Does the Harmonic Oscillator Wave Function Change in Momentum Space?

[frogjg2003]

Homework Statement

I'm trying to prove that the Harmonic oscillator wave function doesn't change (except a phase factor) when I convert from position to momentum space.
$\Phi_{nlm}(\vec p)=(-i)^{2n+l}\Psi_{nlm}(\vec p)$

Homework Equations

$\Phi_{nlm}(\vec p)=\frac{1}{(2\pi)^{3/2}}\int d^3r e^{-i \vec p\cdot\vec r}\Psi_{lmn}(\vec r)$
$\Psi_{nlm}(\vec r)=N_{nl} \alpha^{3/2} e^{-(\alpha r)^2/2} L^{l+\frac{1}{2}}_n((\alpha r)^2) R_{lm}(\alpha\vec r)$
in natural units and α is in GeV. $L_n^{l+\frac{1}{2}}(x^2)$ is the associated Laguerre polynomials, $R_{lm}(\vec r)=\sqrt{\frac{4\pi}{2l+1}} r^lY_{lm}(\hat r)$ and N_nl is a normalization constant. I'm leaving α out of N_nl because I want to keep track of it for later calculations.
$R_{lm}(\vec x+\vec a)=\sum_{\lambda=0}^l \sum_{\mu=-\lambda}^\lambda \sqrt{\binom{l+m}{\lambda+\mu}\binom{l-m}{\lambda-\mu}} R_{lm}(\vec x) R_{(l-\lambda) (m-\mu)}(\vec a)$
$R_{lm}(a\vec x)=a^lR_{lm}(\vec x)$
All integrals are over all space, i.e.
$\int d^3 r=\int_0^\infty dr \int_0^\pi r d\theta \int_0^{2\pi} r\sin{\theta} d\phi$

The Attempt at a Solution

The end result should be $\Phi_{lmn}(\vec p)=(-i)^{2n+l} N_{nl} \alpha^{-3/2} e^{-(p/\alpha)^2/2} L^{l+\frac{1}{2}}_n((\frac{p}{\alpha})^2) R_{lm}(\frac{\vec p}{\alpha})$

I've managed to manipulate it into
$\Phi_{nlm}=\frac{N_{nl}}{(2\pi\alpha)^{3/2}} e^{-q^2/2}\int d^3 x e^{-(\vec x+i \vec q)^2/2} L^{l+\frac{1}{2}}_n(x^2) R_{lm}(\vec x)$
where $x=\alpha r$ and $q=\frac{p}{\alpha}$. I want to further transform it to $\vec y=\vec x-i\vec q$ and integrate over y.
I can't figure out if the differential volume element changes, i.e. if I can just write it as
$\Phi_{nlm}=\frac{N_{nl}}{(2\pi\alpha)^{3/2}} e^{-q^2/2}\int d^3 y e^{-y^2/2} L^{l+\frac{1}{2}}_n((\vec y-i\vec q)^2) R_{lm}(\vec y-i\vec q)$

Assuming I could, I continued and managed to get it into the relatively simple form of
$\Phi_{nlm}=\frac{N_{nl}}{(2\pi\alpha)^{3/2}} e^{-q^2/2}\sum_{\lambda=0}^l\sum_{\mu=-\lambda}^\lambda \sqrt{\frac{4\pi}{2\lambda+1}\binom{l+m}{\lambda+\mu}\binom{l-m}{\lambda-\mu}} (-i)^l R_{(l-\lambda)(m-\mu)}\int d^3y e^{-y^2/2} L^{l+\frac{1}{2}}_n((\vec y-i\vec q)^2) y^\lambda Y_{\lambda\mu}(\hat y)$

For n=0, $L_n^{l+\frac{1}{2}}(x)=1$, so it was trivial to show it was correct. For n=1, I can't get the cross term in $y^2-2i\vec y\cdot \vec q-q^2$ to disappear and it leaves $\delta_{\lambda1}$ terms I can't seem to get rid of.

 

[Oxvillian]

I think that it's not necessary to play about with all the special functions to see this.

If you write out the time-independent Schrodinger equation in p-space, you'll find that it looks exactly the same as it does in x-space, with some annoying factors of $m$, $\omega$ and $\hbar$ shuffled around. So when you solve the energy eigenvalue problem in p-space, you're basically doing exactly the same math as you would be in x-space, and the answers will be of the same form.

It all comes about because of the x-p symmetry of the harmonic oscillator Hamiltonian. That Hamiltonian, roughly speaking, can't tell the difference between x and p.

 

[frogjg2003]

True, but that's the undergrad method. I want to be able to transform wavefunctions that aren't the same in both spaces, and this would is a very good test case.