Atomic and Molecular Structure (Chem 421/521)
Notes #12
These notes begin with a brief introduction to the coupling of angular momenta for multi-electron states. This is followed by a discussion of the considerations, which lead to the "Pauli Principle". Finally, the concept of a "Slater determinant" is introduced.
12.1 Coupling of Angular Momenta
What are the possible (orbital and/or spin) angular momenta states for two electrons in an atom? Since spin behaves as an angular momentum, the coupling rules formally are the same no matter whether we are coupling two angular or two spin momenta. The nomenclature, that is being used, differs, of course.
If we designate the orbital angular momentum for electron 1 as L1 (with eigenvalue ℓ1) and that of electron 2 as L2 (eigenvalue ℓ 2), then the possible values for the total orbital angular momentum are
L = Li L = ℓ 1 + ℓ 2, ℓ 1 + ℓ 2 - 1, ℓ 1 - ℓ 2 - 2,....| ℓ 1 - ℓ 2|
We label the resultant states - called terms - according to total orbital angular momentum in an analogous fashion to the atomic orbital labels, except we use capital letters. Thus, L = 0 gives rise to an S term, L = 1 to a P term, L = 2 to a D term, L = 3 to an F term, L = 4 to a G term, etc. Each term with a particular L has (2L+1) components corresponding to the (2L+1) values of the projection of L onto the z-axis:
Lz = Lz,iML = L, L-1, L-2,...,-L
If there are three angular momenta to be coupled, then you first couple two of them as above, and then add the third to all the terms arising from coupling of the first two momenta.
Spin angular momenta couple in the same fashion (of course). Thus, if particle 1 has spin angular momentum S1 with eigenvalue s1 and particle 2 has spin angular momentum S2 with eigenvalue s2,then the possible values for the total spin angular momentum are
S = Si S = s1+s2, s1+s2-1, s1-s2-2,....|s1-s2|
When the two particles are electrons, s1 = s2 = 1/2, and two states result with S = 0 and S = 1, respectively. Again, each spin state has (2S+1) components corresponding to
Stot,z = Sz,iMS = S, S-1, S-2, ...-S; Stot,z = Sz,i
Thus, an S = 0 state has only one component (MS = 0) and is called a singlet; a S = 1 state has three components (MS = 1, 0 , -1) and is called a triplet. The value of (2S+1) is called the spin multiplicity. The spin multiplicity of a singlet (S = 0) is 1, that of a doublet (S = 1/2) is 2, a triplet (S = 1) is 3, etc; the spin multiplicity is added as a left superscript to the orbital angular momentum labels, e.g., 3P, 1D, 1S.
If there are three spin angular momenta to be coupled, then you first couple two of them as above to give a singlet (S = 0) and a triplet(S = 1); then add the third to all the states arising from coupling of the first two to form a total of two doublets (S = 1/2) and a quartet (S = 3/2).
Collectively, the orbital/spin angular momentum states are known as atomic terms or term states, and the symbols (3P, 1D, etc.) are called atomic term symbols.
The ground state of the H atom has one electron in an s orbital (ℓ 1 = 0, s1 = 1/2); it thus forms a 2S state. If the electron finds itself excited into a 2p orbital (ℓ 1 = 1, s1 = 1/2), it will be in a 2P state.
The He atom has two electrons. If they are both occupying s orbitals, i.e. ℓ 1 = 0, ℓ 2 = 0; s1 = 1/2, s2= 1/2, then we may expect to form 1S and 3S states. If one electron is in an s orbital (ℓ 1 = 0, s1 = 1/2) and the other electron is in a p orbital (ℓ 2 = 1, s2 = 1/2), then we could expect to find 1P and 3P states.
Doubly filled orbitals contribute 0 to both L and S.
Exercise. Show that two electrons in p orbitals may couple their angular momenta to form 1S, 3S, 1P, 3P, 1D, and 3D states. If the two electrons are in the same set of p orbitals (like 2p orbitals in C with ground state electronic configuration 1s22s22p2), then only the 1S, 3P, and 1D states are "Pauli" allowed (see below).
