
#1
Mar1812, 04:55 PM

P: 20

So one of the postulate of quantum mechanics is that observables have complete eigenfunctions. Can someone let me know if I am understanding this properly:
Basically you postulate for example, position kets x> such that any state can be represented by a linear combination of these states (integral), and you postulate an operator x' such that x'x>=xx>. So basically the Hilbert space is the span of x>?...Then you can postulate other kets, like momentum kets p> that also span the Hilbert space, and you postulate an operator p' such that p'p>=pp>. Then any state can be represented as an integral over p>, including the position states. So by postulation, the span of x> equals the span of p>? Then we postulate the Schrodinger equation, and write H' as some combination of the x' and p' operators, and that the eigenfunctions of H' (which can be written in terms of x> or p>) span the same space spanned by x> and p>?...How do we know this? Or do instead the eigenfunctions of H' define the Hilbert space, and there are states that can be represented by linear combinations of x> or p> but aren't actual states? 



#2
Mar1812, 11:22 PM

P: 344

Yeah, that's right. The Hilbert space can be spanned by the eigenvectors of x' or p', which is made explicit by the fact that you can write one operator in terms of the otherspecifically, they're Fourier transforms of each other. The Hamiltonian is then defined in terms of those operators, and by solving for its eigenvalues we find the allowable values of energy in the system.
I'm not sure what you mean by your last line, though, "Or do instead the eigenfunctions of H' define the Hilbert space, and there are states that can be represented by linear combinations of x> or p> but aren't actual states?" Any state that is a linear combination of x>'s or p>'s must itself be a state in the Hilbert space, although they may or may not be eigenstates of the Hamiltonian. The eigenstates of the Hamiltonian do span the Hilbert space, though, so it's possible to build x or p eigenstates out of pure energy states, if that's what you're asking. This is especially easy to see in the case of a free Hamiltonian, where H = p^2/2m, so eigenstates of momentum are also eigenstates of energy and vice versa. 



#3
Mar1812, 11:34 PM

P: 20

That's basically what I'm asking. So is the fact that the energy eigenstates are complete part of the postulate, or is it just a result of the position and momentum eigenstates being complete? If it's the latter, is it obvious?




#4
Mar1812, 11:44 PM

P: 344

Hilbert Space Postulate
That's something I've never been completely clear on, either. It's definitely true that they dothat fact is used in lots of QM calculationsbut I don't know exactly how one proves it. My impression has always been that there's some sort of deep theorem here which says that the eigenstates of any Hermitian operator span the space, or something along those lines, but I don't know exactly what it is. Perhaps somebody betterversed in the technicalities of linear algebra can shed some light on this?




#5
Mar1912, 12:46 AM

P: 20

Ok. So basically if you could measure a particle to be at position b, then the position operator has to have an associated eigenket b>, and whatever states these eigenkets span, H' also spans? I think this would have to be provable..otherwise when you use the variational principle, how would you know if the function you choose can be expressed as a combination of energy eigenstates, which the whole principle relies on?..



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