The Well-Known Result: Rigorous or Not?

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it is a well known result, but it is a rigorious result or not?
 
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You mean Anderson localization?
 


tom.stoer said:
You mean Anderson localization?

yes.
 


In 1D, yes, it is fully rigorous. A quick google brings up references and papers --- too many to list here.
 


genneth said:
In 1D, yes, it is fully rigorous. A quick google brings up references and papers --- too many to list here.

I don't believe this. Even the Bloch's theorem is not dealt with rigorously, so how could the Anderson localization then? The Anderson localization looks like more complicated phenomenon than Bloch waves.
 


jostpuur said:
I don't believe this. Even the Bloch's theorem is not dealt with rigorously, so how could the Anderson localization then? The Anderson localization looks like more complicated phenomenon than Bloch waves.

no, bloch theorem is rigorious!
 


Afaik the Bloch and the (more general) Floquet theorems are rigorous.
 

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