MODULE 9

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MODULE 9
MANY ELECTRON ATOMS  With the exception of
hydrogen the elements of the Periodic Table
consist of atoms that are made up of 2 or more
electrons Thus in our Schrödinger equation for
the energy we have multiple kinetic energy
operators and multiple potential energy operators
to contend with. The simplest many electron
atom is that of helium with Z 2, i.e., two
electrons and one nucleon having 2 charge.
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As with the H atom we neglect the motion of the
atom as a whole through space. Thus we need to
account for the kinetic energy of the two
electrons about the center of mass.
Also there are PE terms for the various Coulombic
attractions and repulsions present.
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The hamiltonian operator in atomic units is
1, 2, ..., n identifies the spatial coordinates
of the n electrons. For He atom n 2, and the
hamiltonian becomes
or
where h(1) and h(2) are one electron
hamiltonians. The third term is a two electron
operator (inter-repulsion of e-) This term
prevents the Schrödinger equation from being
separable into a pair of single variable
equations.
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The 3-particle He atom is impossible to solve
exactly and we need to resort to approximation
methods, such as variation theory. A very obvious
(but drastic) approximation is the one we already
used in our variation treatment, viz., to simply
ignore the term. This led to the concept of
effective nuclear charge This is the
independent electron approximation The h(1)
and h(2) terms in are hamiltonians for a pair of
hydrogenic atoms containing a Z 2 nucleus. The
wavefunctions are the 1s, 2s, 2p, etc atomic
orbitals. Ignoring the 1/r12 term in the full
hamiltonian generates an approximate hamiltonian
operator
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Note that the 1s, 2s, etc atomic orbitals are
eigenfunctions of Happrox but not of the full
hamiltonian. In this approximation (zero
order) the ground state of the He atom can be
written as 1s(1)1s(2), where the (1) and (2)
refer to the individual electrons in the atom.
Both electrons are associated with the same AO
But now we need to worry about
distinguishability of electrons
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Are Electrons Individually Recognizable? In the
macroscopic world we can distinguish individual
objects in a group of identical objects by their
positions relative to a fixed point or, if they
are in motion, by their individual velocity
vectors. Their different trajectories/positions
make them distinguishable. Electrons are also
identical, same rest mass, same intrinsic angular
momentum and same charge. There are no ways to
tag individual electrons. In the
sub-microscopic world of the electron however,
the momentum and position operators fail to
commute and therefore these two observables
cannot be simultaneously specified with arbitrary
precision. If the momentum of an electron is
sharply defined, its position is very uncertain,
and vice versa.
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Thus there is no property of one of the electrons
in an atom that we can use to distinguish it from
another electron. If there are several
electrons occupying different AOs we cannot state
which of the electrons occupies which orbital.
The description of the electronic configuration
of He ground state as 1s(1)1s(2) is satisfactory
because it states that both electrons are
occupying the same AO (1s) and there is no
attempt to distinguish between the electrons
1s2 . However, to write the zeroth-order
wavefunction of the excited state as 1s(1)2s(2)
is not allowed, since to do so implies that
electron (1) is in the 1s AO and electron (2) is
in 2s. This implies that we can somehow
distinguish electron (1) from electron (2) but
we cannot.
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Instead we must use superposition functions such
as This allows for all combinations of electrons
and AOs and does not specify that a given
electron is in a particular AO The wavefunction
of a system of particles is a function of all the
variables that pertain, both spatial and spin.
For electron (1) in a set of n, the variables
are x1, y1 and z1 (space) and s1(spin). For
brevity we denote all of these variables for each
of n electrons as qn. Thus the wavefunction of
a system of n identical particles can be written
as y (q1, q2, qn).
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Define the permutation (or exchange) operator
Pjk as the operator that exchanges all the space
and spin coordinates of the particles (j) and (k)
in the n-particle system. For a two electron
system (He atom), writing the wavefunction as
y(q1, q2) and the exchange operator as P1,2 , we
see thus the operator has exchanged the
coordinates of the electrons. This will not
affect the state of the system since the
electrons are not distinguishable (an electron is
an electron is an electron). It serves only to
change the labels we placed on the electrons.
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Application of the exchange operator a second
time will bring the system back to the original
description, so we can say where 1 is the unit
operator. It can be shown that the eigenvalues
of any operator whos square is the unit operator
are 1 and 1. Thus when a wavefunction is an
eigenfunction of P1,2 then
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If the operation retains the sign of the function
we say that the function is symmetric wrt
interchange of electrons (1) and (2). If the
operation changes the sign of the function we say
that the function is antisymmetric wrt the
interchange. The conclusion applies in the
general case A wavefunction for a system of n
identical particles must be symmetric or
antisymmetric with respect to every possible
interchange of any two of the particles. Since
the particles are identical it is not conceivable
that some interchanges would be symmetric and
others antisymmetric.
