Power of $a^{\dagger}=\sqrt{\frac{\omega m }{2 \hbar}}\left (x-\frac{\hbar}{m\omega}\frac{d}{dx} \right)$

Question

I don't know if a mathematical passage is correct. For the moment I am not entirely interested in formally understanding justification, but only if it is correct operatively. Considerig the ground level autofunction for the harmonic oscillator: $$\phi_0(x)=\left( \frac{m\omega}{\pi \hbar} \right)^{\frac{1}{4}}\exp\left[ -\frac{m\omega}{2\hbar}x^2 \right]$$ I derived that $$\phi_n(x)=\frac{1}{\sqrt{n!}}(a^{\dagger})^n\phi_0(x)$$ Considering that $$a^{\dagger}=\sqrt{\frac{\omega m }{2 \hbar}}\left (x-\frac{\hbar}{m\omega}\frac{d}{dx} \right)$$ It' s true that for example: $$(a^{\dagger})^3= \left( \frac{\omega m }{2 \hbar}\right)^{\frac{3}{2}} \left (x^3-3x^2\frac{\hbar}{m\omega}\frac{d}{dx}+3x\left(\frac{\hbar}{m\omega}\right)^2\frac{d^2}{dx^2} +\frac{d^3}{dx^3}\right)$$

Answer 1

That's almost correct, but not quite. Imagine $(a^\dagger)^2$ is actually acting on a wavefunction (say, $\psi$). Then \begin{align} (a^\dagger)^2 \psi & = \left( \frac{\omega m}{2 \hbar}\right) \left( x - \frac{\hbar}{m \omega}\frac{d}{dx} \right)\left( x - \frac{\hbar}{m \omega}\frac{d}{dx} \right)\psi \\ & = \left( \frac{\omega m}{2 \hbar}\right) \left( x^2 \psi-\frac{\hbar}{m\omega}\frac{d}{dx}(x\psi )-\frac{\hbar}{m\omega}x\frac{d\psi}{dx}+\left( \frac{\hbar}{m\omega}\right)^2\frac{d^2 \psi}{dx^2} \right) \\ &= \left( \frac{\omega m}{2 \hbar}\right) \left( \left(x^2 -\frac{\hbar}{m\omega}\right)\psi-2\frac{\hbar}{m\omega}x\frac{d\psi}{dx}+\left( \frac{\hbar}{m\omega}\right)^2\frac{d^2 \psi}{dx^2} \right) \end{align} The key point is that the $d/dx$ sits outside of the $x$ in the second term in the second line.

Answer 2

Not correct. $x$ and $\frac{d}{dx}$ do not commute. Try to explicitly apply the $a^{\dagger}$ operator twice on some function.