Approximation Methods - kau



A trial function that depends linearly on the variational parameters leads to a secular determinant

- as another example of the variational method, consider a particle in one dimensional box. We should expect it to be symmetric about x = a/2 and to go to zero at the walls.

- one of the simplest functions with this properties is xn ( a-x)n , where n is a positive integer , consequently , let’s estimate Eo by using :

[pic] (1)

Where c1 and c2 are to be the variational parameters. We find after the energy of particle in one dimensional box exactly equal to:

[pic](2)

Now we will use a trial function φ to calculate Emin to particle in one dimensional box.

A trial function can be generally written as:

[pic](3)

Where the Cn are variational parameters and fn are arbitrary known functions

Consider

[pic]

Then :

[pic]

[pic]

[pic](4)

Where

[pic](5)

We will note that :

[pic](6)

So Hij=Hji

Using this result, equation (4) becomes

[pic](7)

Similary we have

[pic] (8)

Where

[pic] (9)

The quantities Hij and Sij are called matrix elements .

By substituting equations 7,8 into equation 10

[pic](10)

We find that :

[pic](11)

Before differentiating E(c1,c2) in equation 11 with respect to c1and c2 , it is convenient to write equation 11 in the form:

[pic](12)

If we differentiate equation 12 with respect to c1 we find that

[pic](13)

Because we are minimizing E with repect to c1 , [pic]=0 and so equation 13 becomes

[pic](14)

Similarly by differentiating E(c1,c2)with repect to c2 instead of c1 we find

[pic] (15)

Equations (14) and (15) constitute a pair of linear algebraic equations for c1 and c2

This equation is not simply solved but if c1 = c2

[pic] (16)

To illustrate the use of equation 16 let us go back to solving the problem of a particle in a one – dimensional box variationally using equation (1)

[pic]

We will set a = 1. In this case ,

f1 = x (1- x) and f2 = x2 (1- x) 2

So, we will solve H11 , H12, H22 , S11, S12 and S22 from the equations (5) , (9)

[pic]

[pic]

[pic][pic]

[pic]

[pic]

- As the same you can get H12 and H22

H12 = H21= [pic] and H22 = [pic]

- Also we can get S11 = [pic]

[pic]

- As the same we can get S12 and S22

S12 = S21 = [pic] and S22 = [pic]

- Substituting the matrix elements Hij and Sij into the secular determinant gives

[pic]

where [pic]= E m/[pic]. The corresponding secular equation is

[pic]- 56[pic]+252=0

whose roots are

[pic]

We choose the smaller root and obtain

[pic]

Compare with Eexact when a =1

[pic]

The excellent agreement here is better than should be expected normally for such a simple trial function. Note that [pic][pic] [pic], as it must be.

Example

Using Equation f1=x (1-x) and f2= x2(1-x) 2, show explicitly that H12=H21

Solution: using the Hamiltonian operator of a particle in a box, we have

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

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[pic]

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