The Variational Principle
Scott Riggs
Expectation Values
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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
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<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
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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
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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
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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
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Expectation Value
Basic Dirac Notation
Variation Principle
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Find an upper bound on the Eg of a
system described by a Hamiltonian.