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Does the operator $a+a^\dagger$ have eigenstates? If yes, what are they?

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    $\begingroup$ No, it has not discrete spectrum on $L^2(\mathbb{R}^d)$. In fact it is proportional to the position operator (or the momentum one, depends on you definition) and those have purely continuous spectrum, so no eigenstates that are square integrable. $\endgroup$
    – yuggib
    Jul 7 '15 at 13:40
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    $\begingroup$ Closely related: physics.stackexchange.com/q/155852/80818 $\endgroup$
    – tok3rat0r
    Jul 7 '15 at 13:42
  • $\begingroup$ @yuggib: Once again, that's an answer ;) $\endgroup$
    – ACuriousMind
    Jul 7 '15 at 13:55
  • $\begingroup$ $\hat x =\sqrt{\frac{\hbar}{2 m \omega}} (a+a^\dagger)$, thus up to scaling - this operator has the eigenvalues and eigenvectors of the position operator $\endgroup$
    – Alexander
    Jul 7 '15 at 13:59
  • $\begingroup$ @Alexander Either you or yuggib should make your comments into answers! $\endgroup$ Jul 7 '15 at 14:01
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No, it has not discrete spectrum (on $L^2(\mathbb{R}^d)$). In fact $a+a^*$ is proportional to the position operator (or the momentum one, depends on your definition of $a$ and $a^*$; by the usual one the position operator $x$ is proportional to the real part $a+a^*$ and the momentum $p$ to the imaginary part $\frac{1}{i}(a-a^*)$). Both position and momentum operators have purely continuous spectrum, so there are no eigenstates that are square integrable (but there are the usual "generalized eigenvectors"; i.e. delta functions for the position operator).

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  • $\begingroup$ In my QM mechanics class, that would be translated into: write down $|x\rangle$ and don't ask any questions! $\endgroup$
    – jinawee
    Jul 11 '15 at 18:21
  • $\begingroup$ @jinawee good ol' physics style :-P $\endgroup$
    – yuggib
    Jul 12 '15 at 8:41
  • $\begingroup$ @jinawee There are places where questions are answered: Dirac's QM book. $\endgroup$ Feb 21 '20 at 12:37

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