Expectation Values/Bra-ket Problem

[The Eggman]

Concerning expectation values...

Also, the derivation in terms of bra-ket rather than wage function would be appreciated.

Where $$\psi$$ is the system state

Knowing that <A>$$\psi$$=<$$\psi$$|A|$$\psi$$>

And A is comprised of a complete eigenvector set $$\phi$$j w/ corresponding eigenvalues aj

How do you derive <A>$$\psi$$=$$\sum$$aj|<$$\psi$$|$$\phi$$j>|^2 ?

Additionally, (unrelated to above)

if |$$\psi$$> is comprised of component states |$$\phi$$>,

And <$$\phi$$|$$\psi$$>=the relevant expansion coefficient (probability)

What is the value of <$$\psi$$|$$\phi$$>? The complex conjugate of the expansion coefficient?

Thanks!

http://en.wikipedia.org/wiki/Expectation_value_(quantum_mechanics )

 

[Tangent87]

If A has a complete set of eigenvectors $$\phi_j$$ with eigenvalues $$a_j$$ then we can write:

$$A=\sum_j a_j|\phi_j><\phi_j|$$

So thus <$$\psi$$|A|$$\psi$$>=$$\sum_j a_j<\psi|\phi_j><\phi_j|\psi>=\sum_j a_j<\psi|\phi_j>(<\psi|\phi_j>)^*=\sum_j a_j|<\psi|\phi_j>|^2$$

 

[Tangent87]

If you're wondering where the first identity comes from then I'll give you a hint:

By completeness: $$1=\sum_j |\phi_j><\phi_j|$$

 

[The Eggman]

Sorry, I still don't totally understand the first identity. How do you derive that? There isn't a scalar product, so is some other operation implicit?

Also, (this is going to be a stupid question) but |$$\phi$$j><$$\phi$$j| cannot be simplified, right?Lastly, can you commute the bra and the ket to form a scalar product by taking it's complex conjugate? ( <$$\phi$$j|$$\phi$$j>*)?

Thanks in advance!

 

[The Eggman]

Oh wait, did you just use the Identity operator? I think I get it now...

 

[Tangent87]

Oh wait, did you just use the Identity operator? I think I get it now...

Yes exactly, it's just the identity operator. And no you can't commute the bra and the ket because <$$\phi$$|$$\phi$$>^*=<$$\phi$$|$$\phi$$>