Dot Product of Momentum and Radial Operators

[gatztopher]

Homework Statement

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

Homework Equations

[tex]
\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}
[/tex]

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

[tex]
d^3r=r^2dr sin\theta d\theta d\varphi
[/tex]

The Attempt at a Solution

After doing some plugging in and integrating, I get
[tex]
\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
[/tex]

And my little roadblock is simply regarding the [tex]\vec{p}\cdot\vec{r}[/tex]. I know that it's [tex]-i\hbar \nabla r[/tex] 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?

Thanks for your help!

 

[gabbagabbahey]

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

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

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

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

 

[gatztopher]

So, I understand that you mean this:
[tex]\vec{p} \cdot \vec{r}=p\hat{z} \cdot r\hat{r}=|p||r|cos\theta[/tex]
But then, that leaves me with [tex]|p||r|cos\theta=|-i\hbar\nabla| |r|cos\theta[/tex] with [tex]\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}[/tex], 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?

 

[gabbagabbahey]

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

 

[gatztopher]

Oh! Overthinking it... thank you!