# Eigenstates of sum of creation and annihilation operators

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

• 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. Jul 7 '15 at 13:40
• Closely related: physics.stackexchange.com/q/155852/80818 Jul 7 '15 at 13:42
• @yuggib: Once again, that's an answer ;) Jul 7 '15 at 13:55
• $\hat x =\sqrt{\frac{\hbar}{2 m \omega}} (a+a^\dagger)$, thus up to scaling - this operator has the eigenvalues and eigenvectors of the position operator Jul 7 '15 at 13:59
• @Alexander Either you or yuggib should make your comments into answers! Jul 7 '15 at 14:01

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).
• In my QM mechanics class, that would be translated into: write down $|x\rangle$ and don't ask any questions! Jul 11 '15 at 18:21