# Harmonic oscillator ladder operator acting on wavefunction appears to be zero

For the Quantum Harmonic Oscillator problem, when trying to derive the relation between different energy levels of a wave function when you apply a ladder operator I seem to be making a faulty assumption, but I'm unsure where.

I started with the definition of the ladder operators

$$a_{\pm}= {1 \over \sqrt {2 \hbar m \omega}}(\mp ip+m\omega x)$$

I applied it to the time independent wave function inside the hamiltonian, and then simplified.

$$\hat H(a_+ (\psi))= \hat H( a_+(\psi))+\hbar \omega \ a_+(\psi)$$

My textbook uses the identity of $$\hat H \psi = E \psi$$ to simplify the expression to

$$\hat H (a_+ \psi)= (E+\hbar \omega)(a_+(\psi))$$

Which is consistent with their definition that the ladder operators can raise or lower energy of a wave function by increments of $$\hbar \omega$$ as long as you don't lower past the ground state.

However, if you use the my second equation

$$\hat H(a_+ (\psi))= \hat H( a_+(\psi))+\hbar \omega \ a_+(\psi)$$

and subtract the hamiltonian from both sides you get

$$0= \hbar \omega \ a_+(\psi)$$

which implies that either $$\hbar\omega=0$$ or $$a_+\psi=0$$

Unless you have a degenerate wave function, $$\omega \neq 0$$ and obviously $$\hbar \neq 0$$

Assuming I haven't made another mistake, the only other conclusion is that somehow

$$a_+(\psi)=0$$

So either I've done something wrong, or this is telling me that for every possible wave function

$$\hbar {d \over dx}\psi=m \omega x \psi$$

Is anyone able to spot what's going on?

• Obviously your second equation is wrong. It's not clear where your coming from but I suspect one of your terms (the first on the rhs) should be $a_+\hat H(\psi)$ rather than $\hat H(a_+(\psi))$: the $E a_+\psi$ factor comes from $a_+\hat H(\psi)= a_+ E\psi = E a_+\psi$. – ZeroTheHero Sep 26 '18 at 0:59

Be careful which side of $$H$$ you put $$a^{\dagger}$$ on, they don't commute.

You wrote that:

$$Ha^{\dagger}\psi=Ha^{\dagger}\psi+\hbar\omega\psi$$

But this is incorrect. What you probably meant was:

$$Ha^{\dagger}\psi=a^{\dagger}H\psi+\hbar\omega\psi$$

Note that on the right hand side of this equation $$a^{\dagger}$$ is to the left of $$H$$.

• That makes sense, but my book seems to say that since $H\psi=E\psi$ and E is just a constant, then $a_+$ and $E$ should commute. Or am I misunderstanding something about H and E in that context? – WindowsNT Sep 26 '18 at 1:08
• Would it be safe to say that $[a_+,H]=-m\omega a_+$? – WindowsNT Sep 26 '18 at 1:10
• Correct, but $H$ only becomes $E$ when it acts on the wave-function directly. If there is a creation operator in the way you must use the commutation relations to get $H$ to act on $\psi$. – Connor Dolan Sep 26 '18 at 1:12
• Ok, I see see my mistake. Does that also mean that $\hat H (a_+(\psi))=E_+$ where $E_+=E+\hbar\omega$ is the energy of the next ladder rung from where E was? – WindowsNT Sep 26 '18 at 1:15
• No, because you still need a creation operator that changes the state. – Connor Dolan Sep 26 '18 at 1:17