Let us take a paraboloid Ex y= x2+ y2 This function has, of course, a minimum at 0 0, but the minimum is of no interest to us.. What we want to find is a minimum of E, but only when x an
Trang 1996 M SLATER–CONDON RULES
Fig M.1. Four Slater–Condon rules (I, II, III, IV) for easy remembering On the left side we see pictorial representations of matrix elements
of the total Hamiltonian ˆ H The squares inside the brackets represent the Slater determinants Vertical lines in bra stand for those spinorbitals, which are different in bra and in ket functions.
On the right we have two square matrices collect-ing the hij’s and ij|ij − ij|ji for i j = 1 N The dots in the matrices symbolize non-zero ele-ments.
and G12= 0 This happens because operators ˆF and ˆG represent the sum of, at most, two-electron operators, which will involve at most four spinorbitals and there will always be an extra overlap integral over the orthogonal spinorbitals.9
The Slater–Condon rules are schematically depicted in Fig M.1
Trang 2N LAGRANGE MULTIPLIERS
METHOD
Imagine a Cartesian coordinate system of n+ m dimensions with the axes labelled
x1 x2 xn+m and a function1 E(x), where x= (x1 x2 xn+m) Suppose
that we are interested in finding the lowest value of E, but only among such x
that satisfy m conditions (conditional extremum): conditional
extremum
Wi(x)= 0 (N.1) for i= 1 2 m The constraints cause the number of independent variables to
be n
If we calculated the differential dE at point x0, which corresponds to an
ex-tremum of E, then we obtain 0:
0=
n+m
j=1
∂E
∂xj
0
dxj (N.2)
where the derivatives are calculated at the point of the extremum The
quanti-ties dxjstand for infinitesimally small increments From (N.2) we cannot draw the
conclusion that the (∂x∂Ej)0are equal to 0 This would be true if the increments dxj
were independent, but they are not Indeed, we find the relations between them by
making differentials of conditions Wi:
n+m
j =1
∂Wi
∂xj
0
dxj= 0 (N.3)
for i= 1 2 m (the derivatives are
calculated for the extremum)
This means that the number of truly
independent increments is only n Let us
try to exploit this To this end let us
mul-tiply each equation (N.3) by a number "i
(Lagrange multiplier), which will be fixed
Joseph Louis de Lagrange (1736–1813), French mathe-matician of Italian origin, self-taught; professor at the Ar-tillery School in Turin, then at the École Normale Supérieure
in Paris His main achieve-ments are in variational cal-culus, mechanics, and also in number theory, algebra and mathematical analysis.
in a moment Then, let us add together all the conditions (N.3), and subtract the
result from eq (N.2) We get
1 Symbol E is chosen to suggest that, in our applications, the quantity will have the meaning of energy.
997
Trang 3998 N LAGRANGE MULTIPLIERS METHOD
n+m
j =1
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
dxj= 0
where the summation extends over n+ m terms The summation may be carried out in two steps First, let us sum up the first n terms, and afterwards sum the other terms
n
j=1
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
dxj+
n+m
j=n+1
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
dxj= 0
The multipliers "ihave so far been treated as “undetermined” Well, we may force them to make each of the terms in the second summation equal zero2
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
= 0 for j = n + 1 n + m (N.4) Hence, the first summation alone is 0
n
j =1
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
dxj= 0
which means that now we have only n increments dxj, and therefore they are
in-dependent Since for any (small) dxj, the sum is always 0, the only reason for this could be that each parenthesis[ ] individually equals zero
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
= 0 for j = 1 n
This set of n equations (called the Euler equations) together with the m
condi-Euler equation
tions (N.1) and m equations (N.4), gives a set of n+ 2m equations with n + 2m unknowns (m epsilons and n+ m components xiof the vector x0)
For a conditional extremum, the constraint Wi(x)= 0 has to be satisfied for
i= 1 2 m and
∂E
∂xj
0
−
i
"i
∂Wi
∂xj
0
= 0 for j = 1 n + m
The xifound from these equations determine the position x0of the condi-tional extremum E
Whether it is a minimum, a maximum or a saddle point, is decisive for the analy-sis of the matrix of the second derivative (Hessian) If its eigenvalues calculated at
x0 are all positive (negative), it is a minimum3(maximum), in other cases it is a saddle point
2 This is possible if the determinant which is built of coefficients (∂Wi
∂xj)0is non-zero (this is what we have to assume) For example, if several conditions were identical, the determinant would be zero.
