This exercise illustrates the variational method of estimating eigenstates of a Hamiltonian.

First consider the eigenstates of a one-dimensional square-well for 0 <= x <= a :

(written in units of \hbar^2/2m)

> H(x)*psi(x) = E*psi(x);-diff(diff(psi(x),x),x) = E* psi(x);

The exact solutions for this problem are given by:

> psi(n,x) = C*sin(n*Pi*x/a); E(n) = n^2*(Pi/a)^2;

The variational method allows us to estimate psi(n,x) and E(n), starting with the lowest eigenvalue (n=1). Suppose we guess psi(1,x) =f(x)==x*(a-x). It will be convenient to keep the normalization arbitrary.

> f:=x->x*(a-x);

E1approx = <f|H|f>/<f|f>

> E1approx:=Int(-f(x)*diff(diff(f(x),x),x),x=0..a)/Int((f(x))^2,x=0..a);

> E1approx:=int(-f(x)*diff(diff(f(x),x),x),x=0..a)/int((f(x))^2,x=0..a);E1exact:=evalf(Pi^2/a^2);

This shows that the approximate eigenvalue is ~1.3% greater than the exact value. (Is it significant that the approximation is greater ?

Comparing the wavefunction shapes, we see why the approximation is fairly good. Setting a=1:

> psi1exact:=x->sin(Pi*x);psi1approx:=x->(4*x*(1-x));

> plot({psi1exact(x),psi1approx(x)},x=0..1);

We can continue this investigation by guessing the next eigenfunctions:

> psi2exact:=x->sin(2*Pi*x);psi2approx:=x->(64/3)*x*(1-x)*(1/2-x);

> plot({psi2exact(x),psi2approx(x)},x=0..1);

The normalization for psi2approx was chosen so that it is comparable to psi2exact.

1. Using the variational method, estimate E2approx and compare it with E2exact.

2. Notice that <psi2approx|psi1approx>=0. Is this important?

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