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The Periodic Table of elements provides abundant
evidence (see later) that for systems of
electrons (atomic states) only the antisymmetric
result is found and we can formulate an important
quantum mechanical postulate that The
wavefunction of a system of electrons must be
antisymmetric with respect to the interchange of
any two electrons. Wolfgang Pauli employed
relativistic quantum field theory to show that
fermions (electrons and other particles with
half-integral spin s 1/2, 3/2,) require
antisymmetric wavefunctions, whereas bosons (s
0, 1, 2, and so on) require symmetric
wavefunctions.
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To demonstrate the effect of the Pauli principle
for identical fermions we start with the
recognition ( from the postulate) that the
antisymmetry requirement is Now assign the same
four coordinates to two (identical) electrons,
I.e. two electrons occupy the same AO. Thus, x1
x2, y1 y2, z1 z2, and s1 s2 thus q1
q2. 2y 0 and the probability of finding the
two electrons in the orbital defined by y is
zero.
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Thus two electrons having the same spin function
(a, say) have zero probability of being found
with the same values of space coordinates, in
other words they cannot co-exist in the same
region of space (AO). This Pauli repulsion
forces electrons of the same spin to keep apart
from one another, e.g., in different orbitals.
In other words the Pauli principle requires
that no two electrons can have the same four
quantum numbers, n, l, ml, and s
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The Helium Atom The ground state of He in the
zeroth order can be written as 1s(1)1s(2), i.e.,
both electrons occupying the same orbital. Now
we must consider the spin state of the two
electrons and at first sight there are 4 spin
combinations The indistiguishability of
identical particles is clearly upheld in the
first pair, since both electrons have the same
spin function. The second two combinations show
a particular electron with a particular spin
function, and this violates the principle. the
first and second are symmetric () and the third
and fourth are neither nor -, with respect to
interchange.
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These last two then are unacceptable. Moreover,
even though the first and second spin
combinations are acceptable in themselves, when
coupled with the symmetric 1s(1)1s(2) spatial
function, we find the Pauli principle forbids the
combinations and since they are both overall
symmetric to electron interchange thereby
violating the Pauli rule.
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The way to proceed is to seek a spin function
that is an antisymmetric superposition. The two
candidates for this are (no conflict with
electron distinguishability) These are the
symmetric () and antisymmetric (-)
eigenfunctions of P1,2, the exchange operator we
combine the antisymmetric one with the symmetric
spatial function 1s(1)1s(2) to make an overall
antisymmetric total function This is the total
wave function for He ground state (in zero order)
including spatial and spin contributions. It is
a single state -- no degeneracies.
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He in an excited electronic state A 1s electrons
is promoted to the n 2 level which could be
described as 1s12s1 or 1s12p1x. In
non-hydrogenic systems the 2s orbital is at a
slightly lower energy than any of the three 2p
orbitals. The two-electron combination might be
written as 1s(1)2s(2) or 1s(2)2s(1), but these
descriptions imply that we can identify electrons
(1) and (2) as being in the designated orbitals.
This clearly violates the concept of
indistiguishability. Moreover, the two product
functions are neither symmetric nor antisymmetric
on electron exchange and thus cannot contribute
to a total wavefunction that must be overall
antisymmetric with respect to electron exchange.
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As before we use a superposition that can satisfy
the symmetry requirements. The sum function is
symmetric and the difference function is
antisymmetric with respect to electron
exchange The symmetric spatial combination can
conform to the Pauli principle of He (1s2s) in a
product with an antisymmetric spin term The
antisymmetric spatial combination can conform to
the Pauli principle in a product with a symmetric
spin term. Thus, four total wavefunctions
satisfy these requirements for the 1s2s excited
state of He.
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And the configuration 1s2s shows 4 distinct
states. To obtain the state energies we need to
operate on the individual wavefunctions with the
full hamiltonian. But we have used the
independent electron approximation (ignored the
inter-electron repulsion term 1/r12) Therefore,
the functions are not eigenfunctions of H(1,2)
and we cannot simply compare the eigenvalues
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However, we can calculate the average energies by
evaluating the expectation values of
with the four functions listed. Note that the
usual denominator is missing since both space and
spin parts are normalized. Nothing in the
operator we are using will interact with any of
the spin contributions in the four wavefunctions,
and since the spin terms are normalized, their
integrals are unity. Therefore, the average
energies will be determined by the spatial
contributions, and since three of the functions
contain the same (antisymmetric) spatial part,
they form a three-fold degenerate set.
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Only ys,a has a different spatial part and
therefore we can anticipate this function to have
a different energy from the three ya,s functions.
