Momentum representation of hydrogen atom

[dingo_d]

Homework Statement

I need to calculate the probability distribution of 1s and 2p state of hydrogen atom in momentum and in coordinate representations.

I have calculated the wave function in coordinate representation, and the dilemma is, do I simply do the Fourier transform for given wave functions?

Since it's just 1s and 2s (l and m=0 so there is no theta and phi dependence) it becomes just radial, and that's just 1D, right?

And when I'm calculating the FT, do I take the limits of integral from $$(-\infty, \infty)$$ or $$(0, \infty)$$? Because when I put put the limits from $$(-\infty, \infty)$$ the integral diverges :\

 

[vela]

You want to evaluate

$$\psi(\mathbf{k})=\frac{1}{\sqrt{2\pi}}\int d^3\mathbf{r}\,\psi(\mathbf{r}) e^{-i\mathbf{k}\cdot\mathbf{r}}$$

Go from there.

 

[dingo_d]

Yeah, but the wave functions $$\psi_{100}$$ and $$\psi_{200}$$ doesn't depend on $$\theta$$ and $$\phi$$, so I just integrate over r.

I was thinking, since the potential I used was Coulomb one: $$-\frac{e^2}{r}$$, it goes from $$(0,\ \infty)$$, so my integral should have the same limits, that way it won't diverge, right?

I found something later, in Schaum, but I didn't see what are his wave functions in coordinate representation...

 

[vela]

You can't just ignore the angles. You still have to integrate over them. The wave function may not depend on the angles, but the integrand as a whole does.

 

[dingo_d]

Oh... Thanks for the hint ^^

EDIT: Oh, and in that case if I take the transform to be:

$$\psi(\mathbf{p})=\frac{1}{\sqrt{2\pi}}\int d^3\mathbf{r}\,\psi(\mathbf{r}) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r}}$$

I need to be carefull about that extra $$\hbar$$ and since it's 3D do I need to take the factor in front:

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

 

[vela]

Yeah, probably. I wasn't paying attention to the factors of $2\pi$ and $\hbar$.

 

[dingo_d]

Ok, thnx ^^

I have one more question, since I need to factor the other coordinates as well, the probability distribution, which I must draw, will be $$P=4\pi r^2|\psi|^2$$, right? (Because in the book I didn't notice the 4\pi part...)

 

[vela]

I don't understand what you are asking.

 

[dingo_d]

Well I need to find the probability in the momentum and coordinate space. And the probability is $$P=|\psi|^2$$ but in my case since I'm dealing with 3D wave function the probability is given over a volume, so the probability is $$P=4\pi r^2|\psi|^2$$, right?

I'm getting nice images :D

 

[vela]

That would be the radial probability density. Multiplied by dr, it would be the probability of finding the electron in a spherical shell of radius r and thickness dr.

 

[dingo_d]

Now I'm all confused :\

I have clculated $$\psi_{100}$$ and $$\psi_{200}$$.

I need the probability density. So I do: $$P=|\psi_{100}|^2 r^2 4\pi$$ right? And when I plot it I get the above picture.

Because just taking $$P=|\psi_{100}|^2$$ will give me diverging line...

What I'm asking, is my formula correct?

 

[vela]

It's impossible to answer your question because you haven't said what exactly you're trying to find. It's kind of like if someone told you he rolled a pair of dice and then asked you, is the probability 1/6? You can't answer that because you don't know what he's talking about exactly.

 

[dingo_d]

the problem is: find the probability density in coordinate and momentum representations for 1s and 2s states of hydrogen atom.

sorry if I'm not understandable :(

 

[vela]

Try posting the complete problem statement exactly as it was given. You're leaving out details which you don't seem to realize are important.

 

[dingo_d]

That is the entire statement XD

 

[vela]

Well, then you'll need to ask for clarification. There are different probability densities depending on what probabilities you're interested in. $|\psi(x,y,z)|^2$ is a probability density; you could integrate out anyone of those variables and get yet another probability density. In other words, you can't plot the probability density because there is more than one. You can plot a probability density, but you need to know exactly which one they're talking about.

(That said, I'd guess the one you plotted, the radial density, is likely the one you want.)

 

[dingo_d]

Oh... I see. I'll mention that to my teaching assistant.

I also think that he thought about radial, because that's the one we have in our books.

I plotted the solution from Schaum (altho I get a different solution when integrating) and the probability distribution in momentum space for 1s state is really low, and 2s is higher, but still lower than those in coordinate space.

Since it's Fourier transform, I predicted that it would be this way (peaked function in one space is flat in other, right?), but I'm kinda surprised at how small it is (by several orders of magnitude)...