MODULE 8

APPROXIMATION METHODS I

Once we move past the two particle systems, the Schrödinger equation cannot be solved exactly.

The electronic inter-repulsion term in the hamiltonian prevents us from separating it into the sum of one-electron operators, and thus exact solutions for systems of three or more particles elude us.

Fortunately approximation methods have been developed which can generate solutions that are very accurate.

Variation theory and Perturbation theory

MODULE 8

The Variation Method

The ground state of an arbitrary system defined by a wavefunction 0 with eigenvalue E0.

When the wavefunction is normalized the denominator is unity, but this is not necessarily the case

0 0 0H E

*0 0

0 *0 0

0 0

0 0

ˆ

ˆ

H dE

d

H

MODULE 8

Now substitute any other function for 0 and calculate the

corresponding energy E

Then according to the variation theorem

and the equality holds only if

ˆ*

*

ˆ

H dE

d

H

0E E

MODULE 8

Thus the energy E of the new system will never be less than that

of the ground state.

From this we see that we can calculate an upper limit (bound) for the value of E0 by using some trial function.

The closer we can get the wavefunction to be like 0, the closer

will E approach the true value E0.

The way to proceed is to choose a trial function that depends on one or more arbitrary parameters (variational parameters).

At this point we minimize Ewith respect to the parameters and

thus obtain the best possible ground state energy that our trial function can provide.

The success of this approach clearly depends on our ability to select a good trial wavefunction ( clever guesswork) .

MODULE 8

For example choose (, , , …).

The energy will also depend on the same parameters

0( , , ,...)E E

MODULE 8

We already know that we can determine the energy of the ground (1s) state of a hydrogen atom exactly without recourse to

approximations.

So this is a good test case for the variation approach-how well does it recover the known value?

In the 1s state, l = 0, and the hamiltonian can be simplified to

4 2 2 21 0/ 32s eE m e

2 22

20

ˆ2 4e

d d eH r

m r dr dr r

MODULE 8

Now we decide on the trial function.

From our knowledge of the exact solution we can expect that the wavefunction will decay to zero as r increases.

One such function is a Gaussian of the form

(r) = exp(-r2)

FIGURE 8.1where can be the variational parameter.

Now substitute trial function and the operator into the numerator and denominator of the basic equation

ˆ*

*

H dE

d

MODULE 8

Now we want to minimize Ewith respect to by differentiating and setting the result equal to zero.

2 2 1/ 2

1/ 2 3/ 20

3( )

2 2e

eE

m

2 2

3/ 2 1/ 20

( ) 3

2 (2 )e

dE e

d m

2 4

3 2 4018

em e

4 4

min 2 2 2 2 2 20 0

40.424

3 16 16e em e m e

E

MODULE 8

This value can be compared to the exact value of E1s from above

As the theorem predicts, Emin > E1s, i.e. it is less negative.

In the normalized trial wavefunction it is convenient to put in terms of the Bohr radius, a0. Thus:

4 4

1 2 2 2 2 2 20 0

10.500

2 16 16e e

s

m e m eE

2

8

9 oa

2 20

1/ 2

(8/9 )3/ 2 3

0

8 1( )

3r ar e

a

MODULE 8

The Gaussian is a very poor approximation to the true exponential, particularly at

values of r less than the Bohr radius.

Nevertheless the trial wavefunction leads to a value for the state energy which is

within 80% of the exact result.

This was with a single variational parameter ().

It might be anticipated that by using more flexible trial functions with more parameters, we could approach the true value even

closer.

MODULE 8

The Helium Atom

Now we use the variational method to find the upper bound of the energy for He, with one nucleon (Z = 2) and two electrons. We do

not know the exact value of this.

The Hamiltonian operator for this is given by:

The resulting Schrödinger equation cannot be solved exactly because of the potential energy term involving r12

If we ignore the inter-electronic repulsion term the ground state wavefunction becomes separable

2 2 22 2

1 20 1 2 0 12

2 1 1 1ˆ ( )2 4 4e

e eH

m r r r

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This equation can be used as a trial function in which Z is a variational parameter.

