The Variational Principle Scott Riggs Expectation Values Basic
The Variational Principle Scott Riggs
Expectation Values ¡ ¡ Basic Statistics Pab = ∫p(x) dx (limits a to b) l The probability that x lies between a & b l 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> = l ∫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 –(h 2/2 m)(d 2 Ψ/dx 2)+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-x) (if a/2 <=x<=a) dΨ/dx = -A = 0 otherwise Ψ’’ = Ad(x)-2 Ad(x-a/2)+Ad(x-2) A is a constant found by < Ψ |Ψ> = 1 = A 2[∫x 2 dx + ∫(a-x)2 dx ]=> A = (a/2)(3/a)1/2
¡ ¡ ¡ Now H = -(h^2/2 m) (Ψ’’) + 0 Use the variation principle <H> = -h 2 A/2 b∫ [Ψ’’] Ψ(x) dx = 12 h 2/2 ma 2 We know from exact E = (Pih)2/2 ma 2 Theorem works because 12>Pi 2
Conclusions Expectation Value ¡ Basic Dirac Notation ¡ Variation Principle ¡ l Find an upper bound on the Eg of a system described by a Hamiltonian.
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