0
$\begingroup$

With operator methods we can set the Hamiltonian of the harmonic oscillator in the following form:

$$\hat{H}=\hbar \omega(A^{\dagger}A+1/2).$$

My question is that how can we know that the lowest eigenvalue for the operator $A^{\dagger}A$ is zero and not a positive number?

$\endgroup$

marked as duplicate by Qmechanic quantum-mechanics Oct 5 '17 at 12:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You are talking of the eigenvalue of $a$ or of the Hamiltonian? The eigenstates of $a$ are not energy eigenstates. Furthermore, there is not an eigenstate of $\mathcal{H}$ with eigenvalue 0. $\endgroup$ – Saramago Oct 5 '17 at 11:27
  • $\begingroup$ There was a typo. Hope you now undertand what I'm asking. $\endgroup$ – Hulkster Oct 5 '17 at 11:41
  • 1
    $\begingroup$ This question is explained in this & this Phys.SE posts and links therein. $\endgroup$ – Qmechanic Oct 5 '17 at 12:23