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

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?

## marked as duplicate by Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 5 '17 at 12:46

• 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 5 '17 at 11:27