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I'm puzzled about what should be the normal ordering of the identity operator (or any proportional operator):

  • looking at it from the "Fock space operators POV",the prescription is to move all the creation operators to the left and the annihilators to the right, but the identity by definition as non of those, so should be left invariant.

But I run into several contractions:

  • normal ordering of any operator should have a vanishing vev, so $$<\, : \mathbb{I} :\, > = 0 \ne\, < \mathbb{I}>$$

  • using the commutator between the $a$ and $a^{\dagger}$: $$:aa^{\dagger}: \, = \,a^{\dagger}a =\, :a^{\dagger}a + \mathbb{I}: = \,a^{\dagger}a + :\mathbb{I}:$$ where I used the commuator to get from the first to the third and the normal ordering otherwise. Identification of the second and fourth gives me again $\,:\mathbb{I}:\,\, = 0$.

  • last the vev of the exponential of any operator should be zero ($<:e^{A}:>\, = 0$), expanding the exponential gives me again $<\,:\mathbb{I}:\,>\, = 0$.

Is this the final result or are there more subtleties?

Qmechanic
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Learning is a mess
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2 Answers2

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Comments to the question:

  1. Under the ordering symbol (such as, e.g. normal ordering $:\ldots:$, time ordering $T(\ldots)$, radial ordering ${\cal R}(\ldots)$, etc) all the operators (super)commute, e.g. $$ : \hat{A}\hat{B}: ~=~ (-1)^{|\hat{A}||\hat{B}|}: \hat{B}\hat{A}:, $$ even if the (super)commutator $[\hat{A},\hat{B}]\neq 0$ is non-vanishing.

  2. Ordering of a single elementary (=non-composed) operator (such as, e.g. $\hat{a}$, $\hat{a}^{\dagger}$, ${\bf 1}$, etc) is superfluous.

Qmechanic
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You can check my answer to a related post that gives an axiomatic definition of normal order as a function defined on the free algebra generated by creation and annihilation operators. According to this formulation ${:}1{:} = 1$. There it is also explained how paradoxes involving the normal order are resolved. The statement "the expectation value of the normal order of any operator (including the identity) on the vacuum is zero" is false according to this formulation. This is consistent with Eq. (82) in the well-known article "Bosonization for beginners and - refermionazation for experts" by Delf and Schöller [Ann. Phys. 7, 225 (1998)], which implies $\langle{:}e^{i\phi}{:}\rangle = 1$.

Sebastiano Peotta
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