The excited state of He (1s2s) will therefore
have two energy states, one of which is triply
degenerate. To find what the energy values are
we need to grind through some integrations see
Appendix I There we find the conclusion that the
two energies are given by where the first two
terms on the RHS are the energies of the 1s and
2s states for He (Z2), respectively.
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The integrals J (the Coulomb integral) and K (the
Exchange integral) are J is always positive
because it describes the Coulombic interaction
between the two electrons (1s1s and 2s2s).
The interaction is repulsive, and thus the
energy of the combination is raised (less
negative). The integral K has the product
functions in the integrand differing by exchange
of electrons, hence its name. As Appendix I
explains K has negative and positive
contributions but is positive overall and not as
large as J.
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The procedure in the Appendix shows that the
triply degenerate level is lower in energy than
the non-degenerate level and the separation is
2K. We can think of the J term as resulting
from homogeneous/time-averaged electron densities
in the charge-clouds of the orbitals We can
think of the K term as being a correction arising
from the inhomogeneities that occur as a
consequence of electrons wanting to avoid each
other.
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For example, if we approximate the two linear
combination spatial terms in our wavefunctions as
follows where a is a function of the radius
vector r1 and b is a function of the radius
vector r2, then as r1 approaches r2 so y- tends
to vanish, i.e., the electrons tend to avoid each
other in the difference combination. This leads
to a Fermi hole. In a magnetic field the
triply degenerate level is split into three
discrete energy states-it is a triplet of states,
or a triplet state.
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Slater Determinants When considering the He
atom (ground and excited states) it was
straightforward to write down antisymmetric total
wavefunctions as multiples of spatial and spin
parts. It was easy because we had to deal with
just two electrons. It is another matter to
construct an asymmetric wave function for N
electrons, by inspection. In 1930s Slater
introduced the concept of using determinants to
construct the required wavefunctions. For
example take the He ground state wavefunction
that we have shown to be
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MODULE 9
This wavefunction is composed of a symmetric
space function and an antisymmetric spin function
so it is overall antisymmetric with respect to
interchange of the pair of electrons-as Pauli
requires. As a determinant this becomes Show
this by expanding the determinant and
rearranging. The individual terms in the
determinant (1sa) are spinorbitals that describe
the spatial and spin state of the associated
electron. The determinant has the individual
spinorbitals along the rows and the associated
electrons in the columns (others reverse this
procedure, but this has no effect since
interchanging columns and rows leaves the
determinant unchanged).
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Determinants such as the above are called Slater
determinants The wavefunction they represent is
a determinantal wavefunction If the labels (1)
and (2) are interchanged in the determinant this
places electron (1) in column 2 and electron (2)
in column 1 This interchanges two columns of
the determinant and changes its sign (see
Barrante, chapter 9) The wavefunction must also
change sign and thus the determinant represents
an antisymmetric function.
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Furthermore, suppose that our two electrons in
the 1s orbital have the same spin, say a. This
changes the determinant as follows Now two
rows (the first and second) are identical This
causes the determinant to vanish (for all
orders) If the Slater determinant vanishes, the
wavefunction it represents also vanishes Thus a
state in which two electrons have identical sets
of space and spin variables does not exist see
Module 9 for shorthand ways of writing SDs
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Now examine the wavefunction for the Li atom
(Z3) using SDs. Firstly is the 1s3
configuration possible? If so The top row
is the principal diagonal Two of the rows (first
and third) are identical, therefore the
determinant vanishes and the 1s3 configuration is
not possible.
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So the next to try is the configuration 1s22s
The 2s electron has been given a spin, but this
is purely arbitrary Thus the ground state of Li
exists as a pair of degenerate states, it is a
doublet of states, or a doublet state. No two
rows are identical and interchanging the labels
(1) and (2) switches the first and second columns
The determinant changes sign, indicating the
wavefunction is antisymmetric, (Pauli requirement)
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Slater Orbitals In Module 8 we applied Variation
Theory to the He ground state in order to
evaluate its energy. In the calculation we used
the atomic number, Z, as a variational parameter
and the trial function was taken as
Minimization of the energy with respect to
variation of Z whereas the experimental result
is 2.9033 au One variational parameter brings us
within 2 of the actual value.
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At first sight this appears to be quite good
agreement However, the ionization energy (IE) of
He is given by the energy difference between He
and He The first term can be calculated
exactly because He is a hydrogenic ion with Z
2, and we can use the variation theory value of
EHe1s as the second term
The experimental value of IE is 2372 kJ mol-1.
