Is this is the correct way to derive the ground state of a harmonic oscillator? In Griffiths Chapter 2, the harmonic oscillator the author assumes that
$$a_{-} \psi_0=0 \tag{1}$$
But we can also express this in a more general form as
$$a_{-}^n \psi = 0 \tag{2}$$
which has the eigenvalue, $E-n\hbar\omega$
$$n = ?$$
Or we could also assume that
$$\psi_0 = a^{n-1}_{-} \psi $$
which is a lot easier,
What I'm trying to say is that, can the wavefunction be derived from the general formulation of the lowest "rung" in eq(2) instead of eq(1)(since $a-{\psi}_0$ is the result of multiple applications of the annihilation operator.
I had given it a try but couldn't get to anything useful. Is this way to derive the wavefunction even logical ?
I think Taylor expansion may do the trick, but the math involved would be very difficult
$${\frac{1}{(\sqrt{2m})^n}\Bigr(\frac{\hbar}{i}\frac{d}{dx} - im\omega x\Bigr) }^n\psi =0$$
Note : I have already seen the straightforward derivation of the ground state.
 A: I gather you are interested in deriving all Hermite functions by the multiple application of the creation operator in coordinate space, or the ground state by laddering down from an excited state.
I assume you are comfortable with Dirac's elegant ladder, in mainstream (not Griffiths) notation,
$$
a|0\rangle=0\\
|n\rangle \equiv \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle, ~~\leadsto\\
\hat H|n\rangle = \hbar \omega (n+1/2)|n\rangle,\\
  a^\dagger|n\rangle = \sqrt{n + 1} | n + 1\rangle \\
          a|n\rangle = \sqrt{n} | n - 1\rangle\\
a^n |n\rangle=\sqrt{n!}|0\rangle ~~~\leadsto \\
a^{n+1}|n\rangle=0,
$$
which appears like what you are after.
In the coordinate representation, $\langle x|0\rangle=\psi_0(x)$,
$$ \left\langle x \mid a \mid 0 \right\rangle = 0  ~~~
  \leadsto \left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\left\langle x\mid 0\right\rangle = 0  ~~~
  \leadsto         \\
        \left\langle x\mid 0\right\rangle = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4} \exp\left( -\frac{m\omega}{2\hbar}x^2 \right)                                                                                                   = \psi_0 (x) ~, $$
solving the above ODE;  hence,
$$\psi_1(x )= \langle x \mid  1 \rangle    =
 \langle x \mid   a^\dagger \mid 0 \rangle = \left(x - \frac{\hbar}{m\omega}\frac{d}{dx}\right) \psi_0 (x) ~,$$
etc, recursively producing all $\psi_n$, utilizing the Hermite function recursion relation
$(x-\partial) (e^{x^2/2}\partial^n e^{-x^2})=- (e^{x^2/2}\partial^{n+1} e^{-x^2})$.
You may, of course, run the recursion backwards
$\left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right) \psi_1 \propto \psi_0$, hence $\left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)^2 \psi_2 \propto \psi_0$, etc, possibly your original question.
