The Variational and WKB Methods
More often than not, it is impossible to find exact solutions to the eigenvalue problem of the Hamiltonian. One then turns to approximation methods, some of which will be described in this and the following chapters. In this section we consider a few exam
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CHAPTER 16
The name of the game is finding the lowest upper bound for a given amount of work. If H happens to be positive definite, Eo~ 0, and we will be able to restrict Eo to the range E(ao, f3o, .. .)~Eo~O. As an example, consider the problem of a particle in a potential V(x)=A.x 4 • Here are the features we expect of the ground state. It will have definite parity, and, since the ground-state function will have no nodes (more nodes-+more wiggles-+ more kinetic energy), it will have even parity. It will be peaked at x = 0 so as to minimize (V). And of course it will vanish as lxl-+oo. A trial function that has all these features (and is also easy to differentiate and integrate) is IJI(X, a)= e-ax'/2
(16.1.3)
where a is a free parameter that determines the width of the Gaussian. The energy as a function of a is
We see here the familiar struggle between the kinetic and potential energy terms. The former would like to see a -+0 so that the wave function is wide and has only large wavelength (small momentum) components, while the latter would like to see a-+oo, so that the wave function is a narrow spike near x=O, where the potential is a minimum. The optimal point, both effects included, is -(6mA.)I/3
ao- - -
fz2
(16.1.4)
The corresponding energy is 3 (6fz4A)l/3 8 m2
E(ao)=--
(16.1.5)
Since His positive definite, we conclude 05;Eo5;E (ao)
(16.1.6)
The best approximation to the wave function of the ground state (among all Gaussians) is IJI(X, a 0 )=exp{-~a 0 x 2 ). The inequality (16.1.6) is of course rigorous, but is utility depends on how close E ( a 0 ) is to Eo. Our calculation does not tell us this. All we know is that since we paid attention to parity, nodes, etc., our upper bound E(a 0 ) is lower than that obtained by someone whose test functions had odd parity and 15 nodes. For instance, if V(x) had been ~mm 2 x 2 instead of A.x 4 , we would have found a 0 =(mmjfz) 112 and E(a 0 ) = fzm/2. Although this is the exact answer, our calculation would not tell us this. The way to estimate the quality of the bound obtained is to try to lower it further by considering a trial function with more parameters. If this produces sub-
stantiallowering, we keep going. On the other hand, if we begin to feel a "resistance" to the lowering of the bound as we try more elaborate test functions, we may suspect that £ 0 is not too far below. In the case of V(x) = ~mro 2x 2 , it will eventually be found that there is no way of going below E ( ao) = liw /2. Our faith in the variational method stems from its performance in cases where the exact answer is known either analytically or experimentally. Let us consider two examples. The first is that of the electron in a Coulomb potential V = -e2 jr. We expect the ground-state wave function to have no angular momentum, no nodes, behave like r 0 as r-+0, and vanish as r-+oo. So we choose lf!(r, (), f/J, a) =exp( -ar 2 ).t We find (upon ignoring the irrelevant angular variables throughout) E( a ) -_
f[
2
2
e ) e -ar2] r 2 d r d -d r2 1e -ar2 ( - fz- 2mr 2 dr dr r
If
e -2ar
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