The separate coupling of orbital and spin angular momenta to form atomic terms is known as the Russell-Saunders coupling scheme. The scheme is valid as long as an individual electron's orbital and spin angular momentum both form "good" quantum numbers, i.e. when the individual electron's orbital and spin angular momentum both commute with the Hamiltonian for that atom. RS coupling is generally ok for the first and second row of the Periodic Table, but the "independence" approximation deteriorates through the third row and rapidly thereafter due to relativistic effects, in particular the presence ofwhat is called "spin-orbit" coupling (L.S). Thus, for heavier atoms, angular momentum and spin no longer form independent quantum numbers; rather, the total angular momentum J = L + S is quantized.
12.2Spin Wave Functions for Two Electrons
The two possible spin wavefunctions for a single electron are and , corresponding to ms = 1/2 and ms = -1/2, respectively. To construct the possible spin functions for two electrons we proceed as follows:
From the coupling scheme outlined above, we know that there is going to be a singlet (S = 0) and a triplet (S = 1) spin state; the latter state has three components corresponding to MS = 1, 0, -1. The total spin operator for two electrons is
S = S1 + S2
and hence the eigenfunctions of S are products of the spin functions for each single electron. The possibilities are therefore (1)(2), (1)(2), (1)(2), and (1)(2), or linear combinations of these functions.
The spin function (1)(2) is "clearly" the component of the triplet state with MS = 1. The spin function (1)(2) is "clearly" the component of the triplet state with MS = -1. The two MS = 0 states must be linear combinations of (2) and (1). Noting that the two triplet spin functions we have ((1)(2) and (1)(2)) are symmetric with respect to interchange of the two electrons, we find (using a |S, Ms> notation)
|1, +1> = (1)(2)|1, -1> = (1)(2)
|1, 0> = (2)-1/2[(2) + (2)]
|0, 0> = (2)-1/2[(2) - (2)]
We note that the three triplet spin functions are symmetric with respect to interchange of the two electrons, and that the single singlet spin function is antisymmetric with respect to interchange of the two electrons.
12.3The Pauli Principle
For the groundstate of the He atom with electronic configuration 1s2 we found a spatial wavefunction and energy
s = 100(1)100(2) E(1s2) = E1s + E1s + J1s1s
For the lowest excited states of the He atom with electronic configuration 1s12s1 we found two spatial wavefunctions and energies
± (1s12s1) = (2)-1/2{100(1)200(2) ± 200(1)100(2)}
E±(1s12s1) = (E1s + E2s+ J1s2s) ± K1s2s
E1s and E2s are the one-electron orbital energies from a hydrogen-like atom; they are negative quantities.Both J's and K's are positive quantities; hence
E+(1s12s1) > E-(1s12s1)
Similar orbital and energy expressions applied to the states arising from the 1s12p1 configuration:
± (1s12p1) = (2)-1/2{100(1)210(2) ± 210(1)100(2)}
E±(1s12p1) = (E1s + E2p+ J1s2p) ± K1s2p E+(1s12p1) > E-(1s12p1)
Overall, E(1s2) < E-(1s12s1) < E+(1s12s1) < E-(1s12p1) < E+(1s12p1) when good wavefunctions (or spectroscopic data) are used.
The wavefunctions s+(1s12s1) and+(1s12p1) are all symmetric with respect to interchange of the two electrons ; - (1s12s1) and - (1s12p1) are antisymmetric with respect to interchange of the two electrons.
We now wish to combine the spatial and the spin functions for the He atom. We found only one wavefunction corresponding to the 1s2 configuration- - can this function be combined with more than one of the spin functions? The spatial part of the 1s2 wavefunction is symmetric with respect to interchange of the two electrons -- the singlet spin function is antisymmetric, whereas the triplet spin functions are symmetric. Thus, we could seemingly generate a total of four wavefunctions, each a product of a spatial and a spin wavefunction. Three wavefunctions would be overall symmetric (triplet) and one would be overall antisymmetric (singlet) to particle interchange. Since spin does not occur explicitly in the electronic Hamiltonian, all these states would be degenerate in the absence of magnetic fields. Is the ground state of the He atom four-fold degenerate? Similar questions may be asked for the excited 1s12s1 configuration, except we now have two spatial wavefunctions to combine with four spin functions. Are there really eight states corresponding to the 1s12s1 electronic configuration? Eight states corresponding to the 1s12p1 electronic configuration?
