A coherent state $|\alpha\rangle$ is an eingenvector of the operator $\hat{a}$, but this is not an observable (i.e., not an hermitian operator). But every vector is eigenvector of a complete set of commuting observables (CSCO), and i've heared in many places that for the one dimension Harmonic Oscillator, all CSCO's have only one observable ($\hat{X},\hat{P},\hat{H}/\hat{N},...$). So how to construct, for an abitrary $|\alpha\rangle$, the observable that has it as eigenvector?
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ Take $|\alpha\rangle\langle \alpha|$. $\endgroup$– Tobias FünkeCommented Jun 2 at 21:23
-
$\begingroup$ This is indeed a hermitian operator, but i do not think it has a physical meaning. I could also choose the identity operator, but i think that exists a more interpretable option. $\endgroup$– QuantumBrachistochroneCommented Jun 2 at 21:33
-
2$\begingroup$ IIRC, the ground state of a linearly (in $X$) perturbed harmonic oscillator is a coherent state, where the $\alpha$ depends on the displacement. $\endgroup$– Tobias FünkeCommented Jun 2 at 21:45
-
$\begingroup$ ...although with that you'd only get real $\alpha$... $\endgroup$– Tobias FünkeCommented Jun 3 at 6:12
Add a comment
|