The expectation value in quantum theory

by aaaa202
Tags: expectation, quantum, theory
aaaa202 is offline
Dec3-12, 06:02 PM
P: 995
Going from the abstract state vector lψ> and the mean-value 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?
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Jorriss is offline
Dec3-12, 07:06 PM
P: 1,025
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.
dextercioby is offline
Dec4-12, 02:42 AM
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P: 11,866
We usually denote the general (abstract, assumed linear and self-adjoint) operators by capitals, A, B, C as to distinguish them from the operators for position xi and momentum pi. 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 \psi|x\rangle \langle x|A|x' \rangle \langle x'|\psi \rangle [/tex]

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