# Dot Product of Momentum and Radial Operators

1. Sep 26, 2009

### gatztopher

1. The problem statement, all variables and given/known data

I need to find the momentum space function for the ground state of hydrogen (l=m=0, Z=n=1)

2. Relevant equations

$$\phi(\vec{p}) = \frac{1}{(2\pi\hbar)^{3/2}}\int e^{-i(\vec{p}\cdot\vec{r})/\hbar}\psi(\vec{r})d^3\vec{r}$$

$$\psi(\vec{r})=Y(\theta,\varphi)R(r)=(\sqrt{\frac{1}{4\pi}})(2(\frac{Z}{a_{0}})^{3/2}e^{-Zr/a_{0}})$$

$$d^3r=r^2dr sin\theta d\theta d\varphi$$

3. The attempt at a solution

After doing some plugging in and integrating, I get
$$\phi(\vec{p})=\frac{4}{\pi}(2a_{0}\hbar)^{-3/2} \int r^2e^{-i(\vec{p}\cdot\vec{r})/\hbar}e^{-r/a_{0}}d^3r$$

And my little roadblock is simply regarding the $$\vec{p}\cdot\vec{r}$$. I know that it's $$-i\hbar \nabla r$$ but I don't know how to calculate that, and as a result, I don't know how to carry out the integral.

One of the hints on the problem was, when using spherical coordinates, to set the polar axis along p. What might that mean?

Last edited: Sep 26, 2009
2. Sep 26, 2009

### gabbagabbahey

No, $\vec{r}$ and $\vec{p}$ are the position and momentum vectors, not operators, and the 'dot' is just an ordinary vector dot product.

The polar axis is usually the z-axis, so the hint is just telling you to choose your coordinate system so that $\vec{p}=p\hat{z}$...which you are free to do, since the integration is over $\vec{r}$, which is independent from $\vec{p}$.

3. Sep 28, 2009

### gatztopher

So, I understand that you mean this:
$$\vec{p} \cdot \vec{r}=p\hat{z} \cdot r\hat{r}=|p||r|cos\theta$$
But then, that leaves me with $$|p||r|cos\theta=|-i\hbar\nabla| |r|cos\theta$$ with $$\nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\hat{\varphi}\frac{1}{rsin\theta}\frac{\partial}{\partial\varphi}$$, and if I carried out that calculation I would have a bunch of unit vectors on my hands, which defies the rule that dot products have scalar results. Where's my mistake?

Last edited: Sep 28, 2009
4. Sep 28, 2009

### gabbagabbahey

No, again $\vec{p}$ and $\vec{r}$ are vectors, not operators...$\vec{p}\cdot\vec{r}\neq-i\hbar\nabla r\cos\theta$...Just substitute $\vec{p}\cdot\vec{r}= pr\cos\theta$ into your integral and integrate...

5. Sep 28, 2009

### gatztopher

Oh! Overthinking it... thank you!