
Re: Little Quantum Mechanics Question
Posted:
Sep 15, 2003 7:36 PM


In article <bk5cjb$n41$1@nntp.itservices.ubc.ca>, israel@math.ubc.ca (Robert Israel) writes: >In article <Gin9b.12$05.8388@news.uchicago.edu>, > <meron@cars3.uchicago.edu> wrote: >>In article <bk3ubi$17hk$1@nntp6.u.washington.edu>, adam@nowhere.man writes: > >>>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. > >OK: in this case H has purely discrete spectrum, and so it does have >an orthonormal basis of eigenvectors. > >>>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} ] > >In physics, maybe, but in math we tend to be more careful.
Yes, true. I admit that physicists tend to be kinda sloppy about these things. Usually they can get away with this, but not always. So, I plead guilty as charged.
>The position operator has a continuous spectrum. As George Jones >observed, x> is not a member of the Hilbert space, and X has >_no_ eigenvectors, let alone an orthonormal basis of them. >Mathematicians (or mathematical physicists) use various devices >to deal with this, e.g. the "projectionvalued measure" form of the >Spectral Theorem.
Yes, George is right on this. > >>>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). > >What??? Please explain yourself. What kind of bases are you >talking about? >In mathematics there definitely is such a requirement. >Every orthonormal basis of a Hilbert space has the same cardinality. >Every Hamel basis of a vector space has the same cardinality (not >the same, in the infinitedimensional case, as that of an orthonormal >basis, of course).
I'm sorry, I should've said that "if you use physicist's sloppy approach and pretend that x> are legit kets, then you just proceed to ignore the little matter of cardinality". When you're sloppy, you've to be consistently sloppy:)
Mati Meron  "When you argue with a fool, meron@cars.uchicago.edu  chances are he is doing just the same"

