# Hydrogen atom:<K>,<V>,momentum distribution

1. Mar 18, 2008

### folgorant

Hello to everybody! This is my first time in PF. I have problems with an QM exercise.

1. The problem statement, all variables and given/known data
About hydrogen atom in the ground state (n=1), evaluate:
<K>, <V> (expectation value of the kinetic energy and potential energy) and the momentum distribution ($$\varphi$$(p)).

3. The attempt at a solution

for now I treat only <K>:
see attachment 009 (<K>=<T>=kinetic energy)
but i know the exact solution so that I evaluate is not right.
where is my mistake??

thanks

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2. Mar 19, 2008

### folgorant

okay. maybe my attachment is not readble.

the wave function for the hydrogen atom in 1s state is:= $$\varphi(r)=\frac{exp(-r/a)}{\sqrt{\pi a^3}}$$

where:

a is the Bohr radius := $$a=\frac{\hbar^2}{m e^2}$$ ;
m is the electron mass ;
$$e=\frac{q^2}{4\pi\epsilon}$$ ;
q is the elementary charge ;

now:

to find <K> (expectation value of kinetic energy) I have to evaluate:

$$\int \frac{p^2}{2m}\left|\varphi(r)\right|^2 r^2 sin\vartheta d\vartheta d\phi dr$$ from 0 to infinity

i.e. = $$\frac{(-i\hbar)^2}{2m\pi a^3} \int \partial^{2}/\partial r^{2}[exp(-2r/a)] r^2 sin\vartheta dr d\vartheta d\phi = \frac{(-i\hbar)^2}{2m\pi a^3} 4 \pi \int \frac{4r^2}{a^2} exp(-2r/a) dr = \frac{(-i\hbar)^2}{2m\pi a^3} 4 \pi a = \frac{(-2i\hbar)^2}{m a^2} = \frac{-2 \hbar^2 m^2 e^4}{m \hbar^4} = \frac{-2me^4}{\hbar^2}$$

but the exact solutions found on Basdevant or Sakurai is:= $$K=\frac{+me^4}{2 \hbar^2}$$

so...where is my mistake??
please, help me to find it!!

3. Mar 19, 2008

### clem

$$\nabla^2$$ is not $$\partial^2/\partial r^2$$ in spherical coordinates.

4. Mar 19, 2008

### folgorant

ops....it's true...

I tried with:

$$<K>=\frac{1}{\pi a^3}\frac{-\hbar^2}{2m} \int \frac{1}{r^2} \partial/\partial r (r^2 \partial/\partial r) exp(\frac{-2r}{a}) r^2 sin\vartheta d \theta d \phi d r$$ from 0 to infinity

.............................
.....................................

then

$$\frac{-2 \hbar^2}{m a^3} \frac{-2}{a} \int exp(\frac{-2r}{a}) (2r-\frac{2r}{a}) dr = = \frac{-2 \hbar^2}{m a^3} \frac{-2}{a} [r^2 exp(\frac{-2r}{a})]^{inf}_{0}$$

and here I can't have a finite solution....
is the integral correct?
how to have a finite solution for $$[r^2 exp(\frac{-2r}{a})]^{inf}_{0}$$ ????

thanks a lot

5. Mar 19, 2008

### malawi_glenn

$$[r^2 exp(\frac{-2r}{a})]^{\infty}_{0} = 0$$

and I dont think your interation is correctly.

you should end up with an integral like:
$$\int ^{\infty}_0(2r-2r^2/a)e^{-2r/a}dr$$

And use the standard integral:
$$\int ^{\infty}_0r^ne^{-\alpha r}dr = (1/\alpha ^{n+1}) \Gamma (n+1)$$

Useful properties of the gamma function:
$$\Gamma (m+1) = m\Gamma (m)$$
$$\Gamma (1) = 1$$
$$\Gamma (1/2) = \sqrt{\pi}$$

Last edited: Mar 19, 2008
6. Mar 19, 2008

### folgorant

I check the solution of the integral with the software "Derive" and it's correct!!!

....?????????????????

7. Mar 19, 2008

### malawi_glenn

$$\int {r^2} \partial/\partial r (r^2 \partial/\partial r) exp(\frac{-2r}{a}) r^2 d r = (-2/a)\int (2r-2r^2/a)e^{-2r/a}dr$$

??

Last edited: Mar 19, 2008
8. Mar 19, 2008

### folgorant

...and I tried also to solve with Gamma function but the integral = 0 !!!!!

9. Mar 19, 2008

### folgorant

yes...it's the same of mine!

10. Mar 19, 2008

### malawi_glenn

ee no..