Because the variational parameter is part of the exponential function this is an example of a non-linear variation

The 1s wavefunctions in the equation are of the form

(remember for 1s, l = 0)

0 1 2 1 1 1 2( , ) ( ) ( )s sr r r r

0

1/ 23/

1 30

( ) jZr a

s j

Zr e

a

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or, in au

laborious math

0 1 2 0 1 2 1 2

0 1 2 0 1 2 1 2

ˆ( , ) ( , )( )

( , ) ( , )

r r H r r drdrE Z

r r r r drdr

42

20

27( )

(4 ) 8em eE Z Z Z

2 27( )

8E Z Z Z

Minimizing E(Z) with respect to Z we find Zmin = 27/16

MODULE 8

Substitution back into equation for E(Z)

The experimental result is –2.9033 and the best-refined calculation provides –2.9037.

Thus the result for Emin is reasonable (it is less negative, i.e.

greater than the experimental value) considering the simplicity of the trial function we have employed.

2

min

272.8477

16E

au

MODULE 8

The use of nuclear charge (Z) as a variational parameter implies that Zmin = 27/16 (or 1.6875) can be interpreted as an effective

nuclear charge.

That the value is significantly less than 2 reflects the fact that each electron is partially screening the nucleon from the other electron,

thereby reducing the net charge seen by the nucleon.

This lessening of the nuclear attraction means that the 1s orbital in helium will be larger than a hydrogenic 1s with Z = 2.

MODULE 8

Linear Combinations and Secular Determinants

improved accuracy can be achieved by increasing the number of variational parameters

employ a trial function that is a linear combination of known functions with variable coefficients.

Such a trial function can be written as

We opt for simplicity by choosing N = 2 and require that cn and fn

are real (not necessary, but keeps it simple).

1

N

n nn

c f

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1 1 2 2c f c f

1 1 2 2 1 1 2 2

1 1 2 2 1 1 2 2

1 1 1 1 1 1 2 2 2 2 1 1 2 2 2 2

2 21 1 1 2 1 2 2 1 2 1 2 2 21

2 21 11 1 2 12 2 1 21 2 22

ˆ ˆ( ) ( )

ˆ ˆ( )( )

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

H d c f c f H c f c f d

c f c f Hc f Hc f d

c f Hc f c f Hc f c f Hc f c f Hc f d

c f Hf d c c f Hf d c c f Hf d c f Hf d

c H c c H c c H c H

ˆij i jH f Hf dwhere

MODULE 8

the operator has to be Hermitian in order to guarantee real eigenvalues, hence

Following the same routine, the denominator of the variational equation becomes

1 2 2 1ˆ ˆ

ij ji

f Hf d f Hf d

H H

2 21 11 1 2 12 2 22

ˆ 2H d c H c c H c H

2 2 21 11 1 2 12 2 222d c S c c S c S

ij ji i jS S f f d

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multiplying out the LHS and taking the partial derivative of E wrt c1, we get

Thus we have E as a function of the variational parameters ci

Prior to differentiating it is better to write it differently

2 21 11 1 2 12 2 22

1 2 2 21 11 1 2 12 2 22

2( , )

2

c H c c H c HE c c

c S c c S c S

2 2 2 21 2 1 11 1 2 12 2 22 1 11 1 2 12 2 22( , )( 2 ) 2E c c c S c c S c S c H c c H c H

2 21 11 2 22 1 11 1 2 12 2 22

1

1 11 2 12

(2 2 ) ( 2 )

2 2

Ec S c S E c S c c S c S

c

c H c H

MODULE 8

Now put the derivative term equal to zero since we are minimizing, and rearrange

Differentiate equation again, but now wrt c2

1 11 11 2 12 12( ) ( ) 0c H ES c H ES

1 12 12 2 22 22( ) ( ) 0c H ES c H ES

These last two equations are a pair of linear homogeneous algebraic equations, containing the two unknowns c1 and c2.

