How to use the completeness relation

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The discussion focuses on demonstrating that if the operator relation (Ωf)* = -Ωf* holds for any real function f, then the expectation value ⟨Ω⟩ must equal zero. Participants explore the completeness relation and its application in quantum mechanics, emphasizing the integral representation of states and the use of orthonormal eigenfunctions. The calculations show that the expectation value can be expressed as a sum over eigenvalues weighted by coefficients from the expansion of f. A participant seeks clarification on the notation and references specific sections in a textbook to better understand the derivation. The conversation highlights the importance of correctly applying quantum mechanical principles to derive meaningful results.
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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>##
 
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Sergio Rodriguez said:
## \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 ##.
Sergio Rodriguez said:
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.,
$$
\langle \psi | \Omega | \psi \rangle = \langle \psi | (1 \times \Omega \times 1) | \psi \rangle = \sum_m\sum_n \langle \psi | m \rangle \langle m | \Omega | n \rangle \langle n | \psi \rangle
$$
 
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.
 
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Could you tell me where in the book this is? I want to explain using the same notation and referring to particular equations.
 
Thank you!

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