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

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

1. Homework Statement
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 ([tex]\varphi[/tex](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??

please help me.

thanks
 

Attachments

Answers and Replies

29
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okay. maybe my attachment is not readble.

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

where:

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

now:

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

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

i.e. = [tex]\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} [/tex]

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

so...where is my mistake??
please, help me to find it!!
 
clem
Science Advisor
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[tex]\nabla^2[/tex] is not [tex]\partial^2/\partial r^2[/tex] in spherical coordinates.
 
29
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ops....it's true...

I tried with:

[tex]<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 [/tex] from 0 to infinity

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

then

[tex]\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} [/tex]

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

thanks a lot
 
malawi_glenn
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[tex][r^2 exp(\frac{-2r}{a})]^{\infty}_{0} = 0[/tex]

and I dont think your interation is correctly.

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

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


Useful properties of the gamma function:
[tex] \Gamma (m+1) = m\Gamma (m) [/tex]
[tex]\Gamma (1) = 1 [/tex]
[tex]\Gamma (1/2) = \sqrt{\pi} [/tex]
 
Last edited:
29
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I check the solution of the integral with the software "Derive" and it's correct!!!

....?????????????????
 
malawi_glenn
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[tex]\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[/tex]

??
 
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29
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...and I tried also to solve with Gamma function but the integral = 0 !!!!!
 
29
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yes...it's the same of mine!
 
malawi_glenn
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ee no..

then

[tex]\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} [/tex]



thanks a lot
 
malawi_glenn
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29
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oh,iI'm sorry but it's only a writing-mistake....in my works I've what you've posted too.....
 
malawi_glenn
Science Advisor
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ok, but I think I know what the error is, you have done things in wrong order:

[tex]\int \psi (r) ^*\Delta\psi (r)r^2dr [/tex]

not

[tex]\int \Delta|\psi (r)|^2r^2dr [/tex]

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.
 
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29
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malawi_glen......you're great....i try to correct it right now!!!!
thanks a lot.....
 
malawi_glenn
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You shold not get 0 now, good luck!
 
29
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yes!!! the result is correct!! thanks a lot!!
 
29
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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 [tex]\psi (r)[/tex]

i.e. [tex]\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 [/tex]

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:

[tex]\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 [/tex]

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

can you help me please???
 
malawi_glenn
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dude [tex] d\mathbf{r} = r^2sin \theta d\theta drd\phi[/tex] or this is even better in this case:
[tex] d\mathbf{r} = r^2 d(\cos \theta )drd\phi[/tex]

so that:

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

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

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:
[tex] \textbf{a} \cdot \textbf{b} = ab \cos \theta [/tex]

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

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
 
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malawi_glenn
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I did som crucial editing, my latex was wrong. Now it shold be more correct.
 
29
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my questions:

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

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

then, I tried to cumpute:
[tex]
\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 =
[/tex]

[tex]\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 =
[/tex]

[tex]
\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 =
[/tex]

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

how can I do??

is what I wrote correct?
 
malawi_glenn
Science Advisor
Homework Helper
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1)
use cos(/theta) as varible instead of theta, is is much easier and you will use this things in the future, so learn this trick now ;)

When you derive the volume element, you have the absolute value when you do the Jacobian etc, so dotn bother about that minus sign.

When theta runs from 0 to pi, cos(theta) goes from 1 to -1, it also quite often this is used in text books, so it's good to know both.

2)
Ok I had no Idea that the spherical harmonic was included in your original wave function, great!

use standard integral tables, maybe you need to do some partial integrations etc.

I dont have time at the moment to check if you have done all steps correct, just try to evaluate the integral. If your answer is wrong, I'll give it a try tomorrow perhaps.

Good Luck!
 
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malawi_glenn
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here is a good integral for you:

[tex] \int_0^{\infty}xe^{-Ax}\sin (bx) dx = 2Ab/((A^2+b^2)^2) [/tex]

[ A > 0 ]
 
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29
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malawi_glenn: starting from [tex]\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 =[/tex]
....and using your integral I get:

[tex]\frac{4\pi}{(2 \pi)^{3/2}} \frac{1}{ \sqrt {\pi a^3}} \frac{2p}{(\frac{1}{a^2} + p^2)^2}[/tex]

..i am a little bit suspicious...can that form be a momentum wave function???
 
malawi_glenn
Science Advisor
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its a function of p right?

What is the answer?

"my" integral? :-) can be found anywhere..

if you plot your phi(p), it looks quite nice.
 
29
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ops, sorry! of course that it could be.....I hadn't seen it in the right way...so.. are you sure the it's the correct solution or could I try to obtain it in other ways??

anyway, thank you very much!!
 

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