Trang 4N LAGRANGE MULTIPLIERS METHOD 999
Example 1 Minimizing a paraboloid going along a straight line off centre. Let us
take a paraboloid
E(x y)= x2+ y2 This function has, of course, a minimum at (0 0), but the minimum is of no
interest to us What we want to find is a minimum of E, but only when x and y
satisfy some conditions In our case there will only be one:
W =1
2x−3
2− y = 0 (N.5)
This means that we are interested in a minimum of E along the straight line
y=1
2x−3
2
The Lagrange multipliers method works as follows:
• We differentiate W and multiply by an unknown (Lagrange) multiplier " thus
getting: "(12dx− dy) = 0
• This result (i.e 0) is subtracted4from dE= 2x dx + 2y dy = 0 and we obtain
dE= 2x dx + 2y dy −1
2ε dx+ ε dy = 0
• In the last expression, the coefficients at dx and dy have to be equal to zero.5In
this way we obtain two equations: 2x−1
2"= 0 and 2y + " = 0
• The third equation needed is the constraint y =1
2x−3
2
• The solution to these three equations gives a set of x y " which corresponds
to an extremum We obtain: x= 3
5, y= −6
5, "= 12
5 Thus we have obtained, not only the position of the minimum (x=3
5, y= −6
5), but also the Lagrange multiplier " The minimum value of E, which has been encountered along the
straight line y=1
2x−3
2is equal to E(35 −6
5)= (3
5)2+ (−6
5)2=9+3625 =9
5
Example 2 Minimizing a paraboloid moving along a circle (off centre). Let us take
the same paraboloid (N.5), but put another constraint
W = (x − 1)2+ y2− 1 = 0 (N.6) This condition means that we want to go around a circle of radius 1, centred at
(1 0), and see at which point (x y) we have the lowest value6of E The example
is chosen in such a way as to answer the question first without any calculations.
Indeed, the circle goes through (0 0), therefore, this point has to be found as the
minimum Beside this, we should find a maximum at (2 0), because this is the point
on the circle which is most distant from (0 0)
Well, let us see whether the Lagrange multipliers method will give the same
result
After differentiation of W , multiplying it by the multiplier ", subtracting the
result from dE and rearranging the terms, we obtain
4 Or added – no matter (in that case we get a different value of ").
5 This is only possible now.
6 Or, in other words, we intersect the paraboloid with the cylindrical surface of radius 1 and the
cylin-der axis (parallel to the axis of symmetry of the paraboloid) is shifted to (1 0).
Trang 51000 N LAGRANGE MULTIPLIERS METHOD
dE= 2x− "(2x − 2) dx+ 2y(1 − ") dy = 0 which (after forcing the coefficients at dx and dy to be zero) gives a set of three equations
2x− "(2x − 2) = 0 2y(1− ") = 0 (x− 1)2+ y2= 1
Please check that this set has the following solutions: (x y ")= (0 0 0) and (x y ")= (2 0 2) The solution (x y) = (0 0) corresponds to the minimum, while the solution (x y)= (2 0) corresponds to the maximum.7 This is what we expected to get
Example 3 Minimizing the mean value of the harmonic oscillator Hamiltonian. This
example is different: it will pertain to the extremum of a functional.8We are often going to encounter this in the methods of quantum chemistry Let us take the energy
functional
E[φ] =
∞
−∞dx φ
∗Hφˆ ≡ φ ˆHφ
where ˆH stands for the harmonic oscillator Hamiltonian:
ˆ
H= − ¯h2 2m
d2
dx2+1
2kx
2
If we were asked what function φ ensures the minimum value of E[φ], such a function could be found right away, it is φ= 0 Yes, indeed, because the kinetic energy integral and the potential energy integral are positive numbers, except in the situation when φ= 0, then the result is zero Wait a minute! This is not what
we thought of We want φ to have a probabilistic interpretation, like any wave function, and therefore φ|φ = 1, and not zero Well, in such a case we want
to minimize E[φ], but forcing the normalization condition is satisfied all the time Therefore we search for the extremum of E[φ] with condition W = φ|φ − 1 = 0
It is easy to foresee that what the method has to produce (if it is to be of any value)
is the normalized ground-state wave function for the harmonic oscillator How will the Lagrange multipliers method get such result?
