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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 7:36 PM
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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 eigen-kets, |n>, form a basis.

>
>>>H|n> = E_n|n> 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_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} ]
>
>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 "projection-valued 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 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).

>
>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 infinite-dimensional 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"




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