# Why is the action of lowering operator on the ground state of a harmonic oscillator to give a 0 wave function?

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 $$a \phi_0 = 0$$

You can do this by direct substitution in the position or momentum basis. Here it is for $x$:

\begin{align} \hat a|0\rangle &= \left[i\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac i {m\omega}\hat p \right) \right] \left[\left(\frac{m\omega}{\pi\hbar}\right) e^{-mwx^2/2\hbar} \right]\\ &\propto \left[x+\frac {\hbar}{m\omega}\frac d{dx} \right]e^{-mwx^2/2\hbar}\\ &\propto xe^{-mwx^2/2\hbar}+\frac {\hbar}{m\omega}(-m\omega x/\hbar )e^{-mwx^2/2\hbar} \\ &= 0 \, . \end{align}

Expanding on J. Murray's answer (which basically is the answer), we can start from the so-called number operator and have $\hat{N} = a^{\dagger}a$. As it was explained Murray's answer, its spectrum is non-negative, because $$ \langle \psi | \hat{N} | \psi \rangle = \langle \psi | a^{\dagger}a | \psi\rangle = || a|\psi\rangle ||^2 \geq 0$$ so every eigenvector $|n\rangle$ must have non-negative eigenvalue $n$ (currently no assumption on $n$. Can be fractional, can be integer, who knows?)

Now an important point is
that $[\hat{N},a]=-a$, which means that if $|n\rangle$ is an eigenvector with eigenvalue $n$, then, applying this commutator we get
$$ \hat{N} a |n\rangle = a(\hat{N}-1)|n\rangle = (n-1)a|n\rangle$$
so $a|n\rangle$ is also an eigenvector of $\hat{N}$ with an eigenvalue $n-1$! This means that if we start with some arbitrary eigenvector, we can create a series of eigenvectors with decreasing eigenvalues, just by repeatedly applying $a$, hurray! The only problem is that we know that this series cannot go on endlessly. It **must** stop otherwise we will get to negative eigenvalues territory, which cannot be true. So the **only** way for this series of lowering eigenvalues to stop is if it exactly hits zero. Then we don't have a state any more, and acting with $a$ on the zero vector just remains zero. So the eigenvalues $n$ have to be the non-negative integers. The state with the lowest possible eigenvalue has to be $a|0\rangle = 0$.

Edit: I just now noticed your last paragraph where you asked for the answer to be in terms of wave-functions. So you can start with $$ \int\! dx \psi^*(x) \hat{N} \psi(x) = \int\! dx |a\psi(x)|^2 \geq 0$$ ensuring that the eigenvalues of $\hat{N}$ are nonnegative, and then have $$ \hat{N} a \psi_n(x) = a(\hat{N}-1)\psi_n(x) = (n-1)a\psi_n(x)$$ telling us that $a\psi_n(x)$ is a wavefunction which is an eigenfunction of $\hat{N}$ with eigenvalue $n-1$. Other than that all progresses as before