# Eigenstates of QM harmonic oscillator in momentum space

In almost every introductory QM book they treat the QM harmonic oscillator. As a result, one finds out that the eigenfunctions in position space are given in terms of Hermite polynomials (in natural units)

$$\psi_n(x) = \frac{1}{\pi^{\frac{1}{4}}\sqrt{2^n (n!)}} e^{-\frac{x^2}{2}} H_n(x)$$.

In 'algebraic' derivations one uses $$|n\rangle$$ to denote the $$n$$-th eigenstate $$\psi_n$$, consequently $$\langle x|n\rangle = \psi_n(x)$$ is the representation of the wave function in the position space. Therefore, we have $$\langle x| n\rangle= \frac{1}{\pi^{\frac{1}{4}}\sqrt{2^n (n!)}} e^{-\frac{x^2}{2}} H_n(x)$$.

I tried to find a similar representation for the wave function in momentum space, $$\langle p | n \rangle$$ but couldn't find anything in my QM literature. I know that I can apply a Fourier transform to obtain the representation in position space (which I did for the ground state). This is quite tedious and becomes infeasible once going to higher $$n$$.

For symmetry reasons, I would expect some quite similar expression as for $$\langle x | n \rangle$$, maybe with some additional complex phase ($$i$$?), coming from the rotation in phase space.

Does anyone know a reference, where they give an expression (and or derivation) for $$\langle p |n \rangle$$ explicitly? Or is it an easy calculation and I have overseen something, making life too complicated?

• Yes, up to a factor they have the same form as in position representation just replacing $x$ for $p$. en.wikipedia.org/wiki/… Commented Apr 26, 2021 at 12:22
• Have you tried to take the Fourier transform of the Schrödinger equation? I think that the Fourier transform of $\psi_n(x)$ obeys a similar differential equation as $\psi_n(x)$, with (very roughly) the replacing $p \leftrightarrow x$. Commented Apr 26, 2021 at 12:30
• You understand the oscillator hamiltonian generates a rotation in phase space taking you from coordinates to momenta and back. This was firmed up by Condon's 1937 paper. Commented Apr 26, 2021 at 13:54

With $$\varphi_n(x)=\frac 1{\sqrt{2^nn! \sqrt{ \pi}}}H_n(x)e^{-x^2/2}$$ being the normalized oscillator wavefunctions, and defining the Fourier transform by
$${\mathcal F}[f](p) =\frac 1{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ipx} f(x)dx$$ you have $${\mathcal F}[\varphi_n](p)= i^n \varphi_n(p).$$

This is most easily derived from the generating function for Hermite polynomials:

We evaluate the Fourier transform of the generating function
$$\exp\left\{2xt -t^2 -\frac 12 x^2\right\}= \sum_{n=0}^{\infty} \frac {t^n}{n!} H_n(x)e^{-x^2/2}$$ to get
$$\sum_{n=1}^\infty \frac {t^n}{n!} \int_{-\infty}^\infty e^{ipx} H_n(x) e^{-x^2/2}\,dx =\frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty e^{ipx} e^{2xt -t^2 -x^2/2}\, dx\nonumber\\ = \exp\left \{-t^2 + \frac 12 (2t+ip)^2\right\} \nonumber\\ = \exp\left\{t^2 +2itp - \frac 12 s^2\right\}\nonumber\\ = \exp\left\{-(it)^2 +2itp - \frac 12 s^2\right\}\nonumber\\ = \sum_{n=0}^{\infty} \frac {t^n}{n!} i^n H_n(p)e^{-p^2/2}.$$
Identifying coefficients of $$t^n$$ then gives the desired result.

• @pcalc You are familiar with the eigenfunctions of the Fourier operator, no? Commented Apr 26, 2021 at 13:42
• Thank you - I wasn't aware that the weighted Hermite-polynomials are eigenfunctions of the Fourier operator! That solves my problem in a very easy and elegant way. [Interestingly, even Mathematica couldn't transform an arbitrary weighted Hermite-polynomial]. Thank's a lot! Commented Apr 27, 2021 at 7:44
• @mike stone A brief remark; shouldn't it be $(-i)^n$, at least if the Fourier transform is given correctly at Wikipedia (en.wikipedia.org/wiki/…) ? Commented Apr 28, 2021 at 9:32
• @pcalc I have not checked this recently, but I did define the sign convention for the FT in my answer. There are many different conventions for Fourier transfoms Commented Apr 28, 2021 at 11:51
• @mikestone would you mind glancing at my answer which gets a factor $(-1)^n$ instead? The phases don't really matter but they are bugging me. Commented Apr 28, 2021 at 13:34

In momentum space the operators of position and momentum are given by $$\hat{x} = i\hbar\frac{\partial}{\partial p},\\ \hat{p} = p$$ So obtaining the solution is rather straigntforward, knowing the solution in position space.

