How to use the completeness relation

[Sergio Rodriguez]

Homework Statement

Show that if ##\left( \Omega f\right) ^* = -\Omega f ^* ##
then ##\left< \Omega \right> = 0 ## for any real function f. where ##\Omega## is an operator

Homework Equations

It's a self test of the completeness relation --Molecular quantum mechanics (Atkins)--
so the equation is
$$ \sum_s \left| s \right> \left< s \right| = 1 $$

The Attempt at a Solution

## \left< \Omega \right> = \left<m\left|\Omega\right|n\right>## for any functions ##f_m## and ##f_n##
if ## f_m = f_n = f ##, then ## \left<\Omega\right> = \left<f\left|\Omega\right|f\right>
##
and as f is a real function ## f^* = f##, so:
##\left( \Omega f\right) ^* = -\Omega f ^* = -\Omega f ##
##\left( \Omega f\right) ^* = \Omega^* f^* = \Omega^* f ##
so ##\Omega^* f = -\Omega f ##
##\Omega^* \left|f\right> = -\Omega \left|f\right>##
##\left<f \left| \Omega^* \right|f\right> = - \left< f\left|\Omega \right|f\right>##

 

[DrClaude]

## \left< \Omega \right> = \left<m\left|\Omega\right|n\right>## for any functions ##f_m## and ##f_n##
if ## f_m = f_n = f ##, then ## \left<\Omega\right> = \left<f\left|\Omega\right|f\right>##

That's not correct. The expectation value of an operator ##\langle \Omega \rangle## over a state ##\psi## is always ##\langle \Omega \rangle = \langle \psi | \Omega | \psi \rangle##. There is thus no need to assume ## f_m = f_n = f ##.

Don't know hot to get in the Bra ket notation and relate it with ##\left<\Omega\right>##

Since you are dealing with a function ##f##, I would approach it using the completeness relation expressed as an integral over a basis set of functions ##\{\phi_n\}##. In any case, the idea is to introduce identity operators that you can then convert using the completeness relation, e.g.,
$$
\langle \psi | \Omega | \psi \rangle = \langle \psi | (1 \times \Omega \times 1) | \psi \rangle = \sum_m\sum_n \langle \psi | m \rangle \langle m | \Omega | n \rangle \langle n | \psi \rangle
$$

 

[Sergio Rodriguez]

I have do this:

As I don't know if f is an eigenfunction of Ω, I expand it as ## f = \sum_n c_nf_n## where ##f_n## are orthonormal eigenfunctions of Ω, so:

##\left<Ω\right> = \left<f\left|Ω\right|f\right> = \left<f\left|1*Ω*1\right|f\right> ##
##\sum_m\sum_n \left<f|m\right>\left<m\left|Ω\right|n\right>\left<n|f\right> = \sum_m\sum_n c_n^2 \left<m|m\right>\left<m\left|Ω\right|n\right>\left<n|n\right>##

As ## f_n ## are orthonormal the double sum become a single sum:
##\sum_n c_n^2 \left<n\left|Ω\right|n\right> = \sum_n c_n^2ω_n## where ##ω_n## are the eigenvalues, so ##\left<Ω\right> = \sum_n c_n^2ω_n ##

The second part is:

##(Ωf)^* = -Ωf^* = -Ωf ## as f is a real function
##f(Ωf)^*= -fΩf ##
##\int f(Ωf)^*\,dτ = \int -fΩf\, dτ##
##\sum_n \int c_nf_n(Ωc_nf_n)^*\,dτ = - \sum_n \int c_nf_nΩc_nf_n\, dτ##
##\sum_n c_n^2ω_n^* \int f_nf_n\,dτ = - \sum_n c_n^2ω_n \int f_nf_n\, dτ## And as ##\int f_nf_n\, dτ = 1##
##\sum_n c_n^2ω_n^* = - \sum_n c_n^2ω_n##
And as ##\sum_n c_n^2ω_n = \left<Ω\right>##
##\sum_n c_n^2ω_n^* = - \left<Ω\right>##

But I don't know how to follow it.

 

[DrClaude]

Could you tell me where in the book this is? I want to explain using the same notation and referring to particular equations.

 

[Sergio Rodriguez]

Thank you!

It's the first chapter. Section 1.20 Matrix Elements. Example 1.9, Self-test 1.9, page 33.