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The expectation value in quantum theory 
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#1
Dec312, 06:02 PM

P: 1,005

Going from the abstract state vector lψ> and the meanvalue of an observable x (operator) given by:
<x> = <ψlxlψ> I want to show how that is done in the position basis: So I take: <x> = <ψlxlψ> And insert completeness in front of the state vector to get the expansion involving the wave function: 1 = ∫lx><xl (1) But when my teacher did this he insisted on using lx'> and furthermore that you actually inserted two different operators ∫lx'><x'l and ∫lx''><x''l both of course represent the unit operator. But I am curious as to why you need to make this primes. Why isnt (1) sufficient? Where does confusion arise and why do you need two "different" unit operators? 


#2
Dec312, 07:06 PM

P: 1,042

It may help to do this. Rather than using a continuous basis, use a discrete basis so the integral is a sum.
Then write out the product of two identity operators I*I where I = (sum)n><n (do it in a small case, such as 3 terms). Write it both ways. Using a different index, and the same index. You'll see you lose cross terms (if the basis is not orthonormal) by only using one index. I hope I understood the question properly and that helps. 


#3
Dec412, 02:42 AM

Sci Advisor
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P: 11,895

We usually denote the general (abstract, assumed linear and selfadjoint) operators by capitals, A, B, C as to distinguish them from the operators for position x_{i} and momentum p_{i}. And then yes, using primes to distinguish between different (but unitarily equivalent) sets of x's and p's expecially when using more then one generalized completion identities.
[tex] \langle \psi A\psi\rangle = \iint dx dx' \langle \psix\rangle \langle xAx' \rangle \langle x'\psi \rangle [/tex] 


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