Can $A_{nm} = |n\rangle \langle m|$ be written in terms of boson operators $b, b^\dagger$?

Asked Mar 13 '22 at 03:07

Active Mar 13 '22 at 03:07

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I am curious whether all operators can be written as a linear combination of product of boson operators $b, b^\dagger$.

More precisely, consider the single harmonic oscillator Hilbert space $H$, whose orthonormal basis is $\{|n\rangle\}_{n\in\mathbb Z_+}$. We have ladder operators $b$ and $b^\dagger$ such that $$b|0\rangle =0, \quad b^\dagger |n\rangle =\sqrt{n+1}|n+1\rangle.$$ Then, can an operator $A_{nm} = |n\rangle \langle m|$ be written by only using $b, b^\dagger$?

If this is true, then one may say that $b, b^\dagger$ are "complete" in the sense that they span the whole algebra of operators on $H$.

asked Mar 13 '22 at 03:07

Laplacian

1,051

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It could, around the vacuum projector: n creators on the left of it, and m annihilators on the right. – Cosmas Zachos Mar 13 '22 at 03:14
@CosmasZachos Then, can the vacuum projector be written in terms of $b,b^\dagger$? – Laplacian Mar 13 '22 at 03:24
Yes. – Cosmas Zachos Mar 13 '22 at 09:16
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According to Louisell and Fan, $|0\rangle\langle 0|=1−+(−1)/2!+...$, with zeros at n=1,2,3,... The elegant formulation is here. – Cosmas Zachos Mar 13 '22 at 10:06
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@CosmasZachos Thanks for noting me the elegant formulation! – Laplacian Mar 13 '22 at 10:39
Actually, at integer eigenvalues, $\operatorname{sinc}(\pi N) \equiv \sin (\pi N)~/(\pi N)$ or any power of it shares the same property. You have more freedom than you think, and can finesse idempotence... – Cosmas Zachos Mar 13 '22 at 23:14

Can $A_{nm} = |n\rangle \langle m|$ be written in terms of boson operators $b, b^\dagger$?

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