Notes#12
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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,i
ML = L, L-1, L-2,...,-L
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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,i
MS = 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).
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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
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S 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
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due to relativistic effects, in particular the presence of what 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.
12.2 Spin Wave Functions for Two Electrons
The two possible spin wave functions 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)
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|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.
12.3 The Pauli Principle
For the ground state of the He atom with electronic configuration 1s2
we found a spatial wave function 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 wave functions 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)
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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 wave functions (or spectroscopic data) are used.
The wave functions 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 wave function corresponding to the 1s2
configuration - - can this function be combined with more than one of
the spin functions? The spatial part of the 1s2 wave function 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 wave
functions, each a product of a spatial and a spin wave function. Three
wave functions 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 wave
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functions 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 wave function (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 wave
functions, 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 wave function, since it has a symmetric
spatial wave function. 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:
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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)]
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Three components forming a 1P term: S = 0; L = 1; (2S+1)(2L+1) = (1)(3) = 3
Electronic configuration 1s12p1; Term 3P
= |-(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
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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 wave function 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 wave functions, 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
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antisymmetric spin function to go with this spatially symmetric wave
function. 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.
12.4 The Slater Determinant
Look at the wave function 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 spinorbitals (spin- and space-labeled orbitals), viz.,
= |s |0,0> = 100(1)100(2)(2)-1/2[ (2) - (2)] =
|100(1)
(2)-1/2|
100(1)|
| = (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 spin orbitals) of the
determinant are explicitly indicated (surrounded by a set of vertical bars
to indicate determinant and the normalization factor is "understood").
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A determinant nicely displays the antisymmetry of the overall
wave function. 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 spinorbital, 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 wave functions for many-electron
atoms (or molecules) as a Slater determinant. Such a wave function
automatically obeys the Pauli Principle. The wave function 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)-1/2 100(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)-1/2 100(2)
100(2)
200(2)| =
|100(3)
100(3)
200(3)
|100(1)100(2)200(3)
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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 wave function 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)
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