Timeline for Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis?
Current License: CC BY-SA 4.0
6 events
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| Jul 13, 2018 at 13:52 | comment | added | Cosmas Zachos | You are asking for a hidebound proof out of a formal wisecrack? The factors are all diagonal matrices. If you truncate at m, of course it works--- if you trusted your normal ordering. But then you are reduced to @Pisanty & Nahmad-Achar 2012. Surely a different question, no? | |
| Jul 13, 2018 at 13:40 | comment | added | lurscher | but can this 'purely formal' expression get arbitrarily close to the operator under some norm or topology sense as we take in more terms in the expansion? | |
| Jun 30, 2018 at 1:21 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jun 29, 2018 at 15:03 | vote | accept | Aharon Brodutch | ||
| Jun 28, 2018 at 15:04 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jun 28, 2018 at 14:58 | history | answered | Cosmas Zachos | CC BY-SA 4.0 |