Timeline for Binomial expansion for noncommutative operator
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2016 at 8:08 | answer | added | user44191 | timeline score: 3 | |
| Nov 18, 2016 at 22:55 | comment | added | user44191 | You will not get an answer in the exact form you specify; the sum of the powers will not necessarily be $n$, as the corresponding algebra is filtered, not graded. Even for $n = 2$, you get: $(A^\dagger - kA)^2 = A^{\dagger 2} - k A^\dagger A - k A A^\dagger + k^2 A^2 = A^{\dagger 2} - 2 k A^\dagger A - k + k^2 A^2$ Which has a term of lower degree (the -k). You'll need to sum over lower degrees. | |
| Nov 18, 2016 at 11:32 | comment | added | Flávio Oliveira Neto | It is the raising operator in quantum mechanics, it takes the state n to n+1. Yes, the commutator is equal to 1. | |
| Nov 18, 2016 at 1:49 | comment | added | BigM | What does the bracket mean? Is it the commutator?Also we are missing the binomial coefficients. | |
| Nov 18, 2016 at 1:32 | answer | added | T. Amdeberhan | timeline score: 0 | |
| Nov 18, 2016 at 1:21 | comment | added | David Handelman | And what is $A^{\dagger}$? Moore-Penrose inverse (this is what the dagger often means, although probably not here)? | |
| Nov 18, 2016 at 1:20 | history | edited | David Handelman | CC BY-SA 3.0 |
non is NOT a word; syntax; TeX corrected
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| Nov 17, 2016 at 22:12 | review | First posts | |||
| Nov 17, 2016 at 22:18 | |||||
| Nov 17, 2016 at 22:07 | history | asked | Flávio Oliveira Neto | CC BY-SA 3.0 |