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

  1. Mar 18, 2008 #1
    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 ([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
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2008 #2
    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!!
     
  4. Mar 19, 2008 #3

    clem

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    [tex]\nabla^2[/tex] is not [tex]\partial^2/\partial r^2[/tex] in spherical coordinates.
     
  5. Mar 19, 2008 #4
    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
     
  6. Mar 19, 2008 #5

    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: Mar 19, 2008
  7. Mar 19, 2008 #6
    I check the solution of the integral with the software "Derive" and it's correct!!!

    ....?????????????????
     
  8. Mar 19, 2008 #7

    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]

    ??
     
    Last edited: Mar 19, 2008
  9. Mar 19, 2008 #8
    ...and I tried also to solve with Gamma function but the integral = 0 !!!!!
     
  10. Mar 19, 2008 #9
    yes...it's the same of mine!
     
  11. Mar 19, 2008 #10

    malawi_glenn

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    ee no..

     
  12. Mar 19, 2008 #11

    malawi_glenn

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    Strange.
     
  13. Mar 19, 2008 #12
    oh,iI'm sorry but it's only a writing-mistake....in my works I've what you've posted too.....
     
  14. Mar 19, 2008 #13

    malawi_glenn

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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.
     
    Last edited: Mar 19, 2008
  15. Mar 19, 2008 #14
    malawi_glen......you're great....i try to correct it right now!!!!
    thanks a lot.....
     
  16. Mar 19, 2008 #15

    malawi_glenn

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    You shold not get 0 now, good luck!
     
  17. Mar 19, 2008 #16
    yes!!! the result is correct!! thanks a lot!!
     
  18. Mar 20, 2008 #17
    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???
     
  19. Mar 20, 2008 #18

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

    malawi_glenn

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    I did som crucial editing, my latex was wrong. Now it shold be more correct.
     
  21. Mar 20, 2008 #20
    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?
     
  22. Mar 20, 2008 #21

    malawi_glenn

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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!
     
    Last edited: Mar 20, 2008
  23. Mar 20, 2008 #22

    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 ]
     
    Last edited: Mar 20, 2008
  24. Mar 20, 2008 #23
    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???
     
  25. Mar 20, 2008 #24

    malawi_glenn

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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.
     
  26. Mar 20, 2008 #25
    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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