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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