I have been working on the quantum harmonic oscillator with ladder operators and I am running into issues with normalising the excited states. There doesn't seem to be a true convention for the ladder operators; I have chosen to use: $A_{\pm}=\frac{1}{\sqrt{2m}}\left(\hat{p}\pm im\omega x\right)$ as it seems simplest to me.
I correctly arrive at wavefunctions proportional to what I want by raising the ground state, but I am having difficulty normalising them. I have tried the normal $\int_{-\infty}^\infty|\psi_n(x)|^2dx = 1$ method, but the answer I get from the second excited state seems to differ from the answer everyone else uses by a sign: I get $$\psi_2 = \left(\frac{m\omega}{4\pi \hbar}\right)^{1/4}\left(1-\frac{2m\omega x^2}{\hbar}\right)\exp\left(-\frac{m \omega x^2}{2\hbar}\right),$$ when all the answers I see are $$\psi_2 = \left(\frac{m\omega}{4\pi \hbar}\right)^{1/4}\left(\frac{2m\omega x^2}{\hbar} - 1\right)\exp\left(-\frac{m \omega x^2}{2\hbar}\right).$$
Now, clearly these yield the same probability distribution, is my answer valid regrading the sign of the wavefunction? If not, some insight into why would be good. I note that in all the material where it is derived, the normalisation constant is derived for a general state $n$ using the ladder operators, but surely the standard method should still work for individual cases.