The answer is on p 198
7 The method does not give us information about the kind of extremum found.
Trang 6O PENALTY FUNCTION METHOD
Very often we are interested in the minimization of a (“target”) function,1i.e in finding such values of variables, which ensure a minimum of the function when some constraints are satisfied Just imagine hiking in the Smoky Mountains: we want to find the point of the lowest ground elevation provided that we hike along
a straight line from, say, Gatlinburg to Cherokee
Suppose the target function for minimization (which corresponds to the eleva-tion of the ground in the Smoky Mountains region) is the funceleva-tion f (x1 x2
xn+m), but the variables xihave to fulfil m equations (“constraints”):
φi(x1 x2 xn+m)= 0 for i = 1 2 m
For such tasks we have at least three possibilities The first is to eliminate m variables (by using the conditions) and express them by others In this way the target function f takes into account all the constraints and depends only on n independent variables Then the target function is to be minimized The second possibility is to use the Lagrange multipliers method (see Appendix N) In both cases there is, however, the complication that the conditions to be satisfied might
be quite complex and therefore solution of the corresponding equations may be difficult to achieve An easier solution may be to choose a penalty method The idea behind the penalty method is quite simple Why go to the trouble of trying
to satisfy the conditions φi= 0, when we could propose the following: instead of function f let us minimize its modification
F= f +
m
i=1
Kiφ2i
where the penalty coefficients Ki> 0 are chosen to be large.2When minimizing
F we admit that the conditions φi= 0 could be non-satisfied, but any attempt to
violate them introduces to F a positive contributionm
i=1Kiφ2i This means that, for minimization of F , it would always be better to explore such points in space (Fig O.1) for whichm
i=1Kiφ2i = 0 If the K’s are large enough, the procedure will force the choice φ2
i = 0, or φi= 0 for i = 1 2 m, and this is what has to
be satisfied during minimization
Note that the task would be much more difficult if φ2
i had more than one mini-mum that corresponds to φi= 0 This penalty method is worth keeping in our tool
1 If we change the sign of the target function, the task is equivalent to maximization.
2 This means a high penalty.
1001
Trang 71002 O PENALTY FUNCTION METHOD
Fig O.1. How does the penalty method work? We have to minimize f (x y), but under the condition that x and y satisfy the equation φ1(x y) = 0 (black line) Function f (x y) exhibits a single minimum
at point B, but this minimum is of no interest to us, because we are looking for a conditional minimum.
To find it we minimize the sum f (x y) + Kφ 2
1 with the penalty function Kφ21 0 allowing any deviation
from the black line φ1(x y) = 0 However, going off this line does not pay, because this is precisely what switches the penalty on As a result, at sufficiently large K we obtain the conditional minimum W This
is what the game is all about.
box, because it is general and easily applicable For the method to work, it has to have a sufficiently large K However, if K is too large, the numerical results might
be of poor quality, since the procedure would first of all take care of the penalty, paying little attention to f It is recommended that we take a few values of K and check whether the results depend on this
As an example of the penalty function method, let us take the docking of two molecules Our goal is to give such values of the atomic coordinates of both mole-cules as to ensure the contacts of some particular atoms of both molemole-cules within some precise distance limits for the contacting atoms The task sounds trivial, until
we try to accomplish it in practice (especially for large molecules) The goal can be rather easily achieved when the penalty function method is used We do the
Trang 8follow-O PENALTY FUNCTION METHOD 1003
we simply add a penalty for not satisfying the desired contacts For a single pair of
the atoms (a contact) the penalty could be set as
