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?
Thank you for your help!
 A: 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.
To add more specifics:

*

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