# 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.

• I really don't understand your notation, what do you mean by $a-\psi_0$? Do you mean $a\psi_0$? The lowering operator acting on the ground state? Otherwise it looks like you're subtracting an operator from a state, which doesn't make sense... Jul 5, 2020 at 18:54
• I couldn't format the $_$ sign so i used $-$ symbol Jul 5, 2020 at 19:01
• The $\psi$'s $_$ messes up the formatting:( Jul 5, 2020 at 19:03
• Also, just to be clear, are you asking if -- once we know the ground state of a harmonic oscillator -- we can find an arbitrary state $\psi_n$ by acting an operator on it? Jul 5, 2020 at 19:13
• why is $n=1$ case not sufficient? Jul 5, 2020 at 19:16

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.