The Variational Principle

Scott Riggs

Expectation Values

Basic Statistics Pab = ∫p(x) dx (limits a to b)

The probability that x lies between a & b Where p(x) dx is the probability that individual

(randomly chosen) lies between x and x+dx

<x> = ∫x |Ψ(x,t)|^2 dx : <x> is the average of measurements performed on particles all in the state Ψ.

Dirac Notation

<f|g> = ∫f(x)*[g(x)] dx <f| bra |g> ket Bracket notation ket is just a vector bra is a linear

combination of vectors. <f|d/dx|g> =

∫f(x)*(d/dx)[g(x)] dx

Variational Principle

Need to calculate the ground state energy for a system described by H, but unable to solve the time-independent Schrödinger equation.

HΨ = EΨ also written –(h2/2m)(d2 Ψ/dx2)+VΨ= EΨ Eg <= <Ψ|H|Ψ> = <H> Says the expectation value of H in the state given

by Ψ is guaranteed to overestimate Eg. If Ψ happens to be the exact ground state of H

then Eg=<H>. Also if Ψ to be an exicted state then <H> > Eg

Example Triangular trial Ψ for an infinite square well. Ψ(x) = Ax (if 0<=x<=a/2) dΨ/dx = A

= A(a-x) (if a/2 <=x<=a) dΨ/dx = -A = 0 otherwiseΨ’’ = Ad(x)-2Ad(x-a/2)+Ad(x-2)

A is a constant found by < Ψ |Ψ> = 1

= A2[∫x2 dx + ∫(a-x)2 dx ]=> A = (a/2)(3/a)1/2

Now H = -(h^2/2m) (Ψ’’) + 0 Use the variation principle <H> = -h2A/2b∫ [Ψ’’] Ψ(x) dx = 12h2/2ma2

We know from exact E = (Pih)2/2ma2

Theorem works because 12>Pi2

Conclusions

Expectation Value Basic Dirac Notation Variation Principle

Find an upper bound on the Eg of a system described by a Hamiltonian.