Thus our calculated value is ca 150 kJ mol-1 in
error, about the strength of some chemical
bonds We need to do better
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MODULE 9
One way to improve is to change our trial
function. In the He1s calculation we chose the
hydrogenic 1s wavefunction, but almost any
function, or combination could be used. In
1930 Slater introduced a new set of orbitals, now
called Slater orbitals which are of the form Y
is the appropriate spherical harmonic z Z/n for
hydrogenic orbitals, but in the Slater-type
orbitals (STOs) it becomes an arbitrary
parameter. The radial parts of STOs have no
nodes, unlike the hydrogenic counterparts for n gt
1.
is a normalizing factor
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MODULE 9
In the calculation of He1s that we did in Module
8 we effectively used the S100. Thus the trial
wavefunction, following the procedure in Module 8
and remembering that l 0 is giving our
earlier values of z 1.6875 and Emin -2.8477
au which gave a value of IP that is about 150
kJ mol-1 in error. A slight improvement can be
gained by also letting n vary (in the normalizing
factor and in the exponent of the r term).
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MODULE 9
Chemists find the concept of orbitals very
satisfactory, especially when we think about
chemical bonds Within the orbital concept,
improvements to Emin can be achieved by using
trial functions of even more flexibility Such as
by defining the spatial function as a product of
one electron orbitals that are completely general
in nature This leads us to a limiting value for
Emin -- the Hartree-Fock limit. This is the
best that can be done with retention of the
orbital concept. For He1s the Hartree-Fock
limit is Emin -2.8617 au.
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MODULE 9
Removing the restriction that the trial function
be a product of one-electron orbitals and
expressing it as a series of variational
parameters can lead to improvements in Emin
Pekeris (1959) used 1078 parameters to obtain
Emin -2.903724375 au in exact agreement with
the true value. In this approach, the emphasis
is on obtaining the exact energy by crunching
numbers and the idea of atomic orbitals has been
abandoned. In modern times, the procedure is
usually to find the H-F orbitals (that generate
the H-F limit) and then correct the energy by
applying a Perturbation Theory treatment (see
later).
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MODULE 9
It is instructive for us to see how the H-F
procedure works in a simple case, viz., the He
atom. Any procedure that is going to be
successful clearly must account for the
interaction between electrons in some way. In
the HF approach the electron-electron repulsion
is treated in an average way each electron is
considered to be moving within the field of the
nucleon and the average field of the remainder of
the electrons. As we shall see this is not
detailed enough, but it will get us started. We
start by writing the two-electron wavefunction as
the product of one-electron orbitals, as outlined
above
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MODULE 9
The functions f(rj) are identical since we allow
both electrons in He 1s2 to be in the same
orbital under the Pauli restriction. The
probability distribution of electron 2 (its
charge-cloud) is given by f(r2)f(r2)dr2 and
thinking classically, we can interpret this as a
charge density. The average potential energy
that is experienced by electron 1 at some value
of r1 because of the presence of the charge-cloud
of electron 2 is (in au) given by
So we can define a one-electron hamiltonian that
includes this averaged potential energy term
(Coulomb operator)
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There is a similar equation for the other
electron but since the wavefunctions are
identical we need only consider one equation.
This is the H-F equation for He 1s2 and solving
this equation will generate the optimum
wavefunction for this atom within the orbital
concept and the H-F approximation. But there
appears to be a problem The operator in the
Schrödinger equation depends on Vav,1 which
depends in turn on f(r2). Thus we need to know
the wavefunction solution to the Schrödinger
equation before we can write down its operator.
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MODULE 9
The way forward uses the self-consistent field
method (Hartree) First we guess a form for
f(r2) and use it to evaluate Vav,1(r1)
using then we solve for f(r1) using After
the first cycle the output f(r) will differ from
the input f(r), unless we had guessed the correct
function in the first place. Use the calculated
f(r) to estimate a new Vav,1(r1) and thence
another hamiltonian and subsequently another
f(r1). (Recall that the two orbital functions
are identical).
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Then the cycle is repeated until the output and
input wavefunctions are sufficiently close, or
are self-consistent. At this point we have the
Hartree-Fock orbitals. As implied above single
functional orbitals do not yield the best energy
and in practice linear combinations of STOs are
employed. The optimum result (five STOs with
different coefficients) for the He ground state,
as stated earlier, was found to give an energy
that was 109 kJ mol-1 too high, It is the best
that can be achieved within the orbital
approximation.
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The way the Coulomb operator was calculated in
the above procedure treats each electron as
though it were moving in the time-averaged field
of the others in the system. However, this is
only partly true because electrons are mutually
repulsive and they tend to avoid each other. In
other words their motions are correlated. The
HF procedure is not constructed to take this into
account and it generates EHF gt Eexpt. The
difference between the HF limit and the true
energy is the correlation energy. In the He
case this was 109 kJ mol-1, only ca 1.5 of the
total electronic energy, but still a large
absolute error.
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