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Topic: Little Quantum Mechanics Question
Replies: 22   Last Post: Sep 22, 2003 12:50 AM

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meron@cars3.uchicago.edu

Posts: 914
Registered: 12/6/04
Re: Little Quantum Mechanics Question
Posted: Sep 15, 2003 2:12 PM
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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 1-D "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 eigen-kets, |n>, form a basis.
>
>H|n> = E_n|n> n=1,2,3,...
>
>and the |n> arn't degenerate.
>
>One can expand |Psi> as:
>
>|Psi> = Sum[ {n,1,Infinity} , c_n|n> ]
>
>where c_n = <n|Psi>
>
>But, also, the postion operator, X , is an observable:
>
>X|x> = x|x>
>
>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 e-kets, or the basis of X e-kets 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"




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