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According to Is there a simple way of finding the eigenstates of the creation and annihilation operator in QM?

The creation operator has no eigenstates. But one postulate of QM says that the state of a system after measuement using an operator must be an eigenstate of that operator. How to make sense of this?

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    $\begingroup$ What is there to make sense of? Who's trying to "measure" the creation operator? $\endgroup$
    – ACuriousMind
    Mar 13, 2022 at 10:56
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    $\begingroup$ The creation operator does not correspond to a physical observable in the same time that the position or spin operators do. So acting on a state with it does not correspond to measuring anything about that state. Rather, it changes the state, $a^\dagger |n \rangle = \sqrt{n+1} |n+1 \rangle$. $\endgroup$ Mar 13, 2022 at 14:56
  • $\begingroup$ This is just a brief comment, but I think you're running into a very common QM misconception: measuring an operator is not at all related to applying the operator to a state. $\endgroup$
    – knzhou
    Mar 14, 2022 at 1:19

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Not all operators are observables. Only self-adjoint operators are observables. A property of self-adjoint operators is that they have real eigenvalues.

The creation operator is not an observable and is not a self-adjoint operator. It's okay for an operator not to have eigenstates and not to be an observable.

Only observables can be measured.

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    $\begingroup$ Not all self-adjoint operators have eigenvectors. That is true in finite dimensions, but false otherwise. Perhaps you have in mind some sort of generalized eigenvalues? $\endgroup$ Mar 13, 2022 at 16:31
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    $\begingroup$ Self-adjoint operators have a spectral resolution, so the projections associated with measurement exist. The continuous spectrum does not have eigenvectors, but measurements of the continuous spectrum always have finite precision anyways. To keep the discussion simple, this subtlety is usually swept under the rug in physics. $\endgroup$ Mar 13, 2022 at 19:01
  • $\begingroup$ Good point, I edited accordingly. $\endgroup$
    – Alwin
    Mar 14, 2022 at 0:51

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