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The Hamiltonian is arbitrary upto a constant anyway. Why don't we just subtract off the vacuum energy? The Hamiltonian was always observable only upto a constant.

Instead, we do normal-ordering, in which we suddenly assume that creation and annihilation operators commute, just for this one step! This procedure is indefensible.

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    $\begingroup$ we suddenly assume that creation and annihilation operators commute Why do you think so? Further: The Hamiltonian was always observable only upto a constant - this only holds if you neglect gravity. $\endgroup$ Jun 28 at 12:35
  • $\begingroup$ @JasonFunderberker But you can still subtract just the "right amount of infinity" to make the resultant Hamiltonian align with the observed cosmological constant. There's still no need for normal-ordering. As for the assumption that they commute, I read this in the book "Student Friendly QFT". Page 61. They use this commutation property in the derivation of the "observed Hamiltonian". And they mention that it's just for this one step. $\endgroup$
    – Ryder Rude
    Jun 28 at 12:42
  • $\begingroup$ Related: What does the ordering of creation/annihilation operators mean? $\endgroup$
    – Qmechanic
    Jun 28 at 13:32

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A normal ordered operatpr has finite matrix elements between any pair of states. Subtracting a constant only gurantees you a finite number for the diagonal matrix element between one specific state and itself.

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  • $\begingroup$ but the hamiltonians look exactly the same in both cases $\endgroup$
    – Ryder Rude
    Jun 28 at 14:12

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