• Position representation $$H = \frac{p^2}{2m} + \frac{m\omega^2x^2}{2} = -\frac{\hbar^2}{2m}\partial_x^2 + \frac{m\omega^2}{2}x^2$$
• Momentum representation $$H = \frac{p^2}{2m} + \frac{m\omega^2x^2}{2} = \frac{1}{2m}p^2 -\frac{m\hbar^2\omega^2}{2}\partial_p^2 = -\frac{\hbar^2}{2\mu}\partial_p^2 + \frac{\mu\omega^2}{2}p^2,$$ where we defined $$\mu=1/(m\omega^2)$$.

Thus, the solutions for the Harmonic oscillator in momentum representation are obtained from those in frequency representation by replacement: $$x \longrightarrow p,\\ m\longrightarrow \mu=\frac{1}{m\omega^2}.$$ Remark
Transformations between representations are discussed in any QM textbooks. Equations for position and momentum operators are perhaps not always spelled... but definitely derived explicitly in the Landau & Livshitz' QM. As noted in the comments, in practice it involves doing basic Fourier transform. However, it is a good idea to internalize it in terms of more general representation theory.

Check "Momentum Representation for the Harmonic Oscillator", https://www.ks.uiuc.edu/Services/Class/PHYS480/qm_PDF/chp4.pdf in p.88

I think the clearest way to see this (ie without having to recall any facts about Fourier transforms) is to remember that the original problem was

$$H = \frac{1}{2m}p^2 + \frac{1}{2}m\omega^2 x^2$$ where $$[x,p]=i$$.

This has eigenfunctions $$\phi_n(x;m,\omega)$$ that depend on the mass and frequency. We want to find the eigenfunctions $$\tilde{\phi}_n(p,m,\omega)$$. To do this note that the Hamiltonian can be written as:

$$H = \frac{1}{2m'}x^2 + \frac{1}{2}m'\omega^2 p^2$$ where $$[p,x]=-i$$ (equivalently, $$[(-p),x]=i$$) and $$m'=(m\omega^2)^{-1}$$. Then going through the same steps as before but mentally swapping $$x,p$$ , one will find that the eigenfunctions are the same as before but with $$m$$ replaced by $$m'$$, since the problem is identical. That is, one should find

$$\tilde{\phi}_n(p,m,\omega)=\phi_n(-p;m',\omega')=(-1)^n\phi_n(p;\frac{1}{m\omega^2},\omega).$$

The phase pre-factor isn't really important since we only care about states up to a phase anyway (however I am a bit annoyed I don't get the $$(i)^n$$ in Mike Stone's answer so if someone can spot a mistake I'd be happy to hear it.)

I would expect them to also be Hermite polynomials for the following reason. The ground state is a Gaussian, and the fourier transform of a Gaussian is also a Gaussian. So we know that $$\psi_0 (p)$$ is a Gaussian just like $$\psi_0(x)$$. Then, from the basic algebra of the creation and annihilation operators, we know that we can find the other states by just acting with $$a^\dagger$$ over and over on the ground state, even if it's in the momentum basis. Now, from memory, the annihilation operator is something like $$a \propto (x - \frac{i}{m \omega} p)$$. If you mutiply by the correct factor of $$m \omega$$ and $$i$$ you can get this to look like the exact same thing with $$x$$ and $$p$$ switched. So now it seems like you can just act with these on the Gaussian $$\psi_0(p)$$ and you'll once again get a tower of Hermite polynomials, but now as functions of $$p$$ instead of $$x$$.

• Thank you! I observed the same for the ground state, but I wasn't sure whether my suspicion for that "rule" for higher $n$ holds true. At least I wasn't sure about the argumentation with the ladder operators Commented Apr 27, 2021 at 7:47