
Re: Little Quantum Mechanics Question
Posted:
Sep 15, 2003 2:12 PM


In article <bk3ubi$17hk$1@nntp6.u.washington.edu>, adam@nowhere.man writes: >Hi sci.math,sci.physics. I was in bed trying to go to sleep >tonight, but I started thinking about something, and I >can't figure it out. So I am asking the newsgroup for some >assistance. Anyway, just for definitness, lets consider the >old 1D "particle in a box" (of length L) problem. > >Here's my problem: > >The state of the system is specified by the ket Psi> > >The Hamiltonian, H, is an observable, i.e. it's a hermetian >operator, and it's eigenkets, n>, form a basis. > >Hn> = E_nn> n=1,2,3,... > >and the n> arn't degenerate. > >One can expand Psi> as: > >Psi> = Sum[ {n,1,Infinity} , c_nn> ] > >where c_n = <nPsi> > >But, also, the postion operator, X , is an observable: > >Xx> = xx> > >So one can expand Psi> as: > >Psi> = Integrate[ f(x)x> , {x,0,L} ] > >So, my question is this. Obviously the state space is >of infinite dimention, because, whether you choose to >use the basis of H ekets, or the basis of X ekets there's >an infinite amount of them... but the n> are infinite, but >COUNTABLY infinite, whereas the x> are uncountable. So, >are there more x> than n>,
In a way, yes.
or what is going on? > Interestingly, once you get to infinitely dimensional spaces, there is no requirement that all the bases have to have same cardinality (though, of course, they all have to be infinite).
Mati Meron  "When you argue with a fool, meron@cars.uchicago.edu  chances are he is doing just the same"