The trivial solution is that both coefficients equal zero, but then the wavefunction would equal zero.

MODULE 8

Non-trivial solutions to such a set of equations do exist

If, and only if, the determinant of the coefficients vanishes, i.e. if

This is called the secular determinant, and expansion of it yields a polynomial (here a quadratic) secular equation in E.

Finding the roots of the quadratic yields two values of E, of which we select the lowest as the approximate energy of our ground

state entity.

11 11 12 12

12 12 22 22

0H ES H ES

H ES H ES

MODULE 8

An example of the linear combination (LC) procedure:

The particle in the one-dimensional box with vertical walls where the potential energy becomes infinitely large, but within the box,

V(x) = 0.

The Hamiltonian for this situation is that for kinetic energy only

The trial functions we choose are from the set

2 2

22

d

m dx

(1 )n nnf x x

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These functions have zero amplitude at the walls and a

single maximum at the center of the box, as needed for the

n = 1 wavefunction.

As usual to keep matters simple we use the first two terms only as our BASIS

SET, thusx

0.0 0.2 0.4 0.6 0.8 1.0

f(x)

0.0

0.1

0.2

0.3

n=3

n=2

n=1

f(x)=xn(1-x)n

2 21 2(1 ) (1 )trial c x x c x x

MODULE 8

Using the outlined procedure we evaluate the H and S integrals to be:

2

11 11

2

12 21 12 21

2

22 22

1;

6 30

1;

30 140

1;

105 630

H Sm

H H S Sm

H Sm

1 ' 1 '

6 30 30 140 01 ' 1 '

30 140 105 630

E E

E E

2

'Em

E

2' 56 ' 252 0E E

MODULE 8

E’ = 51.065 and 4.9347

This is to be compared to the exact energy

2

min

2

4.9387

0.125002 /

Em

h m

2 2 2

2

2

( 1)8 8

0.125000

exact

n h hE when n L

mL m

h

m

Thus the LC technique with a basis set of only two members leads to an energy value within 2 ppm of the exact value - a remarkable

achievement.

MODULE 8

It is often useful to operate with a larger basis set than two

In the general case for a basis set of N-terms we write the set of linear equations as:

1 11 11 2 12 12 1 1

1 21 21 2 22 22 2 2

1 1 1 2 2 2

. . ... .

( ) ( ) ... ( ) 0

( ) ( ) ... ( ) 0

( ) ( ) ... ( ) 0

N N N

N N N

N N N N N NN NN

c H ES c H ES c H ES

c H ES c H ES c H ES

c H ES c H ES c H ES

11 11 12 12 1 1

21 21 22 22 2 2

1 1 2 2

...

...

. . .

...

0

N N

N N

N N N N NN NN

H ES H ES H ES

H ES H ES H ES

H ES H ES H ES

MODULE 8

In the determinant the elements have been inserted using the convention Mrc where r = row number and c = column number.

It can be simplified using the Hermitian conditions that Hij = Hji and

Sij = Sji.

Expansion of the determinant leads to a secular N-order polynomial in E.

For N > 2 we need to use computational methods to obtain the roots.

we select the smallest of the roots for the value of Emin.

MODULE 8

From the root taken we can substitute back to find the values of ci.

Only N-1 of the equations are independent and we can evaluate only the ratios cN/c1.

If we require that the function be normalized to unity viz.

then the value of c1 can be extracted and so all the rest.

2

1

1N

ii

c

MODULE 8

An extension of the LC method is to use a trial function of the form

where fn also contain variational parameters, e.g. 1

N

n nn

c f

2nr

nf e

Now two sets of variational parameters, cn and n.

The minimization of E with respect to the two sets of parameters involves complicated math and needs to be done numerically.

Fortunately algorithms are available for this.

MODULE 8

N Emin / au

1 -0.4244

2 -0.4858

5 -0.49976

10 -0.49998

exact -0.500000

The results obtained by taking different numbers of terms for the ground state of the hydrogen atom are shown in Table