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.

  • 1
    $\begingroup$ 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... $\endgroup$
    – Philip
    Jul 5, 2020 at 18:54
  • $\begingroup$ I couldn't format the $_$ sign so i used $-$ symbol $\endgroup$
    – Tim Crosby
    Jul 5, 2020 at 19:01
  • $\begingroup$ The $\psi$'s $_$ messes up the formatting:( $\endgroup$
    – Tim Crosby
    Jul 5, 2020 at 19:03
  • $\begingroup$ 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? $\endgroup$
    – Philip
    Jul 5, 2020 at 19:13
  • 4
    $\begingroup$ why is $n=1$ case not sufficient? $\endgroup$ Jul 5, 2020 at 19:16

1 Answer 1


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.


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