Wednesday, September 27, 2017

quantum mechanics - Do the ladder operators a and adagger form a complete algebra basis?


It is easy to construct any operator (in continuous variables) using the set of operators {|m|},

where l and m are integers and the operators are represented in the Fock basis, i.e any operator ˆM can be written as ˆM=,mα,m|m|
where α,m are complex coefficients. My question is, can we do the same thing with the set {ak(a)}.


Actually, this boils down to a single example which would be sufficient. Can we find coefficients αk, such that |00|=k,αk,ak(a).

(here |0 is the vacuum and I take a0=I)



Answer




@Accidental reminds you this is a theorem. To actually see it in your terms, use the infinite matrix representation of a,a of Messiah's classic QM, v 1, ChXII, § 5. Specifically, your vacuum projection operator has a 1 in the 1,1 entry and zeros everywhere else.


The operator you chose is freaky to represent, but, purely formally, the diagonal operator for Naa, |00|=(1+N)(1N)2N23N34N4...

would do the trick, once anti-normal ordered.


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