In quantum mechanics of the harmonic oscillator, when we use the operator method to find out the solutions, we find that the action of $\hat{a}$ is to lower the energy of a state by $\hbar\omega$ and the action of $\hat{a}^\dagger$ is to raise the energy by $\hbar\omega$. And so we claim that the action of the lowering operator on the ground state $\phi_0$ would be
$\hat{a}\phi_0=0$ and we use this to calculate the ground-state wave function and so on. But I have neither found not been able to construct a good enough argument as to why $\hat{a}\phi_0=0$. Why not something other than 0? or maybe some other wave function entirely?
It would be very nice if you could provide arguments in terms of wave functions and not the vectors as the Dirac notation confuses me as I am taking a quantum mechanics course for the first time and we have not covered that yet.
Best Answer
The eigenstates of the operator $N = a^\dagger a$ can be labeled by their eigenvalues, i.e. $N \phi_n = n \phi_n$, where $n$ is an integer. Note that $$n = \langle \phi_n,N\phi_n\rangle = \underbrace{\langle\phi_n,a^\dagger a \phi_n\rangle = \langle a \phi_n,a\phi_n\rangle}_{a \text{ and } a^\dagger\text{ are mutually adjoint}} = \Vert a \phi_n\Vert^2 \geq 0 $$
In particular, we have that $$ 0 = \Vert a\phi_0 \Vert^2$$ Since the only element of a Hilbert space with zero norm is the zero vector itself, we have that $$a \phi_0 = 0$$