K(r− r0)2 where r stands for the distance of the atoms, and r0is the optimum (desired)
con-tact distance At a chosen starting geometry the atoms are far from achieving the
optimum distance, and therefore the force field energy is supplemented by a large
distance-dependent penalty The energy is so high that the minimization
proce-dure tries to remove the penalty and relax the system Often this can be done in
only one way: by docking the molecules in such a way as to achieve the proper
contact distance
Trang 9P MOLECULAR INTEGRALS WITH
The normalized 1s spherically symmetric Gaussian Type Orbital (GTO) centred at the point shown by the vector Rpreads as
χp≡
2αp
π
3
exp
−αp|r − Rp|2
The molecular integrals usually involve, at most, four such orbitals: χp χq
χr χs, with corresponding centres Rp Rq Rr Rs, and exponents αp αq αr αs, respectively Since any product of the 1s GTOs represents a (non-normalized) 1s GTO centred between the centres of the individual GTOs (see p 359), let
us denote the centre of χpχq by Rk= αp Rp+α q Rq
α p +α q , and the centre of χrχs by
Rl=α r Rr+α s Rs
α r +α s Then all the integrals needed are as follows:1
overlap integral:
Spq= χp|χq =
4αpαq (αp+ αq)2
3
exp
−α
pαq
αp+ αq|Rp− Rq|2
; (P.1)
kinetic energy integral:
Tpq=
χp
−12
χq
= αpαq
αp+ αq
3− 2αpαq
αp+ αq|Rp− Rq|2
Spq; (P.2)
nuclear attraction integral:2
Vpqα =
χp
|r − R1 α|χq
= 2
(
αp+ αq
π F0
(αp+ αq)|Rα− Rk|2
Spq; (P.3)
electron repulsion integral:
(pr|qs) = (χpχr|χqχs)=
χp(1)∗χq(1) 1
r12χ
∗
r(2)χs(2) dv1dv2
=√2
π
'
αp+ αq√
αr+ αs
'
αp+ αq+ αr+ αs
F0
(αp+ αq)(αr+ αs)
αp+ αq+ αr+ αs |Rk− Rl|2
SpqSrs (P.4)
1S.F Boys, Proc Roy Soc (London) A200 (1950) 542.
2 In order to interpret this integral (in a.u.) as the Coulombic attraction of the electronic charge
χ ∗
p (1)χq(1) by a nucleus (of charge Z, located at Rα ) we have to multiply the integral by −Z.
Trang 10P MOLECULAR INTEGRALS WITH GAUSSIAN TYPE ORBITALS 1s 1005
with F0defined as3
F0(t)=√1
t
√ t
0
exp
−u2 du (P.5) Note that for an atom (all the centres coincide) we have t= 0 and F0(0)= 1
Do these formulae work?
The formulae look quite complex If they are correct, they have to work in several
simple situations For example, if the electronic distribution χ∗p(1)χq(1) centred at
Rkis far away from the nucleus, then we have to obtain the Coulombic interaction
of the charge of χ∗p(1)χq(1) and the nucleus The total charge of the electron cloud
χ∗
p(1)χq(1) is obviously equal to Spq, and therefore Spq
|R α −R k |should be a very good
estimation of the nuclear attraction integral, right?
What we need is the asymptotic form of F0(t), if t→ ∞ This can be deduced
from our formula for F0(t) The integrand is concentrated close to t= 0 For t →
∞, the contributions to the integral become negligible and the integral itself can
be replaced by∞
0 exp(−u2) du=√π/2 This gives[F0(t)]asympt=√π
2 √
t and
Vpqα
asympt= 2
(
αp+ αq
π F0
(αp+ αq)|Rα− Rk|2
Spq
= 2
(
αp+ αq
π
√ π
2 (αp+ αq)|Rα− Rk|2
Spq= Spq
|Rα− Rk| exactly as we expected If χp= χq, then Spq= 1 and we simply get the Coulombic
law for the unit charges It works
Similarly, if in the electronic repulsion integral χp= χq, χr= χs and the
dis-tance|Rk− Rl| = R is large, then we should get the Coulombic law for the two
point-like unit charges at distance R Let us see Asymptotically
(pr|qs)asympt=√2
π
'
αp+ αq√
αr+ αs
'
αp+ αq+ αr+ αs
F0
(αp+ αq)(αr+ αs)
αp+ αq+ αr+ αs |Rk− Rl|2
=√2 π
'
αp+ αq√
αr+ αs
'
αp+ αq+ αr+ αs
√ π 2
(
(αp+α q )(αr+α s )
α p +α q +α r +α s |Rk− Rl|2
= 1
R
which is exactly what we should obtain
3 The values of F(t) are reported in L.J Schaad, G.O Morrell, J Chem Phys 54 (1971) 1965.
... example of the penalty function method, let us take the docking of two molecules Our goal is to give such values of the atomic coordinates of both mole-cules as to ensure the contacts of some... results mightbe of poor quality, since the procedure would first of all take care of the penalty, paying little attention to f It is recommended that we take a few values of K and check whether... Minimizing the mean value of the harmonic oscillator Hamiltonian. This
example is different: it will pertain to the extremum of a functional.8We are often going to encounter