Quantum harmonic oscillator: How can we know that the lowest eigenvalue for the operator $A^{\dagger}A$ is zero and not a positive number?

Asked Oct 05 '17 at 10:45

Active Oct 05 '17 at 12:47

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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?

edited Oct 05 '17 at 12:24

Qmechanic

asked Oct 05 '17 at 10:45

Hulkster

1

This question is explained in this & this Phys.SE posts and links therein. – Qmechanic Oct 05 '17 at 12:23
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. – Saramago Oct 05 '17 at 11:27
There was a typo. Hope you now undertand what I'm asking. – Hulkster Oct 05 '17 at 11:41

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