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

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.