# Normalising the Quantum Harmonic Oscillator

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.

• Since in this simple case only the absolute square of the wavefunction (or an object like $\langle\psi|H|\psi\rangle$) makes physical sense, wavefunctions are determined up to an arbitrary complex phase (like -1 in this case). Which phase you choose is irrelevant, as long as you stick to your convention. There are, of course, cases when the phase can carry a physical significance (e.g. the Berry phase in the Aharonov-Bohm effect), but that's irrelevant at this point. May 7, 2017 at 22:56
• @Goobley Please disregard my initial comment - I hadn't fully grasped the core of the problem. That said, if Griffiths uses $[A_-,A_+]=\hbar \omega$ it is exclusively for didactic purposes; you don't use it in the modern literature as there is a huge inertia that expects $a^\dagger a$ to be the number operator (as opposed to the hamiltonian). May 7, 2017 at 23:22

This isn't really a question about normalization - it's about the phase of the ladder operators. It's important to note that given any ladder operator pair $a,a^\dagger$ such that $$[a,a^\dagger]=1$$ (which is the defining property of the bosonic annihilation and creation operators), you can always define, for any phase $e^{i\varphi}$, a new set of operators $a_\varphi = e^{-i\varphi}a$, $a_\varphi^\dagger = e^{i\varphi} a^\dagger$, and they will also obey the bosonic commutation relation, $$[a_\varphi,a_\varphi^\dagger] = e^{-i\varphi}e^{i\varphi} [a,a^\dagger]=1.$$ Ultimately, it doesn't matter much: it all comes down to convention, and in any case under the normal evolution, in the Heisenberg picture, $a$ will evolve to $a_{\omega t}$ anyways.
However, it's important to note that the usual convention is to take $$a=\sqrt{\frac{m\omega}{2\hbar}} \left( x + i\frac{1}{m\omega} p\right), \tag{*}$$ which differs from your definition by a factor of $i$ (as well as correcting the off-standard normalization $[A_-,A_+]=\hbar \omega$). The phase is not a big problem, so long as you stick to your convention, but it does mean that e.g. if you define the second excited state as $\frac{1}{\sqrt{2!}}{a^\dagger}^2|0⟩$, you will have accumulated a phase of $i^2=-1$, i.e. precisely the minus sign you're talking about. In some ways, the convention $(*)$ is preferable because the eigenfunctions are always real in position space (whereas with your convention the odd-numbered eigenfunctions will be pure imaginary), but that's about the most you can say in its favour - beyond the fact that it has a huge inertia behind it and that it is futile and counter-productive to try and fight that inertia.