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?

  • $\begingroup$ 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
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    $\begingroup$ This question is explained in this & this Phys.SE posts and links therein. $\endgroup$ – Qmechanic Oct 5 '17 at 12:23