Recent content by mrquantum

Surely not! Does that mean lightsabers aren't real? I'm kidding. I was just adding a little whimsy to my question about phase transitions in a partial vacuum. I know spaceships aren't real. A human could never even get into space.

I just found out that if you're exposed to the vacuum of space, you won't explode, but rather you'll die less excitingly from a lack of oxygen. So much for the scientific accuracy of total recall! How is that a vacuum doesn't "pull" matter apart given that lowering pressure can decrease...

What do you mean by finite width? Does all em radiation have the same amplitude? If the beams had the same frequency, how could their phase difference not be the same everywhere?

Okay - let's ignore the term unpolarized light - I'll concede that it's a little ambiguous. If I rephrase my two photon example with two EM beams of identical frequency traveling along an identical path in space (z), but differing in phase by pi, such that when the electric field of one is in...

I don't understand how both can be correct. Is the RF radiation absorbed and re-emitted or does it pass straight through the sample. I'm basically trying to understand how the magnetisation vector (along +z) gets tipped into the xy-plane, or, if you like, how the z component of each nuclei's...

How is it that the electric and magnetic fields in unpolarized light are not canceled out? The simplest example would be two photons of the same energy traveling coherently in time along exactly the same path in space, differing only by a rotation of 180 degrees around the axis of travel...

Am I wrong in thinking that the rf radiation used in nmr is not "absorbed" by protons to cause transitions between spin states but rather it is only there to provide a magnetic field which can rotate the magnetisation vector away from the primary field?

Thanks vela. This isn't actually a homework assignment, I'm just working my way through the textbook. I got the same result as Tom, (Af)* = (Af*)* = -Af = -Af* => a = -a* => a = 0 given hermiticity. This result is actually stated in the book in one of the worked-through examples. I was wondering...

http://www.kinetics.nsc.ru/chichinin/books/spectroscopy/Atkins05.pdf That's a link to the textbook. Self-test 1.9. The question doesn't really specify anything else.

Does that make sense?

Yes.

Sorry, my mistake. Need to show that the expectation value of the operator A is zero. <f|A|f> = 0

I'm assuming A is any real or complex operator for which the relation holds and f is any real function.

I'm stuck on a question in atkins molecular quantum mechanics 4e (self test 1.9). If (Af)* = -Af, show that <A> = 0 for any real function f. I think you are expected to use the completeness relation sum,s { |s><s| = 1. I'm sure the answer is simple but I'm stumped.