How Do You Derive Expectation Values Using Bra-Ket Notation?

[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 \phij w/ corresponding eigenvalues aj

How do you derive <A>\psi=\sumaj|<\psi|\phij>|^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&gt;&lt;\phi_j|

So thus <\psi|A|\psi>=\sum_j a_j&lt;\psi|\phi_j&gt;&lt;\phi_j|\psi&gt;=\sum_j a_j&lt;\psi|\phi_j&gt;(&lt;\psi|\phi_j&gt;)^*=\sum_j a_j|&lt;\psi|\phi_j&gt;|^2

 

[Tangent87]

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

By completeness: <br /> 1=\sum_j |\phi_j&gt;&lt;\phi_j|<br />

 

[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 |\phij><\phij| cannot be simplified, right?Lastly, can you commute the bra and the ket to form a scalar product by taking it's complex conjugate? ( <\phij|\phij>*)?

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>