11. Mar 19, 2008

### malawi_glenn

Strange.

12. Mar 19, 2008

### folgorant

oh,iI'm sorry but it's only a writing-mistake....in my works I've what you've posted too.....

13. Mar 19, 2008

### malawi_glenn

ok, but I think I know what the error is, you have done things in wrong order:

$$\int \psi (r) ^*\Delta\psi (r)r^2dr$$

not

$$\int \Delta|\psi (r)|^2r^2dr$$

if you know what I mean. This has to do with the definition of expectation value and general properties of derivative operators.

\Delta is the Laplacian Operator, which acts on everything to the RIGHT.

Last edited: Mar 19, 2008
14. Mar 19, 2008

### folgorant

malawi_glen......you're great....i try to correct it right now!!!!
thanks a lot.....

15. Mar 19, 2008

### malawi_glenn

You shold not get 0 now, good luck!

16. Mar 19, 2008

### folgorant

yes!!! the result is correct!! thanks a lot!!

17. Mar 20, 2008

### folgorant

momentun distribution of the hydrogen atom in ground state

ok, now I'm trying to evaluate the momentun distribution of the hydrogen atom in ground state.

I should apply the Fourier Transform to the $$\psi (r)$$

i.e. $$\varphi ( \textbf{p} ) = \frac{1}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{-\infty}_{+\infty} exp(-\textbf{r} /a) exp(i \textbf{pr} ) d^{3}r$$

mmm..I'm not good in these -hard math works- ....

my problem is also about the -morevariables- integral...

I don't really know how to start....

Since the wave function depends only from r, I've just tried to do:

$$\varphi ( p ) = \frac{1}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{-\infty}_{+\infty} exp(-r /a) exp(ipr} ) dr = \frac{1}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{-\infty}_{+\infty} exp(\frac{-r}{a} + ipr) ) dr$$

...but the result of the integral is something very strange...I think it's not the good way.....

18. Mar 20, 2008

### malawi_glenn

dude $$d\mathbf{r} = r^2sin \theta d\theta drd\phi$$ or this is even better in this case:
$$d\mathbf{r} = r^2 d(\cos \theta )drd\phi$$

so that:

$$\varphi ( \textbf{p} ) = \frac{1}{(2 \pi)^{3/2}} \int \varphi (\textbf{r}) \exp(i \textbf{p} \cdot \textbf{r}) d^{3}r =$$

$$\frac{1}{(2 \pi)^{3/2}} \int \varphi (\textbf{r}) \exp (i \textbf{p} \cdot \textbf{r}) r^2d(\cos \theta ) drd\phi$$

And you integrate r from 0 to infinity, theta from 0 to pi (but you now have cos(theta) as integration variable, so be careful, and phi from 0 to 2*pi.
Use the forumla for dot-product:
$$\textbf{a} \cdot \textbf{b} = ab \cos \theta$$

But you must use the SPATIAL wave function, just not the radial part. i.e
$$\varphi (\textbf{r}) = \varphi (r) \cdot Y_0^0 (\theta ,\phi )= \frac{\exp(-r/a)}{\sqrt{\pi a^3}}\cdot 1/(\sqrt{4 \pi})$$

I hope you get the picture, I might miss some constants etc, since I have not done this QM for 1 year, but the procedures I remember ;-) I hope

Last edited: Mar 20, 2008
19. Mar 20, 2008

### malawi_glenn

I did som crucial editing, my latex was wrong. Now it shold be more correct.

20. Mar 20, 2008

### folgorant

my questions:

1)
if $$d cos (\theta) / d\theta = -sin(\theta)$$ ,
so $$sin(\theta) d \theta = -cos(\theta)$$....isn't true???

2)
I check for the spatial wave function and the term $$\frac{1}{\sqrt{4\pi}}$$ is included in the $$\varphi(r)=\frac{exp(-r/a)}{\sqrt{\pi a^3}}$$....,no??

then, I tried to cumpute:
$$\varphi ( \textbf{p} ) = \frac{1}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int e^{-r/a -ipr cos \theta } r^2 sin \theta dr d\theta d\phi = \frac{1}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{\infty}_{0}\int^{\pi}_{0} \int ^{2\pi}_{0} e^{-r/a -ipr cos \theta } r^2 sin \theta dr d\theta d\phi =$$

$$\frac{2\pi}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{\infty}_{0}\int^{\pi}_{0} e^{-r/a -ipr cos \theta } r^2 sin \theta dr d\theta =$$

$$\frac{4\pi}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \int^{\infty}_{0} \frac{e^{-r/a } r sin (pr)}{p} dr =$$

but now I can't solve this.....

how can I do??

is what I wrote correct?