1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability density of an electron of hydrogen

  1. Nov 30, 2012 #1
    Hi everyone

    1. The problem statement, all variables and given/known data

    What's the probability density of an electron at a distance r (from hydrogen) which is in the stae n=2, l=1.

    2. Relevant equations
    -



    3. The attempt at a solution

    I think I have to to solve

    [tex] \int |\Psi_{nml}|^2 dV[/tex]

    The solution of the Schrödinger equation is

    [tex] \Psi_{nml} = R(r) Y(\theta,\varphi)[/tex]


    I think I can just ignore my Y and onle calculate

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


    I found in the internet the term for R with n=2 and l=1. But this leads to a horrbile integral. Furthermore I'm not sure about the integration limits. I think the lower limit is zero, but what's the other one? I think it's r because I want to find the probabilty density at r. Is this attempt right and do I really have to solve that integral? (I solved it with wolfram alpha. The solution is pretty long aswell. I think there is an easier way).

    Thanks for your help
     
  2. jcsd
  3. Nov 30, 2012 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You're asked about the probability DENSITY, which means that no integral sign is necessary.
     
  4. Nov 30, 2012 #3
    Oh ok...I'm surprised now actually. What's the approach to calculate the probability density? We haven't heard about that in our lecture :uhh:
     
  5. Nov 30, 2012 #4

    Mute

    User Avatar
    Homework Helper

    ##|\Psi_{mn\ell}(r,\phi,\theta)|^2## is the probability density. That's why when you integrate over it over all space it has to give 1, so that the probability is 1.

    However, ##|\Psi_{mn\ell}(r,\phi,\theta)|^2## is the joint-probability density for the radial position and angular positions. If you only want the radial probability density, then you do need to integrate over the angular variables. If you integrate over the angular variables. If you just neglect the ##Y_{\ell m}(\phi,\theta)##, you'll be off by a constant factor. However, once you've integrated over the angular variables, the

    $$a\int dr r^2 |R(r)|^2$$
    that you wrote down must integrate to 1. A 1-d radial probability density must satisfy ##\int dr \rho(r) = 1##, so what must your probability density be?

    (The "a" in the above formula is just standing in for the constant you get from integrating over the angular variables.)
     
  6. Nov 30, 2012 #5
    so, my probabilty density would be
    [tex] w dr= 4 \pi r^3|R(r)|^2 dr[/tex] ?
     
  7. Nov 30, 2012 #6

    Mute

    User Avatar
    Homework Helper

    Where did the extra factor of ##r## come from? You originally wrote ##r^2|R(r)|^2## in the integral in your earlier post.
     
  8. Dec 1, 2012 #7
    sorry i meant r^2
     
  9. Dec 1, 2012 #8

    Mute

    User Avatar
    Homework Helper

    Ok, that's good then. Two more questions: did your problem not specify which ##m## state the system is in, or that it is at least in a definite m state? If it were in a superposition of the possible m states the calculation would not be as simple, so you might want to double check that, but I suspect the problem is assuming a fixed state of definite m.

    Also, are you sure about that factor of ##4\pi##? If you actually integrate over the ##|Y_{\ell m}(\theta,\phi)|^2## over the solid angle ##d\Omega = d\theta d\phi \sin\theta##, as opposed to just ignoring the spherical harmonic, you do not get a factor of ##4\pi##, I believe.
     
  10. Dec 2, 2012 #9
  11. Dec 2, 2012 #10

    no m was given. Task just says: hydrogen atom in an eigenstate with n l given above.
    To be honest, I'm very unfamiliar with the probabilty density. I think I understood the following: If I integrate the probabilty density I get my probabilty where to find the electron. But let's assume the following: We ignore the angle function Y because we are just interested in the radial ratio. Is my probabilty density now only |R|^2 or a|R|^2 (where a is just a coefficient which comes from the dV in spherical coordinates) ?
     
  12. Dec 2, 2012 #11

    Mute

    User Avatar
    Homework Helper

    Yes, if you integrate the probability density over some region, let's say the region is defined by ##r_1## to ##r_2##, ##\theta_1## to ##\theta_2## and ##\phi_1## to ##\phi_2##, then the result of the integrationis the probability that you will find the electron in that region of space.

    Say that you integrate over all the angles, but only from r = 0 to some r'. That is the probability that you will find the electron at any angle, but within a distance R from the origin. If you differentiate this probability with respect to r', you get the radial probability density - but your probability was just the integral

    $$a\int_0^{r'} dr~r^2 |R(r)|^2,$$
    where the angular components have been integrated out to give some constant a (which you need to find). If you take the derivative of this integral with respect to r' (using the Liebniz rule that I mentioned in a previous thread - it's simpler here because only one limit depends on r'), you should get

    $$\rho(r') = a(r')^2 |R(r')|^2.$$

    (You can relabel r' as r at this point; the only reason they were different before was because I was already using r as the integration variable and did not want you to mix them up).

    Now, to answer the rest of your question:

    You do not ignore the Spherical Harmonic ##Y##. You have to integrate over it. i.e., you have to perform the integral ##a = \int d\theta d\phi \sin\theta |Y_{\ell m}(\theta,\phi)|^2##. This will give you the factor "a" that I have been writing in this formulas. Alternatively, you could note that the integration will give you some factor "a" that you might not want to calculate, but integrating ##\rho(r)## over all space should give you 1, so you could find "a" by figuring out what it needs to be so that the total probability is 1.

    However, you shouldn't need to do that. Do you know what the normalization condition of the spherical harmonics is? That is, do you know

    $$\int_0^\pi d\theta \int_0^{2\pi} d\phi \sin\theta Y_{\ell m}(\theta,\phi) Y_{\ell' m'}^\ast(\theta,\phi) = ?$$

    This is usually covered in books or lectures when the spherical harmonics are introduced, and you can use it to find ##\int d\theta d\phi \sin\theta |Y|^2##.
     
  13. Dec 2, 2012 #12
    I'm not sure, but about what you said, I think the mentioned integral should equal 1? Actually the lecture is a bit of a mess we have to attend. I don't study 'pure' physics. The physisicts here attend two quantum mechanics lectures (total time = 1 year). I have to attend only this one whereas the QM part is about 3 months. So the professor doesn't go into detail at all. It's hard to come by with all the things when you hear it for the first time like the spherical harmonics etc. that's why I'm so clueless. Thanks for your help and patience :redface:
     
  14. Dec 2, 2012 #13

    Mute

    User Avatar
    Homework Helper

    Yes, the angular integral is 1! You might expect this to be the case, as otherwise the radial integral over ##r^2|R(r)|^2## would have to give ##1/a##. (Some people do use a different normalization convention in other fields, in which the ##|Y|^2## integral would give ##4\pi/(2\ell + 1)##, but in quantum mechanics the normalization is typically chosen to be 1).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook