is real) also has a radial node. Now the function 1s(1)2s(1)1s(2)2s(2) will be positive whenever r1 and r2 are either both smaller or both larger than the radial node distance (R in Figure 5-2). But when one r value is smaller than R and the other is greater, corresponding to the electrons being on opposite sides of the nodal surface, the product 1s(1)2s(1)1s(2)2s(2) is negative. These positive and negative contributions to K are weighted by the function 1/r12, which is always positive and hence unable to affect the sign of the integrand. But 1/r12 is smallest when the electrons are far apart. This means that 1/r12 tends to reduce the contributions where the electrons are on opposite sides of the node (i.e., the negative contributions), and so the value of K turns out to be positive (although not as large in magnitude as J , which has no negative contribution at all).

Since the integral K is positive, we can see from Eq. (5-54) that the triply degenerate energy level lies below the singly degenerate one, the separation between them being 2K. (We note in passing that these independent-electron wavefunction energies are not simply the sum of one-electron energies as was the case when we used Happrox, thereby ignoring interelectronic repulsion.)

The experimental observation agrees qualitatively with these results. There are two

state energies associated with the 1s2s conﬁguration. When the atom is placed in an external magnetic ﬁeld, the lower-state-energy-level splits into three levels. The state having the higher energy has a “multiplicity” of one and is called a singlet. The lowerenergy with multiplicity three is called a triplet. (The reference to a “triplet state” should not be construed to mean that this is one state. It is a triplet of states.)

It is possible to use vector arguments similar to those presented in Chapter 4 to under

stand why the triply degenerate level splits into three different levels in the presence of a homogeneous magnetic ﬁeld. Let us ﬁrst consider the case of a single electron. We have already indicated that two spin states are possible, which we have labeled α and β. <img src="./tmp/raw_deb0d1ce85a5c5c4d1ce09953a8c9ef8-166_1.png"/>

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

In a magnetic ﬁeld the angular momentum vectors precess about the ﬁeld axis z, as depicted in Figure 5-3. The z components of the angular momentum vectors are constant but the x and y components are not. Because the allowed z components must in general differ by one atomic unit (stated but not proved in Chapter 4), and because there are but two allowed values (inferred from observations such as the splitting of a beam of silver atoms into two components), and because the two kinds of state are oppositely affected by magnetic ﬁelds, it is concluded that the z components of angular momentum (labeled ms) are equal to + 1 and − 1 a.u. for α and β, respectively. Fol

2

lowing through using orbital angular momentum relations as a model, we postulate an electron spin quantum number s equal to the maximum z-component of spin angular

√

momentum in a.u., 1 , and a spin angular momentum vector s having length

s(s + 1)

2

a.u. The degeneracy, gs = 2s + 1, equals 2, in agreement with the two orientations in Fig. 5-3.

As noted in Chapter 4, half-integer quantum numbers correspond to eigenfunctions

that cannot be expressed as spherical harmonics. We will not pursue the question of detailed expressions for α and β here. (However, see the problems at the end of Chapter 9.)

Now let us turn to the two-electron system. We will assume that the magnetic

moments of the two electrons interact independently with the external ﬁeld. This ignores the fact that each electron senses a small contribution to the magnetic ﬁeld resulting from the magnetic moment of the other electron.

Another factor that could affect the magnetic ﬁeld sensed by the spin moment is the

magnetic moment resulting from the orbital motions of the electrons, although this is not present if both electrons are taken to be in s atomic orbitals (AOs). For two electrons, we can imagine four situations: αα, αβ, βα, and ββ. We pointed out earlier (Section 4-5) that, for a system composed of several moving parts, the total angular momentum is the sum of the individual angular momenta, and the z component is the sum of the individual z components. For the four spin situations listed above, this means that the net z components of spin angular momentum (labeled Ms) are +1, 0, 0, and −1 a.u., respectively. The spin combinations αβ + βα [from the triplet (5-43b)] and αβ − βα Figure 5-3 The angular momentum vectors for α and β precess around the magnetic ﬁeld axis z. The z components of these vectors are constant and have values of + 1 and − 1 a.u. respectively.

2

2 </body></html>

Home

<html><body>Section 5-4 Slater Determinants and the Pauli Principle 5-4 Slater Determinants and the Pauli Principle

1s

(1)α(1)

1s(2)α(2)

1s(3)α(3)

ψ = 1

√ 1s(1)β(1) 1s(2)β(2) 1s(3)β(3)

(5-38)

6

2s(1)α(1)

2s(2)α(2)

2s(3)α(3)

Expanding this according to the usual rules governing determinants (see Appendix 2)

gives

ψ = 1

√ [1s(1)α(1)1s(2)β(2)2s(3)α(3) + 2s(1)α(1)1s(2)α(2)1s(3)β(3)

6

+1s(1)β(1)2s(2)α(2)1s(3)α(3) − 2s(1)α(1)1s(2)β(2)1s(3)α(3) −1s(1)β(1)1s(2)α(2)2s(3)α(3) − 1s(1)α(1)2s(2)α(2)1s(3)β(3)] (5-39)

This can be factored and shown to be identical to wavefunction (5-37) of the preceding section.

A simplifying notation in common usage is to delete the α, β symbols of the spin

orbitals and to let a bar over the space orbital signify β spin, absence of a bar being understood to signify α spin. In this notation, Eq. (5-38) would be written

1s(1) 1s(2) 1s(3)

ψ = 1

√ 1¯s(1) 1¯s(2) 1¯s(3)

(5-40)

6

2s(1)

2s(2)

2s(3)

In the case of our four-electron conﬁguration, the recipe gives

U

1(1) U1(2) U1(3) U1(4)

U2(1) U2(2) U2(3) U2(4)

ψ = 1

√

(5-41)

4! U3(1) U3(2) U3(3) U3(4)

U

4(1)

U4(2) U4(3) U4(4)

Chapter 5 Many-Electron Atoms

U3(3)U4(4)|.

We have indicated the general recipe for writing down a Slater determinant, and we

Suppose we tried to put two electrons into the same space-orbital with the same spin.

This restriction on allowable electronic conﬁgurations is more familiar to chemists as

We showed in Section 5-2 that two space functions having proper space symmetry

√

ψs,a(1, 2) = (1/ 2) 1s(1)2s(2) + 2s(1)1s(2) (1/ 2) α(1)β(2) − β(1)α(2) (5-42)

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

Alternatively, we can multiply the antisymmetric space term by any one of the three

possible symmetric spin terms:



√



 α(1)α(2)

(5-43a)

√

ψa,s(1, 2) = (1/ 2) 1s(1)2s(2) − 2s(1)1s(2) (1/ 2) α(1)β(2)+β(1)α(2)

(5-43b)

 β(1)β(2)

(5-43c)

All four of these wavefunctions satisfy the exclusion principle and each is linearly

independent of the others, indicating that four distinct physical states arise from the conﬁguration 1s2s.

There are a number of important points that can be illustrated using these wave

1s(1)

1s(2)

ψa,s(1, 2) = 1

√

2 2s(1)

2s(2)

(5-44)

1 1s(1) 1s(2)

1¯s(1) 1¯s(2)

ψs,a(1, 2) = 1

√

√

∓ 1

√

(5-45)

a,s

2

2 2¯s(1)

2¯s(2)

2 2s(1)

2s(2)

Next we will investigate the energies of the states as they are described by these

¯

ψ*H ψ dτ

E =

(5-46)

ψ*ψ dτ

¯E = ψ* −1∇2 − 1∇2 − (2/r1) − (2/r2) + (1/r12) ψ dτ

(5-47)

2 1

2 2 <img src="./tmp/raw_deb0d1ce85a5c5c4d1ce09953a8c9ef8-163_1.png"/>

Chapter 5 Many-Electron Atoms

Which of these two state energies should be higher? To determine this requires that

we expand our energy expression (5-47) for each of the two space functions (5-42) and (5-43).

¯E1 = 1

[1s*(1)2s*(2) ± 2s*(1)1s*(2)] −1∇2 − 1∇2 −

1

2

(2/r1)

3

2

2

− (2/r2) + (1/r12) [1s(1)2s(2) ± 2s(1)1s(2)]dv(1)dv(2) (5-48)

(The subscript on ¯

E refers to the degeneracy of whichever energy level we are consider

(5-49)

6We have already shown that, if the 1/r12 term is absent, the energy is equal to E1s + E2s, for He+, which is

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

1s*(1) − 1 ∇2 1s(1)dv(1) +

2s*(1) − 1 ∇2 2s(1) dv(1)

(5-50)

2 1

2 1

Expansion of Eq. (5-48) over (−2/r1 − 2/r2) proceeds analogously to give

1s*(1)(−2/r1)1s(1) dv(1) +

2s*(1) (−2/r1) 2s(1) dv(1)

(5-51)

The ﬁnal term in the hamiltonian, 1/r12, occurs in four two-electron integrals:

1

1s*(1)2s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

2

+

2s*(1)1s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

±

1s*(1)2s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

±

2s*(1)1s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

(5-52)

¯E1 =

1s*(1) − 1 ∇21 1s(1) dv(1) +

1s*(1) [−2/r1] 1s(1) dv(1)

3

2

+ 2s*(1) −1∇2 2s(1) dv(1) + 2s*(1) [−2/r1] 2s(1) dv(1)

2 1

+

1s*(1)2s*(2)(1/r12)1s(1)2s(2) dv(1) dv(2)

±

1s*(1)2s*(2)(1/r12)2s(1)1s(2) dv(1) dv(2)

(5-53)

Notice that, since − 1 ∇2 − 2/r is the hamiltonian for He+, the ﬁrst two integrals of

2

Eq. (5-33) combine to give the average energy of He+ in its 1s state. The second pair gives the energy for He+ in the 2s state. Thus, Eq. (5-53) can be written

¯E1 = E1s + E2s + J ± K

(5-54)

3

The integral J denotes electrons 1 and 2 as being in “charge clouds” described by

Chapter 5 Many-Electron Atoms Figure 5-2 The function produced by multiplying together hydrogenlike 1s and 2s orbitals. R is the radius of the spherical nodal surface.

integral” J be positive. Alternatively, we can argue that the functions 1s1s, 2s2s, and 1/r12 are everywhere positive, so the integrand of J is everywhere positive and J must be positive.

The integral K is called an “exchange integral” because the two product functions in

The experimental observation agrees qualitatively with these results. There are two

It is possible to use vector arguments similar to those presented in Chapter 4 to under

Section 5-5 Singlet and Triplet States for the 1s2s Conﬁguration of Helium

2

lowing through using orbital angular momentum relations as a model, we postulate an electron spin quantum number s equal to the maximum z-component of spin angular

√

momentum in a.u., 1 , and a spin angular momentum vector s having length

s(s + 1)

2

a.u. The degeneracy, gs = 2s + 1, equals 2, in agreement with the two orientations in Fig. 5-3.

As noted in Chapter 4, half-integer quantum numbers correspond to eigenfunctions

that cannot be expressed as spherical harmonics. We will not pursue the question of detailed expressions for α and β here. (However, see the problems at the end of Chapter 9.)

Now let us turn to the two-electron system. We will assume that the magnetic

Another factor that could affect the magnetic ﬁeld sensed by the spin moment is the

2

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