130 likes | 511 Views
The Variational Principle. Scott Riggs. Expectation Values. Basic Statistics P ab = ∫ 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
E N D
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 otherwise Ψ’’ = 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.