Roentgenquantum

Okay thank u

This helped a lot I'm still trying to get familiar to the whole notation and concept of working with operators

Thanks a lot

Hopefully this will be my last dumb question

I will try and answer your questions... I little buisy at the mo, will be back on the comp soon.

 

 

 

Right... well, when you come across something like this [math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math] the fact we can see [math]A^\dagger = A[/math] actually implies something important; if [math]A[/math] is a matrix, then this implies our matrix is hermitian, meaning all the diagonal elements in the matrix are real. Real things are associated to observables in quantum mechanics. This will take you into things like calculating matrices and the hermitian conjugate. You simply transponse your matrix but then complex conjugate it. I could write up some matrices, but I really can't be bothered unless you ask me to. Anyway... coming across something like [math]|\psi><\phi|[/math] is called an outer product, obviously the opposite name-meaning of the inner product. You may come across something like

 

[math]|\psi><\psi|\phi>[/math] this means that the eigenvector has been chosen, then this is an orthogonal vector [math]\phi[/math].

I would have written more, but am in a rush again!!! lol

Hi and thanks for the quick answer.

My whole question is: How to get from [math]\langle \phi|A| \Psi \rangle ^* [/math] to [math] \langle \Psi|A^\dagger| \phi \rangle[/math]. Is this only per definition that way (I already understood that this is the case for matrix elements A, because this is the way for creating the adjoint operator - transpose and conjugate-complex.). Because I tried to review the single steps to get from the left side to the right side.

Or maybe my question is:

If I transpose the scalar product, do I just change the bra and the ket?

Sorry if this sounds confused, but in fact, I am o.O

Thanks in advance, Patrick

Hi!

Since I forgot my PW... and I was too lazy to get a new one, I simply connected with my FB account, and here I am, posting some more questions...

I just reviewed linear operators. And found out that they can have an adjoint.

Well. I asked myself a question while looking at this equation:

[math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math]

,where A is a linear operator, and [math]\phi[/math] and [math]\Psi[/math] are wave functions in the hilbert space. While trying to reproduce the steps I wondered if one is allowed to move bras and kets without forming there conjegate-complex.

Meaning:

[math]| \Psi \rangle \langle \Phi| = \langle \Phi| \Psi \rangle[/math]

Thanks for answering!

EDIT

I just realized that by definition [math] ( \Psi | \phi )= (\phi | \Psi )^* [/math].

Can this be applied to the bra-ket notation? And what happens if an operator is squeezed in between...?

Well, from what I know is the smallest producable time (at the moment) will be at the Europeen XFEL. They will be able to produce "laser" pulses with 100 fs (10^-15 s) puls length and the photon packages have a distance of 0.3 fs. From what I know this is the smallest measurable time yet.