Hey Zack, i forgot to ask you in during the review session but you made the statement that
for a free particle
< k |p| k > = hbar*k or something like that.. My question is, how could you express it like that, since for a free particle the state is non-normalizable and therefore wouldn't the integration over all space go to infinity for a free particle? I could understand perhaps if we were integrating over momentum space, the orthogonality property would kick in and it would seem we might get some sort of spike at where the value of the momentum is since we are talking about an eigenstate, but since this is in cartesian space I'm a little confused on this issue. Thanks in advance.
Zack, could you get the solutions up for homework 9?
ReplyDeleteHey Zack, i forgot to ask you in during the review session but you made the statement that
ReplyDeletefor a free particle
< k |p| k > = hbar*k or something like that.. My question is, how could you express it like that, since for a free particle the state is non-normalizable and therefore wouldn't the integration over all space go to infinity for a free particle? I could understand perhaps if we were integrating over momentum space, the orthogonality property would kick in and it would seem we might get some sort of spike at where the value of the momentum is since we are talking about an eigenstate, but since this is in cartesian space I'm a little confused on this issue. Thanks in advance.
I think you are totally right about that. There is a problem. I suppose one could try to correct it by dividing k | p | k by k | k .
ReplyDelete(I am afraid to put brackets in. How did you do that so nicely?)
I see how that might work. I just put a space between the brackets and the k.
ReplyDelete