Does quantization always work?

[TrickyDicky]

"Does quantization always work?"
My answer to the original question is yes, as quantum physics always works.
If quantization does not work the applied procedure is wrong, incorrectly executed, or both.
Then the discussion went on about tensor products of Hilbert spaces.
I wonder what is the consequence of this for a simple physical system such as the hydrogen atom.
If there is none, as I believe, the math must be wrong or incorrectly applied.

The specific question was about the hydrogen atom Hilbert space. That has been answered:$\mathcal{H}_{Com}\otimes \mathcal{H}_{rel}=L^2(R^6)$ that avoids the problem with the potential mentioned by dextercioby.
Where do you see wrong or incorrectly applied math here?

 

[bhobba]

If quantization does not work the applied procedure is wrong, incorrectly executed, or both.

That depends on what you mean by 'work'. In QM there is an operator order issue:
http://motls.blogspot.com.au/2012/12/ordering-ambiguities-and.html

If its a problem is debatable.

Thanks
Bill

 

[my2cts]

The specific question was about the hydrogen atom Hilbert space. That has been answered:$\mathcal{H}_{Com}\otimes \mathcal{H}_{rel}=L^2(R^6)$ that avoids the problem with the potential mentioned by dextercioby.
Where do you see wrong or incorrectly applied math here?

My comment is intended to provoke a physically more clear statement of the issue.
Concerning the Hilbert space of H, $\mathcal{H}_{p}\otimes \mathcal{H}_{e}=L^2(R^6)$ also does the job.

 

[my2cts]

That depends on what you mean by 'work'. In QM there is an operator order issue:
http://motls.blogspot.com.au/2012/12/ordering-ambiguities-and.html
Bill

It also depends on what is meant by quantization. In any case, the hydrogen problem poses no problems.
Thus, if there is a mathematical problem it must be the math not the atom ;-).

 

[strangerep]

I'm still having a trouble to understand why those projectors can be used in a tensor product decomposition, since the |x_e\rangle\ from that product is not even known, not to mention that the (possibly rigged) Hilbert space in which it "lives" is also not known.

Isn't it? One can take the (generalized) momentum eigenstates for a free electron, giving a rigged Hilbert space expressed in momentum basis. Then perform a Fourier transform...