From a detailed study of the absorption and emission spectra of the He atom, it was concluded that all the states spectroscopically observed were overall antisymmetric with respect to interchange of the two electrons. We express this observation in the
"Pauli Principle": The total wavefunction (spatial and spin) must be antisymmetric with respect to interchange of any pair of electrons.
Electrons are spin 1/2 particles and the Pauli Principle actually applies more generally to all half-integral spin particles (fermions). Particles (including photons) with integral spin (including 0) have wavefunctions, which are symmetric with respect to interchange of any two particles (bosons).
Thus, we find that the ground state configuration 1s2 must have a singlet (antisymmetric) spin wavefunction, since it has a symmetric spatial wavefunction. The ground state of the He atom is not degenerate after all.
For the excited configurations, 1s12s1 and 1s12p1, we get four (not eight) possibilities: the spatially symmetric + functions must be combined with the antisymmetric singlet spin function; the spatially antisymmetric - functions must be combined with the symmetric triplet spin functions. We can thus write:
Electronic configuration 1s2; Term 1S
= |s |0, 0> = 100(1)100(2)(2)-1/2[(2) - (2)]
One component forming a 1S term: S = 0; L = 0; (2S+1)(2L+1) = (1)(1) = 1
Electronic configuration 1s12s1; Term 1S
= |+(1s12s1)>|0,0> =
(2)-1/2{100(1)200(2) + 200(1)100(2)}(2)-1/2[(2) - (2)]
One component forming a 1S term: S = 0; L = 0; (2S+1)(2L+1) = (1)(1) = 1
Electronic configuration 1s12s1; Term 3S
= |-(1s12s1)>|1,+1> = (2)-1/2{100(1)200(2) - 200(1)100(2)}(2)
= |-(1s12s1)>|1,0> =
(2)-1/2{100(1)200(2) - 200(1)100(2)}(2)-1/2[(2) + (2)]
= |-(1s12s1)>|1,-1> = (2)-1/2{100(1)200(2) - 200(1)100(2)}(2)
Three components forming a 3S term: S = 1; L = 0; (2S+1)(2L+1) = (3)(1) = 3
Electronic configuration 1s12p1; Term 1P
= |+(1s12p11)>|0,0> = (2)-1/2{100(1)211(2) + 211(1)100(2)}(2)-1/2
[ (2) - (2)]
= |+(1s12p01)>|0,0> = (2)-1/2{100(1)210(2) + 210(1)100(2)}(2)-1/2
[ (2) - (2)]
= |+(1s12p-11)>|0,0> = (2)-1/2{100(1)21-1(2) + 21-1(1)100(2)}(2)-1/2
[ (2) - (2)]
Three components forming a 1P term: S = 0; L = 1; (2S+1)(2L+1) = (1)(3) = 3
Electronic configuration 1s12p1; Term3P
= |-(1s12p11)>|1,+1> = (2)-1/2{100(1)211(2) - 211(1)100(2)}(2)
= |-(1s12p11)>|1,0> = (2)-1/2{100(1)211(2) - 211(1)100(2)}(2)-1/2
[ (2) + (2)]
= |-(1s12p11)>|1,-1> = (2)-1/2{100(1)211(2) - 211(1)100(2)}(2)
= |-(1s12p01)>|1,+1> = (2)-1/2{100(1)210(2) - 210(1)100(2)}(2)
= |-(1s12p01)>|1,0> = (2)-1/2{100(1)210(2) - 210(1)100(2)}(2)-1/2
[ (2) + (2)]
= |-(1s12p01)>|1,-1> = (2)-1/2{100(1)210(2) - 210(1)100(2)}(2)
= |-(1s12p-11)>|1,+1> = (2)-1/2{100(1)21-1(2) -211(1)100(2)}(2)
= |-(1s12p-11)>|1,0> = (2)-1/2{100(1)21-1(2) - 21-1(1)100(2)}(2)-1/2
[ (2) + (2)]
= |-(1s12p-11)>|1,-1> = (2)-1/2{100(1)21-1(2) - 211(1)100(2)}(2)
Nine components forming a 3P term: S = 1; L = 1; (2S+1)(2L+1) = (3)(3) = 9
Furthermore, the optical spectrum breaks up into two parts: Lines are only seen connecting states, which are spatially symmetric to states which are also spatially symmetric; or connecting states which are spatially antisymmetric to states which are spatially antisymmetric. This is a consequence of the mathematical form of the transition moment integral, if = <i|erk|f>. For two electrons, we get
if = <i|er1 + er2|f> = (-e) <i|r1 +r2|f
The (r1 +r2) operator is symmetric with respect to interchange of the two electrons. If you try to connect a spatially symmetric state to a spatially antisymmetric state in the He atom, +-= (-e)<|r1 +r2|->, then the full integrand becomes antisymmetric to exchange of the two electrons. Hence, the transition moment integral must change sign, if you interchanged the labels of the two electrons -- which is "nonsense" and the integral must equal zero. "+" states only connect to "+" states"----
"-" states only to "-" states in electronic spectroscopy of atoms. That means that singlets only connect to singlets-- and triplets only to triplets. Hence we are guided to another selection rule for optical spectroscopy, the ‘no change in spin multiplicity’ rule: Sif = 0.
