Emuberable and Demumerable sets

[sweet springs]

Hi. I have a question about numbers of basis in quamutum mechanics space.

Hamiltonian of harmonic oscillator is observable and have countably infinite sets |En>s

Together with position or momentum basis identity equation is,

|state>=\int|x><x|state>dx=\int|p><p|state>dp=\Sigma_n\ |E_n><E_n|state>

The same state is expressed as both enumerable and denumerable infinite sets.

Is it OK? Denumerable sets should be interpreted correctle as enumerable or vice versa?

 

[bhobba]

Sure - that's fine.

To really understand it though you need to investigate Rigged Hilbert Spaces, but that requires considerable background in analysis:
http://physics.lamar.edu/rafa/webdis.pdf

At he beginning level simply accept you can have both.

Thanks
Bill

 

[mathman]

Hi. I have a question about numbers of basis in quamutum mechanics space.

Hamiltonian of harmonic oscillator is observable and have countably infinite sets |En>s

Together with position or momentum basis identity equation is,

|state>=\int|x><x|state>dx=\int|p><p|state>dp=\Sigma_n\ |E_n><E_n|state>

The same state is expressed as both enumerable and denumerable infinite sets.

Is it OK? Denumerable sets should be interpreted correctle as enumerable or vice versa?

Enumerable is essentially a mathermatics term meaning that the set is ordered. Denumerable means countbly infinite, but not necessarily ordered.

 

[bhobba]

Enumerable is essentially a mathermatics term meaning that the set is ordered. Denumerable means countbly infinite, but not necessarily ordered.

You are of course correct.

But reading between the lines I am pretty sure he is asking about the existence of continuous basis in separable spaces which is a bit strange until you are used to it.

Thanks
Bill

 

[dextercioby]

But of course, continuous bases do not exist in separable spaces.

 

[aleazk]

This paper may be useful for you, it requires some concepts from functional analysis, though:

Mathematical surprises and Dirac’s formalism in quantum mechanics

 

[sweet springs]

Thanks ALL for your advise. I wil learn it.

 

[bhobba]

But of course, continuous bases do not exist in separable spaces.

Again of course. The RHS formalism is sneaky in making it look like it does by allowing a continuous spectrum.

To the OP its tied up with the so called Nuclear Spectral Theorem which is very difficult to prove - in fact I haven't even seen a proof - the one in the standard text by Gelfland that I studied ages ago is in fact incorrect - don't you hate that sort of thing

However the following details an approach useful in QM based on so called Hilbert-Schmidt Riggings:
http://mathserver.neu.edu/~king_chris/GenEf.pdf

Thanks
Bill

 

[dextercioby]

I feel like a teacher, every once in a while, we get to write here the same things. :) https://www.physicsforums.com/threa...spectral-theorem-for-rhs.750266/#post-4727570