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.

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    $\begingroup$ 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. $\endgroup$ Commented May 7, 2017 at 22:56
  • $\begingroup$ @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). $\endgroup$ Commented May 7, 2017 at 23:22

1 Answer 1


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.

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    $\begingroup$ Thanks! This makes a lot of sense, I hadn't got a true grip on the concept of the phase here, this is my second course in QM and is more focused on the the theory of how to use it than it is with doing the maths, sometimes I don't haven't immediately linked the concepts when they actually pop out of the maths. I like the phase argument and it is good to be aware of the usual convention. $\endgroup$
    – Goobley
    Commented May 7, 2017 at 23:13

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