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