The "Pauli Principle" restricts the number of particles, which can occupy a particular state: Imagine two electrons of the same spin in the same orbital. A suitable spin wavefunction is then (1)(2), which is symmetric to interchange. If the spin function is symmetric, we must have an antisymmetric spatial function (Pauli Principle), i.e., something of the form = a(1)b(2) - b(1)a(2). But if the orbitals are also the same (a = b), then is identically zero. Hence, two electrons cannot have the same space and spin wavefunctions, a principle we normally express as: "No two electrons can be in the same state" or (for atoms) "No two electrons can have the same quantum numbers".
Why isn't the electronic ground state configuration for Li 1s3, with spatial part = 1s(1)1s(2)1s(3)? Because we need an antisymmetric spin function to go with this spatially symmetric wavefunction. And it turns out to be mathematically impossible to write a spin function (different from zero) for three electrons using only and , which is antisymmetric with respect to interchange of any two electrons. Hence the 1s3 assumption must be wrong.
12.4The Slater Determinant
Look at the wavefunction for the ground state of the He atom:
= |s |0,0> = 100(1)100(2)(2)-1/2[ (2) - (2)]
We can write this wavefunction as a determinant composed of spin-orbitals (spin- and space-labeled orbitals), viz.,
= |s |0,0> = 100(1)100(2)(2)-1/2[ (2) - (2)] =
|100(1) 100(1)|
(2)-1/2| | =(2)-1/2det(100(1)100(2)
|100(2)100(2)|
The last expression
(2)-1/2det(100(1)100(2))|100(1)100(2)|
is convenient shorthand notation, where only the diagonal elements (i.e., the electronic configuration in terms of spinorbitals) of the determinant are explicitly indicated (surrounded by a set of vertical bars to indicate determinant and the normalization factor is "understood").
A determinant nicely displays the antisymmetry of the overall wavefunction. Interchanging the labels of two electrons corresponds to interchanging two rows, and a determinant automatically changes sign if two rows are interchanged. If the two electrons were in the same spin-orbital, then two columns would be the same in the determinant; a determinant vanishes if it has two identical columns (or rows).
Thus, we generally write the wavefunctions for many-electron atoms (or molecules) as a Slater determinant. Such a wavefunction automatically obeys the Pauli Principle. The wavefunction for the ground state configuration of the Li atom (1s22s1, 2S term) may thus be written
|s2s11/2,1/2> =
|100(1) 100(1) 200(1)
(6)-1/2100(2)100(2)200(2)| =
|100(3) 100(3) 200(3)
(6)-1/2det(100(1)100(2)200(3)|100(1)100(2)200(3)
The other spin component for this 2S term is simply
|s2s11/2,-1/2> =
|100(1) 100(1) 200(1)
(6)-1/2100(2)100(2)200(2)| =
|100(3) 100(3) 200(3)
|100(1)100(2)200(3)
and, of course, the two components (|s2s11/2,1/2> and |s2s11/2,-1/2>) are energetically degenerate.
For an N electron system, a suitable wavefunction is the Slater determinant expressed in spin-orbitals (atomic or molecular) as
= (N!)-1/2det((1)2(2)3(3)4(4)5(5)...N(N) = |(1)2(2)3(3)4(4)5(5)...N(N)
1