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I am trying to compute the $\Psi(\vec{p})$ for an electron in the ground state of an Hydrogen atom, H(1s) in order to compare the Bohr model with the results of 1982 Weigold et al.

I have found that the function in the position space representation is $$\Psi(\vec{r})=\sqrt{\frac{1}{\pi a_{0}^{3}}}\exp(-\rho/a_0),$$ being $a_0$ the Bohr radius and $\rho$ the radial component using spherical coordinates. My concern is how to calculate the probability distribution in the momentum space representation? I am trying to do it using Fourier analysis: $$\Psi(\vec{p})=\frac{1}{(2\pi \hbar)^{3/2}}\int{e^{-i\frac{\vec{p}\vec{r}}{\hbar}}\psi(\vec{r})dr},$$ where $\vec{r}=(\rho,\varphi,\theta)$ and $\vec{p}=(0,0,p)_{x,y,x}. $

And my question is:

  • Can I just use $\vec{p}=p(cos\theta,0,-sin\theta)$? (I don't think so because I guess the Fourier Transform is not the same for spherical coordinates?)
  • So, if using $\rho=\sqrt{x^2+y^2+z^2}\ $ in $\Psi(\vec{r})$ is correct way? (I'm trying to avoid this because I can't solve the integral)
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To find this Fourier transform, you start with $$\Psi(\vec{p})=\frac{1}{(2\pi \hbar)^{3/2}}\int{e^{-i\vec{p}\cdot\vec{r} / \hbar }\psi(\vec{r})\mathrm d\vec r},$$ as you've written, but you need to be a bit more careful after that.

To begin with, you can't assume that $\vec p$ is along the $z$ axis, because $\vec p$ has been given to you as the argument of $\Psi(\vec p)$, and you can't change it. What you can do, however, is set up the coordinate system for the $\vec r$ integration so that its axes simplify your life, i.e. with $z$ along the direction of $\vec p$.

(The reason you can do this, by the way, is because the ground state $\psi(\vec r)$ is spherically symmetric. If it's not symmetric, such as e.g. anything that's not an $s$ state, then you need to use a decomposition of the plane wave into partial (spherical) waves, like the one in Jackson, 2nd ed., p. 471, eq. (10.43).)

Once you've done that, then yes, you can write $\vec p = (0,0,p)$, and you can further use spherical polar coordinates $\vec r = r(\sin(\theta) \cos(\phi), \sin(\theta)\sin(\phi), \cos(\theta))$ for your position vector, write \begin{align} \Psi(\vec{p}) & = \frac{1}{(2\pi \hbar)^{3/2}} \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-ipr\cos(\theta) / \hbar }\psi(\vec{r}) \:r^2\sin(\theta)\:\mathrm d\phi\mathrm d\theta\mathrm dr \\& = \frac{1}{(2\pi \hbar)^{3/2}}\sqrt{\frac{1}{\pi a_{0}^{3}}} \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-ipr\cos(\theta) / \hbar }e^{-r/a_0} \:r^2\sin(\theta)\:\mathrm d\phi\mathrm d\theta\mathrm dr, \end{align} and integrate away.


Alternatively (and this is really for any future visitors who chance upon here looking for an actual answer for the title question), you can cheat and look up the answer, which is found in e.g.

The Momentum Distribution in Hydrogen-Like Atoms. B. Podolsky and L. Pauling. Phys. Rev., 34, 109 (1929).

(Not you, Marc, by the way. Go and do the integral.)

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