The expectation value in quantum theory

[aaaa202]

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 isn't (1) sufficient? Where does confusion arise and why do you need two "different" unit operators?

 

[Jorriss]

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]

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 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.

\langle \psi |A|\psi\rangle = \iint dx dx&#039; \langle \psi|x\rangle \langle x|A|x&#039; \rangle \langle x&#039;|\psi \rangle