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Say in my Hamiltonian I had a term

$$Q[a_j,a^\dagger_j]_\pm$$

where $Q$ is a constant. Suppose I didn't realise that this quantity equals 1 and calculated a normal order. Of course you get 0 However, the normal order of 1 should be 1 or else any operator could be multiplied by 1s and get 0 at the end. What is a correct way of normal ordering any operators?

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  • $\begingroup$ Hi Tom, it seems to me that you are operating under a faulty assumption that there is a unique preferred way of quantizing any classical theory. I've been there. The truth is – operator orderings are pure quantum-mechanical features, they haven't any classical counterparts since all commutators are proportional to $\hbar$ and thus can be neglected in the classical regime. There could be many QFTs corresponding to the same classical theory, all related by reordering of different operators. Which one we choose to describe nature is decided by internal consistency and agreement with observations. $\endgroup$
    – Prof. Legolasov
    Commented Mar 17, 2017 at 5:19

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Well, it depends on context. E.g.:

  1. Assume we are given a classical model that we want to quantize, i.e. to construct a corresponding quantum theory. Be aware that quantization is not unique. The classical theory doesn't know about operator ordering so this introduce ambiguity. To parametrize our ignorance we should allow the possibility of an arbitrary constant term. Often the constant can later be fixed by other consistency requirements, cf. e.g. my Phys.SE answers here and here.

  2. If we are just doing operator manipulations in some consistent operator formulation of a quantum theory, then there are no ambiguities. The commutation relations dictate the outcome of any operator rearrangement.

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  • $\begingroup$ Normal order of constant being one is consistent with the vacuum expectation of any operator being 0 (I.e. Vacuum expectation of normal ordered Hamiltonian). $\endgroup$
    – Tom
    Commented Mar 15, 2017 at 23:48
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic ♦
    Commented Mar 16, 2017 at 12:20

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